Class 11 Chemistry

NCERT Solutions For Class 11 Chemistry Chapter 8 Redox Reactions Long Questions And Answers

Question 1. In the following redox reactions, identify the oxidation half-reactions and reduction half-reactions along with the oxidants and reductants

1. \(\mathrm{Cl}_2(g)+2 \mathrm{I}^{-}(a q) \rightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_2(g) \)

2. \( \mathrm{Sn}^{2+}(a q)+2 \mathrm{Fe}^{3+}(a q) \rightarrow \mathrm{Sn}^{4+}(a q)+2 \mathrm{Fe}^{2+}(a q)\)

3. \( 2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3(a q)+\mathrm{I}_2(s) \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6(a q)+2 \mathrm{NaI}(a q)\)

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4. \(\mathrm{Fe}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{H}_2(g)\)

5. \( \mathrm{H}_2 \mathrm{~S}(a q)+\mathrm{Cl}_2(g) \rightarrow \mathrm{S}(s)+2 \mathrm{HCl}(a q)\)

6. \(2 \mathrm{FeCl}_2(a q)+\mathrm{Cl}_2(g) \rightarrow 2 \mathrm{FeCl}_3(a q)\)

7. \(2 \mathrm{Hg}^{2+}(a q)+\mathrm{Sn}^{2+}(a q) \rightarrow \mathrm{Hg}_2^{2+}(a q)+\mathrm{Sn}^{4+}(a q)\)

Answer:

1.

Oxidation half-reaction: 2l-(aq)→I₂(s) + 2e

Reduction half-reaction: Cl₂(g) + 2e→2Cl-(aq)

Oxidant: Cl₂(g); Reductant: l-(aq)

2.

Oxidation half-reaction: Sn²⁺(aq)→Sn⁴⁺(aq) + 2e

Reduction half-reaction: Fe³⁺(aq) + e→Fe²⁺(aq)

Oxidant: Fe³⁺(aq); Reductant: Sn2+(aq)

3.

Oxidation half-reaction:\(\mathrm{S}_2 \mathrm{O}_3^{2-}(a q) \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}(a q)+2 e\)

Reduction half-reaction: I2(s) + 2e→2l-(aq)

Oxidant: I₂(s); Reductant: Na₂S₂O₃(aq)

4.

Oxidation half-reaction: Fe(s)→Fe²⁺(aq) + 2e

Reduction half-reaction: 2H+(aq) + 2e→H2(g)

Oxidant: H+(aq); Reductant: Fe(s)

Redox Reactions Class 11 Long Questions and Answers

5.

Oxidation half-reaction: S2-(aq)→S(s) + 2e

Reduction half-reaction: Cl₂(g) + 2e→CI-(aq)

Oxidant: Cl₂(g); Reductant: H2S(aq)

6.

Oxidation half-reaction: Fe²⁺(aq)→Fe³⁺(aq)

Reduction half-reaction: Cl₂(g) + 2e→2Cl-(aq)

Oxidant: Cl₂; Reductant: FeCl₂

7.

Oxidation half-reaction: Sn²⁺(aq)→Sn⁴⁺(aq) + 2e

Reduction half-reaction: \(2 \mathrm{Hg}^{2+}(a q)+2 e \rightarrow \mathrm{Hg}_2^{2+}(a q)\)

Oxidant: Hg²⁺(aq); Reductant: Sn²⁺(aq)

Question 2. A compound is composed of three elements A, B, and C. The oxidation numbers of A, B, and C in the compound are +1, +5, and -2, respectively. Which one of the following formulas represents the molecular formula of the compound?

1. A₂BC₄
2. A₂(BC₃)₂.

Answer:

The algebraic sum of the oxidation numbers of all atoms present in a molecule is equal to zero.

In an A₂BC₄ molecule, the algebraic sum ofthe oxidation numbers of all the constituent atoms

= 2 × (+1) +1 × (+5) + 4 × (-2) = -1

In A₂(BC₃)₂ molecule, the algebraic sum of the oxidation numbers of all the atoms

= 2 × (+ 1) + 2[5 + 3 ×  (-2)] = 0

Therefore, A₂(BC₃)₂ represents the molecular formula ofthe compound.

Question 3. Among the reactions given below, identify the redox reactions and also mention the oxidant and the reductant in each case

1. \(\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_2(\mathrm{~g})\)

2. \(2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3(a q)+\mathrm{I}_2(s) \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6(a q)+2 \mathrm{NaI}(a q)\) →\(\)

3. \(\left(\mathrm{NH}_4\right)_2 \mathrm{~S}(a q)+\mathrm{Cu}\left(\mathrm{NO}_3\right)_2(a q)\) →\(\mathrm{CuS}(s)+2 \mathrm{NH}_4 \mathrm{NO}_3(a q)\)

4. \(\mathrm{Cu}_2 \mathrm{~S}(s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{SO}_2(g)\)

5. \(\mathrm{Ca}(\mathrm{OH})_2(a q)+\mathrm{H}_2 \mathrm{SO}_4(a q) \rightarrow \mathrm{CaSO}_4(s)+2 \mathrm{H}_2 \mathrm{O}(l)\)

6. \(\mathrm{H}_2 \mathrm{~S}(g)+\mathrm{HNO}_3(a q)\) →\(\mathrm{H}_2 \mathrm{SO}_4(a q)+\mathrm{NO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)\)

Answer:

1.

The Oxdination number of Fe decreases from +3 to 0 while that of C increases from +2 to +4. so in this reaction, Fe₂O₃ undergoes reduction and Co undergoes Oxdination. Hence, it’s a redox reaction in which Fe₂O₃ acts as an Oxdiant and Co as Reductant.

2.

The oxidation number of S increases from +2 to +2.5 while that of decreases from 0 to -l. So, in this reaction, Na₂S₂O₃ undergoes oxidation and I₂ undergoes reduction. Hence, it is a redox reaction in which Na₂S₂O₃ acts as a reductant and I₂ as an oxidant.

3. In this reaction, there occurs no change in oxidation number for any element. So, it is not a redox reaction.

4.

The oxidation number of Cu decreases from +1 to 0 while that of S increases from -2 to +4. Also, the oxidation number of 0 decreases from O to -2. So, in this reaction, Cu₂S undergoes oxidation as well as reduction and O₂ undergoes reduction. Hence, it is a redox reaction in which Cu₂S serves as both oxidant and reductant and O₂ acts as an oxidant.

5. In this reaction, no change in oxidation number for any element takes place. So, it is not a redox reaction.

6.

The oxidation number of S increases from -2 to +6 while that of N decreases from +5 to +4. So, in this reaction, H₂S undergoes oxidation and HNO₃ undergoes reduction. Hence, it is a redox reaction in which H₂S acts as a reductant and HNO₃ acts as an oxidant

Hence, it is a redox reaction in which H₂S acts as a reductant and HNO₃ acts as an oxidant.

Question 4. Identify the following half-reactions as oxidation half¬ reactions and reduction half-reactions:

1. \(\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow 2 \mathrm{Cr}^{3+}\)

2. \(\mathrm{Cr}(\mathrm{OH})_4^{-}(a q) \rightarrow \mathrm{CrO}_4^{2-}(a q)\)

3. \(\mathrm{IO}_3^{-}(a q) \rightarrow \mathrm{IO}_4^{-}(a q)\)

4. \(\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{Cl}^{-}(a q)\)

5. \(\mathrm{MnO}_4^{-}(a q) \rightarrow \mathrm{MnO}_2(s)\)

Answer:

1.

The oxidation number of Cr changes from +6 to +3. This indicates that the reaction involves the reduction of Cr₂O₇^-2 Hence, it is a reduction half-reaction.

2.

The reaction involves the oxidation of Cr(OH)₄ because the oxidation number of Cr increases from +3 to +6. So, this reaction represents an oxidation half-reaction.

3.

This reaction represents an oxidation half-reaction since the oxidation number of I increases from +5 to +7 in the reaction.

4.

This reaction represents a reduction half-reaction because the oxidation number of Cl decreases from +1 to -1 in the reaction.

5.

This is a reduction half¬ reaction because the oxidation number of Mn decreases from +7 to +4.

Question 5. Give an example of a disproportionation reaction. Calculate the volume of /0.225(M) KMnO₄ solution that can completely react with 45mL of a 0.125(M) FeSO₄ solution in an acid medium.
Answer:

⇒  \(\left[\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)+e\right] \times 5\)

Reduction reaction:

⇒ \(\mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e\) → \(\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_2 \mathrm{O}(l) \times 1\)

Net reaction:

⇒ \(5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}(a q)\) → \(5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_2 \mathrm{O}(l)\)

5 mol of FeSO₄ = 1 mol of MnO₄

or, 1000mL of 5 M FeSO₄ = 1000 ml limit of 1M KMnO₄

or, 1mL of 5M FeSO₄ s lmL of lM KMnO₄

or, 1mL of 1M FeSO₄ \(\equiv \frac{1}{5} \mathrm{~mL}\) of 1M KMno4

or, 45mL of 0.125M FeSO₄ = 45 × 0.125 ,\(\times \frac{1}{5} \mathrm{~mL}\) of 1M KMnO₄ \(\equiv \frac{9 \times 0.125}{0.225} \mathrm{~mL}\) = 5mL of0.225M KMnO₄

= The volume of KMnO₄ required = 5mL

NCERT Class 11 Chemistry Chapter 8 Redox Reactions Long Questions & Answers

Question 6. Identify the following reactions as disproportionation and comproportionation reactions

1. \(\mathrm{Ag}^{2+}(a q)+\mathrm{Ag}(s) \rightarrow 2 \mathrm{Ag}^{+}(a q)\)

2. \(2 \mathrm{H}_2 \mathrm{O}_2(l) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(g)\)

3. \({O}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})\)

4.  \(4 \mathrm{KClO}_3(s) \rightarrow \mathrm{KCl}(s)+3 \mathrm{KClO}_4(s)\) → \(2 \mathrm{MnO}_4^{-}(a q)+\mathrm{MnO}_2(s)+4 \mathrm{OH}^{-}(a q)\)

5. \(2 \mathrm{NH}_4 \mathrm{NO}_3(s) \rightarrow \mathrm{N}_2(g)+4 \mathrm{H}_2 \mathrm{O}(g)+\mathrm{O}_2(g)\)

6. \(\mathrm{IO}_3^{-}(a q)+5 \mathrm{I}^{-}(a q)+6 \mathrm{H}^{+}(a q) \rightarrow 3 \mathrm{I}_2(s)+3 \mathrm{H}_2 \mathrm{O}(l)\)

Answer:

1.

In this reaction, the resulting species, Ag+, exists in an oxidation state (+1) that lies between the oxidation states of Ag²⁺(+2) and Ag(0). Hence, it is a comproportionation reaction.

2.

In this reaction, H₂O₂ (oxidation number of 0 is -1 ) decomposes to form H₂O (oxidation number of O is -2) and O₂ (oxidation number of O is zero).

The oxidation state -1 lies between 0 and -2. So, this reaction represents a disproportionation reaction.

3.

In this reaction, the oxidation number of Cl decreases (+5 to-1 ) as well as increases (+5 to +7). This means KClO₃ undergoes both oxidation and reduction in the reaction. Hence, this reaction is a disproportionation reaction.

4.

In this reaction, the oxidation number of Mn increases (+6 to +7) and decreases (+6 to +4) as well. ‘Hus’ means MnO^– undergoes both oxidation and reduction in the reaction. Therefore, this reaction is a disproportionation reaction.

5.

NH₄NO₃ molecule has two N-atoms, out of which one has an oxidation number of -3 and the other has an oxidation number of +5—the decomposition of NH₄NO₃ results in N₂(g).

So, in this reaction, the oxidation number of one N-atom increases from -3 to 0 while the oxidation number of other N-atom decreases from +5 to 0. Since the oxidation number 0 lies between the oxidation numbers -3 and +5, the reaction represents a comproportionation reaction.

6.

In this reaction, 10J (oxidation number of I = +5) reacts with I- (oxidation number of = -1 ) to form I₂ (oxidation number of = 0 ). Since the oxidation number 0 lies between -1 and +5, the reaction represents a comproportionation reaction.

Question 7. Determine the equivalent masses of the following underlined compounds by both oxidation number and electronic methods

1. SO₂ + 2H₂O→H₂SO₄

2. HNO₃→NO₂ + H₂O

3. HNO₃ + 3H⁺→ NO + 2H₂O

4. MnO₂ + 4H⁺→ Mn²⁺→ + 2H₂O

5.  \(\mathrm{KMnO}_4+\mathrm{FeSO}_4+\mathrm{H}_2 \mathrm{SO}_4\) → \(\mathrm{K}_2 \mathrm{SO}_4+\mathrm{MnSO}_4+\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}\)

Answer:

1. Oxidation number method: \(\stackrel{+4}{\mathrm{SO}_2}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4\)

In this reaction, the oxidation number of S increases from +4 to +6. The change in oxidation number per molecule of SO₂ = 6 – 4

= 2 units.

∴ Equivalent mass of SO₂

⇒ \(\frac{\text { Molecular mass of } \mathrm{SO}_2}{2}=\frac{64}{2}=32\)

Electronic method: SO, + 2H₂O→4H⁺+ SO₄ ^2-+ 2e

Number of electrons lost by a molecule of SO₂ for its oxidation = 2.

∴ Equivalent mass of SO₂ \(=\frac{64}{2}=32\)

Class 11 Chemistry Chapter 8 Redox Reactions NCERT Long Questions

2. Oxidation number method: \(\stackrel{+5}{\mathrm{HN}} \mathrm{O}_3 \rightarrow \stackrel{+4}{\mathrm{~N}} \mathrm{O}_2+\mathrm{H}_2 \mathrm{O}\)

The change in oxidation number per molecule of HNO₃ = 5-4 = unit:

∴ Equivalent mass of HNO₃

⇒ \(\frac{\text { Molecular mass of } \mathrm{HNO}_3}{1}=\frac{63}{1}=63\)

Electronic method: NO^–₃ + 2H⁺ + e→ NO₂+ H₂O.

The number of electrons involved in the die reduction of NO^–₃ is 1.

∴ Equivalent mass of HNO₃

⇒ c\(\frac{\text { Molecular mass of } \mathrm{HNO}_3}{1}=\frac{63}{1}=63\)

3. Oxidation number method: \(\mathrm{H}^{+5} \mathrm{~S}_3+3 \mathrm{H}^{+} \rightarrow \stackrel{+2}{\mathrm{~N} O}+2 \mathrm{H}_2 \mathrm{O}\)

In this reaction, the oxidation number of N decreases from +5 to +2. So, the change in oxidation number per molecule of HNO₃ = 5-2

= 3 units.

∴ Equivalent mass of HNO₃

⇒ \(=\frac{\text { Molecular mass of } \mathrm{HNO}_3}{1}=\frac{63}{3}=21\)

Electronic method:  \(\mathrm{NO}_3^{-}+4 \mathrm{H}^{+}+3 e \rightarrow \mathrm{NO}+2 \mathrm{H}_2 \mathrm{O}\)

The number of electrons involved in the reduction of 1 molecule of HNOg = 3

∴ Equivalent mass of HNO₃

⇒ \(\frac{\text { Molecular mass of } \mathrm{HNO}_3}{1}=\frac{63}{3}=21\)

4. Oxidation number method: \(\stackrel{+4}{\mathrm{MnO}_2}+4 \mathrm{H}^{+} \rightarrow \stackrel{+2}{\mathrm{M}}{ }^{2+}+2 \mathrm{H}_2 \mathrm{O}\)

In this reaction, the oxidation number decreases from +4 to +2. So, the change in oxidation number per molecule of MnO₂ = 4-2

= 2 units.

∴  Equivalent mass of MnO₂

⇒ \(=\frac{\text { Molecular mass of } \mathrm{MNO}_2}{1}=\frac{87}{2}=43.5\)

  Electronic method: \(\mathrm{MnO}_2+4 \mathrm{H}^{+}+2 e \rightarrow \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O}\)

The number of electrons involved in the reduction of a molecule of MnO₂ =-2

⇒ \(=\frac{\text { Molecular mass of } \mathrm{MnO}_2}{1}=\frac{87}{2}=43.5\)

5. Oxidation number method:

⇒ \(\stackrel{+7}{\mathrm{~K}} \mathrm{nO}_4+\stackrel{+2}{\mathrm{~F}} \mathrm{eSO}_4+\mathrm{H}_2 \mathrm{SO}_4\)

⇒ \(\mathrm{~K}_2 \mathrm{SO}_4+\stackrel{+2}{\mathrm{MnSO}_4+\stackrel{+3}{\mathrm{Fe}} \mathrm{e}_2}\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}\)

In this reaction, KMnO₄ undergoes reduction because the oxidation number of Mn decreases from +7 to +2. So, the change in oxidation number of Mn= 7-2

=5 units.

∴  Equivalent mass of KMnO₄

⇒ \(\frac{\text { Molecular mass of } \mathrm{KMnO}_4}{\begin{array}{c}
\text { Change in oxidation number per molecule } \\
\text { of } \mathrm{KMnO}_4 \text { because of its reduction }
\end{array}}\)

⇒  \(\frac{158}{5}\)

= 31.6

In the reaction, FeSO₄ undergoes oxidation because the oxidation number of Fe increases from +2 to +3. So, the change in oxidation number of Fe = 3-2

= 1 unit.

∴  Equivalent mass of FeSO₄

⇒ \(\frac{\text { Molecular mass of } \mathrm{FeSO}_4}{\begin{array}{c}
\text { Change in oxidation number per molecule } \\
\text { of } \mathrm{FeSO}_4 \text { because of its oxidation }
\end{array}}\)

⇒ \(\frac{151. 85}{1}\)

= 151.85

∴  Electronic method:

MnO₄ + 8H⁺+ 5e→ Mn²⁺  + 4H₂O

Equivalent mass of KMnO₄

⇒ \(=\frac{\text { Molecular mass of } \mathrm{KMnO}_4}{\begin{array}{c}
\text { Number of electrons gained in reduction of } \\
\text { one molecule of } \mathrm{KMnO}_4
\end{array}}\)

=  \(\frac{158}{5}\)

31.6

Fe²⁺ →Fe³⁺ + e

∴  Equivalent mass of FeSO₄

⇒ \(=\frac{\text { Molecular mass of } \mathrm{FeSO}_4}{\begin{array}{c}
\text { Number of electrons lost in oxidation of } \\
\text { one molecule of } \mathrm{FeSO}_4
\end{array}}\)

=\(\frac{151.85}{1}\)

Question 8. Determine the equivalent mass of Br₂(l) [Molecular mass =159.82 in the given reaction:

⇒ \(2 \mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}(a q)+\mathrm{Br}_2(l)\)→ \(2 \mathrm{Mn}^{2+}(a q)+2 \mathrm{BrO}_3^{-}(a q)+2 \mathrm{H}_2 \mathrm{O}(l)\)

In the given reaction, the oxidation half-reaction is —

⇒ \(\mathrm{Br}_2(l)+6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{BrO}_3^{-}(a q)+12 \mathrm{H}^{+}(a q)+10 e\)

The number of electrons involved in the oxidation of one molecule of Br₂ = 10 Equivalent mass of Br₂.

⇒ \(=\frac{\text { Molecular mass of } \mathrm{Br}_2}{\begin{array}{c}
\text { Number of electrons involved per } \\
\text { molecule of } \mathrm{Br}_2 \text { in its oxidation }
\end{array}}\)

=\(\frac{159.82}{10}\)

= 15.982

Question 9. \(\mathrm{MnO}_4^{2-}\) undergoes a disproportionation reaction in an acidic medium but MnO₄ docs do not. Give reason.
Answer:

The oxidation number of Mn in MnO₄ is +7, which is the highest oxidation number that Mn can possess. So, it does not undergo the disproportionation reaction.

Again, in the case of \(\mathrm{MnO}_4^{2-}\) the oxidation number of Mn is +6. Therefore, Mn in \(\mathrm{MnO}_4^{2-}\) can increase its oxidation number to +7 or decrease it to some lower value. So, \(\mathrm{MnO}_4^{2-}\) undergoes a disproportionation reaction as given below—

In the above reaction, the oxidation number of Mn increases from +6 in \(\mathrm{MnO}_4^{2-}\) to +7 in MnO₄ and decreases to +4 in MnO₄.

Question 10. What amount of K₂Cr₂O₇? (in mmol) is required to oxidize 24 mL 0.5 M Mohr’s salt?
Answer:

The number of mmol of Mohr’s salt in 24 mL 0.5MMohr’s salt solution =24 x 0.5 = 12.

So, the balanced redox reaction is

⇒ \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+6\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot \mathrm{FeSO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}+7 \mathrm{H}_2 \mathrm{SO}_4\)

⇒ \(\mathrm{~K}_2 \mathrm{SO}_4+6\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4+3 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+43 \mathrm{H}_2 \mathrm{O}\)

So, from the balanced equation, we see that 6mmol Mohr’s salt gets oxidized by 1mmol K₂Cr₂O₇.

12 mmol Mohr’s salt gets oxidized by \(\frac{1}{6} \times 12\) = 2 mmol K₂Cr₂O₇.

Question 11. Explain with reaction mechanism why the reaction between 03 and H₂O₂ is written as—

⇒ \(\mathrm{O}_3(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}_2(l) \rightarrow \mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})\)

Answer:

The reaction between O₃ and H₂O₂ is as follows

First step: O₃(g)→O₂(g) + O(g)’

Second step: H₂O₂(g) + O(g)→H₂O(g) +O₂(g)

In the first step, ozone produces nascent oxygen which is H₂O₂ in the second step. So, the overall reaction is as follows H₂O₂ + O₃→H₂O + O₂ + O₂

So, in the overall reaction, O₂ Is written twice because a total of two molecules of O₂ are produced during the reaction.

Question 12. 12.53 cm3 0.051M SeO₂ reacts completely with 25.5 cm3 0.1 M CrSO₄ to produce Cr₂(SO₄)₃. What is the change in the oxidation number of Se in this redox reaction?
Answer:

Let the oxidation number of Se in the newly produced compound be x.

The redox reaction is as follows:

⇒ \({\left[\mathrm{Se}^{4+}+x e \longrightarrow \mathrm{Se}^{4-x}\right] \times 1}\)

⇒\({\left[\mathrm{Cr}^{2+} \longrightarrow \mathrm{Cr}^{3+}+e\right] \times x}\)

⇒ \(\mathrm{Se}^{4+}+x \mathrm{Cr}^{2+} \longrightarrow \mathrm{Se}^{4-x}+x \mathrm{Cr}^{3+}\)

Now, 12.53 cm3 0.051M SeO₂ = 12.53 x 0.051 = 0.64 mmol SeO₂

25.5 cm3 0.1 M CrSO₄= 25.5 ×  0.1 = 2.55 mmol CrSO₄ However according to the balanced equation, 1 mol SeO₂ gets reduced by x mol CrSO₄.

2.55 mmol CrSO₄ is reduced by \(\frac{2.55}{x}\) mmol SeO₂

But 0.64 mmol SeO₂ gets reduced

⇒ \(\text { So, } \frac{2.55}{x}=0.64 \quad \text { or, } x=4\)

The change in oxidation number of Se -atom = 4- (4- x) = x = 4.

Question 13. 30 ml 0.05 M KMnb4 is required for the complete oxidation of 0.5 g oxalate in an acidic medium. Calculate ) tl,e percent amount of oxalate in that salt sample:
Answer:

⇒ \(2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\)

According to the equation, 2 mol MnO-4 = 5 mol C₂O₄

⇒ \(1 \mathrm{~mol} \mathrm{MnO}_4^{-} \equiv \frac{5}{2} \mathrm{~mol} \mathrm{C}_2 \mathrm{O}_4^{2 \mathrm{ris}}\)

Again, 1000 mL 0.05(M)KMnO4 =→ 0.05 mol KMnO₄

30 mL 0.05(M)MnO4 \(\Rightarrow \frac{0.05 \times 30}{1000}\)

= 1.5×10-3 mol KMnO₄

Now, 1 mol \(\mathrm{MnO}_4=\frac{5}{2} \mathrm{~mol} \mathrm{C}_2 \mathrm{O}_4\)

⇒ \(1.5 \times 10^{-3} \mathrm{~mol} \mathrm{MnO}_4=\frac{5}{2} \times 1.5 \times 10^{-3} \mathrm{~mol} \mathrm{C}_2 \mathrm{O}_4^2\)

⇒ \(=\frac{5}{2} \times 1.5 \times 10^{-3} \times 38 \mathrm{~g} \mathrm{C}_2 \mathrm{O}_4^{2-}=0.33 \mathrm{~g} \mathrm{C}_2 \mathrm{O}_4^{2-}\)

⇒ \(\text { Percentage of } \mathrm{C}_2 \mathrm{O}_4^{2-} \text { in the sample }=\frac{0.33 \times 100}{0.5}=66 \%\)

Long Answer Questions for Redox Reactions Class 11 Chemistry

Question 14. What will be the nature of the suit formed when 2 mol Nil Is added to the pigeon’s solution of mol pyrophosphoric? Clive equation?

Answer:

From the structure of pyrophosphoric acid, it Is clear that it contains four replaceable hydrogen atoms.

So, the reaction between and 2 mol NaOH will be as follows:

⇒ \(-\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{H}_2 \mathrm{P}_2 \mathrm{O}_7+2 \mathrm{H}_2 \mathrm{O}\)

There are two replaceable 11-atoms in Na₂H₂P₂O₇ fonned due to the above reaction. So, it is an acidic salt.

Question 15. Oxidation million of the elements A, It mid-C are 12,1 to mid -2 respectively. Which one will lie the formula of the compound containing these three elements?

⇒ \(\mathrm{A}_2\left(\mathrm{BC}_2\right)_2, \mathrm{~A}_3\left(\mathrm{~B}_2 \mathrm{C}\right)_2, \mathrm{~A}_3\left(\mathrm{BC}_4\right)_2\)

Answer:

The total oxidation number of all the elements in a compound should be zero (0).

In A₂(BC₂) molecule, the sum of oxidation numbers of all the atoms

=2 ×(+ 2) + 2 × (+ 5) + 4 × (-2) = + 6

In the A₂(B₂C)₂ molecule, the sum of oxidation numbers of all the atoms

= 3 × (+ 2) + 4 × (+ 5) + 2 × (-2) = + 22

In A₃(BC₄)₂ molecule, the sum of oxidation numbers of all the atoms

= 3 × (+ 2) + 2 × (+ 5) + 8 ×(-2) = 0.

∴ The correct formula of the compound will be A₃(BC₄)₂

Question 16. In an acidic medium, for the reduction of each NO₃ ion in the given reaction, how many electrons will be required? NO₃ NH₂OH
Answer:

NO₃— NH₂OH; For equalizing the number of O -atoms on both sides, two H₂O molecules are added to the right side (having a lesser number of O -atoms) and two H+ ions are added to the left side for each molecule of H₂O added.

⇒ \(\mathrm{NO}_3^{-}+4 \mathrm{H}^{+} \longrightarrow \mathrm{NH}_2 \mathrm{OH}+2 \mathrm{H}_2 \mathrm{O}\)

For equalizing the number of H -atoms on both sides, three additional H+ ions are required on the left side

so we get, \(\mathrm{NO}_3^{-}+7 \mathrm{H}^{+} \longrightarrow \mathrm{NH}_2 \mathrm{OH}+2 \mathrm{H}_2 \mathrm{O}\)

To balance the charge on both sides, 6 electrons are added to the left side ofthe equation.

⇒ \(\mathrm{NO}_3^{-}+7 \mathrm{H}^{+}+6 e \longrightarrow \mathrm{NH}_2 \mathrm{OH}+2 \mathrm{H}_2 \mathrm{O}\)

Hence, for the reduction of each NO₃ ion into an NH₂OH molecule, 6 electrons are required.

Question 17. Identify the redox reactions among the following:

1. 2CuSO₄ + 4KI→2CuI + I₂ + 2K₂SO₄

2. BaCl₂ + Na₂SO₄→BaSO₄ + 2NaCl

3. 2NaBr + Cl₂→2NaCl + Br₂

4. NH₄NO₂→+N₂ + 2H₂O

5. CuSO₄ + 4NH₃→[Cu(NH₃)₄]SO₄

6. 3I₂ + 6NaOH→NaIO₃ + 5NaI + 3H₂O

Answer:

In this reaction, the oxidation number of Cu decreases (+ 2 → +1) and the oxidation number of I₄ increases (-1→ 0) i.e. reduction of CuSO₄ and oxidation of KI take place. Thus, it is a redox reaction.

1.  \(\stackrel{+2}{\mathrm{CuSO}_4}+\stackrel{-1}{\mathrm{KI}} \rightarrow \stackrel{+1}{\mathrm{CuI}}+\stackrel{0}{\mathrm{I}} 2+\mathrm{K}_2 \mathrm{SO}_4\)

This reaction does not involve any change in the oxidation number of any element i.e., in this reaction, oxidation or reduction does not take place. Hence, it is not a redox reaction

2.  \(\stackrel{+2}{\mathrm{BaCl}_2}+\stackrel{+1}{\mathrm{Na}_2} \mathrm{SO}_4 \rightarrow \stackrel{+2}{\mathrm{BaSO}}{ }_4+2 \mathrm{NaCl}^{-1}\)

In this reaction, the oxidation number of bromine increases from -1 to 0 and the oxidation number of chlorine decreases from 0 to -1. In this case, NaBr gets oxidized whereas Cl₂ gets reduced. Hence, this reaction is a redox reaction.

3. \(2 \mathrm{NaBr}+\stackrel{0}{\mathrm{C}} \mathrm{l}_2 \rightarrow 2 \mathrm{Na}{ }^{-1} \mathrm{Cl}+\stackrel{0}{\mathrm{Br}}{ }_2\)

In NH₄NO₂, the oxidation number of N in NH₄ increases from -3 to 0, while the oxidation number of N in NO₂ decreases from +3 to 0, i.e., in this reaction, simultaneous oxidation of NH+ and reduction of NO₂ in the compound occur. So, this reaction is a redox reaction.

4. \(\stackrel{-3}{\mathrm{NH}_4}{\stackrel{+3}{\mathrm{~N}} \mathrm{O}_2 \rightarrow \stackrel{0}{\mathrm{~N}}}_2+2 \mathrm{H}_2 \mathrm{O}\)

This reaction does not involve any change in the oxidation number of any element. So, it is not a redox reaction.

5.  \(\left(\stackrel{+2}{\mathrm{CuSO}_4}+4 \mathrm{NH}_3\right) \rightarrow\left[\stackrel{+2}{\mathrm{Cu}}\left(\mathrm{NH}_3\right)_4\right] \mathrm{SO}_4\)

In the given reaction, the oxidation number of iodine increases from 0 to +5 and decreases from 0 to -1 i.e., both oxidation and reduction occur in this reaction. Thus, it is a redox reaction.

6. \(3 \stackrel{0}{\mathrm{I}}_2+6 \mathrm{NaOH} \rightarrow \mathrm{NaIO}_3^{+5}+5 \mathrm{NaI}^{-1}+3 \mathrm{H}_2 \mathrm{O}\)

Question 18. Which of the following reactions are disproportionation reactions and comproportionation reactions?

1. 4KClO₃→KCl + 3KClO₄

2. 3K₂MnO₄ + 2H₂O→2KMnO₄ + MnO₂ + 4KOH

3. KIO₃ + 5KI + 6HCI→3I₂ + 6KCI + 3H₂O

4. 2C₆H₅CHO + NaOH C₆H₅COONa + C₆H₅CH₂OH

5. Ag²⁺ + Ag→2Ag⁺

Answer:

1.

In KClO₃, the oxidation number of Cl = + 5. The oxidation numbers of Cl in KCl and KClO₄ are -1 and + 7 respectively. So in this reaction, KClO₃ undergoes simultaneous oxidation and reduction producing KClO₄ and KCl respectively. Hence, it is a disproportionation reaction.

2.

The oxidation number of Mn in K₂MnO₄ = + 6. On the other hand, the oxidation numbers of Mn in the products KMnO₄ and MnOa are + 7 and + 4 respectively. Therefore, in the reaction, K₂MnO₄ is oxidized and reduced at the same time forming KMnO₄ and MnO₄. Thus, it is a disproportionation reaction.

3.

The oxidation numbers of iodine in KIO₃ and KI are + 5 and -1 respectively and the oxidation number of iodine in 12 is zero. This oxidation number lies between + 5 and -1, which is an intermediate oxidation state. So, it is a comproportionation reaction.

4.

In this reaction, the oxidation number of carbon in the benzene ring does not change. However, the oxidation number of carbon atoms in the —CHO group (the oxidation number of carbon in —CHO is +1 ) changes to a higher oxidation number of +3 in the —COONa group and a lower oxidation number of -1 in the —CH₂OH group. Here, —the CHO group gets simultaneously oxidized & reduced. Hence, it is a disproportionation reaction.

5.

The oxidation numbers of the reactants Ag²⁺ and Ag are + 2 and O respectively and the oxidation number of the product is +1. This oxidation number is intermediate between the oxidation numbers +2 and O . so it is a comproportionation reaction.

Question 19. Identify the redox reaction(s) and also the oxidants B and the reductants from the following reaction(s).

1. \(2 \mathrm{MnO}_4^{-}+5 \mathrm{SO}_2+6 \mathrm{H}_2 \mathrm{O}\) → \( 5 \mathrm{SO}_4^{2-}+2 \mathrm{Mn}^{2+}+4 \mathrm{H}_3 \mathrm{O}^{+} \)

2. \(\mathrm{NH}_4^{+}+\mathrm{PO}_4^{3-} \longrightarrow \mathrm{NH}_3+\mathrm{HPO}_4^{2-} \)

3. \( \mathrm{HClO}+\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}+\mathrm{S}\)

Answer:

In this reaction, the oxidation number of Mn decreases from (+7→+2) and the oxidation number of S increases from (+4→+6). So this reaction causes a reduction of MnO₄ and oxidation of SO₂. Hence, the reaction is a redox reaction. Here Mnt)ÿ acts as an oxidant and SO₂ as a reductant.

This reaction does not involve any change in the oxidation number of any element. So it is, not a redox reaction.

This reaction involves an increase in the oxidation number of S from -2 to 0, and a decrease in the oxidation number of, Cl from +1 to -1. So it is a redox reaction. Here HCIO is an oxidising agent and H₂S is a reducing agent.

Question 20. Determine the equivalent masses of Na₂S₂O₃.5H₂O +2 and KBrO₃ In the following reactions.

⇒ \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}\)

⇒ \(\mathrm{BrO}_3^{-}+6 \mathrm{H}^{+}+6 e \longrightarrow \mathrm{Br}^{-}+3 \mathrm{H}_2 \mathrm{O}\)

Answer:

In this reaction, two \(\mathrm{S}_2 \mathrm{O}_3^{2-}\) ions are oxidised by losing 2 electrons. So, for the oxidation of one \(\mathrm{S}_2 \mathrm{O}_3^{2-}\) ion, one electron is given up

Equivalent mass of Na₂S₂O₃.5H₂O

⇒ \(=\frac{\text { Molecular mass of } \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \cdot 5 \mathrm{H}_2 \mathrm{O}}{1}=248\)

This reaction produces Br- from BrO₃. In the reduction of each molecule of KBr03, 6 electrons are accepted.

In the given reaction, the equivalent mass of KBrO₃

⇒ \(\frac{\text { Molecular mass of } \mathrm{KBrO}_3}{6}=\frac{167}{6}=27.8\)

NCERT Solutions Class 11 Chemistry Chapter 8 Long Questions & Answers

Question 21. Determine the equivalent weights of the underlined compounds in the following two reactions:

1. \(\mathrm{FeSO}_4+\mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{SO}_4\)

→ \(\mathrm{K}_2 \mathrm{SO}_4+\mathrm{MnSO}_4+\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}\)

2. MnO₂+HCl  → MnCl₂+ Cl₂+H₂O

K = 39, Mn = 55, O = 16

Answer:

In this reaction, the decrease in oxidation number of Mn =(7-2) = 5 units. Equivalent weight of Mn.

1.  \(\frac{\text { Molecular weight of } \mathrm{KMnO}_4}{\text { Change in oxidation number }}\)

= \(\frac{158}{5}\)

= 31.6

⇒ \(\stackrel{+4}{\mathrm{MnO}}{ }_2 \longrightarrow \stackrel{+2}{\mathrm{MnCl}}{ }_2\)

Here the oxidation number of Mn decreases by (4-2)

= 2 unit.

∴ Equivalent weight of MnO₂

⇒ \(=\frac{\text { Molecular weight of } \mathrm{MnO}_2}{\text { Change in oxidation number }}=\frac{87}{2}=43.5\)

Question 22. Balance the following equation with the help of the oxidation number method.

⇒ \(\mathrm{Fe}_3 \mathrm{O}_4+\mathrm{CO} \rightarrow \mathrm{FeO}+\mathrm{CO}_2\)

Answer:

In Fe3O₄, the oxidation number of each Fe -atom =+2.67. In the reaction, the oxidation number of each Fe -atom decreases from + 2.67 to 2. So, decrease in oxidation number. of eachFe -atom = +0.67

Therefore for three Fe -atoms, the total decrease in oxidation number 3 × (+ 0.67) =+ 2 unit.

On the other hand, the oxidation number of C increases from +2 to +4. Hence, an increase in the oxidation number of C= 2 units.

Therefore, in the reaction, Fe₃O4 and CO will react with each other in the molar ratio of 2: 2 or 1: 1 .

Again 1 molecule of Fe₃O₄ will produce 3 molecules of F₂O. Hence, the balance equation will be

Fe₃O₄ + CO → 3FeO + CO₂.

Question 23. Balance by ion-electron method: MnO₂ + HCl→+Mn²⁺ + Cl₂ + H₂O
Answer:

Or MnO₂ + 4H⁺ + 4Cl→ Mn²⁺ + 2Cl + Cl₂+ 2H₂O

Therefore the balanced chemical equation is:

MnO₂ + 4H +  4Cl → MnCl₂ + Cl₂ + 2H₂O

Question 24. In a basic medium, balance the half-reactions below:

1. \(\mathrm{Cr}(\mathrm{OH})_3 \rightarrow \mathrm{CrO}_4^{2-}\)
2. \(\mathrm{Cl}_2 \mathrm{O}_7 \rightarrow 2 \mathrm{ClO}_2^{-}\)

Answer:

In order to equalize the number of O -atoms, one H₂O molecule is added to the right side (because it has an excess O -atom) and for this 1 molecule of H₂O, two OH Now, we get Cr(OH)₂ + 2OH^– ions are added to the left side. CrO^2- +H₂O. In this equation, five H -atoms are on the left side, and two H atoms are present on the right side. To equalize the number of H -atoms on both sides, three OH-(aq) and three H₂O molecules are added to the left side and right side respectively. Finally, we get

⇒ \( \mathrm{Cr}(\mathrm{OH})_3+2 \mathrm{OH}^{-}+3 \mathrm{OH}^{-} \rightarrow \mathrm{CrO}_4^{2-}+\mathrm{H}_2 \mathrm{O}+3 \mathrm{H}_2 \mathrm{O}\)

Or, \(\mathrm{Cr}(\mathrm{OH})_3+5 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_4^{2-}+4 \mathrm{H}_2 \mathrm{O}\)

For balancing the charge, 3 electrons are added to the right side. Therefore, the final half-reaction becomes,

⇒ \(\mathrm{Cr}(\mathrm{OH})_3+5 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_4^{2-}+4 \mathrm{H}_2 \mathrm{O}+3 e\)

2. To balance the number of 0 -atoms on both sides, 3 molecules of H₂O(Z) are added to the left side (because there are excess O -atoms on this side) and for these three H₂O(l) molecules, six OH- (aq) ions are added to the right side. Then we get

⇒ \(\mathrm{Cl}_2 \mathrm{O}_7+3 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{ClO}_2^{-}+6 \mathrm{OH}^{-}\)

To balance the charge, electrons are added to the left side. Hence, the balanced equation is

⇒ \(\mathrm{Cl}_2 \mathrm{O}_7+3 \mathrm{H}_2 \mathrm{O}+8 e \longrightarrow 2 \mathrm{ClO}_2^{-}+6 \mathrm{OH}^{-}\)

Question 25. Balance the following reaction in acidic and alkaline medium: \(\mathrm{SO}_3^{2-}(a q) \longrightarrow \mathrm{SO}_4^{2-}(a q)\) 
Answer:

Addlemedium:

⇒ \(\mathrm{SO}_3^{2-}(a q) \longrightarrow \mathrm{SO}_4^{2-}(a q)\)

In Tills reaction, this left side of the equation is deficient in one () -atom. So the number of O -atoms on both sides is equalized by adding one H2O(aq) molecule to the left side.

⇒ \(\mathrm{SO}_3^{2-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{SO}_4^{2-}(a q)\)

To equalize the number of H -atoms on both sides, two H+ ions are added to the right side. So we get—

⇒ \(\mathrm{SO}_3^{2-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{SO}_4^{2-}(a q)+2 \mathrm{H}^{+}(a q)\)

To balance the charge on both sides, 2 electrons are added to the right side, we get—

⇒ \(\mathrm{SO}_3^{2-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{SO}_4^{2-}(a q)+2 \mathrm{H}^{+}(a q)+2 e\)

Basic medium: \(\mathrm{SO}_3^{2-}(a q) \longrightarrow \mathrm{SO}_4^{2-}(a q)\)

To balance the number of O -atoms on both sides, two OH”(a<7) ions and one H20(Z) molecule are added to the left side and right side respectively. This gives—

⇒ \(\mathrm{SO}_3^{2-}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{SO}_4^{2-}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

To balance the change on both sides, we add 2 electrons to the right side. Thus, the balance equation is—

⇒ \(\mathrm{SO}_3^{2-}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{SO}_4^{2-}(a q)+\mathrm{H}_2 \mathrm{O}(l)+2 e\)

Question 26. Determine the values of x and y in the following balanced equation:

⇒ \(5 \mathrm{H}_2 \mathrm{O}_2+x \mathrm{ClO}_2+2 \mathrm{OH}^{-} \longrightarrow x \mathrm{Cl}^{-}+y \mathrm{O}_2+6 \mathrm{H}_2 \mathrm{O}\) 
Answer:

Oxidation half-reaction:

⇒  \(\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-} \longrightarrow \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+2 e\)……………………….(1)

Reduction half-reaction:

⇒\(\mathrm{ClO}_2+2 \mathrm{H}_2 \mathrm{O}+5 e \longrightarrow \mathrm{Cl}^{-}+4 \mathrm{OH}^{-}\) ……………………….(2)

Multiplying equations (1) and (2) by 5 and 2 respectively and adding the equations, we get—

⇒ \(5 \mathrm{H}_2 \mathrm{O}_2+10 \mathrm{OH}^{-}+2 \mathrm{ClO}_2+4 \mathrm{H}_2 \mathrm{O}\)→ \(2 \mathrm{Cl}^{-}+5 \mathrm{O}_2+10 \mathrm{H}_2 \mathrm{O}+8 \mathrm{OH}^{-}\)

∴ 5H₂O₂ + 2ClO₂ + 2OH^–→2Cl^–+ 5HO₂ + 6H₂O ……………………….(3)

Comparing equation (3) with the given equation, we get x = 2 and y – 5.

Question 27. In the given reaction determine the equivalent weight of AS₂S₃:
Answer:

⇒ \(\mathrm{As}_2 \mathrm{~S}_3+7 \mathrm{ClO}_3^{-}+7 \mathrm{OH}^{-}-7\) → \(2 \mathrm{AsO}_4^{3-}+7 \mathrm{ClO}^{-}+3 \mathrm{SO}_4^{2-}+6 \mathrm{H}_2 \mathrm{O}\)

The increase in oxidation number of each As -atom =5-3 = 2 units.

So, the total increase in oxidation number of two As -atoms = 2 ×2

= 4 units.

Two increases in the oxidation number of each S-atom =+6-(-2)

Therefore, the total Increase in oxidation number of three S -atoms = 3 × 8

= 24 units.

Hence, the total Increase In oxidation number for each As 2S,t molecule In Its oxidation Is a (24 + 4) = 20 unit.

Equivalent weight of AS₂S₃ In the given reaction

⇒ \(\frac{\text { Molecular weight of } \mathrm{As}_2 \mathrm{~S}_3}{\text { Increase in oxidation number }}=\frac{M}{28}\)

Redox Reactions Chapter 8 Long Answer Solutions Class 11

Question 28. Determine the equivalent mass of Fe₂O₄ in the given Mn = 6-4 = 2 units. reaction: FeO₄ + KMn0_4↓Fe₂O₃ + MnO₂ (Assume that the molecular mass of Fe₃O₄ =M
Answer:

⇒ \(\stackrel{+\mathrm{F} / 3}{\mathrm{Fe}} \mathrm{e}_3 \mathrm{O}_4 \rightarrow \stackrel{+3}{\mathrm{~F}} \mathrm{e}_2 \mathrm{O}_3\stackrel{+\mathrm{F} / 3}{\mathrm{Fe}} \mathrm{e}_3 \mathrm{O}_4 \rightarrow \stackrel{+3}{\mathrm{~F}} \mathrm{e}_2 \mathrm{O}_3\)

In this reaction, the increase in oxidation number for each Pc -atom \(=3-\frac{8}{3}=\frac{1}{3} \text {. }\)

So, the total increase in oxidation number for 3 Fe -atoms

⇒ \(=3 \times \frac{1}{3}=1 \text { unit. }\)

Therefore, the equivalent mass of Fe3O4

⇒ \(=\frac{\text { Molecular mass of } \mathrm{Fe}_3 \mathrm{O}_4}{\text { Increase in oxidation number }}=\frac{M}{1}=M\)

Question 29. What is the ratio of equivalent weights of MnO₄ in acidic, basic & neutral mediums?
Answer:

The reactions that the MnO₄ ion undergoes in acidic, basic, and neutral mediums are as follows. Acidic Medium

⇒ \(\mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e \rightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_2 \mathrm{O}(l)\)

• Equivalent weight \(E_1=\frac{M}{5}\) where M = KMn04.
• Basic Medium: MnO₄ (aq) + e→MnO₄ (aq)
• Equivalent weight \(E_2=\frac{M}{1}\)

Neutral Medium:

⇒ \(\mathrm{MnO}_4^{-}(a q)+2 \mathrm{H}_2 \mathrm{O}(l)+3 e \rightarrow \mathrm{MnO}_2(s)+4 \mathrm{OH}^{-}(a q)\)

Therefore \(E_1: E_2: E_2=\frac{1}{5}: 1: \frac{1}{3}=3: 15: 5\)

Question 30. MnO₄ reacts with A^x+ to form AO₃^{– ,}  Mn²⁺, and O2. One mole of MnO₄^– oxidizes 1.25 moles of A^x+ to AO₃^–. What is the value of x?
Answer:

⇒  \(\mathrm{MnO}_4^{-}+\mathrm{A}^{x+} \rightarrow \mathrm{AO}_3^{-}+\mathrm{Mn}^{2+}+\mathrm{O}_2\)

The change in oxidation number of = 5 unit (+7→+2) and that of A = (5- x) unit [+x→+5 in AO_3^-3]

1 mole of MnO₃ reacts completely with 1.25 mole of Ax+

Therefore, 1×5 = 1.25(5 -x)

Solving for x gives x = +1

Thus, the value of x = +1

Question 31. 20 mL solution of 0.1 (M) FeSO₂ as completely oxidized using a suitable oxidizing agent. What is the number of electrons exchanged?
Answer:

20 mL of 0.1(M) FeSO₄

=\(\frac{0.1}{1000} \times 20\)

=\(2 \times 10^{-3} \mathrm{~mol} \text { of } \mathrm{Fe}^{2+}\)

Fe²⁺ is oxidized to Fe³⁺, leaving 1 electron.

Hence, the number of electrons exchanged by 2 x 10-3 mol of Fe²⁺ is

2 × 10^-3× 6.022 × 10²³

= 1.2044  ×  10²¹ electrons

Question 32. What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?

1. Kl₃
2. CH₃COOH

Answer: The chemical structure of KI₃ is K+[1-1→1]¯

In Kl₃-1 ion forms a coordinate Bond with I₂. The Osditaion Number Of Each 1 atom in 12 molecules is zero, and the oxidation number of K+ id +1, therefore from the oxidation number of one I- will be -1

2. The C-1 atom is linked to one O-atom and one OH group, For O -atom and OH -group, the oxidation numbers are 2 and I respectively, As C-1 and C-2 are the atoms of the name element, the covalent linkage between them makes no change In oxidation number of either atom.

So, the oxidation number of C- 1 would be +3 since the total oxidation number of one O-atom and one Oil -group is -3. The total oxidation number of three OH-atoms linked to the C-2 atom Is +3. So, the oxidation number of C-2 would be -3.

Question 33. Justify the following reactions are redox reactions:

1. \(\mathrm{CuO}(s)+\mathrm{H}_2(g) \rightarrow \mathrm{Cu}(s)+\mathrm{H}_2 \mathrm{O}(g)\)

2. \( \mathrm{Fe}_2 \mathrm{O}_3(s)+3 \mathrm{CO}(g) \rightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_2(g)\)

3. \(4 \mathrm{BCl}_3(g)+3 \mathrm{LiAlH}_4(s)\) → \(2 \mathrm{~B}_2 \mathrm{H}_6(g)+3 \mathrm{LiCl}(s)+3 \mathrm{AlCl}_3(s)\)

4. \(2 \mathrm{~K}(s)+\mathrm{P}_2(g) \rightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}(s)\)

 5. \( 4 \mathrm{NH}_3(g)+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_2 \mathrm{O}(g)\)

Answer:

1.

In the reaction, the oxidation number of Cu decreases (+2→0) indicating the reduction of CuO, and the oxidation number of 2 increases (0 →+1), indicating the oxidation of H₂. Hence, it is a redox reaction.

2.

In the reaction, the oxidation number of Fe decreases (+3→+0) and that of C increases
→+4). Therefore, the reaction involves the reduction of Fe₂O₃ and the oxidation of CO. Hence, it is a redox reaction.

3.

The reaction involves the reduction of BC1₃ because the oxidation number of B decreases from +3 to -3 and the oxidation of LiAH₄ as the oxidation number increases from -1 to +1. So, it is a redox reaction.

4.

In the reaction, the increase in the oxidation number of K (0→+1) and the decrease in the oxidation number of F (0 to -1) indicate that the former undergoes oxidation and the latter reduces. Hence, the given reaction is a redox reaction.

5.

In the reaction, NH₃ undergoes oxidation because the oxidation number of N-atom increases (-3 to +2), while O₂ undergoes reduction, as is evident from the decrease in the oxidation number of O (0→-2). So, it is a redox reaction.

Redox Reactions Long Question Answers for Class 11 Chemistry

Question 34. Calculate the oxidation number of sulfur, chromium and nitrogen in  \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) & NO₃^–. Suggest the structure of these compounds. Count for the
Answer:

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\): If the oxidation number of Cr in \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\)

Be x, then 2(x)+ 7(-2)

=-2

∴ x = +6

The structure of \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) is

Therefore, there is no fallacy.

NO₃^–:

According to a conventional method, the oxidation number of N in

NO₃^–: x+ 3(-2) = -1 or, x = +5

Where x = +3 -1. oxidation number of N in NO₃^–)

However, according to the chemical bonding method, the structure of NO₃^– is

If the oxidation number of N is -0 = 0 the above structure is x,

Then, x+ (-1) + (-2) + (-2)

⇒ \(\begin{gathered}
x+(-1)+(-2)+(-2)=0 \\
\quad\left(\text { for } \mathrm{O}^{-}\right)(\text {for }=0)(\text { for } \rightarrow 0)
\end{gathered}\)

or, x = +5

So, in both conventional and chemical bonding methods. the oxidation number of N in NO-₃ is +5. Therefore, there is no fallacy.

Question 35. Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5
Answer:

Carbon (c):

Nitrogen (N):

Question 36. While sulfur dioxide and hydrogen peroxide can act as oxidizing as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. why?
Answer:

A species can act as both oxidant and reductant am one of its constituent atoms has an intermediate value of oxidation number. So, in a reaction the atom can increase or decrease its oxidation number i.e., it can act as an oxidant as well as a; reductant.

1. In SO₂, the oxidation number of S is +4. The highest and lowest oxidation numbers of S are +6 and -2 respectively. Therefore, the S-atom in SO₂ can increase its oxidation number in a reaction in which SO₂ acts as a reductant and decrease its oxidation number in a reaction in which SO₂ plays the role of an oxidant. Hence, SO₂ can act as an oxidant as well as a reductant.

2. In H₂O₂, the oxidation number of O is  -1 The highest and lowest oxidation numbers of oxygen are -2 and O respectively. Therefore, the oxygen atom in H₂O₂ is capable of increasing or decreasing its oxidation number. In the reaction in which H₂O₂ acts as an oxidant, the oxidation number of oxygen decreases from -1 to -2 and in the reaction in which it acts as a reductant, the oxidation number of oxygen increases from -1 to 0. Hence, H₂O₂ can act both as an oxidant and a reductant.

3. In O₃, the oxidation number of oxygen is zero. Oxygen can show two oxidation numbers, -1 and -2. So, the oxidation number of oxygen O₃ can reduce to -1 or -2, but it can never increase. Hence, O₃ can act only as an oxidant.

4. In HNO₃, the oxidation number of nitrogen is +5. It is the maximum oxidation number that nitrogen can exhibit. So, the only opportunity for nitrogen in HNO₃ is to decrease its oxidation number. Hence, HNO₃ can act only as an oxidant.

Question 37. Consider the reactions:

1. 6CO₂(g) + 6H₂O(l)→ C₆H₁₂O₆(aq) + 6O₂(g)

2. O₃(g) + H₂O₂(l) → H₂O(Z) + 2O₂(g)

Why it is more appropriate to write these reactions

1. 6CO₂(g) + 12H₂O(l) → C₆H₁₂O₆(aq) + 6H₂O(l) +6O₂(g)

2. O₃(g) + H₂O₂(l) → H₂O(l) +O₂(g) + O₂(g)

Also, suggest a technique to investigate the path of the above (1) and (2) redox reactions

Answer:

The reaction shown by the equation 1 takes place in the photosynthesis process. From the equation 1, it may seem that the reaction involves only the consumption of H₂O. However, if we look at the steps that are supposed to be involved in the photosynthesis reaction, it becomes evident that consumption as well as formation of H₂O takes place in the photosynthesis reaction. The proposed steps of the photosynthesis reaction are

1. Decomposition of H₂O into H₂ and O₂

12H₂O(Z)→12H₂(g) + 6O₂(g) …………………………(1)

2. Formation of C₆H₁₂0O₆ and H₂O due to reduction of CO₂(g) by H₂(g) produced in step.

6CO₂(g) + 12H₂(g)→C₆HI₂O₆(s) + 6H₂O(l)…………………………(2)

Combining equations (1) and (2) gives the complete reaction for the photosynthesis process.

6CO₂(g) + 12H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g) +6H₂O(l)…………………………(3)

Thus, equation (3) will be more appropriate for representing the photosynthesis reaction because it gives the actual stoichiometry of the reactants and the products involved in the given reaction.

From the equation 2, the source of O₂ formed in the reaction is not obvious. One may think O₂ is formed from O₃ or H₂O₂ or both O₃ and H₂O₂. The detailed steps of this reaction as shown below reveal that O₂ is formed from both O₃ and H₂O₂.

……………..(1)

Therefore, it is appropriate to represent the reaction by equation (1).

To investigate the paths of the reaction 1 and 2, we adopt the tracer technique method. In this method, we use H₂O¹⁸ instead of H₂O for reaction 1 and H₂O₂¹⁸ instead of H₂O₂ for reaction 2.

Question 38. Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a compound of a lower oxidation state is formed if the reducing agent is in excess, and a compound of a higher oxidation state is formed if the oxidizing agent is in excess. Justify this statement by giving three illustrations.
Answer:

1. The given statement can be justified from the examples of j reactions mentioned below.

The reaction of C (reductant) with O₂ (oxidant) may result in CO or CO₂ or a mixture of CO and CO₂. However, if this reaction is initiated with the excess amount of C, the only product that forms is CO. On the other hand, if O₂ is taken in excess in the reaction, only CO₂ is formed. In CO, the oxidation state of C is +2, and in CO₂, it is +4.

Thus, we see that taking an excess amount of reductant leads to the formation of a compound lower oxidation state. Conversely, a compound of a higher oxidation state is formed when the oxidant is taken in excess.

The reaction of P₄ (reductant) with Cl₂ (oxidant) results in PCl₃ when P₄ is taken in excess, while it results in PCl₅ when Cl₂ is taken in excess.

The oxidation state of PCl₃ is +3 and that in PC1₅ is +5. Thus, an excess amount of reductant produces an oil compound lower oxidation state and an excess amount of oxidant produces a compound with a higher oxidation state.

The same thing happens when Na (reductant) is reacted with O₂ (oxidant).In the presence of excess Na, the resulting compound is Na₂O, in which the oxidation state of oxygen is -2 in the presence of excess O₂, the resulting compound is Na202, in which the oxidation state of oxygen is -1. → ↑

[In this reaction, the mass of Na is 46g and that of oxygen is 64 g]

NCERT Class 11 Chemistry Redox Reactions Long Answer Solutions

Question 39. How do you count for the following observations? Although alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet In the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.

When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colorless pungent-smelling gas HC1, but if the mixture contains bromide then we get red vapour of bromine. Why?
Answer:

The reaction in acid medium:

Multiplying equation (1) by 6 and equation (2) by 5 and then adding them together, we have

The reaction in alkaline or neutral medium:

Multiply equation (3) by 2 and then adding it to equation (4), we have

Even though toluene oxidizes to benzoic acid in the presence of acidic or alkaline KMnO₄, the manufacture of benzoic acid from toluene is usually carried out by using alcoholic KMnO₄ as an oxidant.

This is because ofthe following advantages:

The use of alcoholic KMnO₄ is cost-effective because carrying out the reaction in the presence of it does not require adding either acid or alkali in the reaction medium. In a neutral medium, OH- ions are produced during the reaction.

Both KMnO₄ and toluene are soluble in alcohol and they form a homogeneous mixture. This facilitates the reaction and contributes towards speeding up the reaction.

When concentrated H₂SO₄ is added to an inorganic mixture containing chloride, HC₁, which has a pungent smell, is produced.

Question 40. Identify the substance oxidized, reduced, oxidizing agent, and reducing agent for each of the following reactions

1. \(2 \mathrm{AgBr}(s)+\mathrm{C}_6 \mathrm{H}_6 \mathrm{O}_2(a q) \rightarrow 2 \mathrm{Ag}(s)+2 \mathrm{HBr}(a q)+\mathrm{C}_6 \mathrm{H}_4 \mathrm{O}_2(a q)\)

2. \(\mathrm{HCHO}(l)+2\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}(a q)+3 \mathrm{OH}^{-}(a q)\)

→ \(2 \mathrm{Ag}(s)+\mathrm{HCOO}^{-}(a q)+4 \mathrm{NH}_3(a q)+2 \mathrm{H}_2 \mathrm{O}(l)\)

3. \(\mathrm{HCHO}(l)+2 \mathrm{Cu}^{2+}(a q)+5 \mathrm{OH}^{-}(a q) \rightarrow \mathrm{Cu}_2 \mathrm{O}(s)+\mathrm{HCOO}^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l)\)

4. \(\mathrm{N}_2 \mathrm{H}_4(l)+2 \mathrm{H}_2 \mathrm{O}_2(l) \rightarrow \mathrm{N}_2(g)+4 \mathrm{H}_2 \mathrm{O}(l)\)

5.\(\mathrm{Pb}(s)+\mathrm{PbO}_2(s)+2 \mathrm{H}_2 \mathrm{SO}_4(a q) \rightarrow 2 \mathrm{PbSO}_4(s)+2 \mathrm{H}_2 \mathrm{O}(l)\)

Answer:

Question 41. Consider the reaction:

1. 2S₂O₃(aq) + I₂(s) →S₄O₆^2-aq) + 2I^–(aq)

2. S₄O₃^2-(aq) + 2Br₂(Z) + 5H₂O(l) →  2SO₄^2-(aq) + 4Br^–(aq)+10H⁺(aq)

Answer:

The standard reduction potential for Br₂/2Br^– system is greater than that for I2/2I- system

⇒ \(E_{\mathrm{Br}_2 / 2 \mathrm{Br}}^0=1.09 \mathrm{~V}\)

And \(E_{1_2 / 21^{-}}^0=0.54 \mathrm{~V}\)

This indicates that Br₂ is a stronger oxidizing agent than I. The average oxidation number of S in

⇒ \(\mathrm{S}_2 \mathrm{O}_3^{2-}\) is +2. and that in \(\mathrm{S}_4 \mathrm{O}_6^{2-}\) is 2.5, while the oxidation number of S in SO₂– is +6.

The oxidation number per S-atom changes by 0.5 units in the reaction

⇒  \(\mathrm{S}_2 \mathrm{O}_3^{2-} \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}\) while in the reaction \(\mathrm{S}_2 \mathrm{O}_3^{2-} \rightarrow \mathrm{SO}_4^{2-}\) This change occurs by 4 units.

Being a stronger oxidizing agent, Br₂ is capable of increasing the oxidation number of S in

⇒ \(\mathrm{S}_2 \mathrm{O}_3^{2-}\) To the maximum oxidation number of 6, thereby leading to the formation of S₂O₃^2-ion.

On the other hand, I₂, being a weaker oxidizing agent, increases the oxidation number of Sin S₂O₃^2- to an oxidation number of 2.5 and results in the formation of

⇒ \(\mathrm{S}_4 \mathrm{O}_6^{2-}\) ion.

Question 42. Justify giving reactions that among halogens, fluorine is the best oxidant, and among hydrohalic compounds, hydroiodic acid is the best reductant
Answer:

The standard redox potentials (or standard reduction potentials) ofthe redox couples formed by halogens are

Are doxcouple consists of a reduced form and an oxidised form. The larger the value of E° for a redox couple, the greater the tendency of its oxidized form to get reduced and the smaller the tendency of its reduced form to oxidize. The reverse is true when the value of E° for a redox couple is small.

Combining this idea with standard electrode potentials of the redox couples given, we can infer that the tendency of oxidized forms (i.e., F₂, Cl₂, Br_2, and I₂) to get reduced or the strength of oxidizing power of the oxidized forms follows the order: F₂ > Cl₂ > Br₂ > I₂, and the tendency of reduced forms {i.e., F-, Cl-, Br- and I-) to get oxidized or the strength of reducing power ofthe reduced forms follows the order:

I¯> Br¯> Cl¯ > F¯

As the oxidizing power of F₂ is the highest among the halogens, it is capable of oxidizing other halides to the corresponding halogens. No other halogen except F₂, has this ability

⇒ \(\mathrm{F}_2(g)+2 \mathrm{Cl}^{-}(a q) \rightarrow 2 \mathrm{~F}^{-}(a q)+\mathrm{Cl}_2(g)\)

⇒ \(\mathrm{F}_2(g)+2 \mathrm{Br}^{-}(a q) \rightarrow 2 \mathrm{~F}^{-}(a q)+\mathrm{Br}_2(l)\)

⇒ \(\mathrm{F}_2(g)+2 \mathrm{I}^{-}(a q) \rightarrow 2 \mathrm{~F}^{-}(a q)+\mathrm{I}_2(s)\)

⇒ \(\mathrm{Cl}_2(g)+2 \mathrm{Br}^{-}(a q) \rightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{Br}_2(l)\)

⇒ \(\mathrm{Cl}_2(g)+2 \mathrm{I}^{-}(a q) \rightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_2(s)\)

⇒ \(\mathrm{Br}_2(l)+2 \mathrm{I}^{-}(a q) \rightarrow 2 \mathrm{Br}^{-}(a q)+\mathrm{I}_2(s)\)

Therefore, F₂ has the strongest oxidizing power among the halogens. The oxidation of a hydrohalic acid produces its halogen. The tendency of a hydrohalic acid to get oxidized or the reducing power of a hydrohalic acid is high when the halide ion of the hydrohalic acid exhibits a greater tendency to get oxidized. As the tendency ofhalide ions to get oxidized follows the order I¯ > Br¯ > Cl¯ > F-, the reducing power of hydrohalic acids will follow the order HI>HBr>HCl>HF.

The following reactions confirm this:

HI or HBr can reduce H₂SO₂ to SO₂, but HC1 or HF cannot reduce it.

⇒ \(2 \mathrm{HI}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{I}_2+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}\)

⇒ \(2 \mathrm{HBr}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Br}_2+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}\)

I^– can reduce Cu²⁺ to Cu⁺, but Br^– cannot.

⇒ \(2 \mathrm{Cu}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightarrow \mathrm{Cu}_2 \mathrm{I}_2(s)+\mathrm{I}_2(s)\)

⇒ \(\mathrm{Cu}^{2+}(a q)+2 \mathrm{Br}^{-}(a q) \rightarrow \text { No reaction }\)

Therefore, we can conclude that HI is the strongest reducing agent among the hydrohalic acids.

Question 43. Why does the following reaction occur? XeO₆^-4 (aq) + 2F^–(aq) + 6H⁺(aq) →  XeO₃(g) + F₂(g) + 3H₂O(l), What conclusion about the compound Na₄XeO₆ (of which XeO₃ is a part) can be drawn from the reaction?
Answer:

The oxidation numbers of Xe in XeO₆^-4 and XeO₃ are +8 and +6 respectively. Thus, in the reaction the oxidation number of Xe decreases, and hence XeO₆^-4– undergoes reduction and acts as an oxidising agent. On the other hand, the oxidation number of F increases from -1 to 0.

Therefore, in the reaction fluorine undergoes oxidation and hence it acts as a reductant. As the reaction is spontaneous and XeO₆^-4  oxidizes F^–, it can be concluded that Na₄XeO₆ has stronger oxidizing power than F₂.

Class 11 Chemistry Chapter 8 Long Questions Redox Reactions

Question 44. Consider the reactions:

1. H₃PO₂(aq) + 4AgNO(aq) + 2H₂O(l) → H₃PO₂(aq) + 4Ag(s) + 4HNO₃(aq)

2. H₃PO₂(aq) + 2CuSO₄(aq) + 2H₂O(Z)→ H₃PO₄(aq) + 2Cu(s) + H₂SO₄(aq)

3.  C₆H₅CHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH^–(aq) → C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃(aq) + 2H₂O(l)

4. C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^– →(aq) No change observed

What inference do you draw about the behavior of Ag⁺ and Cu²⁺ from these reactions?
Answer:

1. The reaction involves the reduction of the Ag⁺ ion to Ag and the oxidation of H₃PO₂ to H₃PO₄. Thus, in this reaction, the Ag+ ion behaves as an oxidant. It oxidizes H₃PO₂ to H₃PO₄.

2. The reaction involves the reduction of Cu²⁺ ion to Cu and the oxidation of H₃PO₂ to H₃PO₄. Thus, in this reaction, the Cu²⁺ ion behaves as an oxidant. It oxidizes H₃PO₂ to H3PO₄.

3.The reaction involves the oxidation of C₆H₅CHO to C₆H₅COOH and the reduction of [Ag(NH₃)₂]⁺ to Ag. Thus, in this reaction, the Ag⁺ ion acts as an oxidant. It oxidizes C₆H₅CHO to C₆H₅COOH.

This reaction indicates that the Cu²⁺ ion is not capable of oxidizing C₆H₅CHO. This explains why the Cu²⁺ ion is a weaker oxidizing agent than the Ag⁺ ion.

Question 45. What sorts of information can you draw from the following reaction?
Answer:

Let us first analyze whether (CN)₂ in the reaction gets oxidized or reduced or simultaneously oxidized as well as reduced. To know this the knowledge of the oxidation states of C in (CN)₂, CN, and CNO- are required.

The oxidation state of C in (CN)₂ is

⇒ \(+3\left[(\stackrel{+3-3}{(\mathrm{CN}})_2\right]\) is +2 \(\mathrm{CN}^{-}\left[\mathrm{CN}^{-}\right] \text {and }+4 \text { in } \mathrm{CNO}^{-}\left[\mathrm{C}^{+3-32} \mathrm{CN}^{-}\right]\)

Now let us consider the given equation

⇒  \((\stackrel{+3}{\mathrm{C}} \mathrm{N})_2(g)+2 \mathrm{OH}^{-}(a q) \xrightarrow[+2]{\mathrm{C}} \mathrm{N}^{-}(a q)+\stackrel{+4}{\mathrm{C}} \mathrm{NO}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

From this equation, we see that (CN)₂ reduced to the oxidation number of C reduces (+3→+2) in this change, and It gets oxidized to CNO^– because the oxidation number of C Increases (+3→+4) In this change. Thus, from this reaction, we have the following information: CD In an alkaline medium cyanogen gas dissociates into cyanide Ion (ON-) and cyanate Ion (CNO^–).

It Is a redox reaction. More particularly, it is a disproportionation reaction because (CN)₂ undergoes oxidation and reduction simultaneously. Cyanogen is a pseudohalogen. It acts as a halogen on reacting with alkalis.

Question 46. The Mn³⁺ ion Is unstable In solution and undergoes disproportionation to give Mn²⁺, MnO₂, and H⁺ ions. Write a balanced ionic equation for the reaction.
Answer:

Mn³⁺ is unstable in an aqueous medium and undergoes a disproportionation reaction, forming Mn³⁺, MnO₂ and H⁺. So, the reaction is—

Here the oxidation reaction is

⇒ \(\mathrm{Mn}^{3+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{MnO}_2(s)+4 \mathrm{H}^{+}(a q)+e \quad \cdots[1]\)

Adding equation (1) and equation (2), we have

⇒ \(2 \mathrm{Mn}^{3+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{MnO}_2(s)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}^{+}(a q)\)

This is the balanced equation for the disproportionation reaction that the Mn³⁺ ion undergoes in an aqueous medium.

Question 47. Consider the elements: Cs, Ne, I, and F

1. Identify the element that exhibits only a negative oxidation state.
2. Identify the element that exhibits only a positive oxidation state.
3. Identify the element that exhibits both positive and negative oxidation states.
4. Identify the element that exhibits neither the negative nor the positive oxidation state

Answer:

1. Being an element of the highest electronegativity, F always shows a negative oxidation state. It exhibits only a -1 oxidation state

2. Cs is an alkali metal and shows a strong electropositive character. As a result, it always shows a positive oxidation state. It exhibits only a +1 oxidation state.

3. Like other halogens, I also show a -1 oxidation state. In addition, it shows positive oxidation states, +1, +3, +5, and +7 when it forms compounds with more electronegative elements.

4. Ne is an inert element and does not tend to gain or lose electrons. As a result, it shows neither a positive oxidation state nor a negative oxidation state.

Question 48. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess chlorine is removed by treating it with sulfur dioxide. Present a balanced equation for this redox change taking place in water.
Answer:

The reaction that occurs when SO₂ is used to remove excess Cl₂ in drinking water is

⇒ \(\mathrm{Cl}_2(a q)+\mathrm{SO}_2(a q) \rightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{SO}_4^{2-}(a q)\)

Oxidation reaction: \(\stackrel{+4}{\mathrm{SO}_2}(a q) \xrightarrow{+6} \mathrm{SO}_4^{2-}(a q)\)

To balance O-atoms, we add 2H₂O to the left-hand side and 4H+ to the right-hand side.

⇒ \(\mathrm{SO}_2(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{SO}_4^{2-}(a q)+4 \mathrm{H}^{+}(a q)\) ……………….(1)

Balancing charges on both sides, we have

Reduction reaction: Cl₂(aq) + 2e→2Cl^–(aq)………………..(2)

Adding equation (1) to equation (2), we have

⇒ \(\mathrm{Cl}_2(a q)+\mathrm{SO}_2(a q)+ 2 \mathrm{H}_2 \mathrm{O}(l)\) → \(\mathrm{SO}_4^{2-}(a q)+2 \mathrm{Cl}^{-}(a q)+4 \mathrm{H}^{+}(a q)\)

This is the balanced equation for the reaction.

Question 49. Refer to the periodic table given in your book and now answer the following questions:

1. Select the possible non-metals that can show a disproportionation reaction.
2. Select three metals that can show a disproportionation reaction.

Answer:

1. Non-metals such as P4(s), Cl₂(g), and Br₂(Z) undergo disproportionation reaction.

⇒ \(\mathrm{P}_4(s)+3 \mathrm{NaOH}(a q)+3 \mathrm{H}_2 \mathrm{O}(l)\) → \(\mathrm{PH}_3(g)+3 \mathrm{NaH}_2 \mathrm{PO}_2(a q)\)

⇒ \(\mathrm{Cl}_2(g)+6 \mathrm{NaOH}(a q)(\mathrm{Hot})\) → \(5 \mathrm{NaCl}(a q)+\mathrm{NaClO}_3(a q)+3 \mathrm{H}_2 \mathrm{O}(l)
\)

⇒ \(\mathrm{Br}_2(l)+6 \mathrm{NaOH}(a q)(\mathrm{Hot})\)→ \(5 \mathrm{NaBr}(a q)+\mathrm{NaBrO}_3(a q)+3 \mathrm{H}_2 \mathrm{O}(l)\)

Three metals that can show disproportionation reactions are copper, gallium and manganese.

⇒ \(2 \stackrel{+1}{\mathrm{C}} u^{+}(a q)\rightarrow \stackrel{0}{\mathrm{Cu}}(s)+\stackrel{+2}{\mathrm{Cu}^{2+}}(a q)\)

⇒  \(3 \mathrm{Ha}^{+}(a q)\rightarrow \stackrel{+3}{\mathrm{Ga}}^{3+}(a q)+2 \stackrel{\ominus}{\mathrm{Ga}}(s)\)

⇒ \(2 \mathrm{Mn}^{3+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l)\) → \(\mathrm{MnO}_2(s)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}^{+}(a q)\)

Redox Reactions Chapter 8 NCERT Long Answer Questions Class 11

Question 50. In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen?
Answer:

Reaction:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

4 × 17 g = 68 g 5 × 32 g

= 160 g 4 × 30 g = 120 g

Therefore, 160 g O₂ is required to oxidize 68 g NH₃.

Therefore, 160 g O₂ is required to oxidize 68 g NH₃.

20g O₂ is required to oxidise \(\frac{68}{160} \times 20 \mathrm{~g}\)

= 8.5 g of NH₃

So, here O₂ is the limiting reagent. The amount of nitric oxide produced depends upon the amount of oxygen taken and not on the amount of NH3 taken

According to the above equation,

160 g O₂ produces 120 g NO

⇒ \(20 \mathrm{~g} \mathrm{O}_2 \text { produces } \frac{120}{160} \times 20 \mathrm{~g} \mathrm{NO}=15 \mathrm{~g} \mathrm{NO}\)

Thus the reaction between 10 g NH₃ and 20gO₂ produces a maximum amount of 15 g NO.

NCERT Solutions For Class 11 Chemistry Chapter 5 States Of Matter Gases And Liquids Very Short Question And Answers

Question 1 Arrange O₂, CO₂, Ar, and SO₂ gases to present a sample of air in order of their increasing pressures.
Answer:

⇒ \(d_1=\frac{P M}{R T}, d_2=\frac{P M}{4 \times 8 R T}=\frac{1}{32} \frac{P M}{R T}=\frac{d_1}{32}\)

Question 2. Why is it not possible to liquefy an ideal gas?
Answer: Because of the absence of intermolecular forces of attraction in an ideal gas, such a gas cannot be liquefied

Question 3. Why is the nib of the fountain pen split?
Answer: The split part of the nib of a fountain pen acts like a capillary tube. The ink moves towards the tip of the die nib by capillary action against gravitation through the split par.

Read and Learn More NCERT Class 11 Chemistry Very Short Answer Questions

Question 4. At a constant temperature, gas A (volume VA and pressure PA) is mixed with gas B (volume VR and pressure PB). What will the total pressure ofthe gas mixture be?
Answer:

Total pressure ofthe gas mixture \(P=\frac{P_A V_A+P_B V_B}{V_A+V_B}\)

States of Matter Chapter 5 Class 11 Very Short Questions

Question 5. For which of the following gas mixtures is Dalton’s law of partial pressures applicable? NO + O₂, CO₂ + CO , CO + O_2, CH₄ + C₂H₆,CO + H₂
Answer: Dalton’s law is applicable in case NO reacts with O₂ to form NO₂. CO reacts with O₂ to form CO₂. Hence, Dalton’s law does not apply to 1 and 3.

Question 6. In a mixture of A, B, and C gases, the mole fractions of A are 0.25 and 0.45, respectively. If the total pressure of the mixture is P, then find the partial pressure of B in the mixture.
Answer:

In the gas mixture, the mole of B =1- (0.25 + 0.45) = 0.3.

∴ The partial pressure of B in the mixture = 0.3 p.

Question 7. Why are diffusion rates of N₂O and CO₂ gases the same under identt of conditions?
Answer: N₂O and CO₂ have the same molar mass. So, diffusion rates will be equal at a certain temperature and pressure.

Question 8. At constant temperature and pressure, the rate of diffusion of H₂ gas is A/15 times. Find the value of n.
Answer:

⇒ \(\frac{r_{\mathrm{H}_2}}{r_{\mathrm{C}_n \mathrm{H}_{4 n-2}}}=\sqrt{\frac{16 n-2}{2}}=\sqrt{8 n-1}=\sqrt{15}\)

Question 9. Besides the lower layer, CO₂ is also found in the upper layer of the atmosphere, although it is heavier than O₂ or N₂—Explain.
Answer: The rate of diffusion of a gas is not influenced by the gravitational force. Hence, CO₂ diffuses throughout the atmosphere instead of residing only at the lower layer of the atmosphere.

Question 10. The molecular masses of A, B, and C are 2, 4, and 28, respectively. Arrange them according to their increasing rates of diffusion.
Answer:

Molar masses of the three gases follow the order: MA < MB < Mc So, at a certain temperature and pressure, diffusion rates will be in the order: rC<rB<rA.

NCERT Solutions Class 11 Chemistry Chapter 5 Very Short Q&A

QuestionWhichwhich type of gas molecules are the energy and translational kinetic energy equal?
Answer: For monatomic gases (He, Ne, etc.) the total kinetic energy and translational kinetic energy are equal.

Question 12. Which type of velocity does a gas molecule with average kinetic energy possess?
Answer: The average kinetic energy of a gas molecule \(=\frac{1}{2} m c_{r m s}^2.\). So, a gas molecule with average kinetic energy has the root-mean-square velocity.

Question 13. On what factors does the total kinetic energy of the molecules in a gas depend?
Answer: The total kinetic energy of the molecules in a gas depends on the absolute temperature as well as the amount of the gas.

Question 14. Between H₂ and CO₂ gas, which one has the value of compressibility factor greater than 1 at ordinary temperature and pressure?
Answer: At ordinary temperature and pressure, the compressibility factor of H₂ is greater than 1.

Question 15. Which types of intermodular forces of attraction act between the modules in liquid HF?
Answer: HF is a polar covalent molecule. Hence, the forces of attraction between HF molecules in their liquid state are dipole-dipole attraction and H -H-H-bonding.

Question 16. What do you mean by the pressure of the gas?
Answer: Since E₂>E₁, T₂>T₁. This is because, with the increase in the absolute temperature of a gas, the average kinetic energy of the gas molecules increases

Very Short Questions and Answers for Class 11 Chemistry Chapter 5

Question 17. Mention the variables and constant quantities in Charles’ law
Answer: Variables; Volume ( V) and absolute temperature (T) and Constants: Mass (m) and pressure (P) of the gas.

Question 18. What is the numerical value of N/n?[N and n are the number of gas molecules and number of moles of gas]
Answer: Number of gas molecules (N) = several moles of the gas (n) x Avogadro’s number or\(\frac{N}{n}\) = Avogadro’s number = 6.022 x1023.

Question 19. The equation of state of a real gas is P(V-b) = RT. Can this gas be liquefied?
Answer: For the given gas, the value of van der Waals constant a = 0, indicates the absence of intermolecular forces of attraction in the gas. So this gas cannot be liquefied.

Question 20. At 25°C, the vapour pressure of ethanol is 63 torrs. What does it mean?
Answer: At 25°C, the pressure exerted by eth vapour vapour in equilibrium with the liquid ethanol is 63 torrs.

Question 21. Why is the equilibrium established in the evaporation of a liquid in a closed vessel at a constant temperature called dynamic equilibrium?
Answer: This is so called because the process of evaporation and condensation does not stop at equilibrium and they keep on occurring equally and take

Question 22. Which one between water and ethanol has greater surface tension at a particular temperature?
Answer: As intermolecular forces of attraction are stronger in water than in ethanol, the surface tension of water is greater than that of ethanol.

Question 23. Plot density vs pressure for a fixed mass of an ideal gas at a constant temperature.
Answer: Since docs for a given mass of gas at a constant temperature.

Chapter 5 States of Matter Very Short Answer Solutions

Question 24. What are the molar volumes of nitrogen and argon gases at 273.15K temperature and 1 atm pressure? [consider both the gases behave ideally]
Answer: At 273.15K and late pressure, both nitrogen and argon behave ideally. Hence, at this temperature and pressure, the molar volume of each of them will be 22.4L

Question 25. What is the value of the surface tension of a liquid at its critical temperature?
Answer: The surface tension of a liquid at critical temperature is zero because the surface of separation between the liquid and vapour disappears at this temperature.

Question 26. Among the following properties of a liquid, which one increases with the increase in temperature? Surface tension, viscosity and vapour pressure
Answer: Vapour pressure increases with an increase in temperature

Question 27. At a given temperature, the viscosity of liquid A is greater than that of liquid B. Which of these two liquids has stronger intermolecular forces of attraction?
Answer: The stronger the intermolecular forces of attraction of a liquid, the higher the viscosity will be. Therefore, the inter¬ molecular forces of attraction of A will be greater than B

Question 28. How does the average velocity or the root mean square velocity of gas molecules depend on temperature and pressure?
Answer: The average kinetic energy (c) and the root mean square velocity (c₂) of gas molecules are given by \(\bar{c}=\sqrt{\frac{8 R T}{M}}\) and \(c_{r m s}=\sqrt{\frac{3 R T}{M}}\)

NCERT Class 11 Chemistry Chapter 5 Gases and Liquids Short Q&A

Question 29. Why does oil spread over water when it is poured over water
Answer: The surface tension of water is greater than that of oil. Also, the density of oil is less than that of water. When oil is poured over water, the higher surface tension of water causes oil to spread over water.

Question 30. The vapour pressures of benzene and water at 50°C are 271 and 92.5 torr, respectively. Which one would you expect to have stronger intermolecular forces of attraction?
Answer: A liquid with strong intermolecular forces of attraction has low vapour pressure. Thus, intermolecular forces of attraction are stronger in water.

Question 31. Give two examples where capillary action occurs.
Answer: Due to capillary action, water from the soil reaches the leaves of a tree through its sWaterwater comes out through the pores of a clay pot and evaporates. As a result, the water in the pot gets cooled.

Question 32. Comment on the validity of Boyle’s law for the following reaction: N₃O₄(g) 2NO₂(g)
Answer: The number of molecules of N₂O₄ and NO₂ varies with the pressure at constant temperature. Hence, the mass ofthe gas mixture does not remain constant. Thus, Boyle’s law is not applicable here.

Question 33. The normal boiling point of diethyl ether is 34.6°C What will be its vapour pressure at this temperature?
Answer: At the normal boiling point of a liquid, the vapour pressure of the liquid is equal to the atmospheric pressure. Therefore, the vapour pressure of diethyl ether is at 34. Temperatureature is 1 atm or 760 torr.

Question 34. The normal boiling points of ethanol and benzene are 78.3°C and 80°C, respectively. Is the vapour pressure of ethanol lower than, greater than or equal to the vapour pressure of benzene?
Answer:

At the normal boiling point of a liquid, the vapour pressure of the liquid is equal to the atmospheric pressure. Therefore, the vapour pressure of ethanol at 78.3°C and that of benzene at 80°C are the same and equal to 1 atm.

Question 35. The normal boiling points of two gases A and B are -150°C and -18°C, respectively. Which one of the two gases will behave more like an ideal gas at STP?
Answer: The very low boiling point of A implies the very weak intermolecular forces of attraction in gas A. Hence, gas A will show more ideal behaviour at STP.

Very Short Answer Solutions for States of Matter Class 11

NCERT Solutions For Class 11 Chemistry Chapter 5 States Of Matter Gases And Liquids Fill In The Blanks

Question 1. 0.1 atm = _________________  Pa =___________________
Answer:

1. 1.013 × 10⁴

2. 76

Question 2. The plot of pressure (P) versus temperature (T) for an  ideal gas will be a straight line passing through the origin _____________  and ______________________ remain constant
Answer: Mass and volume

Question 3. At a given temperature and pressure, the density of CO₂ gas is greater than N2 gas, because  it the ___________________ greater than CO₂
Answer:  Molecular mass

Question 4. The density of CO₂ at ___________________  standard atmospheric pressure is 1.788 g-L^-1
Answer: 27°C

Gases and Liquids Chapter 5 Class 11 Very Short Answer Questions

Question 5. The velocity of the gas molecules having average kinetic  energy is called  ___________________ 
Answer: Root mean square velocity

Question 6. At a particular temperature, the average kinetic energies of the S02 molecule and H2 molecule are___________________ 
Answer: Equal

Question 7. The most probable velocity of He atoms at 127°C is_____________
Answer: 40.78 m.s^-1

Question 8. The volume of 1 mole of a gas at STP is smaller than 22.4 L. The compressibility factor of that gas at STP is
___________________  one.
Answer: Less than

Question 9. In van der Waals equation__ _________________ effective size ofthe gas molecules.
Answer: ‘b’

NCERT Class 11 Chemistry Chapter 5 States of Matter Short Q&A

Question 10. At low pressure and a particular temperature, the pressure of 1 mol of an ideal gas is ___________________  a real gas.
Answer: Greater than

Question 11. Due to the ___________________  becomes spherical.
Answer: Surface tension

Question 12. The aqueous solution of soaps can spread like a thin film because ___________________  the surface water tension of the aqueous solution of soap is
Answer: Less

Question 13. At a particular temperature, the viscosity coefficient of ethyl alcohol is  ___________________ than dimethyl ether
Answer: Greater.

NCERT Class 11 Chemistry Equilibrium Long Question And Answers

Question 1. Equilibria involving physical and chemical changes are dynamic in nature—why?
Answer:

In a system, if two processes occur simultaneously at the same rate, then the system is said to be in a state of dynamic equilibrium. When a chemical reaction remains at equilibrium, the forward and the backward reactions occur simultaneously at the same rate.

Due to this, the equilibrium established in a chemical reaction is regarded as a dynamic equilibrium.

Similarly, when a physical process remains at equilibrium, the two opposite processes occur simultaneously at the same rate.

Read and learn More NCERT Class 11 Chemistry

For this reason, a physical equilibrium is also dynamic. For example, at equilibrium established during the evaporation of a liquid in a closed container, the evaporation of the liquid and the condensation of its vapor take place simultaneously at the same rate.

Long Answer Questions for Class 11 Chemistry Equilibrium

Question 2. The equilibrium established in the dissolution of a solid in a liquid or a gas in a liquid is dynamic. Explain
Answer:

If a solid solute is continuously dissolved in a suitable solvent, eventually a saturated solution of the solute is obtained. In this solution, an equilibrium exists between dissolved solute and undissolved solute.

At this equilibrium, the solute particles from undissolved solid solute get dissolved at the same rate as the solute particles from the solution get deposited on the surface of undissolved solid solute. Since these two processes occur at the same rate, the equilibrium established between dissolved solute and undissolved solute is dynamic.

When a gas is brought in contact with a suitable solvent, the gas keeps on dissolving in the solvent, and eventually, a saturated solution of the gas is formed. At this state, the gas over the liquid surface remains in equilibrium with the dissolved gas in the liquid.

In this state of equilibrium, the gas molecules from the gas phase enter the liquid phase at the same rate as the dissolved gas molecules from the liquid enter the gas phase. Two opposite processes therefore occur at the same rate. For this reason, the equilibrium that forms when a gas remains in equilibrium with its saturated solution is dynamic.

Question 3. Write the expressions of K_c and K_p for the following reactions:

1. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

2. 2BrF₅(g)⇌  Br₂(g) + 5F₂(g)

3. 3O₂(g) ⇌ 2O₃(g)

4. 4HCl(g) + O₂(g) ⇌ 2Cl₂(g) + 2H₂O(g)

5. P₄(g) + 3O₂(g) ⇌ P₄O₆(s)

6. CO(g) + 2H₂(g)⇌  CH₃OH(l)

Answer:

1.  \(K_c=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]} ; K_p=\frac{p_{\mathrm{SO}_3}^2}{p_{\mathrm{SO}_2}^2 \times p_{\mathrm{O}_2}}\)

2.  \(K_c=\frac{\left[\mathrm{Br}_2\right]\left[\mathrm{F}_2\right]^5}{\left[\mathrm{BrF}_5\right]^2} ; K_p=\frac{p_{\mathrm{Br}_2} \times p_{\mathrm{F}_2}^5}{p_{\mathrm{BrF}_5}^2}\)

3.  \(K_c=\frac{\left[\mathrm{O}_3\right]^2}{\left[\mathrm{O}_2\right]^3} ; K_p=\frac{p_{\mathrm{O}_3}^2}{p_{\mathrm{O}_2}^3}\)

4.  \(K_c=\frac{\left[\mathrm{Cl}_2\right]^2\left[\mathrm{H}_2 \mathrm{O}\right]^2}{[\mathrm{HCl}]^4\left[\mathrm{O}_2\right]} ; K_p=\frac{p_{\mathrm{Cl}_2}^2 \times p_{\mathrm{H}_2 \mathrm{O}}^2}{p_{\mathrm{HCl}}^4 \times p_{\mathrm{O}_2}}\)

5.  \(K_c=\frac{\left[\mathrm{P}_4 \mathrm{O}_6(s)\right]}{\left[\mathrm{P}_4\right]\left[\mathrm{O}_2\right]^3}=\frac{1}{\left[\mathrm{P}_4\right]\left[\mathrm{O}_2\right]^3}\)

∵ [P₄O₆(s)]=1]

⇒  \(K_p=\frac{1}{p_{\mathrm{P}_4} \times p_{\mathrm{O}_2}^3}\)

6.  \(K_c=\frac{\left[\mathrm{CH}_3 \mathrm{OH}(l)\right]}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^2}=\frac{1}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^2}\)

∵ [CH₃OH(l)]=1

⇒ \(K_p=\frac{1}{p_{\mathrm{CO}} \times p_{\mathrm{H}_2}^2}\)

Question 4. Write the expression of K_c the following reactions:

1. 2Ag⁺(aq) + Cu(s) ⇌ Cu²⁺(aq) + 2Ag(s)

2. NH₄NO₃(s) ⇌ N₂O(g) + 2H₂O(g)

3. Au⁺(aq) + 2CN^–(aq) ⇌ [Au(CN)₂]^–(aq)

4. 3Cu(s) + 2NO₃^–(aq) + 8H⁺(aq) ⇌ 3Cu²⁺(aq) + 2NO(g) + 4H₂O(l)

5. Cl₂(g) + 2Br^–(aq) ⇌ Br₂(l) + 2Cl^–(aq)

Answer:

1.  \(K_c=\frac{\left[\mathrm{Cu}^{2+}(a q)\right][\mathrm{Ag}(s)]^2}{\left[\mathrm{Ag}^{+}(a q)\right]^2[\mathrm{Cu}(s)]}=\frac{\left[\mathrm{Cu}^{2+}(a q)\right]}{\left[\mathrm{Ag}^{+}(a q)\right]^2}\)

∵  [Ag(s)]=1 and [Cu(s)]=1

2.  \(K_c=\frac{\left[\mathrm{N}_2 \mathrm{O}(g)\right]\left[\mathrm{H}_2 \mathrm{O}(g)\right]^2}{\left[\mathrm{NH}_4 \mathrm{NO}_3(s)\right]}\)

=\(\left[\mathrm{N}_2 \mathrm{O}(g)\right]\left[\mathrm{H}_2 \mathrm{O}(g)\right]^2\)

∵ [NH₄NO₃(s)]=1

3. \(K_c=\frac{\left[\mathrm{Au}(\mathrm{CN})_2^{-}(a q)\right]}{\left[\mathrm{Au}^{+}(a q)\right]\left[\mathrm{CN}^{-}(a q)\right]^2}\)

4.  \(K_c=\frac{\left[\mathrm{Cu}^{2+}(a q)\right]^3[\mathrm{NO}(g)]^2\left[\mathrm{H}_2 \mathrm{O}(l)\right]^4}{[\mathrm{Cu}(s)]^3\left[\mathrm{NO}_3^{-}(a q)\right]^2\left[\mathrm{H}^{+}(a q)\right]^8}\)

⇒ \(\frac{\left[\mathrm{Cu}^{2+}(a q)\right]^3[\mathrm{NO}(g)]^2}{\left[\mathrm{NO}_3^{-}(a q)\right]^2\left[\mathrm{H}^{+}(a q)\right]^8}\)

∵ Cu(s)=1 and H₂O(l)]=1

5.  \(K_c=\frac{\left[\mathrm{Br}_2(l)\right]\left[\mathrm{Cl}^{-}(a q)\right]^2}{\left[\mathrm{Cl}_2(g)\right]\left[\mathrm{Br}^{-}(a q)\right]^2}=\frac{\left[\mathrm{Cl}^{-}(a q)\right]^2}{\left[\mathrm{Cl}_2(g)\right]\left[\mathrm{Br}^{-}(a q)\right]^2}\)

 ∵ [Br₂(l)]=1

NCERT Solutions Class 11 Chemistry Equilibrium Long Questions & Answers

Question 5. Establish the relation between Kp and Kc for the following: aA + bB₂⇌ IL + mM [where the terms have their significance]
Answer:

⇒  \(K_c=\frac{[L]^l \times[M]^m}{[A]^a \times[B]^b}\)

Where [A],[B],[L] and [M] are Molar Concentrations Of A, B, L, and M At Equilibrium

⇒ \(K_p=\frac{p_L^l \times p_M^m}{p_A^a \times p_B^b}; \text { where } p_A, p_B, p_L \text { and } p_M\)

Partial pressures of A, B, L, and M respectively at equilibrium. Now, p_A = [A]RT, p_B = [B]RT, p_L = [L]RT and P_M = [M]RT.

Putting the values of p_A, p_B, p_L, and PM in the expression of, K_p we have—

⇒ \(K_p=\frac{[L]^l \times[M]^m}{[A]^a \times[B]^b} \times(R T)^{(l+m)-(a+b)}\)

Or \(K_p=K_c \times(R T)^{(l+m)-(a+b)}\)

Question 6. Find the relation between K_p &K_c for the given reactions:

1. NH₃(g) +HCl(g) ⇌NH₄Cl(s)

2. C(s) + CO₂(g) + 2Cl₂(g)⇌2COCl₂(g)

3. ½N₂(g) + O₂(g) ⇌NO₂(g)

4. C(s) + H₂O(g) ⇌CO(g) + H₂(g)

5. Fe(s) + H₂O(g)⇌ FeO(s) + H₂(g)

6. CH₄(g) + H₂O(l) ⇌CO(g) + 3H₂(g)

7. CO(g) + 2H₂(g)⇌ CH₃OH(Z)

Answer:

1. \(\Delta n=0-(1+1)=-2 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-2}\)

2. \(\Delta n=2-(1+2)=-1 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-1}\)

3.  \(\Delta n=1-\left(\frac{1}{2}+1\right)=\left(-\frac{1}{2}\right) ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-\frac{1}{2}}\)

4.  \(\Delta n=(1+1)-1=+1 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)\)

5.  \(\Delta n=1-1=0 ; K_p=K_c(R T)^{\Delta n}=K_c\)

6.  \(\Delta n=(1+3)-1=+3 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^3\)

7. \(\Delta n=0-(1+2)=(-3) ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-3}\)

Question 7. At a constant temperature, the equilibrium constant of the reaction

1. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is K, and that of the reaction

2. ½ N₂(g) + H₂(g) ⇌ NH₃(g) is R.

Explain this difference in K values even though the reactants and products in both the reactions are same.
Answer:

The value of the equilibrium constant of a reaction depends upon the mode of writing the balanced equation of the reaction.

1. N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Equilibrium constant

⇒ \(K=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}\)

2. \(\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g}) ;\)

Equilibrium constant

⇒ \(K_1=\frac{\left[\mathrm{NH}_3\right]}{\left[\mathrm{N}_2\right]^{1 / 2} \times\left[\mathrm{H}_2\right]^{3 / 2}}\)

∴ \(K_1^2=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}\) = K

⇒ \(K_1=\sqrt{K}\)

In both reactions 1 and 2 reactants and products are the same, but in the equations of these two reactions, the stoichiometric coefficients of reactants and products are different, resulting in different values of the corresponding equilibrium constant.

Question 8. If the value of the equilibrium constant of the reaction 2SO₂(g) + O₂(g)→2SO₃(g) is K, then what will be the lues of equilibrium constants of the following reactions?

1. 4SO₂(g) + 2O₂(g) ⇌4SO₃(g)
2. 2SO₃(g) ⇌ 2SO₂(g) + O₂(g)
3. SO₂(g) + ½O₂(g) ⇌ SO₃(g)
4. SO₃(g) ⇌ SO₂(g) + ½O₂(g)

Answer:

⇒ \(2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) 2 \mathrm{SO}_3(g) ; K=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}\)

1.  \(\text { (I) } 4 \mathrm{SO}_2(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g})\rightleftharpoons 4 \mathrm{SO}_3(\mathrm{~g})\)

Equilibrium constant = \(\frac{\left[\mathrm{SO}_3\right]^4}{\left[\mathrm{SO}_2\right]^4\left[\mathrm{O}_2\right]^2}=K^2\)

2.  2SO₃(g) ⇌ 2SO₂(g) + O₂(g)

Equilibrium constant =\(\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}=\frac{1}{K}\)

3.  SO(g)+½O₂(g)⇌ SO₃(g);

Equilibrium constant = \(\frac{\left[\mathrm{SO}_3\right]}{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{1 / 2}}=\sqrt{\mathrm{K}}\)

4.  SO₃(g) ⇌ SO₂(g) +½ O(g)

Equilibrium constant = [SO₂][O₂]^½/[SO₃]

= \(\frac{1}{\sqrt{K}}\)

Equilibrium Chapter 7 Long Answer Questions Class 11

Question 9. A few reactions and their equilibrium constants are given:

1. A ⇌ B + C; K_c = 2;
2. C ⇌ B +D; K_c  = 3
3. D ⇌ B +E; K_c  = 4.

Find the equilibrium constant ofthe reaction, A ⇌3B +E

Answer:

1. \(K_c=2=\frac{|B|[C \mid}{|A|}\)

2. \(K_c=3=\frac{[B \| D]}{[C]}\)

3. \(K_c=4=\frac{[B][E]}{[D]}\)

For the reaction A ⇌ 3B + E

Equilibrium constant K_c=  \(\frac{[B]^3[E]}{[A]}\) ……………………..(1)

Multiplying equilibrium constants of reactions (1),(2)and (30), we have 2× 3× 4

= \(\frac{[B \mid[C]}{\mid A]} \times \frac{[B][D]}{[C]} \times \frac{[B] \mid E]}{[D]}\)

Or,  24 = \(\frac{[B]^3[E]}{[A]}\)…………………….(24)

From the equations (1) and (2), we get  K_c =  24.

Therefore, the equilibrium constant for the reaction A ⇌ 3B + E is 24

Question 10. If the values of Kc for the reactions, A + 2B ⇌ C and C ⇌ 2D are 2 and 4, respectively, then what will be the value of  K_c for the reaction, 2D ⇌ A + 2B

Answer:

A + 2B ⇌ C; K_c = 2 = \(\frac{[C]}{[A][B]^2}\) ………………(1)

C ⇌ 2D ; K_c = 4 =  \(\frac{[D]^2}{[C]}\) ………………(2)

Adding equations 11 1 and [2), A + 26 s=± 2D ………………(3)

The equilibrium constant for equation (3)= Product of The equilibrium constants for reactions (1) and (2)

⇒ \(=\frac{[D]^2}{[A][B]^2}=2 \times 4=8\)

The equilibrium constant For the reaction 2D ⇒ A + 2B, the equilibrium constant = reciprocal of the equilibrium constant for reaction (3)

⇒  \(\frac{1}{8}=0.125\)

Question 11. For the reaction A + B C + D, the equilibrium constant is K. What would be the value of the reaction quotient (Q) when the reaction just starts and when it reaches equilibrium?
Answer:

For the given reaction, the equilibrium constant,

⇒ \(K=\frac{[C]_{e q} \times[D]_{e q}}{[A]_{e q} \times[B]_{e q}} \text { where }[A]_{e q},[B]_{e q},[C]_{e q}\)

[D]_eq are the equilibrium concentrations of A, B, C, and D respectively.

For the given reaction, the reaction quotient (Q) at any moment during the reaction.

⇒ \(Q=\frac{[C] \times[D]}{[A] \times[B]} ; \text { where }[A],[B],[C] \text { and }[D]\)

Are the molar concentrations of.A, B, C, and D respectively at the moment under consideration.

1. Initially: [C] = 0, [D] = 0. Therefore, Q = 0.

2.  At The Equilibrium state;

⇒ \([A]=[A]_{e q},[B]=[B]_{e q}\)

⇒ \([C]=[C]_{e q} \text { and }[D]=[D]_{e q}\)

Hence, Q = K

Question 12. Consider the reaction, A(g) + 2B(g) 2D(g) + 3E(g) + heat, and state how the following changes at equilibrium will affect the yield of the product, D(g)— Temperature is increased, Pressure is increased at a constant temperature, Some amount of E (g) is removed from the reaction system at constant temperature and volume, and Some amount of D(g) is added to the reaction system, keeping temperature and volume constant.
Answer:

The reaction is exothermic. So, the increasing temperature at the equilibrium of the reaction will cause the equilibrium to shift to tire left. This decreases the yield of D(g). As a result, the concentration of D(g) at the new equilibrium will be lower than that at the original equilibrium.

The reaction involves an increase in the number of gas molecules [Δn = (2 + 3)-(l + 2) =+2] in the forward direction. Therefore, at a constant temperature, if the pressure is increased at the equilibrium of the reaction, then, according to Le Chatelier’s principle, the equilibrium will shift to the left. As a result, the yield of D(g) will decrease and the concentration of D(g) at the new equilibrium will be lower than that at the original equilibrium.

At constant temperature and volume, if some amount of E(g) [product] is removed from the reaction system, then, according to Le Chatelier’s principle, the equilibrium of the reaction will shift to the right. As a result, the yield of D(g) will increase, and the concentration of D(g) will increase. Thus, the concentration of D(g) at the new equilibrium will be more than that at the original equilibrium.

At constant temperature and volume, if some amount of D(g) [product] is added to the reaction system at equilibrium, then, according to Le Chatelier’s principle, the equilibrium will shift to the left. As a result, the yield of D(g) will decrease. However, the concentration of D(g) at the new equilibrium will be greater than that at the original equilibrium.

Question 13. Consider the following reactions; What will be the effect on the following if the temperature is increased?

1. The equilibrium constant
2. The position of equilibrium
3. The yield of products.

Answer:

1. The reaction is endothermic as ΔH > 0.

As the reaction is endothermic, the value of its equilibrium constant will increase with the rise in temperature

According to Le Chatelier’s principle, a rise in Na₂CO₃ is a salt of strong base NaOH and weak acid H₂CO₂.

In its aqueous solution, Na₂CO₃ dissociates completely and forms Na⁺ and CO₃ ions. In water, COl- is a stronger base than HaO, and hence it reacts with water, forming OH^– and HCO₂ ions. the temperature at the equilibrium of an endothermic.

Reaction shifts the equilibrium to the right. As a result, the yields of the products increase.

2. The reaction is exothermic as ΔH < 0.

As the reaction is exothermic, the value of its equilibrium constant will decrease with the temperature rise. According to Le Chatelier’s principle, increasing temperature at the equilibrium of an exothermic reaction shifts the equilibrium to the left. Consequently, the yields of the products decrease.

Class 11 Chemistry Equilibrium Long Answer Solutions

Question 14. For these reactions, predict the effect on the following if pressure is increased at equilibrium Position of equilibrium, Yield of the products.
Answer:

The reaction involves no change in volume. Therefore, pressure has no effect either on the position of the equilibrium or on the yield of the product.

The reaction involves a decrease in volume in the forward direction. So, at a constant temperature increasing pressure at the equilibrium of the reaction favors the shifting of equilibrium to the right, thereby increasing the yield of the product.

The reaction involves a decrease in volume in the reverse direction. So, at constant temperature, increasing pressure at the equilibrium of the reaction shifts the equilibrium to the left, thereby decreasing the yield ofthe product

Question 15. At constant temperature, in a closed vessel, the following equilibrium is established during the decomposition of

NH₄Cl(S); NH₄Cl(s)⇌NH₃(g) + HCl(g).

If pressure of. the equilibrium mixture =P and value of equilibrium =K , then show that

⇒ \(P=2 \sqrt{K_p} \text {. }\)

Answer 

Total number of moles at equilibrium = x + x = 2x

∴ \(p_{\mathrm{NH}_3}=\left(\frac{x}{2 x}\right) P ; p_{\mathrm{HCl}}=\left(\frac{x}{2 x}\right) P\)

∴ \(K_p=p_{\mathrm{NH}_3} \times p_{\mathrm{HCl}}=\left(\frac{x}{2 x}\right) P \times\left(\frac{x}{2 x}\right) P=\frac{p^2}{4}\)

∴ \(K_p=\frac{P^2}{4} \quad \text { or, } P^2=4 K_p\)

∴ \(P=2 \sqrt{K_p}\)

Question 16. A weak tribasic acid, H3A, in its aqueous solution, ionizes in the following three steps: 

1. H₃A(aq) + H₂O(l) ⇌ H₃O+(aq)+ H₂A^–(aq)
2. H₂A^–(aq) + H₂O(l) ⇌ H₃O+(aq) + HA^2-(aq)
3. HA^2-(aq) + H₂O(l) ⇌ H₃O+(aq) + A^3-(aq)

If the ionization constants of the steps are K₁, K₂, and K₃ respectively, then determine the overall ionization constant of HgA.

Answer:

Step – 1:  \(K_1=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{H}_2 \mathrm{~A}^{-}\right]}{\left[\mathrm{H}_3 \mathrm{~A}\right]} \text {; }\)

Step – 2: \( K_2=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{HA}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{~A}^{-}\right]}\)

Step – 3: \(K_3=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{3-}\right]}{\left[\mathrm{HA}^{2-}\right]}\)

The equation for the overall ionisation of H₃A is:

⇒ \(\mathrm{H}_3 \mathrm{~A}(a q)+3 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{A}^{3-}(a q)+3 \mathrm{H}_3 \mathrm{O}^{+}(a q)\)

If the overall ionization constant of H₃A is K, then

⇒ \(K=\frac{\left[\mathrm{A}^{3-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]^3}{\left[\mathrm{H}_3 \mathrm{~A}\right]}\)

Multiplying ionization constants of steps 1,2 and 3

We have, \(K_1 \times K_2 \times K_3=\frac{\left[\mathrm{A}^{3-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]^3}{\left[\mathrm{H}_3 \mathrm{~A}\right]}\)

Therefore, K = K₁ × K₂ × K₃

Question 17. Between 0.1(M) and 0.01(M) acetic acid solutions which one will have a higher pH, and why?
Answer:

In an aqueous solution of a weak monoprotic add

Example CH₃COOH )

⇒  \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\sqrt{c \times K_a}\)

Where c = molar concentration of the solution and Ka = ionization constant of the weak acid.

Therefore, in 0.1(M) CH₃COOH solution,

⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\sqrt{0.1 \times K_a} \mathrm{~mol} \cdot \mathrm{L}^{-1}\) and in 0.01(M) CHgCOOH solution,

[H₃O⁺] \(\sqrt{0.01 \times K_a} \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

Hence, the concentration of H₃O⁺ ions in 0.1(M) CH₃COOH solution will be more than that in 0.01(M) CH₆COOH solution.

Therefore, the pH of 0.01 (M) CH₃COOH solution will be greater than that of O.l (M) CH₃COOH solution.

Long Answer Solutions for NCERT Class 11 Chemistry Equilibrium

Question 18. What color change does a blue or red litmus paper exhibit when it is put separately in each of the following aqueous solutions? CH₃COONa₂CH₃COONH₄, NH₄Cl
Answer:

An aqueous solution of CH₃COONa is alkaline because in solution CH₃COO^– ions resulting from the complete dissociation of CH₃COONa undergo hydrolysis, thereby increasing the concentration of OH^–. So, if a red litmus paper is dipped into this solution, it will turn blue

In its aqueous solution, CH₃COONH₄ dissociates completely forming NH⁺(aqr) and CH₃COO^–(aq) ions. Both these ions undergo hydrolysis in water. The important fact is that in water the strength of CH₃COO^– ion as a base and the strength of NH₂ ion as an acid are the same. Consequently, the concentration of OH^– ions produced due to hydrolysis of CH₃COO^– ions:

⇒ \(\left[\mathrm{CH}_3 \mathrm{COO}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(a q)+\mathrm{OH}^{-}(a q)\right]\)

Almost the same as the concentration of H₃O⁺ ions produced due to hydrolysis of NH^– ions:

⇒ \(\left[\mathrm{NH}_4^{+}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_3(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)\right]\)

As a result, an aqueous solution of CH₃COONH₄ is neutral. So, if a litmus paper is dipped into this solution, it will not exhibit any color charge.

In its aqueous solution, NH₄Cl dissociates completely forming NH₂(aq) and Cl^–(aq). NH₂ ions so formed undergo hydrolysis and thereby increase the concentration of H₃O⁺ ions in the solution. As a result, an aqueous solution of NH₄Cl is acidic. Therefore, if a blue litmus paper is dipped into this solution, its color will change to red.

Question 19. Arrange the following aqueous solutions in order of their increasing pH values: Na₂CO₃, CH₃COONH₄, and CuSO₄.
Answer:

Na₂CO₃ is a salt of strong base NaOH and weak acid H₂CO₂. In its aqueous solution, Na₂CO₃ dissociates completely and forms Na⁺ and CO₃^2- ions. In water, CO₃^2- is a stronger base than HaO, and hence it reacts with water, forming OH^– and HCO^–₃ ions.

This results in an increase in the concentration of OH ions in the solution of Na₂CO₃ and makes the solution alkaline.

In its aqueous solution, CuSO₄ dissociates completely, forming [Cu(H₂O)₆]₂⁺ and SO^– ions. SO^–ion, being a conjugate base of strong acid H₂SO₄, is a very weak base and cannot react with water.

In [Cu(H₂O)g]²⁺, the charge density of the Cu²⁺ ion is very high, and consequently, H₂O molecules bonded to the Cu²⁺ ion get polarized.

This causes the weakening of the O — H bonds in the H₂O molecule. Consequently, the O — H bond gets ionized easily, thereby forming an H+ ion. This H+ ion is then accepted by H₂O, which forms the H₃O⁺ ion.

⇒ [Cu(H₂O)6]²⁺(aq) + H₂O(l) ⇌ [Cu(H₂O)₅OH]⁺(aq) + H₃O⁺(aq)

Therefore, the hydrolysis of CuSO₄ in its solution increases the concentration of H₃O⁺ ions, thereby making the solution acidic. So, the increasing order of pH values of the given aqueous solutions: CuSO₄ < CH₃COONH₄ < Na₂CO₃

Question 20. Which of the following are buffer solutions? Give reasons:

1. 100 mL 0.1 (M)NH₃ + 50 mL 0.1(M) HCI
2. 100 mL 0.1 (M)CH₃COOH + 50 m L 0.2 (M) NaOH
3. 100 mL 0.1 (M) CH₃COOH +100 mL 0.15 (M) NaOH
4. 100 mL 0.1 (M) CH₃COONa +25mL 0.1(M) HCI
5. 50 mL 0.2 (M) NH₄Cl + 50 mL 0.1 (M) NaOH

Answer:

1. 100 mL of 0.1(M) NH₃ = O.Olmol of NH₃ and 50mL of 0.1(M) HCI = 0.005mol of HCl. In the mixed solution, 0.005 mol of HCl reacts completely with the same amount of NH₃, forming 0.005 mol of NH₄Cl, and 0.005 mol of NH₃ remains in the solution. Therefore, the mixed solution contains weak base NH₃ and its salt NH₄Cl. Hence, it is a buffer solution.

2. 100mL of 0.1(M) CH₃COOH = 0.01 mol of CH₃COOH and 50mL of 0.2 (M) NaOH = 0.01 mol of NaOH. In the mixed solution, 0.01 mol of CH₃COOH reacts completely with 0.01 mol of NaOH, forming 0.01 mol of CH₃COONa. Therefore, the mixed solution is a solution of CH₃COONa. Hence, it is not a buffer solution

3.100mL of 0.1 (M) CH₃COOH = 0.01 mol of CH₃COOH and 100mL of 0.15(M) NaOH = 0.015 mol of NaOH. In the mixed solution, 0.01 mol CH₃COOH reacts completely with 0.01 mol of NaOH, forming 0.01 mol of CH₃COONa, and 0.005 mol of NaOH remains in the solution. Therefore, the mixed solution is a solution of NaOH. Hence, it cannot be a buffer solution.

4. 100mL of 0.1(M) CH₃COONa = O.Olmol of CH₃COONa and 25mL of 0.1(M) HCI = 0.0025 mol of HCI. In the mixed solution, 0.0025 mol of HCl reacts completely with the same amount of CH₃COONa, forming CH₃COOH and NaCl.

The number of moles of CH₃COONa remaining in the solution is (0.01- 0.0025) = 7.5 × 10^-3 mol. Therefore, the mixed solution contains CH₃COOH and its salt CH₃COONa. Hence, it is a buffer solution.

5. 50mL of 0.2 (M) NH₄Cl EE 0.01 mol of NH₄Cl and 50mL of 0.1 (M) NaOH = 0.005 mol of NaOH . In the mixed solution, 0.005mol of NaOH reacts completely with the same amount of NH₄Cl, thereby forming NH₃ and NaCl. The number of moles of NH₄Cl left behind in the solution is (0.01-0.005) = 0.005 mol. Therefore, the mixed solution contains NH₃ and NH₄Cl. Hence, it is a buffer solution.

Equilibrium Chapter 7 NCERT Long Answer Questions

Question 21. The buffer capacity of a buffer solution consisting of a weak acid, HA, and its salt, NaA becomes maximum when its pH is 5.0. At this pH, what is the relation between the molar concentrations of HA and NaA? Abo finds the value of a of HA.
Answer:

In the case of a buffer solution consisting of a weak acid and its salt, the buffer capacity of the solution is maximum when pH = pK_a.

It is given that the buffer capacity of the given buffer is maximum when its pH = 5. Therefore, the pK_a ofthe weak acid present in the buffer is 5.

Now, the pH of a buffer solution consisting of a weak acid and its salt is given by

⇒  \(p H=p K_a+\log \frac{[\text { salt }]}{[\text { acid }]}\)

Since, \(p H=p K_a=5, \log \frac{[\text { salt }]}{[\text { acid }]}=0\) , [salt] = [acid]

∴ [NaA]=[HA]

Question 22. At a given temperature, if the solubility product and the
solubility of a sparingly soluble salt M₂X₃ are Ksp and S, respectively, then prove that S \(S=\left(\frac{K_{s p}}{108}\right)^{1 / 5}.\)
Answer:

In its saturated solution, M₂X₃ ionizes partially, forming M₃ (aq) and X₂ (aq) ions. Eventually, the following equilibrium is established.

⇒ \(\mathrm{M}_2 \mathrm{X}_3(s) \rightleftharpoons 2 \mathrm{M}^{3+}(a q)+3 \mathrm{X}^{2-}(a q)\)

If the solubility of M₂X₃ in its saturated solution is S mol.L^-1 , then in the solution [M³⁺] = 2S mol.L^-1 and [X^2-] = 3S mol.L^-1

Therefore, the solubility product of M₂X₃

⇒ \(K_{s p}=\left[\mathrm{M}^{3+}\right]^2\left[\mathrm{X}^{2-}\right]^3=(2 S)^2(3 S)^3=108 S^5\)

∴ \(S=\left(\frac{K_{s p}}{108}\right)^{\frac{1}{5}}\)

Question 23. Find out the equilibrium constant for the reaction, XeO₄(g) + 2HF(g)⇌ XeO₃F₂(g) + HaO(g) Consider K₁ as the equilibrium constant for the reaction, XeF₆(g) + HaO(g) ⇌ XeOF₄(g) + 2HF(g) and K₂ as the equilibrium constant for the reaction, XeO₄(g) + XeF₆(g) ⇌ XeOF₄(g) + XeO₃F₂(g).
Answer:

According to the given condition

⇒ \(K_1=\frac{\left[\mathrm{XeOF}_4\right] \times[\mathrm{HF}]^2}{\left[\mathrm{XeF}_6\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]} \text { and } K_2=\frac{\left[\mathrm{XeOF}_4\right] \times\left[\mathrm{XeO}_3 \mathrm{~F}_2\right]}{\left[\mathrm{XeO}_4\right] \times\left[\mathrm{XeF}_6\right]}\)

∴ \(\text { For } \mathrm{XeO}_4(g)+2 \mathrm{HF}(g) \rightleftharpoons \mathrm{XeO}_3 \mathrm{~F}_2(g)+\mathrm{H}_2 \mathrm{O}(g)\)

Equilibrium constant ,\(K=\frac{\left[\mathrm{XeO}_3 \mathrm{~F}_2\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{XeO}_4\right] \times\left[\mathrm{HF}^2\right.}\)

⇒ \(\text { Now, } \frac{K_2}{K_1}=\frac{\left[\mathrm{XeO}_3 \mathrm{~F}_2\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{XeO}_4\right] \times[\mathrm{HF}]^2}=K\)

∴ K= \(\frac{K_2}{K_1}\)

Class 11 Chemistry Long Answer Equilibrium Questions and Answers

Question 24.  At a particular temperature, for the reaction, aA + bB dC + cD, the equilibrium constant is K. Find out the equilibrium constants for the following reactions at the same temperature.

1. m a A +  m b B m d C+m c D

2. \(\frac{1}{m} a A+\frac{1}{m} b B\frac{1}{m} d C+\frac{1}{m} c D\)

Answer:

For the reaction, aA + bB ⇌ dC+cD the equilibrium constant

⇒ \(K=\frac{[C]^d \times[D]^c}{[A]^a \times[B]^b}\)

1. Now,  For the reaction maA + mbB ⇌ mdC + mcD equilibrium constant,

⇒ \(K_1=\frac{[C]^{m d} \times[D]^{m c}}{[A]^{m a} \times[B]^{m b_3}}\)

= \(\left\{\frac{[C]^d \times[D]^c}{[A]^a \times[B]^b}\right\}^m=K^m\)

2. For the reaction

⇒  \(\frac{1}{m} a A+\frac{1}{m} b B \rightleftharpoons \frac{1}{m} d C+\frac{1}{m} c D\) equilibrium constant,

⇒ \(K_2=\frac{[C]^{d / m} \times[D]^{c / m}}{[A]^{a / m} \times[B]^{b / m}}\) \(=\left\{\frac{[C]^d \times[D]^c}{[A]^a \times[B]^b}\right\}^{1 / m}\)

=\((K)^{1 / m}\)

Question 25. State the nature of aqueous solutions containing the following ions with reason:

1. NH⁺₄,
2. F^–
3. Cl^–.

Answer:

1. NH₄⁺:

NH₄⁺ ion is the conjugate acid of a weak base, NH₃. In an aqueous solution, NH₂ is stronger than H₂O [weak Bronsted acid]. Hence, in an aqueous solution, NH₂ ion reacts with water to produce H₃0+ ions, leaving the die solution acidic.

NH⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)

2. F^–:

F^– ion is the conjugate base of a weak acid, HF. F^– is stronger titan H₂O [weak Bronsted base] in aqueous solution. For this reason, in an aqueous solution, the F^– ion reacts with water to give OH^– ions, and as a result, the solution becomes basic.

⇒  \(\mathrm{F}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HF}(a q)+\mathrm{OH}^{-}(a q)\)

3. Cl^–:

Cl^– is the die conjugate base of strong acid, HCl. Hence, it is very weak and cannot react with water. Consequently, the die aqueous solution of Cl^– is neutral.

NCERT Class 11 Equilibrium Long Questions

Question 26. The solubility of zinc phosphate in water is S mol. L^-1 . Derive the mathematical expression of the solubility product of the compound.
Answer:

The following equilibrium is established by zinc phosphate, [Zn3(P04)2] in its saturated aqueous solution.

⇒ \(\mathrm{Zn}_3\left(\mathrm{PO}_4\right)_2(s) \rightleftharpoons 3 \mathrm{Zn}^{2+}(a q)+2 \mathrm{PO}_4^{3-}(a q)\)

∴ \(K_{s p}=\left[\mathrm{Zn}^{2+}\right]^3 \times\left[\mathrm{PO}_4^{3-}\right]^2\)

As given, the solubility of Zn₃(PO₄)₂ is S mol. L^-1 .

Hence, \(\left[\mathrm{Zn}^{2+}\right]=3 S \mathrm{~mol} \cdot \mathrm{L}^{-1} \text { and }\left[\mathrm{PO}_4^{3-}\right]=2 S \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

∴ K_sp = (3S)³ × (2S)²= 108 S⁵; this is the mathematical expression of the solubility product of zinc phosphate.

Question 27. When H₂S gas Is passed through an acidified solution of Cu²⁺ and Zn²⁺, only CuS is precipitated—why?
Answer:

H₂S is a very weak acid. In its aqueous solution, only a small fraction of it ionizes to produce H₃O⁺ and S^2- ions

⇒\(\left[\mathrm{H}_2 \mathrm{~S}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{S}^{2-}(a q)\right]\)

In an acidified solution of H₂S, due to the common ion (H₃O⁺) effect, the ionization of H₂S is further reduced, and as a result, the concentration of S^2- ions becomes extremely low.

When H₂S gas is passed through an acidified solution of Cu²⁺ and Zn²⁺ ions, the concentration of S^-2 ions becomes so low that only the product of the concentrations of Cu²⁺ and S^2- ions exceeds the solubility product of CuS. However, the product of the concentrations of Zn²⁺ and S^2- ions lies well below the value of the solubility product of ZnS. This is why only CuS is precipitated in preference to ZnS.

Question 28. Why is the aqueous solution of Cu(NO₃)₂ acidic?
Answer:

Being a strong electrolyte, Cu(NO₃)₂ dissociates almost completely in its aqueous solution to form [Cu(H₂O)₆]²⁺ and NO₂ ions. In the solution, H₂O also ionizes slightly to form H₃O⁺ and OH^– ions

Cu(NO₃)₂(aq) + 6H₂O(l) → [Cu(H₂O)g]₂+(aq) + 2NO(aq)

2H₂O(l) ⇌  + H₃O⁺(aq) + OH

NO^–₃ is the conjugate base of a strong acid, HNO₃. Hence it is a very weak base. So, NO₂ ions cannot react with water. On the other hand, since the charge density of Cu²⁺ ion in the cationf[Cu(H₂O)₆]⁺² is very high, the H₂O molecules attached to it get polarised, and their O—H bonds become weaker.; These O —H bonds dissociate to produce H+ ions, which are accepted by H₂O to give H₃O⁺ ions.

Cu(H₂O)₆]2+(aq) + H₂O(l) ⇌  [Cu(H₂O)₅OH]⁺(aq) + H₃O⁺(aq)

As a result, the concentration of H₃O⁺ ions in the aqueous solution of Cu(NO₃)₂ becomes higher than that of OH- ions (from H₂O), and consequently, the solution becomes acidic.

Equilibrium Chapter 7 Long Answer Explanation Class 11

Question 29. Despite being a neutral salt, the aqueous solution of Na₂CO₃ is alkaline—why?
Answer:

1. Na₂CO₃, being a strong electrolyte, dissociates almost completely in aqueous solution, forming Na⁺ and CO₃ ions:

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3(a q) \rightarrow 2 \mathrm{Na}^{+}(a q)+\mathrm{CO}_3^{2-}(a q)\)

2. Water, being a very poor electrolyte, ionizes slightly to form H3O+ and OH- ions:

⇒ \(2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

3. Hydrated Na+ ion (Na⁺(aq) ] is a very weak acid and cannot react with water.

4. CO₃^2- the conjugate base of weak acid HCO₃^–, is a stronger base than H₂O (which is a Bronsted base). Therefore, it reacts with water in aqueous solution, forming the following equilibrium.

⇒ \(\mathrm{CO}_3^{2-}(a q)+\mathrm{H}_2 \mathrm{O}(a q) \rightleftharpoons \mathrm{HCO}_3^{-}(a q)+\mathrm{OH}^{-}(a q)\)

HCO₃^– so formed undergoes ionization

⇒ \(\mathrm{HCO}_3^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{CO}_3^{2-}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)]\)

But, due to the common ion (CO₂^-3) effect, this ionization occurs to a very small extent.

As a result, [OH^–] in solution is much higher than [H⁺]. So, aqueous solution of Na₂CO₃ is alkaline.

NCERT Class 11 Chemistry Equilibrium Very Short Question And Answers

Question 1. Give examples of two systems involving solid-vapour equilibrium.
Answer:  The equilibrium, solid vapour is found to be established when naphthalene or iodine undergoes sublimation in a closed container.

Question 2. Which of the following is the strongest Bronsted base? CIO^–, ClO₂^–, ClO₃^–, ClO₄^–.
Answer: CIO^– is the strongest Bronsted base

Question 3. The equilibrium established in the evaporation of a liquid at a given temperature is due to the same rate of two processes. What are these two processes?
Answer: The rate of evaporation of the liquid will be equal to the rate of condensation of its vapours at the equilibrium.

Equilibrium Chapter 7 Class 11 Very Short Answer Questions

Read and Learn More NCERT Class 11 Chemistry Very Short Answer Questions

Question 4. In a soda water bottle, CO₂ gas remains dissolved in water under high pressure. Write down the equilibrium established in this case.
Answer: CO₂(gas) CO₂ (solution)

Question 5. pH of an aqueous 0.1 (M) CH₃COOH solution is 2.87. State whether the pH of the solution will decrease or increase if CH₃COONa is added to this solution.
Answer: pH will increase.

Question 6. Which of the following mixtures will act as buffer solution(s)? 

1. 50 mL 0.1 (M) NH₃ + 100 mL 0.025(M) HCl 
2. 100mL 0.05(M) CH₆COOH + 50mL 0.1 (M) NaOH

Answer: 1. Will act as a buffer solution.

Question 7. At a constant temperature and pressure, what is the value of ΔG for a reaction at equilibrium?
Answer: Zero

Question 8. For a chemical reaction, K_c> 1. Will the value of ΔG° for this reaction be negative or positive?
Answer: Negative

Question 9. Find the relation between K and Kc for the following system: 2NaHCO₃(s) Na₂CO₃(s) + CO₂(g) + H₂O(g)
Answer: K_p=K_c(RT)₂

Question 10. The values of the equilibrium constant for a particular reaction at 25°C and 50°C are 0.08 and 0.12 respectively. State whether it is an exothermic or an endothermic reaction.
Answer: Exothermic

NCERT Solutions Class 11 Chemistry Equilibrium VSAQ

Question 11. If ‘mol-L^-1′ and ‘atm’ are the units of concentration and pressure respectively, then what will be the value of K_p/K_c for the reaction, N₂O₄(g) ⇌ 2NO₂(g) at 300K?
Answer: 24.63L.atmmol^-1

Question 12. Give an example of a reaction for which, K_p = K_c = K_x.
Answer: N₂(g) + O₂(g) ⇌ 2NO(g)

Question 13. Consider the reaction: A(s) + 2B₂(g) ⇌ AB₄(g); ΔH<0. At equilibrium, what would be the effect of increasing temperature on the concentration 
Answer: Increases

Question 14. For the reaction, N₂O₄(g) ⇌ 2NO₂(g) ΔH > 0 Mention two factors whose change at the equilibrium of the reaction will increase Mention two factors whose change at the equilibrium of the reaction will increase the yield of NO₂(g).
Answer: Decrease in temperature, increase in pressure at constant temperature

Question 15. Give an example of a reaction for which, K_p /Kc Is independent of temperature.
Answer: H₂(g) + I₂(g) ⇌ 2HI(g)

Question 16. At a given temperature, if the total pressure of the equilibrium A(s) ⇌ B(g) + C(g) is P, then what is the value of K?
Answer: K_p = P²/2

Question 17. What are the conjugate acid and conjugate base of HPO₄^2-?
Answer:

Conjugate acid: H₂PO₄^–

Conjugate base: PO₄^3-

Very Short Answer Questions for Class 11 Chemistry Equilibrium

Question 18. What are the conjugate acid and conjugate base of H₂O?
Answer:

Conjugate acid: H₃O⁺

Conjugate base: OH^–

Question 19. Kb for NH₃ = 1.8 × 10^-5 at 25°C. What is the value of Ka for NH₄ ions at the same temperature?
Answer: 5.55 × 10^-10

Question 20. For pure water, pK_w=12 at a certain temperature. At this temperature, what is the molar concentration of H₂O⁺ In a neutral aqueous solution?
Answer: 10^-6(M)

Question 21. At a certain temperature, pK_w for pure water is 12. At the same temperature, the concentration of H₃O⁺ in aqueous solution is 10^-8 mol L^-1. Is this solution acidic or basic?
Answer: Basic

Question 22. pH of a sample of pure water is x at a certain temperature. Find the value of pK_w at that temperature.
Answer: 2X

Question 23. pK_a values for two acids, HA and HB, are 4 and 6 respectively. Which one is the stronger acid?
Answer: HA

Question 24. pK_a values for two acids, HA, and HB are 4 and 5 respectively. If each of the acid solutions has a concentration of 0.1(M), then in which solution the molar concentration of H₃O⁺ ions is greater?
Answer: HA

Question 25. At 25 °C, pKa for weak acid, HA = x, and pK_b for A^–, the conjugate base of HA = y. Find the value of (x + y) at this temperature.
Answer: (x+y)=14

Class 11 Chemistry Equilibrium VSAQ Solutions

Question 26. Give examples of two salts, in case of which the pH value of their aqueous solutions is independent of their concentrations.
Answer: CH₃

Question 27. For a weak acid, HA, pK_a = 5. What is the effective range of pH for a buffer comprised of HA and its salt, NaA?
Answer: 4 to 6

Question 28. Give examples of two buffer solutions prepared by mixing two salt solutions of polybasic acid.
Answer: (NaH₂PO_4, Na₂HPO₄)and (NaH₆P₃, Na₂CO₃)

Question 29. In a buffer solution comprising a weak base (B) and its ion (BH⁺), [B] = [BH⁺]. If Kb for the weak base =10^-5, then find the pH value of the buffer solution.
Answer: 9

Question 30. 100 mL of 0.05(M) NaOH solution is added to a 100 mL of 0.1(M) NaH₂PO₄ solution. Will the mixed solution act as a buffer?
Answer: Yes

NCERT Class 11 Equilibrium Chapter 7 VSAQ

NCERT Class 11 Chemistry Equilibrium Fill In The Blanks

Question 1. At a given temperature for a reversible reaction, K< 1. The rate constant of the forward reaction is___________________ than that of the reverse reaction.
Answer: Lower

Question 2. If K<1 for a reaction at a particular temperature, then the value ΔG° is___________________
Answer: Positive

Question 3. For a gaseous reaction, K_p > K_c An increase in pressure at constant temperature will ___________________ the product.
Answer: Decrease

Question 4. If C(s) is added the concentration of to the reaction system C(s) + CO₂(g) ⇌ 2CO(g) at constant temperature and volume, then the concentration of CO₂(g) will___________________.
Answer: Remain Unchanged

Equilibrium Very Short Answer Solutions Class 11

Question 5. For the reaction, COCl₂(g) ⇌  CO(g) + Cl₂(g); ΔH > 0, an increase in temperature at equilibrium will increase the concentration of___________________
Answer: CO and Cl₂

Question 6. The addition of inert gas to the reaction system, PCl₅(g)⇌ PCl₃(g) + Cl₂(g) at constant___________________does do not affect the state of equilibrium.
Answer: Temperature, Volume

Question 7. At a particular temperature, pKw for pure water =12. Its pH will be___________________
Answer: 6

Question 8. The first and second ionization constants of H2S in its aqueous solution are x and y respectively. So, x is _ than y.
Answer: Greater

Question 9. The conjugate base of [Al(H₂O)₆]³⁺ is___________________.
Answer: [Al(H₂O)₅OH]²⁺

Question 10. The conjugate acid and conjugate base of HPO₂^– are respectively.
Answer: \(\mathrm{H}_2 \mathrm{PO}_4^{-}, \mathrm{PO}_4^{3-}\)

NCERT Class 11 Chemistry Chapter 7 Very Short Answer Q&A

Question 11. At 25°C, in an aqueous solution of HA, Ka for HA = 1(T6. Kb for A+ ion is___________________
Answer: 10^-8

Question 12. If the solubility of Ag3P04 in its saturated aqueous solution is S(M), then, its solubility product will be ___________________
Answer: 27 S⁴

Question 13. The solubility of Ag₂CrO₄ ___________________in an aqueous solution of than its solubility in pure water.
Answer: Lower

Question 14. Addition of CH₃COONa to an aqueous solution of CH₆COOH ___________________AgNO₃ is the pH value.
Answer: Increases

NCERT Class 11 Chemistry Equilibrium Multiple Choice Questions

Question 1. Some reactions, their equilibrium constants are as follows:

⇒  \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) ; K_1\)

⇒  \(\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) ; K_2 \)

⇒ \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2(\mathrm{~g}) ; K_3\)

The relation among K₁, K₂ and K₃ is

1. \(K_3 \times K_2^3=K_1^2\)
2. \(K_1 \sqrt{K_2}=K_3\)
3. K₂ × K₃ = K₁
4. K₃ = k₁ × k₂

Answer: 4. K₃ = k₁ × k₂

Question 2. At a given temperature, the reaction, A(g) 20(g), is In equilibrium In a closed flask. At the same temperature, the reaction, C(g) D(g) + O(g) Is in equilibrium in another closed flask. The values of equilibrium constants of these two reactions are Kp and Kp respectively and the total pressures of the equilibrium mixtures are P₁ and P₂ respectively. If: Kp₂ =1:4 and P₁ P₂ = then the ratio of the degree of dissociation of (g) and C(g) are (assume the degrees of dissociation of body care very small compared to 

1. 0.15
2. 0.5
3. 1.0
4. 1.5

Answer: 2. 0.5

Question 3. At a given temperature, the reaction, A₂(g) + B₂(g) 2A₂(g) was started with 0.4 mol of A₂(g) and 0.0 mol of H₂(g) in a flask of volume 2L. When the reaction achieved equilibrium, it was found that the reaction mixture contained 0.5 mol of AO. (The equilibrium constant ( Kp) for the reaction is

1. 8.30
2. 4.76
3. 10.27
4. 6.49

Answer: 2. 4.76

Equilibrium Chapter 7 Class 11 MCQs

Question 4. At a given temperature, if the degree of dissociation of N₂O₄ in the following reaction N₂O₄(g) 2NO₂(g) is a, and the total pressure of the equilibrium mixture is P, then it can be shown that the equilibrium constant for the reaction, K_p= a₂P (assuming a is very small compared to 1). Which ofthe following comments is true for this relation

1. K_p increases as a increases
2. K_p increases as P increases.
3. value of Kp does not depend on P but depends on a
4. The value of Kp depends neither on P nor on a

Answer: 4. The value of Kp depends neither on P nor on a

Question 5. For the reactions

⇒ \( \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g),\)

⇒ \(\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) \& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2}\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \text {, }\)

If the equilibrium constants are K₁, K₂ and K₃ respectively, then the equilibrium constant for the reaction \(4 \mathrm{NH}_3(g)+5 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\) is

1. \(\frac{K_2 K_3^2}{K_1}\)

2. \(\frac{K_2^2 K_3^2}{K_1}\)

3.\(\frac{K_1^3 K_2^2}{K_3}\)

4. \(\frac{K_2^2 K_3^6}{K_1^2}\)

Answer: 4. \(\frac{K_2^2 K_3^6}{K_1^2}\)

Question 6. For a hypothetical reaction, K_c = 0.9 and K_p = 538. Which of the following equations pan represents the reaction properly at 25°C-

1. \(A(g) \rightleftharpoons 2 C(s)+D(g)\)
2. \(B(g) \rightleftharpoons C(l)+D(l)\)
3. \(A(l)+2 B(g) \rightleftharpoons 2 C(g)\)
4. \(A(g)+B(s) \rightleftharpoons 3 C(g)\)

Answer: 4. \(A(l)+2 B(g) \rightleftharpoons 2 C(g)\)

Question 7. When a mixture containing N₂ and H₂ in the molar ratio 1:3 heated in the presence of a catalyst in a closed vessel, die following equilibrium is established:

⇒ \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\)

At equilibrium, if the mole fraction of NH₃ is 0.6 and the total pressure ofthe equilibrium mixture is 10 atm then Kp for the reaction, \(2 \mathrm{NH}_3(g) \rightleftharpoons \mathrm{N}_2(g)+3 \mathrm{H}_2(g)\)

1. 1.33 atm^-2
2. 0.75 atm^-2
3. 1.333 atm^-2
4. 0.75 atm^-2

Answer: 1. 1.33 atm^-2

Question 8. The reaction, \(A(g)+4 B(g) \rightleftharpoons 2 C(g)+3 D(g)\) is carried out in a closed vessel of volume 2L by taking 3 mol of A(g) and 4 mol of 5(g). At equilibrium, if the amount of C(g) be1 mol, then Kc for the reaction is

1. 0.056
2. 0.038
3. 0.084
4. 1.24

Answer: 3. 0.084

Question 9. The total pressure at the equilibrium of the reaction, XY(g) X(g) + Y(g) is P. If the equilibrium constant P for the reaction is Kp and \(K_p=\frac{P}{8}\), then the percent dissociation of XY is

1. 30.49%
2. 33.33%
3. 41.90%
4. 19.26%

Answer: 2. 33.33%

Question 10. At 500K, for the reaction, PCl₅(g) PCl₃(g) + Cl₂(g) , the equilibrium constant, Kp = 0.52 . In a closed container, these three gases are mixed. If the partial pressure of each of these gases is 1 atm, then in the reaction system.

1. The number of moles of PCl₅ will increase
2. The number of moles of PCl₃ will increase
3. The reaction will attain equilibrium when 50% of the reaction gets completed
4. The reaction will attain equilibrium when 75% of the reaction is completed

Answer: 1. The number of moles of pcl₅ will increase

Question 11. For the reaction: \(2 A(g)+B(g) \rightleftharpoons 3 C(g)+D(g)\), two moles of each A and B were taken into a flask which of the following relation between the concentration terms is true when the system attains equilibrium

1. [A]=[B]
2. [A]<[B]
3. [A]=[B]
4. [A]>[B]

Answer: 2. [A]<[B]

Multiple Choice Questions for Class 11 Chemistry Equilibrium

Question 12. The reaction, 2A(g) + 5(g) C(s); ΔH < 0, is in equilibrium in a closed vessel. Which of the following changes at equilibrium’ will increase the yield of C (s)

1. Temperature is increased
2. At constant volume and temperature, some amount of 5(g) is added to the reaction system
3. At constant volume and temperature, some amount of c (s) is removed from the reaction system noriw
4. Pressure is decreased at a constant temperature

Answer: 2. At constant volume and temperature, some amount of 5(g) is added to the reaction system.

Question 13. At 300 K, the reaction A(g) + 5(g) C(s), 1 is in equilibrium in a closed, vessel. At the beginning of the reaction, the partial pressures of A and B gases are 0.2 and 0.3atm respectively and the total pressure of the equilibrium mixture is 0.3atm. Kc, for the reaction, is—

1. 6.06 ×10⁴ L².mol^-2
2. 2.59 × 10³ L².mol^-2
3. 3.03 ×10⁴ L².mol^-2
4. 8.2 ×10² L².mol^-2

Answer: 3. 3.03 × 10⁴ L².mol^-2

Question 14. At 300K, for the reaction, \(\mathrm{AB}_3(\mathrm{~g}) \rightleftharpoons \mathrm{AB}_2(\mathrm{~g})+\frac{1}{2} \mathrm{~B}_2(\mathrm{~g}),\) Kp = 1.66. At the same temperature, ΔG° for the reaction, 2AB₂(g) + B₂(g) 2AB₃(g) is

1. +2.19kJ
2. -2.52KJ
3. +3.85KJ
4. -3.26kJ

Answer: 2. -2.52KJ

Question 15. At a given temperature, when a reversible reaction is carried out in the absence of a catalyst, the ratio of the rate constants for the forward and reverse reactions is found to be 8.0. At the same temperature, if the reaction is carried out in the presence  of a catalyst, then the ratio will be

1. >8.0
2. <8.0
3. =8.0
4. <8.0

Answer: 3. =8.0

Question 16. At a given temperature, a closed vessel contains NH₃ gas and solid NH₄HS. The pressure of NH₃ gas in the vessel is 0.50 atm. On dissociation, NH₄HS produces NH₃ and H₂S gases. The total pressure in the flask at equilibrium is 0.84 atm. The equilibrium constant for the dissociation reaction (Kp) of NH₄HS is

1. 0.30 atm^-2
2. 0.16 atm^-2
3. 0.11 atm^-2
4. 0.22 atm^-2

Answer: 3. 0.11 atm^-2

Question 17. N₂O₄ is dissociated to 33% and 40% at total pressures P₁ and P₂ atm respectively. Hence, the ratio of P₁ to P₂ is

1. \(\frac{7}{3}\)
2. \(\frac{8}{3}\)
3. \(\frac{8}{5}\)
4. \(\frac{7}{4}\)

Answer: 3. \(\frac{8}{5}\)

Question 18. The reaction, A(g) + 25(g) 2C(g) + D(g) was studied using an initial concentration of 5 which was 1.5 times that of. The equilibrium concentrations of A and C were found to be equal. So, Kc for the equilibrium is

1. 4
2. 0.32
3. 2.73
4. 8.17

Answer: 2. 0.32

Question 19. The equilibrium constants Kp₁ and Kp₂ for the reactions x ⇌ 2y and x ⇌ p + q, respectively are in the ratio of 1: 9 . If the degree of dissociation and Z is equal then the ratio of total pressure at this equilibrium is

1. 1:36
2. 1:1
3. 1:3
4. 1:9

Answer: 1. 1:36

Question 20. If the equilibrium constant for Mutrotion or-D-glucose is the percentage ofthe a – form in the equilibrium mixture is

1. 64.5
2. 35.7
3. 53.7
4. 44.8

Answer: 2. 35.7

NCERT Solutions Class 11 Chemistry Equilibrium MCQs

Question 21. The reaction, C(s) + CO₂(g) ⇌ 2CO(g), is at equilibrium in a closed vessel under a given set of conditions. If the degree of dissociation of CO₂ at equilibrium is a and the total pressure of the equilibrium mixture and the value of equilibrium constant are P and Kp respectively, then a

1. \(\frac{K_p}{\sqrt{2 P}}\)

2. \(\frac{1}{2} \sqrt{\frac{K_p}{p}}\)

3. \(\frac{\sqrt{K_p}}{P}\)

4. \(\sqrt{\frac{P}{K_p}}\)

Answer: 4. \(\sqrt{\frac{P}{K_p}}\)

Question 22. At a given temperature, the equilibrium constant, Kc, for the reaction, A + B C is 10. At the same temperature, the reaction is allowed to occur in a closed vessel of volume 1L. At a particular moment during the reaction, if the amounts of A, B, and C in the reaction system are 0.1, 0.4, and 0.3 mol respectively, then

1. The reaction is in equilibrium at that moment
2. The reaction will occur to a greater extent towards the left to attain equilibrium
3. The reaction will occur to a greater extent towards the right to attain equilibrium
4. The reaction will occur to a greater extent towards the left for achieving equilibrium and concentrations of reactants and product will be the same at a new equilibrium

Answer: 3. The reaction will occur to a greater extent towards the right to attain equilibrium

Question 23. A mixture containing N₂ and H₂ in a mole ratio of 1: 3 is allowed to attain equilibrium when 50% ofthe mixture has reacted. If P is the pressure at equilibrium, then the partial pressure of NH₃ formed is

1. p/3
2. p/2
3. p/9
4. p/5

Answer: 1. p/3

Question 24. If the concentration of OH^– ions in the reaction Fe(OH)₃(s) ⇌ Fe³⁺(aq) + 3OH^–(aq) is decreased by 1/4 times, then equilibrium concentration of Fe³⁺ will be increased by

1. 64 times
2. 4 times
3. 8 times
4. 16 times

Answer: 1. 64 times

Question 25. The equilibrium constant (K_p) for the decomposition of gaseous H₂O, \(\mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g)\) is related to degree of dissociation (a ) at a total pressure (P) as-

1. \(\kappa_p=\frac{\alpha^3 p^{1 / 2}}{(1+\alpha)(2+\alpha)^{1 / 2}}\)
2. \(K_p=\frac{\alpha^3 p^{3 / 2}}{(1-\alpha)(2+\alpha)^{1 / 2}}\)
3. \(K_p=\frac{\alpha^{3 / 2} p^2}{(1-\alpha)(2+\alpha)^{1 / 2}}\)
4. \(K_P=\frac{\alpha^{3 / 2} p^{1 / 2}}{(1-\alpha)(2+\alpha)^{1 / 2}}\)

Answer: 4. \(K_P=\frac{\alpha^{3 / 2} p^{1 / 2}}{(1-\alpha)(2+\alpha)^{1 / 2}}\)

Question 26. 2 mol of PCl₅(g) is heated at a given temperature in a closed vessel of volume 2L. As a result, PCl₅(g) dissociates and forms PCl₃(g) and Cl₂(g). When the dissociation reaction reaches equilibrium, it Is found that 50% of PClg(g) has dissociated. Kc for the reaction is

1. 0.15
2. 0.30
3. 0.25
4. 0.5

Answer: 4. 0.5

Question 27. At a given temperature, the reaction, SO₂Cl₂(g) SO₂(g) + CI₂(g), is in a state of equilibrium in a closed vessel. At constant temperature and volume, if some amount of He gas is added to the reaction system, then —

1. The concentration of SO₂(g) will increase
2. The concentration of SO₂Cl₂(g) will increase
3. The concentrations of SO₂(g) , Cl₂(g) , SO₂cl₂(g) will remain the same
4. The value of the equilibrium constant will decrease

Answer: 3. The concentrations of SO₂(g) , Cl₂(g) , SO₂cl₂(g) will remain the same

Question 28. The reaction, C(s) + CO₂(g);⇌2CO(g), is in a state of equilibrium in a closed vessel at a constant temperature. The equilibrium of the reaction will shift towards the left and get re-established if at constant temperature and volume some amount of—

1. C(s) is removed from the reaction system
2. CO₂(g) is added to the reaction system
3. CO₂(g) is removed from the reaction system
4. CO(g) is removed from the reaction system

Answer: 3. CO₂(g) is removed from the reaction system

Class 11 Chemistry Equilibrium Multiple Choice Questions and Answers

Question 29. The pair of compounds which cannot exist together in solution is

1. NaHCO₃ and NaOH
2. Na₂CO₃ and NaHCO₃
3. Na₂CO3 and NaOH
4. NaHCO₃ and NaCl

Answer: 1. NaHCO₃ and NaOH

Question 30. Equimolar solutions of the following were prepared in water separately. Which of the solutions will have the highest pH

1. Srcl₂
2. Bacl₂
3. MgC₂
4. Cacl₂

Answer: 2. Bacl₂

Question 31. A student wants to prepare a saturated solution of Ag⁺ ion. He has only three samples of Ag— AgCl (K_sp = 1.8 × 10^-18), AgBr (K_sp = 5 × 10^-13), and Ag₂CrO₄ (K_sp = 2.4 × 10^-12). Which compound should he take’ to obtain maximum [Ag⁺] 

1. Agcl
2. AgBr
3. Ag₂CrO₂₄
4. None of these

Answer: 3. Ag₂CrO₄

Question 32. The correct relationship between the pH of picomolar solutions of sodium oxide (pH₂), sodium sulfide (pH₂), sodium selenide (pH₃), and sodium telluride (pH₄) is

1. pH₁>pH₂>pH₃>pH₄
2. pH₁ <pH₂<pH₃ < pH₄
3. pH₁<pH₂<pH₃<pH₄
4. pH₁>pH₂=pH₃>pH₄

Answer: 1. pH₁>pH₂>pH₃>pH₄

Question 33. Solubility product constant (K_sp) of salts of types MX₁, MX₂ & M₃X at temperature K_sp are 4.0 × 10^-8, 3.2 ×10^-14 & 2.7×10^-13 respectively. Solubilities (mol dm^-3) of the salts at temperature T are in the order—

1. M₃>MX₂>M₃X
2. M₃X>MX₂>MX
3. MX₂>M₃X>MX
4. MX>M₃X>MX₂

Answer: 4. MX>M₃X>MX₂

Question 34. If the solubilities of AgCl in H₂O,0.01(M) CaCl₂ , 0.01(M) NaCl and 0.05(M) AgNO₃ are S₁, S₂, S₂ Ansd S₂ respectively, then—

1. S₁>S₂>S₃>S₄
2. S₁ > S₂ = S₃ > S₄
3. S₁ > s₃ > S₂ > S₄
4. S₄>S₂>S₃>S₁

Answer: 3. S₁ > s₃ > S₂ > S₄

Question 35. The degree of hydrolysis of a salt of weak acid and weak base in its 0.01(M) solution is found to be 50%. If the molarity of the solution is 0.2(M), the percentage hydrolysis of the salt should be

1. 100%
2. 50%
3. 25%
4. 10%

Answer: 2. 50%

Question 36. The first and second dissociation constants of an acid H₂A are 1.0×10^-5 and 5.0×10^-10 respectively. The first and second dissociation constants of an

1. 5.0 ×10^-5
2. 5.0 ×10¹⁵
3. 5.0 ×10^-15
4. 0.2×10⁵

Answer: 3. 5.0 ×10^-15

Question 37. If three salts P₂X, QY₂, and RZ₃ have the same solubilities in water then the correct relation among their Ksp values is—

1. \(K_{s p}\left(\mathrm{P}_2 \mathrm{X}\right)=K_{s p}\left(\mathrm{QY}_2\right)<K_{s p}\left(\mathrm{RZ}_3\right)\)
2. \(K_{s p}\left(\mathrm{P}_2 \mathrm{X}\right)>K_{s p}\left(\mathrm{QY}_2\right)=K_{s p}\left(\mathrm{RZ}_3\right)\)
3. \(K_{s p}\left(\mathrm{P}_2 \mathrm{X}\right)=K_{s p}\left(\mathrm{QY}_2\right)=K_{s p}\left(\mathrm{RZ}_3\right)\)
4. \(K_{s p}\left(\mathrm{P}_2 \mathrm{X}\right)>K_{s p}\left(\mathrm{QY}_2\right)>K_{s p}\left(\mathrm{RZ}_3\right)\)

Answer: 1. \(K_{s p}\left(\mathrm{P}_2 \mathrm{X}\right)=K_{s p}\left(\mathrm{QY}_2\right)<K_{s p}\left(\mathrm{RZ}_3\right)\)

Question 38. The pH of the solution was obtained by mixing 20 mL of 0.01 (M) Ca(OH)₂ and 30 of 0.1 (M) HCl solution is

1. 6.32
2. 9.85
3. 11.3
4. 4.74

Answer: 3. 11.3

Equilibrium MCQs Chapter 7 NCERT Class 11

Question 39. The pH of an aqueous solution of MCI is 3,0 and that of an aqueous solution of NaOH is 12. The pH of the solution obtained by mixing 100mL of NaOH solution with 500 ml, of HC1 solution’ is

1. 6.71
2. 10.92
3. 12.05
4. 3.08

Answer: 2. 10.92

Question 40. At 25°C, the pH of 0.1(M) aqueous solution of NH₃ is 11.13. At the same temperature, the pH of a ‘solution containing 0.1 (M) of NH₄Cl and 0.01 (M) of NH₃ is-

1. 4.74
2. 6.25
3. 8.26
4. 9.34

Answer: 3. 8.26

Question 41. 800ml. 0.1 (M) HCl solution is mixed with 200ml. 0.5(M) CH₃NH₂ solution. In the resulting solution, traction of H₃O⁺ ions is = fix 10 is

1. 3×10^-5(M)
2. 1.25 ×10^-4(M)
3. 8×10^-11 (M)
4. 7.2×10^-10(M)

Answer: 3. 8 × 10^-11 (M)

Question 42. A solution of a weak acid ( Kn = 10^-5) has a molarity of (M/5). lol of this solution is neutralized completely with a NaOH solution of molarity (M/20). At the neutra¬ lisation point, concentration of H(Of ions (mol-L^-1 ) is

1. 4.39 × 10^-5
2. 1.25 × 10^-6
3. 7.02× 10^-8
4. 1.58 × 10^-9

Answer: 4. 1.58 × 10^-9

Question 43. At 25°C, K for PbCI, is 1.6 × 10^-5 in water. At the same temperature, the amount of PbCl₂ (molar mass = 278.19 g-mol^-`1 ) that remains dissolved in 100 mL of a saturated solution of PbCL, is

1. 0.28g
2. 0.44g
3. 0.17g
4. 0.35g

Answer: 2. 0.44g

Question 44. At 25°C, the solubility product for Gd(OH)₂ in water is 1.2 ×10^-14 What would be the pH of tut aqueous solution of 0.01 (M) Cd²⁺ ions when Cd(OH)₂ starts precipitating

1. 4.29
2. 5.60
3. 8.04
4. 7.56

Answer: 3. 8.04

Question 45. At 25°C the solubility product of a salt AB0 in water is 4.0 × 10^-15 If 0.1 mol of A^2- ions are added to 1 1. of a saturated solution of the salt (assuming the volume of the solution does not change on the addition of A^2- ions), then

1. The solubility product of AB₂ will increase
2. The solubility product of AB₂ will decrease
3. The cone, of b- ions in the solution will be 2 × 10^-7 mol L^-1
4. The solubility of AB₂ in solution will be 4 × 10^-10. mol L^-1

Answer: 3. Cone, of b- ions in solution will be 2 ×10^-7 mol L^-1 

Question 46. At 25°C, Kp for Al(OH)₃.( in water is 2 x 10-33 aqueous solution of ph = 13, the solubility of AI(OH)₃ is 2 ×1 0^-20 . The value of x is

1. 10
2. 15
3. 22
4. 30

Answer: 2. 15

Question 47. At 25°C, K(l for a weak acid, HA in water, is 10- V₂ ml. of 0.1 (M) NaOH solution is added to K, mL of 0.1 (M) solution of HA. How many times would Vl he of K, so that the pH of the solution is 6

1. 2 times
2. 1.5 times
3. 1.1 times
4. 1.4 times

Answer: 3. 1.1 times

MCQs on Equilibrium Class 11 Chemistry

Question 48. Which of the relations are correct for the given physical change;

⇒ \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_4 \cdot 3 \mathrm{H}_2 \mathrm{O}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\) 

1. \(K_p=p_{\mathrm{H}_2 \mathrm{O}}^2\)
2. Kc = [H₂O(g)]₂
3. Kp – K_c(RT)₂
4. Kc = K_p(RT)₂

Answer: 1. \(K_p=p_{\mathrm{H}_2 \mathrm{O}}^2\)

Question 49. At a given temperature, Kc = 6.3 x 10-e for the reaction S-(g) 4S₂(g). At the same temperature, if the lead Ion is started with 2 moles of S0(g) and 0.2 mol of S₂(g) In a closed vessel of volume 1 L, then which of the following comments are true regarding this reaction—

1. At the beginning of the reaction, Q_p = 8.0 × 10^-4 
2. The reaction will occur to a greater extent towards the right To attain equilibrium
3. The reaction will occur to a greater extent towards the left to attain equilibrium
4. The concentration of S₈ at equilibrium is greater than 2 mol. L^-1 

Answer: 1. At the beginning of the reaction, Q_c = 8.0 × 10^-4 

Question 50. 4NH₃(g) + 5O₂(g)⇌4NO(g) + 6H₂O(g), is in equilibrlum In a closed container of volume 1L at a given temperature. If the reaction is started with 1 mol NH₃(g) and 1 mol of O₂(g) and the number of mol of H₂O(g) at equilibrium is 0.6 mol, then at equilibrium-

1. [NH₃] = [NO]
2. [NO]<[O₂]
3. [NO] > [NH₃]
4. [O2] < [H₂O]

Answer: 2. [NO]<[O₂]

Question 51. The reaction, fe(s) + H₂O(g) ⇌ CO(g) + H₂(g); ΔH > 0, is in equilibrium. At equilibrium

1. If the temperature is increased, the partial pressure of H20(g) will decrease
2. The concentration of H₂(g) will decrease if an inert gas is added at constant temperature and volume
3. the concentration of CO(g) will increase if pressure is increased at a constant temperature
4. The equilibrium will move towards the right if an inert gas is added at constant temperature and pressure.

Answer: 1. If the temperature is increased, the partial pressure of H₂O(g) will decrease

Question 52. The reaction, 2NOCl(g)⇌ 2NO(g) + Cl₂(g) ; ΔH>0, Is In equilibrium. Winch of the following changes at equilibrium will decrease The yield of NO(g) —

1. At constant temperature and volume, some amount of nocl(g) is lidded to the reaction system u
2. At constant temperature and volume, some amount of Cl₂(f> ) is added to the reaction system
3. Temperature is decreased at equilibrium
4. At constant temperature and pressure, some airfoil of the gas is added to the reaction system

Answer: 2. At constant temperature and volume, some amount of Cl₂(f> ) is added to the reaction system

Question 53. The vapor density of the equilibrium mixture of NO₂ and N₂O₄ is found to be 40 for the given equilibrium N₂O₄(g) 2NO₂(g). For the given equilibrium

1. 1 Mole percent of nO₂ present in the mixture is 59%
2. 1 Mole percent of nO₂ present in the mixture is 26%
3. The degree of dissociation of n₂O₄ is 0.45
4. The degree of dissociation of n₂O₄ is 0.15

Answer: 2. 1 Mole percent of nO₂ present in the mixture is 26%

Question 54. The vapor pressure of liquid methanol at 50°C is 55.5 kPa. These are correct for the equilibrium reaction attained in a closed vessel of 5 L at 50°C for the following equilibrium CH₃OH(Z) CH₃OH(g)

1. Kp = 55.5kPa
2. Kc = 0.021 mol- L^-1 
3. K = 0.555
4. K = 0.555kPa

Answer: 1. Kp = 55.5kPa

Question 55. The reactions in which the yield of the products cannot be increased by the application of high pressure are—

1. 2SO₂(g) + O₂(g) ⇌2SO₃(g)
2. NH₄HS(s)⇌ NH₃(g) + H₂S(g)
3. N₂O₄(g)⇌2NO₂(g)
4. N₂(g) + 3H₂(g)⇌2NH₃(g)

Answer: 2. NH₄HS(s) NH₃(g) + H₂S(g)

NCERT Class 11 Chemistry Chapter 7 Multiple Choice Questions

Question 56. Aqueous solutions of which of the following compounds on dilution do not suffer any change in pH value—

1. PhCOONH₄
2. NH₄CN
3. HCOONa

Answer: 1. PhCOONH₄

Question 57. Which can act as an acid as well as a base

1. SO₄^2-
2. HS^–
3. HCO^-3
4. HSO^-4

Answer: 2. HS^–

Question 58. Which mixtures (in molar ratio) can act as buffer

1. H₂CO₃ + NaOH (3:2)
2. H₂CO₃ + NaOH (3: 4)
3. NH₃ + HCl (5: 4)
4. NH₃ + HCl (4: 5)

Answer: 1. H₂CO₃ + NaOH (3:2)

Question 59. If equal volumes of the given solutions are mixed, precipitation of AgCl (Ksp = 1.8 x 10-11) will occur only with

1. 10-4(M)Ag+ and 10-4(M)Cl^–
2. 10-5(M)Ag+ and 10-5(M)Cl^–
3. 10-6(M)Ag+ and 10-6(M)Cr
4. 10-10(M)Ag+ and l0-10(M)Cl^–

Answer: 1. 10-4(M)Ag+ and 10-4(M)Cl^–

Question 60. Which ofthe following is true regarding H₃PO₄

1. Ka = Ka₁ X Ka₂ X Ka₃
2. Ka₁<Ka₂<Ka₃
3. ka₁ >ka₂ > ka₃
4. ka₁=ka₂=ka₂

Answer: 1. Ka = Ka₁ X Ka₂ X Ka₃

Question 61. Select the buffer solutions-

1. 0.8(M) H₂S+0.8(M)KHS
2. 2(M) C₆H₅NH₂+2(M)C₆H₅NH₃Br
3. 3(M)H₂Co₃+3(M)KHCO3
4. 0.05(M) KClO₄+0.05(M) HClO₄

Answer: 1. 0.8(M) H₂S+0.8(M)KHS

Question 62. The solubility Of BaSO₆ will be almost the same in

1. 0.1(M) H2O₄
2. 0.1(M) Ba(OH)₂
3. 0.1(M)BA(NO3)₂
4. 0.2(M) HCl

Answer: 1. 0.1(M) H₂SO₄

Question 63. The special that can act both and as Bronsted adds and as Bronsted base in water are-

1. H₂PO₂-4
2. PO^-3₄
3. HCO₂₃
4. [Fe(H₂O)₆]³⁺

Answer: 1. H₂PO₂

Question 64. Among the following salts, whose aqueous solutions will turn blue litmus paper red-

1. NaHCO₃
2. FeCl₃
3. KCN
4. C₆H₅NH₃Cl

Answer: 2. NaHCO₃

Question 65. At a given temperature, the first and the second ionization constants of the acid, H2A, are 10 × 10^-5 and 5.0 × 10^-10  respectively. Which of the following comments are true regarding this add—

1. The concentration of A²⁺ Ions In 0.01 (M) aqueous solution of  H₂A Js 0,01 (Ml
2. The overall Ionisation constant for H₂A is 5.0 ×10^-15
3. In 0,01 aqueous solution of H₂A, the molar concentration of H30* Ions Is twice that of A²⁺ ions.
4. In 0,01 (M) aqueous solution of H₂A, (H₃O+) ×  [HA-)

Answer: 2. The overall Ionisation constant for H₂A is 5.0 ×10^-15

Question 66. At 25C, Kb for CN^– (the conjugate base of HCN) is 2.5 × 10^-5, If 25rnf, of 0,01 (M) aqueous NaOH solution Is added to 50ml, of0,0 1 (M) HCN solution, then-

1. The pH of the solution is 1 1.2
2. The pH of the respiting solution is 9.4
3. At 20*c, the ionization constant for him is 4 × 10^-18
4. At 23*c, the ionization constant for him is 2.5 × 10^-5

Answer: 2. Ph of the respiting solution is 9.4

Equilibrium Multiple Choice Questions for Class 11

Question 67. At a given temperature, If the solubility products for MX, MA_2, and M., B2 In the water are 10^-22, and 10^-33 respectively and their solubilities are S₁, S_2, and S₃ mol-1 respectively, then

1. S₁<S₃
2. S₂>S₃
3. S₂>S₁
4. S₂=S₃

Answer: 1. S₁<S₃

Question 68. At 25-C temperature, the solubility products for BaCrO₄ and SrCrO₄ salts are 2,4 × 10¹⁰ and 3.6 × 10^-6 respectively. If an aqueous solution of K₂CrO₄ is added drop by drop to an aqueous solution containing Ba²⁺ and Sr²₄ ions with concentrations of 10^-4 and 10^-3M) respectively, then—

1. BaCrO₄ will be precipitated first
2. SrCrO₄ will be precipitated first
3. The concentration of sr²⁺ ions will be 6.6 × 10^-8mol-L^-1 when ba²⁺ ions start precipitating
4. The concentration of ba²⁺ ions will be 6.6 × 10^-8 mol-L^-1
5. when sr²⁺ ions start precipitating.

Answer: 1. BaCrO₄ will be precipitated Erst

Question 69. In a buffer solution composed of NaCN and HCN [pKa = 9.4), [NaCN] = 0.2(M) and [HCN] = 0.4(M). An aqueous solution contains Zn²⁺, Ca²⁺, Mn²⁺, and Cr³⁺ ions, each of which has a concentration of 0.1(M). If 500mL of the buffer solution is added to 500mL of this aqueous solution, then the ions that will precipitate in the resulting solution are

1. Zn⁺
2. Ca²⁺
3. Mn²⁺
4. Cr³⁺

Answer: 1. Zn⁺

Question 70. At 25°C, pKb(NH₃) = 4.74, pK_a(HF) = 3.14 and pK_a(HCN) = 9.4. Hence

1. An aqueous solution of nh₄f is acidic
2. Aqueous solution of nh₄cn is acidic
3. The pH of an aqueous solution of nh4cn is greater than that of an aqueous solution of nh4f
4. Ph values of the solutions of both nh4cn and nh4f are independent of the concentrations ofthe solutions

Answer: 1. Aqueous solution of nh4f is acidic

Question 71. Which of the following comments is true

1. If pK_avalues for the acids HA and HB are 4 and 5 respectively, then the concentration of OH- ions in 0.1(M) aqueous solution of HB will be greater than that in 0.1(M) aqueous solution of HA
2. pH of pure water at 0°C is smaller than that at 25°C
3. The degree of hydrolysis of NH₄F in its 0.1 (M) and 0.2(M) aqueous solutions is the same at a particular temperature
4. pH of an acid is 5, implying that the acid is weak.

Answer: 1. If pKa values for the acids HA and HB are 4 and 5 respectively, then the concentration of OH- ions in 0.1(M) aqueous solution ofHB will be greater than that in 0.1(M) aqueous solution of HA

Class 11 Chemistry Equilibrium Chapter MCQs

Question 72. A certain buffer solution contains equal concentrations of A- and HA. Kb for A- is 10-10. Hence

1. K_a for HA is 10^-3
2. k_a for HA is 10^-4
3. pH of the buffer is 4
4. PH of the better is 9

Answer: 2. ka for HA is 10-4

Question 73. A buffer solution containing NH3 and NHÿCl has a pH value of 9. pKb for NH3 is 4.7. If in the buffer solution, the total concentration of buffering reagents is 0.6 mol I.-1, then the amount of—

1. NH₃ in the solution is 3.4g.L^-1
2. NH₄C1 in the solution is 8.9g.L^-1
3. NH₄Cl in the solution is 21.4g.L^-1
4. NH₃ in the solution is 17.5g.L^-1

Answer: 1. NH₃ in the solution is 3.4g.L^-1

NCERT Class 11 Chemistry Chemical Thermodynamics Long Question And Answers

Question 1. Classify the following systems into open, closed, or isolated:

1. Living cell,
2. A gas is enclosed in a cylinder fitted with a movable piston.
3. The walls ofthe container and the piston are impermeable and thermally insulated.
4. The substances present in a soda water bottle. The chemicals participated in a chemical reaction occurring in a closed glass container.
5. Hot tea is kept in a thermos flask.

Read and learn More NCERT Class 11 Chemistry

Answer:

1. An open system: A living cell exchanges matter and energy with its surroundings to maintain itself.
2. A closed system: The walls of the container and the piston are impermeable and thermally insulated. So, the system cannot exchange matter and heat with its surroundings.  However, as the piston is movable, the system can do work or work can be done on it if the pressure on the die piston is decreased or increased. So, the system can exchange energy in the form of work with its surroundings,
3. A closed system: Here, the components present in the bottle constitute the system. As the bottle is closed, the system cannot exchange matter with its surroundings. However, it can exchange heat (energy) with its surroundings.
4. A closed system: Here, the chemicals constitute the system. As the reaction container is closed, the system is unable to exchange matter with its surroundings. However, it can exchange heat (energy) with its surroundings,
5. An isolated system: The walls of a thermos flask are made up of insulating materials. Again, the mouth of the flask is closed. So, the system can exchange neither matter nor energy with its surroundings.

Question 2. Identify the following an extensive or intensive property: Enthalpy, internal energy, pressure, viscosity, heat capacity, density, electric potential, specific heat capacity, molar volume, surface tension, universal gas constant, vapour pressure, number of moles, refractive index, entropy.
Answer:

An extensive property ofa system depends upon the mass ofthe substance present in the system. Its value increases as the amount of substance in the system increases. Enthalpy, internal energy, heat capacity, number of moles, and entropy are extensive properties.

An intensive property of a system is independent of the amount ofthe substance present in the system. Pressure, viscosity, density, electric potential, specific heat capacity, molar volume, surface tension, universal gas constant, vapour pressure, and refractive index are intensive properties.

Chemical Thermodynamics Long Answer Questions Class 11

Question 3. Thermodynamic state functions are path-independent quantities. Explain with an example.
Answer:

A state function is a path-independent quantity. This means that when a system undergoes a process, the change in any state function depends only on the initial and final states of the system, and not on the path of the process.

To make it clear, let us consider the following process in which the mol of an ideal gas changes its state:

2 atm, 4L, 273K→1 atm, 8L, 273K

We can make this change by either of the following two processes. However, in each of these processes, the change in a state function, viz., P or V is the same.

Question 4. Why is the change in any state function in a cyclic process zero? Is the change in any function both reversible and irreversible cyclic processes zero?
Answer:

1. The value of any state function of a system depends only on the present state of the system. In a cyclic process, the initial and the final states of a system are the same, and so are the values ofa state function at these two states.
2. Hence, the change in a state function will be zero in a cyclic process. The change in any state function depends only on the initial and the final states ofthe system.
3. It does not depend on the path followed to carry out the change. This means that a state function undergoes the same change in a process with the specified initial and final states irrespective of whether the process is carried out reversibly or irreversibly.
4. Now, in a cyclic process, the change in any state function is always zero. Hence, for both reversible and irreversible cyclic processes, the change in any state function will be zero.

Question 5. 1mol of an ideal gas participates in the process as described in the figure.

1. What type is the overall process?
2. Is this an isothermal process?
3. Mention the isobaric and isochoric steps in this process.

Answer:

1. It is a cyclic process because the system returns to its initial state after performing a set of consecutive processes.
2. This is not an isothermal process, because the temperature of the system does not remain constant throughout the process although the initial and the final temperatures are the same.
3. Step BC represents an isobaric process because the pressure ofthe system remains constant in this step. Step CA represents an isochoric process because the volume ofthe system remains constant in this step.

Question 6. Are the following changes reversible or irreversible? Give proper explanations:

1. Melting of ice at 0°C and 1 atm pressure
2. The pressure ofa gas enclosed in a cylinder fitted with a piston is 5 atm.
3. The gas is expanded against an external pressure of 1atm.

Answer:

1. At 0°C and 1 atm pressure ice remains in equilibrium with water. If the temperature ofthe system is increased by an infinitesimal amount, ice melts into the water slowly. Again, if the temperature of the system is decreased by an infinitesimal amount, water freezes into ice slowly.
2. Thus, an infinitesimal increase or decrease in temperature causes a change in the direction of the process. Hence, the melting of ice at 0°C and 1 atm pressure can be considered as a reversible process,
3. An irreversible process: The external pressure is considerably less than the pressure of the gas. So, the gas will expand rapidly without maintaining thermodynamic equilibrium during the process. Hence, this expansion will occur irreversibly.

Question 7. In the process A→ B→ C, the change in internal energy of the system in the steps A→ B and B→ C are -x kJ.mol¯1 and y kj-mol^-1, respectively. What will be the change in the internal energy ofthe system in step C→ A?
Answer:

A→ B→ C

Given, ΔU_{A→ B} = -x KJ. mol^-1 and ΔU_{B→ C} = -x KJ. mol^-1

∴ ΔU_C→A  = -ΔU_A→C = (x-y) kJ. mol^-1

[Since U is a state function, its change in die forward direction of a process is the same as that in the backward direction but opposite in sign].

Question 8. A certain amount of a gas participates in the cyclic process ABCD (follow figure). Calculate the total work done g in the process
Answer:

Pressure volume work, w = -P_ex(V₂-V₁)=-P_exΔV

In step: AB: P_ex = x atm, ΔV= (2y-y)L=yL

∴ w₁ = -P_ex ΔV = -xy L . atm

In steps BC and DA. work done is zero because the system remains constant in these steps. Instep CD: P_ex = 0.5x atm and AV = (y- 2y) L = -y L.

∴ ω₂ = -P_exΔV = —0.5x (-y L) = 0.5.xy} L . atm

So, the total work done in the process, ω = ω₁ + ω₂

= (—xy + 0.5xy) L .atm =-0.5xy L.atm

= -0.5xy x 101.3 J [since 1Latm = 101.3 J] =-50.65xy J

NCERT Solutions Class 11 Chemistry Chemical Thermodynamics

Question 9. A particular amount of gas participates separately in the two processes given below: Process-1 For which process, the work done is maximum?
Answer:

⇒ Process-1:

Step-1: ω₁ = -P_ex-ΔV = -P₁(V₂– V₁)

⇒ Step-2: ω₂ = 0 [Since the volume of  the system is constant]

⇒ Total work, ω = ω₁ + ω₂ = -P₁(V₂– V₁)

⇒ So, |ω| = |P₁(V₂-V₁)|

⇒ Process-2:

Step-1: ω₁ = 0 [∴ the volume of the system is constant]

⇒ Step-2: ω₂ = —P_ex ΔV = -P₂(V₂-V₁)

⇒ Total work, ω’ = ω₁+ ω₂ = -P₂(V₂– V₁)

⇒ So, |ω’| = |P₂(V₂-V₁)| [Since P₂ < P₁ , |ω’| < |ω’| ]

Question 10. For an ideal gas, the isothermal free expansion and adiabatic free expansion are the same processes— Explain For chemical changes, why is the change in enthalpy more useful than the change in internal energy?
Answer:

According to the first law of thermodynamics, ΔU = q+ω. In the free expansion of a gas, w = 0. Again, in an adiabatic process, q = 0 Hence, in an adiabatic free expansion of an ideal gas, the change in internal energy ΔU =q + 0=0.

Also, in an isothermal process, the change in the internal energy of an ideal gas is zero. Again for free expansion of an ideal gas, w = 0. So, in an isothermal free expansion of an ideal gas, ΔU = q + w or, 0 = q+ 0 or q = 0. Therefore, it can be concluded that both processes are the same.

Question 11. The heat required to raise the temperature of 1 mol ofa gas by 1°C is q at constant volume and q’ at constant pressure. Will q be better than, less than or equal to q’? Explain
Answer:

The heat required to raise the temperature of1 mol of gas by 1 ° at constant pressure is greater than that required at constant volume. At constant pressure, the heat absorbed by a gas is used up in two ways.

One part of it is used by the gas for doing external work, and the remaining part is utilised for increasing the temperature of the gas.

At constant volume, the heat absorbed by a gas is completely utilised for increasing the temperature of the gas as no external work (P-Vwork) is possible at constant volume. Therefore, q’ must be greater than q.

Question 12. A 0.5 mol sample of H₂(g) reacts with a 0.5 mol sample of Cl₂(g) to form 1 mol of HCl(g). The decrease in enthalpy for the reaction is 93 kj. Draw an enthalpy diagram for this reaction.

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{g}) ; \Delta H=-93 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The reaction is associated with a decrease in enthalpy. So, it is an exothermic reaction. In such a reaction, the total enthalpy of the product(s) (2Hp) is less than that of the reactant(s)(£flfi). Therefore, in the enthalpy diagram for the reaction, ZHp lies below ZHR.

Question 13. A 1 mol sample of N₂ (g) reacts with 1 mol of O₂(g) to form 2 mol of NO₂(Og), where the increase in enthalpy is 180.6kj. Draw An enthalpy diagram for this reaction.
Answer:

⇒ \(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) ; \Delta H=+180.6 \mathrm{~kJ}\)

In the reaction, the enthalpy increases. So, it is an endothermic reaction. In such a reaction the total enthalpy ofthe product(s)(ΣH_R) is greater than that ofthe reactant(s) (ΣH_p). Therefore, in the enthalpy diagram for the reaction, ΣH_p lies above ΣH_R

ΣH_p

ΣH  =  +180.6 kJ

ΣH_R

Class 11 Chemistry Chemical Thermodynamics Long Answer Solutions

Question 14. Identify the exothermic and endothermic changes:
Answer: 

⇒ \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{NO}_2(g)+57.0 \mathrm{~kJ}\)

Answer:

⇒  \(\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{NO}_2(\mathrm{~g})+57.0 \mathrm{~kJ}\).

Heat is realed in this reacton,. so it’s an exothermic reaction.

H₂O(s) + 6.02 kJ→H₂O(l) . Heat is absorbed in this reaction. So, it is an endothermic process.

⇒ \(\mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+\mathrm{H}_2(g)-130 \mathrm{~kJ}\)

⇒ \(\text { i.e. } \mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g)+130 \mathrm{~kJ} \rightarrow \mathrm{CO}(g)+\mathrm{H}_2(g)\)

Question 15. Write down the thermochemical equations for the following reactions:

1. A 1mol sample of methane gas reacts with 2 mol of oxygen gas to form lmol of carbon dioxide and 2mol of water. In this reaction, 890.5 kl of heat is produced.
2. A 1mol sample of carbon (graphite) reacts with 1 mol ofoxygen to form 1 mol of carbon dioxide gas. The heat evolved in this reaction is 393.5 kj.
3. 6 mol of carbon dioxide gas reacts with 6 mol of water to form 6 mol of oxygen gas and lmol of glucose. The heat absorbed in this reaction is 2200 kj.

Answer:

⇒ \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)+890.5 \mathrm{~kJ}\)

⇒  \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \( \Delta H=-890.5 \mathrm{~kJ}\)

⇒ \(6 \mathrm{CO}_2(g)+6 \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(s)+6 \mathrm{O}_2(g)-2800 \mathrm{~kJ}\)

⇒ \(\text { i.e., } 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s})+6 \mathrm{O}_2(\mathrm{~g}) \text {; }\)

⇒  \( \Delta H=-2800 \mathrm{~kJ}\)

Question 16. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l); \dot{2} \Delta \mathrm{H}=-285.8 \mathrm{~kJ}\)  What will be the value of ΔH for the reaction: 2H₂O(l)→ 2H₂(g) + O₂(g)?
Answer:

⇒  \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) \dot{2} \Delta H=-285.8 \mathrm{~kJ}\)

If this equation is written in the reverse manner, we have

⇒ \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) ; \Delta H=+285.8 \mathrm{~kJ}\)

Multiplying this equation by 2, we have

⇒ \(2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) ; \Delta H=4571.6 \mathrm{~kJ}\)

Chemical Thermodynamics Long Questions and Answers Class 11

Question 17. Why is ΔH = ΔU for the following two reactions? Explain.

1. NaOH(aq) + HCl(aq)→ NaCl(aq) + H₂O(l)
2. CHΔ(g) + 2O₂(g) -> CO₂(g) + 2H₂O(g)

Answer: This reaction occurs In a solution. For a reaction occurring in a solution, All = AH

For this reaction An (total number of moles of gaseous products – total number of moles of gaseous reactants) = (1 +2)-(l + 2) = 0. So, Δ7 = ΔU according to the relation ΔH = ΔU + ΔT.

Question 18. Give an example of a reaction for each of the following relations between AH and ΔH: 

1. ΔH<ΔH
2. ΔH > ΔH
3. ΔH = AH.

Answer:

In a reaction, if a gaseous substance (either as a reactant or as a product or both) participates, the change in enthalpy in the reaction at constant pressure and temperature is given by ΔH= AU + ΔnRT. K An (total number of moles of gaseous products – total number of moles of gaseous reactants) >0, < 0 or =0, then AH > AH, AH < AH or All = AH respectively.

⇒ \( 2 \mathrm{H}_2 \mathrm{O}(g) \rightarrow 2 \mathrm{H}_2(g)+\mathrm{O}_2(\mathrm{~g})\)

⇒ \(\Delta n=(2+1)-2=+1 \text {. So, } \Delta H>\Delta U\)

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\Delta n=0-(2+1)=-3 \text {. So, } \Delta H<\Delta U\)

⇒ \(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g)\)

⇒ \(\Delta n=2-(1+1)=0 . \text { So, } \Delta H=\Delta U\)

Chemical Thermodynamics Chapter 6 Long Answer Questions

Question 19. In which of the following reactions At 25°C, does the standard enthalpy change correspond to the standard enthalpy of formation of H₂O(Z)? Give reasons.

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{3}{2} \mathrm{O}_3(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)\(2 \mathrm{H}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
Answer:

1. The standard enthalpy change in this reaction does not indicate the standard heat of formation of H₂O(Z) because, the standard state of oxygen at 25 °C, is O₂(g) and not O₃(g).
2. The standard enthalpy change in this reaction refers to the standard heat of formation of H₂O(Z) because 1 mol of H₂O(Z) is formed from its stable constituent elements.
3. The standard enthalpy change in this reaction does not indicate the standard heat of formation of H₂O(Z). This is because the stable state of hydrogen at 25 °C, is not H(g).

Question 20. Which one of the given reactions indicates the formation reaction ofthe compound produced in the reaction?

1. S (monoclinic) + O₃(g)→ SO₃(g)
2. C (graphite, s) + 2H₂(g)→CH₄(g)
3. N₂(g) + O₂(g)→2NO(g)

Answer:

In the formation reaction of a compound, 1 mol of the compound is formed from its constituent elements. S(s, monoclinic) +O₂(g)→SO₂(g), at 25 °C this reaction does not represent the formation of S02(g) because, the stable form of sulphur is S(s, rhombic) at 25°C.

2. C(s, graphite) + 2H₂(g)→CH₄(g), at 25 °C reaction represents the formation reaction of CH₄(g) because lmol of CH₄(g) is formed from its stable constituent elements.

3.  N₂(g) + O₂(g)→2NO(g), at 25°C this reaction does not represent the formation reaction of NO(g) because 2 mol of NO(g) are formed in the reaction.

Question 21. The standard heats of combustion of CH₄(g) and C₂H₆(g) are -890 kj.mol^-1 and -1560 kj. mol^-1 respectively. Why is the calorific value of C₂H₆(g) lower than that of CH₄ (g)?
Answer:

The enthalpy of combustion of a compound is always negative. So, the standard enthalpy of combustion of the given compound, ΔHº_C = -Q kj .mol-1. The thermochemical equation for this combustion reaction

⇒ \(\mathrm{C}_x \mathrm{H}_y(l)+\left(x+\frac{y}{4}\right) \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H_c^0=-Q \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 22. Identify whether the enthalpy of the initial state is greater than, less than or equal to that of the final state in the following changes: solid;→ liquid; vapour→liquid →vapour
Answer: Solid→Liquid:

It is an endothermic process. In this process

⇒ \(\Delta H>0 \text {, i.e., } H_{\text {liquid }}-H_{\text {solid }}>0 \text { or, } H_{\text {liquid }}>H_{\text {solid }} \text {. }\)

Therefore, the enthalpy of the final state will be greater than that ofthe initial state.

Vapour→Solid: It is an exothermic process. So, in this process

Question 23. What docs Mi signify in each of the following equations?

1. HCl(g) + 5H₂O(l)→HCl(5H₂O); ΔH = -64 kJ
2. HCl(g) + aq→HCl(aq); ΔH = -75 kJ
3. HCl(5H₂O)+20H₂O(l)→HCl(25H₂O); ΔH=-8.1 kJ

Answer:

In this process, 1 mol of HCl dissolves in 5 mol of water, forming a solution of definite concentration. The enthalpy change in such a process is known as the integral heat of the solution. So, ΔH, in the process Indicates the integral heat of the solution.

In this process, 1 mol HCl dissolves in a large amount of water, forming an infinitely dilute solution. The enthalpy change in such a process is called the heat of solution. So, vH, in process 2, indicates the heat of the solution.

This is a dilution process because a solution with a definite concentration is diluted by adding solvent to it. So, ΔH, in this process, indicates the heat of dilution.

Change in enthalpy remains the same whether a reaction is carried out in one step or several steps under similar reaction conditions Explain the rearms.

Question 24. Given (at 25°C and 1 atm pressure):

1. C (s, diamond) + O₂(g)→CO₂(g); ΔH°=-393.5 kj-mol^-1
2. C (s, graphite) +O₂(g)→CO₂(g); ΔH°=-391.6 kj-mol^-1

Find the standard heat of transition from graphite to diamond.
Answer:

⇒ \(\mathrm{C}(s, \text { diamond })+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒  \(\Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{C}(s, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒  \(\Delta H^0=-391.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

[Subtracting equation 1 From equation We get C(s, graphite)→ C(s, diamond); AH° = 1.9 kj. mol^-1 So, the heat of transition for this process is 1.9 kj.mol^-1

Class 11 Chemical Thermodynamics Long Question and Answers

Question 25. At 25°C, if standard enthalpies of formation o/MX(s), M+(ag) and X-(aq) are -x, y and -zkj-mol^-1 respectively, then what will the heat of reaction before the reaction M+(aq) + X~(aq)-+ MX(s).
Answer:

M+(aq) + X-(aq)→MX(s)

The standard heat of reaction,

⇒ \(\Delta H^0=\Delta H_f^0[\mathrm{MX}(s)]-\Delta H_f^0\left[\mathrm{M}^{+}(a q)\right]-\Delta H_f^0\left[\mathrm{X}^{-}(a q)\right]\)

=\((-x-y+z) \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

Question 26. At 25°C, the bond dissociation energy of N₂(g) is 946 kj.mol-1. What does it mean? What would be the standard atomisation enthalpy of N₂(g) at 25°C?
Answer:

At 25°C, the standard bond dissociation energy of N₂(g) is 946 kj.mol-1. This means that the energy required to break 1 mol of N=N bonds completely in the gaseous state to form gaseous nitrogen atoms is 948 kj. At 25 °C, the standard state of nitrogen is N₂(g). Now, the formation of 1 mol of N(g) takes place by the following process \(\frac{1}{2} \mathrm{~N}_2(g) \rightarrow \mathrm{N}(g)\)

Since 1 mol of N(g) is produced from N₂(g) in the process [1], the enthalpy change in this process at 25 will be equal to the standard atomisation enthalpy of nitrogen.

In process [1], change in enthalpy \(=\frac{1}{2}\) x bond dissociation energy of N=N

= \(\frac{1}{2} \times\) 946 = 473 k.mol-1 Therefore, at 25 °C the standard atomisation enthalpy of nitrogen is 473 kJ.mol-1.

Chemical Thermodynamics Long Answer NCERT Solutions

Question 27. A—B bonds present in AB3(g) molecule undergo stepwise dissociation by the following sequence of steps.

1. AB₃(g)→AB₂(g) + B(g) 
2. AB₂(g)→AB(g) + B(g)
3. AB(g)→A(g) + B(g)

Answer:

Bond energy of A — B bond in AB₃(g) molecule = the average bond dissociation energy of three A— B bonds in AB₃(g) molecule. If the standard enthalpy change in step is ΔH° kj.mol-1, then the bond dissociation energy of the A-B bond in AB₃(g) molecule

= \(\frac{x+\Delta H^0+z}{3} \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \text {. }\)

Now, the bond dissociation energy of the A-B bond = kJ.mol^-1

⇒ \(y=\frac{x+\Delta H^0+z}{3} \quad \text { or, } \Delta H^0=3 y-(x+z) \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

Therefore, the standard enthalpy change in step (2)is [3y-(x+z)kJ.mol^-1

Question 28. The following changes are performed on 1 mol of N₂ gas

1. Pressure Is decreased at a constant temperature
2. Volume is decreased at a constant temperature
3. What will be the sign of S_sys In These changes?

Answer:

If the pressure of a gas is decreased at a constant temperature, the volume of the gas increases. At a larger volume of a gas, the gas molecules get a greater chance to move about. As a result, the randomness of the molecules increases, which increases the entropy ofthe gas. Hence, A = +ve.

At constant temperature, the decrease in the volume of a gas reduces the availability of space for the movement of gas molecules. This results in a decrease in the randomness of the molecules; consequently, the entropy ofthe gas decreases. Hence, Δ = -ve

Question 29. The following two reactions occur spontaneously. What will be the signs of AS and AS_surr in these two reactions?

1. A(s)→ B(s) + C(g); ΔH > 0
2. 2X(g)→ X₂(g); ΔH < 0

Answer:

At a given temperature and pressure, the change in entropy of the surroundings, \(\Delta S_{\text {surr }}=-\frac{\Delta H}{T}\)…………………(1)

In this reaction, ΔS_sys = + ve is the gaseous substance produced in the reaction. Since, in this reaction, ΔH > 0; according to equation ΔS_surr = – ve.

In this reaction, Δ S_sys = -ve as the number of gas particles decreases. Again, in this process, AH < 0. So, according to the equation, ΔS_surr > 0

NCERT Chemical Thermodynamics Long Answer Solutions Class 11

Question 30. For a reaction ΔH > 0, and another ΔH < 0. For both the reactions ΔSsys < 0. Which one is likely to occur spontaneously? Which one always occurs nonspontaneous? define

1. Gibbs free energy
2. The standard energy of formation of a substance
3. The standard free energy change in a chemical reaction.

Answer:

For the reaction with ΔH>0, ΔS_surr. is -ve. Again, for this reaction ΔS_sys < 0, and hence ΔS_sys, + ΔS_surr < 0 i.e.  ΔS_univ < 0. So, this reaction will be non-spontaneous.

For the reaction with ΔH < 0,  ΔS_surr is -ve. Again, for this reaction ΔS_surr < 0, Therefore,

ΔS_univ = (ΔS_sys, + ΔS_surr may be Positive or negative depending on the magnitudes of S_sys and S_surr If |ΔS_sys|< |ΔS_surr| then the reaction will be spontaneous.

NCERT Class 11 Chemistry Chemical Thermodynamics Multiple Choice Questions

Question 1. Which of the following statements is true

1. Entropy increases when water vaporizes
2. Randomness decreases in the fusion of ice
3. Randomness increases in the condensation of water vapor
4. Randomness remains unchanged during the vaporization of water

Answer: 1. Entropy increases when water vaporizes

Vaporization of water (water → water vapor) involves an increase in the entropy of the system because the molecular randomness in water vapor is greater than that in water.

Question 2. Identify the correct statement in a chemical reaction

1. The entropy always increases
2. The change in entropy along with a suitable change in enthalpy decides the tire rate of reaction
3. The enthalpy always decreases
4. Both tire enthalpy and tire entropy remain constant

Answer: 2. The change in entropy along with a suitable change in enthalpy decides the tire rate of reaction.

For a reaction occurring at a constant temperature and pressure, ΔG<0, ie., ΔH-TΔS<0 the change in entropy (Δs) and (flng with the change in enthalpy (ΔH) determines the spontaneity of a reaction.

Question 3. The condition for the spontaneity of a  process is—

1. Lowering of entropy at constant temperature & pressure
2. Lowering of Gibbs free energy of tire system at constant temperature and pressure
3. Increase in entropy of tire system at constant temperature and pressure
4. Increase in Gibbs free energy of the universe at constant temperature and pressure

Answer: 2. Lowering of Gibbs free energy of tire system at constant temperature and pressure

The condition of spontaneity for a reaction to occur at constant t and p is ag < 0.

Chemical Thermodynamics MCQs Class 11

Question 4. P-v work done by an ideal gaseous system at constant volume is the internal energy of the system)

1. – ΔP/P
2. Zero
3. -VΔp
4. -Δe

Answer: 2. Zero

As the system’s volume remains constant in the process, the system cannot do any external work.

Question 5. Mixing of two different ideal gases under an isothermal reversible condition wall leads to

1. Increase in Gibbs free energy of the system
2. No change in the entropy of the system
3. Increase in entropy of the system
4. Increase in enthalpy of the system

Answer: 3. Increase in entropy of the system

The molecular randomness in a gas mixture is larger than that in the individual gases because a gas mixture contains a larger number of molecules than that in the individual gases. Consequently, the mixing of two gases will lead to an increase in the entropy of the system.

Question 6. For an isothermal expansion of an ideal gas, the correct of the thermodynamic parameters will be

1. ΔU =0,Q = 0,ω≠0 and ΔH≠ 0
2. ΔU ≠0,Q ≠ 0,ω≠0 and ΔH= 0
3. ΔU =0,Q ≠ 0,ω=0 and ΔH≠ 0
4. ΔU =0,Q≠ 0,ω≠0 and ΔH= 0

Answer: 4. ΔU =0,Q≠ 0,ω≠0 and ΔH= 0

For an ideal gas undergoing an isothermal process, ‘ah = 0. When a gas undergoes expansion it absorbs heat from its surroundings and performs external work. So, in such a process, q≠0 and w≠0.

We know, Δh = Δu+ Δ(pv). For an ideal gas, pv = nrΔt.

∴ Δh = Δu+Δ(nrt) = Δu+nrΔt

In an isothermal process of an ideal gas, aΔu = 0, at- 0, and hence ΔH = 0.

Question 7. The change in entropy (ds) is defined as

1. \(d s=\frac{\delta q}{t}\)
2. \(d s=\frac{d h}{t}\)
3. \(d s=\frac{\delta q_{r e v}}{t}\)
4. \(d s=\frac{(d h-d g)}{t}\)

Answer: 3. \(d s=\frac{\delta q_{r e v}}{t}\)

If a system undergoing a reversible process at tk absorbs an amount of heat, 6qrev, then the change in entropy of the system in the process, \(d s=\frac{\delta q_{r e v}}{t}.\).

Question 8. Δh for cooling 2 mol ideal monoatomic gas from 225 °c to 125°c at constant pressure will be?

1. 250R
2. -500R
3. 500R
4. -250R

Answer: 2. ΔH=nCp(T₂-T₁)

Question 9. For a spontaneous process, correct statement(s) is (are)

1. \(\left(\Delta G_{s y s}\right)_{T, P}>0\)

2. \(\Delta S_{\text {sys }}+\Delta S_{\text {surr }}>\)

3. latex]\left(\Delta G_{s y s}\right)_{T, P}<0[/latex]

4. \(\left(\Delta U_{s y s}\right)_{T, V}>0\)

Answer: 3. \(\left(\Delta G_{s y s}\right)_{T, P}<0\)

For a spontaneous process at constant pressure and temperature,

ΔS >0 or, ΔS + ΔS_surr> 0

Now, ΔG = -TΔS_univ. For a spontaneous process

ΔS_univ > 0 , and hence ΔG < 0

Question 10. Given: C+O₂→CO₂; ΔH⁰ = -xkj; 2CO+O₂→2CO₂; ΔH⁰=-ykj The heat of formation of carbon monoxide will be

1. Y+2x/2
2. Y+2x
3. 2x-y
4. 2x-y/2

Answer: 1. Y+2x/2

C+O₂→CO₂; ΔH°=-xKJ; 2CO+O₂→2CO₂;ΔH°=-yK

Subtracting equation [2] from the equation obtained by multiplying equation [1] by 2, we have,

2CO+O₂→2CO₂;ΔH⁰=-yK

∴ Heat of formation of co \(=\left(\frac{y-2 x}{2}\right) \mathrm{kj} \cdot \mathrm{mol}^{-1}\)

Question 11. The enthalpy of vaporization of a certain liquid at its boiling point of 35°c is 24.64 kl-mol^-1. The value of change in entropy for the process is—

1. 704 J.k^-1 Mol^-1
2. 80J.k^-1 Mol^-1
3. 24.64j.k^-1. Mol^-1
4. 7.04 j.k^-1. Mol^-1

Answer: 2. 80J.k^-1. Mol^-1

⇒ \(\Delta S_{v a p}=\frac{\Delta H_{v a p}}{T_b}\)

= \(=\frac{24.64 \times 10^3 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{(273+35) \mathrm{K}}\)

⇒  \( =80 \mathrm{~j} \cdot \mathrm{k}^{-1} \cdot \mathrm{mol}^{-1}\)

Question 12. ΔH and ΔS of a certain reaction are -400 kj. mol^-1 and -20kj.mol^-1.k^-1 respectively. The temperature below which the reaction is spontaneous is

1. 100K
2. 20°C
3. 20K
4. 120°C

Answer: 3. 20K

For a spontaneous reaction, Δg = ΔH- TΔs <0

⇒ \(T \Delta S>\Delta H\)

Or, \(T>\frac{\Delta H}{\Delta S}\)

Or, \(T>\frac{400}{} \mathrm{~K}_1 \text { or, } T>20 \mathrm{~K}\)

NCERT Solutions Class 11 Chemistry Chemical Thermodynamics MCQs

Question 13. For the reaction x²y⁴(f)→2xy(g) at 300 k the values of a u and as are 2kcal and 20 cal.k-1 respectively. The value of Δg for the reaction is-

1. -3400 Cal
2. 3400 Cal
3. -2800 Cal
4. 2000 Cal

Answer: 3.  ΔH = ΔU+Δn_gRt or, ΔH = 2000 + 2 × 2 × 300

Δn_g = 2-0 = 2, R = 2 cal -k^-1– mol^-1 ]

ΔH = 3200 cal

ΔG = ΔH – TΔS

= 3200- (300 ×  20)

= -2800 cal

Question 14. For the reaction 2SO₂(g) + O(g)⇌ 2SO₃(g) at 300 k, the value of ΔG⁰ 0the equilibrium constant value for the reaction ax that temperature is (r is gas constant)—

1. 10 Atm^-1
2. 10 Atm
3. 10
4. 1

Answer: 3. 10

2SO₂(g) + O₂(g)⇌ 2SO₃(g)

Now, ΔG = -RTlnk

Or, -690.9R = -RTlnk

∴ K = 10¹

Question 15. The condition for the reaction to occur spontaneously is

1. ΔH must be negative
2. ΔS must be negative
3. (ΔH- tΔs) must be negative
4. (Δ H+ tΔs) must be negative

Answer: 3. (ΔH- tΔs) must be negative

At constant temperature and pressure, for a spontaneous process of the reaction ΔG < 0. According to Gibbs’s equation ΔG = ΔH- TΔS. Therefore the condition spontaneity is, ΔH- TΔS < 0.

Question 16. During a reversible adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio \(\frac{c_p}{c_v}\) for the gas is-

1. \(\frac{3}{2}\)
2. \(\frac{7}{2}\)
3. \(\frac{5}{3}\)
4. \(\frac{9}{7}\)

Answer: 1. \(\frac{3}{2}\)

For adiabatic reversible process, \(t p^{\frac{(1-\gamma)}{\gamma}}=\text { constant }\)

Or, \(p t^{\frac{\gamma}{(1-\gamma)}}=\text { constant }\)

Again as mentioned, P ∝ T³

Or, PT^-3= constant

Comparing equations (1) and (2), it may be written as

⇒ \(\frac{\gamma}{1-\gamma}=-3\)

∴ \(\gamma=\frac{3}{2}\)

Question 17. The heat of neutralization of a strong base and a strong add is 13.7 kcal the heat released when 0.6 mol HCl solution is added to 0.25 mol of NaOH is-

1. 3.425kcal
2. 8.22kcal
3. 11.645kcal
4. 13.7kcal

Answer: 1. 3.425kcal

⇒  \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{O}(l)\) ‘

⇒  \(\Delta H^0=-13.7 \mathrm{kcal}\)

⇒ \(0.25 \mathrm{~mol} \mathrm{HCl}(a q)+0.25 \mathrm{~mol} \mathrm{NaOH}(a q)\)

⇒ \(0.25 \mathrm{~mol} \mathrm{NaCl}(a q)+0.25 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\Delta H^0=-13.7 \times 0.25=-3.425 \mathrm{kcal}\)

∴ Amount of heat released = 3.425 kcal

Question 18. The entropy change involved in the isothermal reversible expansion of 2 mol of an ideal gas from a volume of 10 cm³ to a volume of 100 cm3 at 27°cis —

1. 38.3j.mol^-1.k^-1
2. 35.8j.mol^-1.k^-1
3. 32.3j.mol^-1. K^-1
4. 42.3j.mol^-1.k^-1

Answer: 1. 38.3j.mo^-1.k^-1

For isothermal reversible expansion,

⇒ \(\delta s=n r \ln \frac{v_2}{v_1}=2 \times 8.314 \ln \frac{100}{10}=38.3 \mathrm{~j} \cdot \mathrm{mol}^{-1} \cdot k^{-1}\)

Question 19. Among the following expressions which one is incorrect-

1. \(w_{r e v, i s o}=-n R T \ln \frac{V_f}{V_i}\)

2. \(\ln K=-\frac{\Delta H^0-T \Delta S^0}{R T}\)

3. \(K=e^{-\Delta G^0 / R T}\)

4. \(\frac{\Delta G_{\text {sys }}}{\Delta S_{\text {total }}}=-T\)

Answer: 2. \(\ln k=-\frac{\delta h^0-t \delta s^0}{r t}\)

ΔG° = -RTInk and ΔG° = ΔH₀– TΔS₀

⇒  \(\Delta H^0-T \Delta S^0=-R T \ln K \text { and } \ln K=-\left(\frac{\Delta H^0-T \Delta S^0}{R T}\right)\)

Question 20.  A cylinder filled with 0.04mol of ideal gas expands reversibly from 50ml to 375ml at a constant temperature of 37.0°c. As it does so, it absorbs 208 j heat, q and w for the process will be (a = 8.314 j.mol^-1.k^-1 )

1. q = +208 J, w = +208 J
2. q = +208 J, w = -208 J
3. q = -208 J , w = -208 J
4. q = -208 J , w = +208 J

Answer: 2. Q = +208 J, w = -208 J

The process is isothermal and the system is an ideal gas. So, in this process, a u = 0. Given that q = +208J.

∴ Au = q + w or, 0 = 208 + w

∴ W = -208J .

Question 21. For complete combustion of ethanol, \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)\)

The amount of host produced ao measured in a bomb calorimeter, u 1364,47kJ mol 1 at 25°C, assuming ideality the enthalpy of combustion, ΔH_c(  – mol^-1 j for the reaction will be (R= 8.3m J. k^-1. Mol^-1)

1. -1350.50
2. -1366.95
3. -1361.95
4. -1460.50

Answer: 2. -1366.95

In a bomb calorimeter, a reaction occurs under constant volume. Hence, q_v = ΔU. For the given reaction, an = 2-3 = -1 we know, ΔH = ΔU+ΔnRT

∴ ΔH = [- 1364.47 × 8.314 × 10^-3× 298] kj . Mol^-1

=-1366.95 kj- mol^-1

Multiple Choice Questions for Class 11 Chemistry Chemical Thermodynamics

Question 22. The following reaction is performed at 298k. 2No(g) + O₂(g/ v= 2nO₂(g) the standard free energy of formation is 86.6 kJ/mol at 298k, what is the standard free energy of formation of NO₂(g) at 298k (kp – 1.6 × 10¹²) 

1. 8660- \(\frac{\ln \left(1.6 \times 10^{12}\right)}{r(298)}\)

2. 0.5[2 × 86600 -R(298) in(1.6 × 10¹²)

3. R(298)ln(1.6 × 10¹²)- 86600

4. 86600 + R(298) In (1.6 × 10¹²)

Answer: 2. 0.5[2 × 86600 -R(298) in(1.6 × 10¹²)

Given: \(T=298 \mathrm{~K}, \Delta G_f^0(\mathrm{NO})=86.6 \mathrm{~kJ} / \mathrm{mol},\)

⇒ \(\Delta G_f^0\left(\mathrm{NO}_2\right)=?, K_p=1.6 \times 10^{12}\)

⇒ \( 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_0(g) \rightleftharpoons 2 \mathrm{NO}_0(g)\)

∴ \(\Delta G_r^0=2 \Delta G_f^0\left(\mathrm{NO}_2\right)-2 \Delta G_f^0(\mathrm{NO})-\Delta G_f^0\left(\mathrm{O}_2\right)\)

⇒ \(\text { or, } \Delta G_r^0=2 \Delta G_f^0\left(\mathrm{NO}_2\right)-2 \times 86600\)

⇒ \(Again,\Delta G_r^0=-R T \ln K_p
or, 2 \Delta G_f^0\left(\mathrm{NO}_2\right)-2 \times 86600=-R(298) \ln \left(1.6 \times 10^{12}\right)\)

⇒ \(\text { or, } \Delta G_f^0\left(\mathrm{NO}_2\right)=\frac{2 \times 86600-R(298) \ln \left(1.6 \times 10^{12}\right)}{2}\)

= 0.5[2 × 86600- (298)Ln(1.6× 10¹²)

Question 23. The standard Gibbs energy change at 300k for the reaction 2az±b + c is 2494.2 j. At a given time, the composition of the reaction mixture is [a] \([a]=\frac{1}{2},[b]=2 \text { and }[c]=\frac{1}{2}\) reaction proceeds in the [a = 8.314)/k/mol, c=2.718] 

1. Forward direction because q<kc
2. Reverse direction because q<kc
3. Forward direction because q > kc
4. Reverse direction because q > kc

Answer: 4. Reverse direction because q > kc

⇒ \(2 A \rightleftharpoons B+C\)

Given: T=300K, ΔG° = 2494.2J,

R = 8.314 J. K^-1.mol^-1

Now, ΔG° = -2.303 RTlogK_c

or, 2494.2 = -2.303 × 8.314 × 300 ×  logk_c

or, logK_c = -0.4342

∴ K_c = 0.3679

⇒ \(Q_c=\frac{[B][C]}{[A]^2}=\frac{2 \times \frac{1}{2}}{(1 / 2)^2}\)

= 4

Question 24. The heats of carbon and carbon monoxide combustion are -393.5 and -283.5 kj.mol^-1 respectively. The heat of formation (in kj) of carbon monoxide per mole is

1. 110.5
2. 676.5
3. -676.5
4. -110.5

Answer: 4. -110.5

C(s) + O₂(g) →CO₂(g) , ΔH° = -393.5 kj.mol^-1 ………………(1)

⇒ \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g), \Delta H^0=-283.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)  ………………(2)

Subtracting equation (2) from equation (1) we get,

⇒ \(\mathrm{C}(s)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}(g)\)

ΔH° =[- 393.5- (-283.5)] kj.mol^-1

=-110.0 kj-mol^-1

Therefore, the heat of the formation of CO(g)

= -110.0 kj-mol^-1

Question 25. Given, C(graphite) + O₂(g) → CO₂(g), Δ_rH⁰= -393.5 kj .mol^-1

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l), \delta_r H^0=-285.8 \mathrm{~kj} \cdot \mathrm{mol}^{-1} \)

⇒  \(\mathrm{CO}_2(\mathrm{~g})+2 \mathrm{h}_2 \mathrm{O}(l) \rightarrow \mathrm{CH}_4(g)+2 \mathrm{O}_2(\mathrm{~g})\)

Based on the above thermochemical equations, the value of arh° at 298 k for the reaction, c(graphite) + 2h₂(g) → CH₄(g) , will be—

1. -748 Kj.mol^-1
2. -144.0kj.mol^-1
3. +74.8 kj.mol^-1
4. +144.0kj.mol^-1

Answer: 1. -748 Kj.mol^-1

⇒ \(\mathrm{C}(\text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta_r H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)  …………………..(1)

⇒ \(\mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{CH}_4(g)+2 \mathrm{O}_2(g)\) …………………..(2)

⇒ \(\mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g})\)

⇒ \(\Delta_r H^0=+890.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) …………………..(3)

(1) eqn. +2 x eqn. (2) no. + eqn. (3) no.

⇒  \(\mathrm{C}(\text { graphite })+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_4(\mathrm{~g})\)

⇒  \(\Delta_r H^0=[-393.5-2 \times 285.8+890.3] \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

= \(-74.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 26. The combustion of benzene (z) gives CO₂(g) and h₂O(z). Given that the heat of combustion of benzene at constant volume is -3263.9 kj -mol^-1 at 25°c, the heat of combustion (in kj .mol^-1 ) of benzene at constant pressure will be [r = 8.314 j.k^-1.mol^-1 )

1. -4152.6
2. -452.46
3. 3260
4. -3267.6

Answer: 4. -3267.6

⇒ \(\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta n=6-\frac{15}{2}=-\frac{3}{2}\)

⇒ \(\Delta H=\Delta U+\Delta n R T\)

⇒ \(\Delta H=\left[-3263.9-\frac{3}{2} \times 8.314 \times 10^{-3} \times 298\right] \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

=\(-3267.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 27. If \(\frac{1}{2} a \rightarrow b, \quad \delta h=+150 \mathrm{~kj} \cdot \mathrm{mol}^{-1} ; \quad 3 b \rightarrow 2 c+d\)   ah=-125 kj mol^-1 ; e + a-*2d , ΔH= + 350 k. Mol^-1 then ah of the reaction b + d→ £ + 2c will be

1. 525Kj.mol^-1
2. -175Kjmol^-1
3. -325Kj.mol^-1
4. 352Kj.mol^-1

Answer: 2. -175Kjmol^-1

½A→ B: ΔH = + 150kJ.mol^-1

3B→2c + D; ΔH = -125 kj. mol^-1

E + A →2D; ΔH = +350 kj . mol^-1

Subtracting equation [3] from (2 x equation (1) + equation (2), we have

B + D→E + 2c; ΔH = 2 × 150- 125- 350

= -175 kJ.

Question 28. Which is the correct option for free expansion of an ideal gas under adiabatic conditions—

1. q=0, ΔT≠0, W=0
2. q≠0, ΔT≠0, W=0
3. q=0, ΔT≠0, W=0
4. q=0, ΔT<0, W≠0

Answer: 3. q=0, ΔT≠0, W=0

For an adiabatic process, q = 0, and for free expansion of a gas, w = 0.

∴ Δu=q + w = 0 + 0 = 0. For an ideal gas undergoing an isothermal process, ah = 0. So, the given process is isothermal and hence at = 0.

Question 29. The enthalpy change for the reaction, 4h(g)→2h2(g) is -869.5 kl. The dissociation energy of the H – Hbond is

1. -434.8 kJ
2. -869.6kJ
3. +434.8 kJ
4. +217.4kJ

Answer: 3. +434.8 kJ

We know, bond formation enthalpy =(-)x bond dissociation enthalpy. Now, bond formation enthalpy for 2 mol h —h bonds = -869 kJ

Bond dissociation enthalpy for h— h bond \(=\frac{1}{2} \times 869=434.5 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

Question 30. If the enthalpy change for the transition of liquid water to steam is 30 kj.mol-1 at 27°c, the entropy change in ] mol^-1 k^-1 for the process would be

1. 10
2. 1.0
3. 0.1
4. 100

Answer: 4. 100

Question 31. In which of the given reactions, standard reaction entropy change (as0) is positive and standard Gibbs energy change (ag°) decreases sharply with increasing temperature—

1. \(\mathrm{c} \text { (graphite) }+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})\)
2. \(\mathrm{co}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)
3. \(\mathrm{mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{mgo}(\mathrm{s})\)
4. \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

Answer: 1. \(\mathrm{C} \text { (graphite) }+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})\)

In the reaction, C(graphite) \(+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})\) number of gas molecules increases. As a result, the system’s entropy increases (ΔS° > 0).

We know, ΔG = ΔH – TΔS …………….(1)

for the given reaction ΔH° < 0, as it is a combustion reaction. Since at ΔH° < 0 and ΔS° < 0 according to the reaction (1), ΔG ° decreases with rise in temperature.

Question 32. For the reaction, X₂O₄(l)→2XO₂(g), ΔU=2.1 kcal, as = 20 cal.k^-1 at 300 k. Hence, ΔG is—

1. 2.7 kcal
2. -2.7 kcal
3. 9.3kcal
4. -9.3kcal

Answer: 2. -2.7 kcal

ΔH= ΔU +ΔnRT

For the given reaction, an = 2.

∴ ΔH = 2.1 + 2 × 1.987 × 10^-3 × 300 =3.2922 kcal and

ΔG = ΔH- TΔS= 3.2922 -300 × 20 × 10^-3 =-2.7 kcal

Chemical Thermodynamics Chapter 6 NCERT MCQs

Question 33. The heat of combustion of carbon to CO₂ is -393.5 kj/mol. The heat released upon the formation of 35.2g of CO₂ from carbon and oxygen gas is

1. -315 Kj
2. +315 Kj
3. -630 Kj
4. -3.15kj

Answer: c(s) + O₂(g)→CO₂(g); afh = -393.5 kj.mol^-1

The heat released on formation of 44g CO₂ =-395.5 kj-mol^-1

The heat released by the formation of 35.2g of CO₂

⇒ \(-\frac{393.5 \mathrm{~kj} \cdot \mathrm{mol}^{-1}}{44 \mathrm{~g}} \times 35.2 \mathrm{~g}=-315 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

Question 34. For a sample of perfect gas when the pressure is changed  isothermally from p_i to p_f the entropy change is given by

1. \(\Delta S=n R T \ln \left(\frac{p_f}{p_i}\right)\)

2. \(\Delta S=n R \ln \left(\frac{p_i}{p_f}\right)\)

3. \(\Delta S=n R T \ln \left(\frac{p_f}{p_i}\right)\)

4. \(\Delta S=R T \ln \left(\frac{p_i}{p_f}\right)\)

Answer: 2. \(\Delta S=n R \ln \left(\frac{p_i}{p_f}\right)\)

For the expansion of n mol of an ideal gas, the change in entropy of the reversible isothermal process is

⇒ \(\Delta S=n R \ln \frac{V_f}{V_l}\)

At constant temperature, for n mol of an ideal gas \(\frac{V_f}{V_i}=\frac{p_i}{p_f}\)

Therefore \(\Delta S=n R \ln \left(\frac{p_i}{p_f}\right)\)

Question 35. Consider the following liquid-vapor equilibrium. Liquid ⇌ vapour which of the following relations is correct

1. \(\frac{d \ln P}{d \mathrm{~T}^2}=-\frac{\Delta H_v}{T^2}\)

2. \(\frac{d \ln P}{d \mathrm{~T}}=\frac{\Delta H_v}{R T^2}\)

3. \(\frac{d \ln G}{d T^2}=\frac{\Delta H_v}{R T^2}\)

4. \(\frac{d \ln P}{d T}=-\frac{\Delta H_v}{R T}\)

Answer: 2. \(\frac{d \ln P}{d \mathrm{~T}}=\frac{\Delta H_v}{R T^2}\)

For liquid-vapor equilibrium, the relationship between equilibrium pressure (P), the heat of vaporization (ΔH_v) and temperature (T) will be

InP = \(-\frac{\Delta H_v}{R T}+Z\)

Z = constant

Differentiating the equation concerning, we get

⇒ \(\frac{d \ln P}{d T} \frac{\Delta H_\nu}{R T^2}\)

Question 36. The correct thermodynamic conditions for die spontaneous reaction at all temperatures is—

1. ΔH < 0 and ΔS > 0
2. ΔH < 0 and ΔS < 0
3. ΔH < 0 and ΔS = 0
4. ΔH > 0 and ΔS < 0

Answer: 1. ΔH < 0 and ΔS > 0

At constant temperature and pressure, the reaction is to be spontaneous if ΔG _< 0. According to the Gibbs free energy ag = ΔH – TΔS, ΔG will be negative at any temperature if ΔH  < 0 and ΔS  > 0

Question 37. A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 l to a final volume of 4.50 l. The change in internal energy a u of the gas in joules will be

1. -500 Jk^-1
2. -505 J
3. +505 J
4. 1136.25j

Answer: 2. -505 J

Work done of irreversible process,

w = \(-P_{e x} \Delta V=-2.5(4.5-2.5)=-5 \mathrm{~L} \cdot \mathrm{atm}\)

=-5 × 101.3)J

=-506.5J

Since, it is an insulated system, q = 0.

From the first law ofthermodynamics

Δ U = q+w

Or, ΔU = 0-506.5 J

=-506. 5J

Question 38. For a given reaction, ah = 35.5 kj. mol^-1 and as = 83.6 j.mol^-1. The reaction is spontaneous at (assume that ah and as do not vary with temperature)-

1. T > 425 k
2. All temperatures
3. T> 298 k
4. T< 425 k

Answer: 1. T > 425 k

ΔG = ΔH-TΔS = 35.5 ×  103- T× 83.6

Reaction is to be spontaneous if Δ < 0.

Thus, 35.5 ×10³– T ×  83.6 < 0

Therefore, T × 83.6 > 35.5 × 10³ or, T> 424.64 K

Question 39. The bond dissociation energies of x₂, y₂, and xy are in the ratio of 1 : 0.5: 1 . Ah for the formation of xy is -200 kj.mol^-1. The bond dissociation energy of x² will be

1. 200Kj.mo1^-1
2. 100Kj.mol^-1
3. 800 Kj.mol^-1
4. 400Kj.mol^-1

Answer: 3. 800 Kj.mol^-1

⇒ \(\frac{1}{2} \mathrm{x}_2+\frac{1}{2} \mathrm{y}_2 \rightarrow \mathrm{xy}\)

⇒  \(\Delta \mathrm{H}_{\text {reaction }}=\sum(\mathrm{BE})_{\text {reactant }}-\sum(\mathrm{BE})_{\text {product }}\)

BE = Bond energy

If the bond energy of x² is a kj.mol^-1 then the bond energy of y² and xy are 0.5a and a kj.mol^-1 respectively.

∴ \(-200=\frac{a}{2}+\frac{0.5}{2} a-a=-0.25 a \quad \text { or, } a=800\)

Therefore, bond energy of x₂ = 800kj.mol^-1

Question 40. Which of the following is an intensive property

1. Enthalpy
2. Entropy
3. Specific heat
4. Volume

Answer: 3. Specific heat

Intensive property: specific heat; extensive property-: enthalpy, entropy, volume.

Question 41. Which of the tires following is not a thermodynamic function—

1. Internal energy
2. Work done
3. Enthalpy
4. Entropy

Answer: 2. Workdone

Thermodynamic functions are internal energy, enthalpy, entropy, pressure, volume, temperature, free energy, and number of moles.

Question 42. For the adiabatic process, which is correct-

1. At = 0
2. As = 0
3. Q = 0
4. Qp = 0

Answer: 3. Q = 0

For the adiabatic process, no exchange of heat takes place between the system and surroundings. i.e., Q = 0.

Class 11 Chemistry Chemical Thermodynamics MCQs

Question 43. The enthalpy of formation of CO(g), CO₂(g), N₂O(g), and N₂O₄(g) is -110, -393, +811′ and kj/mol respectively for the reaction N₂O₄(g) + 3CO(g)→N₂O(g) + 3CO₂(g) ahr (kj/mol) is –

1. -212
2. +212
3. +48
4. -48

Answer: 4. -48

N₂O₄(g) + 3CO(g)→ N₂O(g) + 3CO₂(g)

ΔH_reaction =∑ Heat of formation of products heat of formation of reactants

-∑ Heat of formation of reactants

Question 44. The bond dissociation energy of CH₄ is 360 kj/mol and C₂H₆ is 620 kj/mol. Then bond dissociation energy of the C- C bond is—

1. 170 Kj/mol
2. 50Kj/mol
3. 80Kj/mol
4. 220Kj/mol

Answer: 3. 80Kj/mol

Dissociation energy of-methane = 360 kj.mol^-1

∴ Bond energy of c—h bond \(=\frac{360}{4}=90 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

The bond energy of ethane,

1× B.E. (C-C) + 6 × B.E. (C-H) = 620 KJ.mol^-1

Or, B.E.(C—C) + 6 × 90 = 620

Or, B.E. (C—C) + 540 = 620

Or, B.E. (C—C) = 620-540

Or, B.E. (C—C) = 80 KJ.mol^-1

Bond dissociation of c —C bond = 80 kj. mol^-1

Question 45. Which thermodynamic parameter is not a state function-

1. Q at constant pressure
2. Q at constant volume
3. W at adiabatic
4. W at isothermal

Answer: 4. W at isothermal

H and u are state functions but w and q are not state functions.

From the equation, Δh = ΔU+Δpv

At constant pressure, Δh = ΔU+pΔV

At constant volume, ΔH = ΔU+ VΔp

At constant pressure, Δp = 0, ΔH = qp so, it is a state function.

At constant volume, ΔV = 0, ΔU = q_V so, it is a state function.

Work done in any adiabatic process is a state function.

ΔU = q- w

∴ (Δq – 0)

ΔU = -w work done in the isothermal process is not a state function.

W = -q

∴ ΔT = 0, q ≠ 0

Question 46. For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, a u and w correspond to—

1. Au< 0, w = 0
2. Alt < 0 , w < 0
3. Al/> 0, w = 0
4. A17 > 0 , w> 0

Answer: 1. Au< 0, w = 0

For adiabatic conditions, PV^ϒ  = constant

⇒ \(p_1 v_1^\gamma=p_2 v_2^\gamma ; v_2=\frac{1}{2} v_1\)

⇒ \(p_2=p_1\left(\frac{v_1}{v_2}\right)^\gamma \text { [for diatomic gas, } \gamma=1.4 \text { ] }\)

⇒ \(p_2=p_1\left(\frac{v_1 \times 2}{v_1}\right)^{1.4} p_2=p_1(2)^{1.4}=(2)^{1.4} p\)

Question 47. For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, a u and w correspond to—

1. Au< 0, w = 0
2. Au< 0, w <0
3. Au> 0, w = 0
4. Au> 0, w>0

Answer: 1. Au< 0, w = 0

Bomb calorimeter is commonly used to find the heat of combustion of organic substances which consists of a sealed combustion chamber, called a bomb.

If the process is rim in a sealed container then no expansion or compression is allowed, so w = 0 and au = q ΔU< 0, w = 0.

Question 48. The heat is liberated when 1.89 g of benzoic acid is burnt in a bomb calorimeter at 25° c and it increases the temperature of 18.94 kg of water by 0.632°c. The specific heat of water at 25°c is 0.998 cal.g^-1 °c^-1, and the value of the heat of combustion of benzoic acid is

1. 881.1kcal
2. 771.124kcal
3. 981.1kcal
4. 871.2kcal

Answer: 2. 771.124kcal

Given:

Weight of benzoic acid = 1.89 g;

The temperature of the bomb calorimeter =25°c=298k;

The mass of water (m) = 18.94 kg = 18940 g;

Increase in temperature (ΔT) = 0.632°c and specific heat of water (s) = 0.998 cal- g°.C^-1

We know that heat gained by water or heat liberated by benzoic acid (q) = msΔT

= 18940 ×  0.998 × 0.632 = 11946.14 cal

Since 1.89 g of acid liberates 11946.14 cal of heat, therefore heat liberated by 122 g of acid = \(\frac{11946.14 \times 122}{1.89}\)

= 771126.5 cal

= 771.12 kcal

(Where 122 g is the molecular weight of benzoic acid)

Question 49. The heat is liberated when 1.89 g of benzoic acid is burnt in a bomb calorimeter at 25°c and it increases the temperature of 18.94 kg of water by 0.632°c. The specific heat of water at 25°c is 0.998 cal.g^-1.°C^-1, and the value of the heat of combustion of benzoic acid is

1. 881.1kcal
2. 771.124kcal
3. 981. kcal
4. 871.2kcal

Answer: 2. 771.124 kcal

NCERT Class 11 Chemistry Chemical Thermodynamics Multiple Choice

Question 50. What is the entropy change in 2 mol n2, when its temperature is taken from 400 k to 800 k, at constant pressure

1. 30J/k
2. 60J/k
3. 40J/k
4. 20J/k

Answer: 3. 40J/k

⇒ \(\Delta S=n C_P \ln \frac{T_2}{T_1}\)

Or, \(R T \ln \frac{4}{2}=3 R T \ln \frac{x}{2}\)

Question 51. 1 Mole of an ideal gas expands isothermally reversible from 2 liters to 4 liters and 3 moles of the same gas expand from 2 liters to x liter and do the same work, what is ‘x’-

1. (8)^1/3
2. (4)^2/3
3. 2
4. 4

Answer: 2. (4)^2/3

⇒ \(w=n R T \ln \frac{V_2}{V_1} \text { or, } R T \ln \frac{4}{2}=3 R T \ln \frac{x}{2}\)

Or, \(\ln 2=\ln \left(\frac{x}{2}\right)^3 \text { or, }\left(x^3=16\right) \text { or, } x=(16)^{\frac{1}{3}}=4^{\frac{2}{3}}\)

Question 52. Which of the following are extensive properties

1. Volume and enthalpy
2. Volume and temperature
3. Volume and specific heat
4. Pressure and temperature

Answer: 1. Volume and enthalpy

Extensive properties depend upon the quantity of the matter contained in the system,

Example: Volume and enthalpy, etc.

Intensive properties depend only upon the nature of the substance and are independent of the amount of the substance present in the system

Example: Temperature, pressure-specific heat, etc.

Question 53.  H₂O(l) → H+(aq)+OH-(aq); ΔH°=57.32 kj .mol^-1 H₂(g) + ½ (g)→H₂O(Z); ΔH°=-285.8 kj . mol^-1 at 25°C. If Δ⁰H₂[H⁺+(aq)] = 0, then the standard heat of formation (kj.mol^-1) for OH^–(aq) at 25^°C ls

1. -142.9
2. -228.48
3. -343.12
4. -253.71

Answer: 2. -228.48

Question 54. For a reaction at T K, ΔH> 0 and ΔS > 0. If the reaction attains equilibrium at a temperature of T₁K, (assume ΔH and ΔS are independent of temperature) then

1. T<T₁
2. T>T₁
3. T=T₁
4. T>T₁

Answer: 2. T>T₁

Question 55. The change in entropy for 2 mol ideal gas in an isothermal reversible expansion from 10 mL to 100 mL at 27°C is—

1. 26.79 J. K^-1
2. 38.29 J.K^-1
3. 59.07 J-K^-1
4. 46.26 J-K^-1

Answer: 2. 38.29 J.K^-1

Question 56. Which of the statements is true

1. A reaction, in which a 77 < 0 is always spontaneous;
2. A reaction, in which ah > 0 can never occur spontaneously
3. For a spontaneous process in an isolated system,
4. For a spontaneous process in an isolated system,

Answer: 3. For a spontaneous process in an isolated system,

Question 57. For the reaction, CaCO₃(s)→CaO(s) + CO₂(g), ΔH⁰= +179.1 kl-mol-1 and ΔSO = 160.2 If ΔH⁰ and ΔS⁰ are temperature independent, then the temperature above which the reaction will be spontaneous is equal to—

1. 1008 K
2. 1200 K
3. 845 K
4. 1118 K

Answer: 4. 1118 K

Question 58. When a gas (molar mass =28 g-mol^-1) of mass 3.5g is burnt completely in the presence of excess oxygen in a bomb calorimeter) the temperature of the calorimeter increases from 208 K to 298.45 K. The heat of combustion at constant volume for the gas (Given; the heat capacity of the calorimeter 2.5 k K4)

1. 4.5 kJ .mol^-1
2. 8.0 kJ. mol^-1
3. 9.0 KJ .mol^-1
4. 9.5 kJ mol^-1

Answer: 3. 9.0 KJ .mol^-1

Question 59. For the reaction, 2NH₃(g)→ N₂(g)+3H₂(g)-

1. ΔH< 0, ΔS> 0
2. ΔH> 0, ΔS > 0
3. ΔH > 0, ΔV <0
4. ΔH < 0, Δ5 < 0

Answer: 2. AH> 0, AS > 0

Question 60. Which of the following pairs is true for the process C₆H₆(g)[1atm, 80.1°C]→C₆H₆(l)[1 atm, 80.1°C]

1. ΔG< 0, ΔS> 0
2. ΔG< 0, ΔS< 0
3. ΔG = 0, ΔS < 0
4. ΔG = 0, ΔS > 0

Answer: 3. ΔG = 0, ΔS < 0

Question 61. The internal energy change when a system goes from state P to Q is 30kJ> mol^-1 If the system goes from P to Q by a reversible path and returns to state P by an irreversible path, what would be the net change in internal energy

1. 30kj
2. <30kJ
3. zero
4. >30kJ

Answer: 3. Zero

MCQs on Chemical Thermodynamics Class 11 Chemistry

Question 62. If at normal pressure and 100°C the changes in enthalpy and entropy for the process, H₂O(l)→H₂O(g), are ΔH and ΔS respectively, then ΔH-ΔU is—

1. 5.6 kj. mol^-1
2. 6.2 kj – mol^-1
3. 3.1 kj-mol^-1
4. 4.8 kj-mol^-1

Answer: 3. 3.1 kj-mol^-1

Question 63. At 25°C, the standard heat of formation for Br₂(g) is 30.9 kj.mol^-1. At this temperature, the heat of vaporization for Br₂(l) is—

1. <30.9 kj. mol^-1
2. 30.9 kj .mol^-1
3. >30.9 kj. mol^-1
4. Cannot Be Predicted

Answer: 2. 3.1 kj-mol^-1

Question 64. At 25°C, when 0.5 mol of HCl reacts completely with 0.5 mol of NaOH in a dilute solution, 28.65 kj of heat is liberated. If at 25°C Δ _f H⁰[H₂O(l)]=  then Δ _f H⁰OH^– (aq) is—

1. -314.45 kj. mol^-1
2. -285.8 kj.mol^-1
3. -228.5 kj. mol^-1
4. -343.1 kJ. mol^-1

Answer: 2.  -285.8 kj.mol^-1

Question 65. On combustion, C_xH_Y(l) forms CO₂(g) and H₂O(l). At a given temperature and pressure, the value of \(\left(\frac{\Delta H-\Delta U}{R T}\right)\) in this combustion reaction is—

1. \(\frac{x}{5}\)
2. \(\frac{x+y}{3}\)
3. \(\frac{y}{4}\)
4. \(\frac{x-y}{4}\)

Answer: 3. \(\frac{y}{4}\)

Question 66. An ideal gas is compressed isothermally at 25°C from a volume of 10 L to a volume of 6 L. Which of the following is not true for this process—

1. q<0
2. w>0
3. ΔU = 0
4. ΔH > 0.

Answer: 4. ΔH > 0.

Question 67. At 27°C, for the reaction aA(g) + B(g)→2C(g) PΔV = -2.5 kj . The value of ‘a’ is

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 68. When 1 mol of an ideal gas is compressed in a reversible isothermal process at T K, the pressure of the gas changes from 1 atm to 10 atm. In the process, if the work done by the gas is 5.744 kJ, then T is—

1. 400k
2. 300k
3. 420k
4. 520k

Answer: 2. 300k

Question 69. For 0.5 mol of an ideal gas, 15 cal of heat is required to raise its temperature by 10 K at constant volume. The molar heat capacity for the gas at constant pressure is-

1. 3cal.K^-1 . mol^-1
2. 4 cal.k^-1mol^-1
3. 5cal.k^-1 .mol^-1
4. 4.5cal.k^-1.mol^-1

Answer: 3. 5cal.k-1 .mol^-1

Question 70. According to the enthalpy diagram given below, the standard heat of formation (kJ mol^-1) of CO(g) at 25°C is

1. -283.0
2. -110.5
3. +283.0
4. +110.5

Answer: 2. -110.5

Question 71. 1 mol of an ideal gas is enclosed in a cylinder fitted with a frictionless and weightless piston. The gas absorbs x kJ heat and undergoes expansion. If the amount of expansion work done by the gas is x kJ, then the expansion is—

1. Adiabatic
2. Cyclic
3. Isothermal
4. It cannot Be Predicted

Answer: 3. Isothermal

Question 72. Given (at 25°C)

⇒ \(\mathrm{Ca}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CaO}(\mathrm{s}) ; \Delta H^0=-635.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒  \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \)

⇒  \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(\mathrm{~s}) ; \)

⇒  \(\Delta H^0=-65.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The heat of formation (in kj.mol^-1 ) for Ca(OH)₂(s) is –

1. -855.4
2. -673.9
3. -986.6
4. -731.7

Answer: 3. -986.6

Question 73. At 25°C, the standard heats of formation of H₂O(g) , H₂O₂(g), H(g) and O(g) are -241.8, -135.66, 218 and 249.17 k).mol^-1 respectively. The bond energy (in kj.mol-1 ) of O —O bond in H₂O₂(g) molecule is—

1. 179.23
2. 160.19
3. 142.60
4. 157.16

Answer: 3. 142.60

Question 74. Given (at 25°C):

⇒ \(\mathrm{C}(\mathrm{s} \text {, graphite })+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})\)

⇒ \(\Delta H^0=-110.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{O}(\mathrm{g}) ; \Delta H^0=+249.1 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \)

⇒ \(\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{C}(\mathrm{g})+\mathrm{O}(\mathrm{g}) ; \Delta H^0=1073.24 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The standard enthalpy change for the process, C(s, graphite)→C(g) at 25°C is—

1. +934.64 KJ.Mol^-1
2. 713.64 KJ.Mol^-1
3. 962.64 KJ.Mol^-1
4. 652.64 KJ.Mol^-1

Answer: 2. 713.64 KJ.Mol^-1

Question 75. At 25°C, for the reaction, H+{aq) + OU-(aq)yH₂O(l) , AHO = -57.3 kj . mol^-1. If the ionization enthalpy of HCN in water is 45.2 kj.mol^-1, then the standard heat of reaction (in kj mol^-1 ) for the reaction, HCN(aq) + NaOH(at jr)-NaCN(ag) + H₂O(Z), in dilute aqueous solution is

1. -113.5
2. -12.1
3. -102.5
4. -35.7

Answer: 2. -12.1

Question 76. The temperature of a bomb calorimeter changes from 25°C to 32.7°C when wg of naphthalene mass (molar = 128 g. mol^-1 ) is burnt completely in the calorimeter. If the heat of combustion at constant volume for naphthalene is -5152 kj mo^-1 then w is (heat capacity of the calorimeter = 8.19 kj K^-1)

1. 0.87 g
2. 1.91 g
3. 2.37 g
4. 1.57g

Answer: 4. 1.57g

Question 77. In which of the following processes the change in entropy for the system is zero

1. Irreversible adiabatic processes
2. Reversible adiabatic process
3. A spontaneous process occurring in an isolated system
4. Isothermal expansion of an ideal gas

Answer: 2. Reversible adiabatic process

Question 78. A system undergoes the process: A→ B → C→ D. In this process, the change in a state function (.X) of the system is x. In steps A→B and B→C of the process, if the changes in X are y and z respectively, then the change in X in step D→C is

1. x-y-z
2. x-z+y
3. y+z-x
4. y-z-x

Answer: 3. y+z-x

Question 79. At 27°C, ΔH = + 6 kJ for the reaction A + 2B-3C. In the reaction, if ΔASuniv = 2 J . K^-1 , then ΔS_sys (in J . K^-1 ) is

1. +2
2. +3
3. +20
4. +22

Answer: 4. +22

Chemical Thermodynamics Multiple Choice Questions for Class 11

Question 80. An LPG cylinder contains 14 kg of butane. A family requires 2 X 104 kj of heat for their cooking purpose every day. By how many days will the butane in the cylinder be used up (Given: heat of combustion for butane = -2658 kj-mol^-1 )

1. 15 days
2. 20 days
3. 32 days
4. 40 days

Answer: 3. 32 days

Question 81. For a reaction involving 1 mol of Zn and 1 mol of H₂SO₄ in a bomb calorimeter

1. ΔH> 0 , w> 0
2. ΔU> 0 , w> 0
3. ΔU< 0 , w> 0
4. ΔU< 0 , w> 0

Answer: 4. ΔU< 0 , w> 0

Question 82. Assuming that water vapor is an ideal gas, the internal energy change (ALT) when mol of water is vaporized at bar pressure and 100°C, will be (Given: at bar and 373K, molar enthalpy of vaporization of water is 41 kj. mol^-1, R = 8.3 J.mol^-1 . K^-1 )

1. 4.100 KJ. mol
2. 3.7904 KJ. mol-1
3. 37.904 kj. mol-1
4. 41.00 kj. mol-1

Answer: 3. 37.904 kj. mol-1

Question 83. At 25°C and 1 atm pressure, ΔH and pressure-volume work for the reaction, 2H₂(g) + O₂(g)y₂H₂O(g) are —483.7 kj and 2.47 kj respectively. In this reaction the value U is-

1. -483.7 kJ
2. -481.23 KJ
3. -400.23 Kj
4. -492.6 KJ

Answer: 2. -481.23 KJ

Question 84. An ideal gas’s initial state of 1 mol is (P₁, V₂, T₁ ). The gas is expanded by a reversible isothermal process and also by a reversible adiabatic process separately. If the final volume of the gas is the same in both of the processes, and changes in internal energy in the isothermal and adiabatic processes are ΔU₁ and ΔU₂ respectively, then

1. ΔU₁=ΔU₂
2. ΔU₁<ΔU₂
3. ΔU₁>ΔU₂
4. Cannot Be Predicated

Answer: 3. ΔU₁>ΔU₂

Question 85. At constant pressure, the amount of heat required to raise the temperature of 1 mol of an ideal gas by 10-C is x kj. If the same increase in temperature were carried out at constant volume, then the heat required would be

1. > xKJ
2. <x Kj
3. = x KJ
4. > xKJ

Answer: 1. > xKJ

Question 86. The enthalpy of fusion of ice at 0- C and 1 atm is 6.02 kj – mol^-1. The change in enthalpy (J K^-1)of the surroundings when 9 g of water is frozen at 0°C and 1 atm pressure is

1. +11.02
2. -11.02
3. -20.27
4. +23.09

Answer: 2. -11.02

Question 87. At a given pressure and a temperature of 300 K, Δ_Surr and Δ_sys for a reaction are 8.0 J. K^-1 and 4.0 J . K^-1surr respectively. ΔG for this reaction is-—

1. -3.0 KJ
2. -3.6 KJ
3. 3.0 kJ
4. -4.2 kJ

Answer: 2. -3.6 KJ

Question 88. In a reversible process, if changes in the entropy of the system and its surroundings are ΔS₁ and ΔS₂ respectively, then

1. ΔS₁ +ΔS₂ >0
2. ΔS₁ +ΔS₂<0
3. ΔS₁+ΔS₂= 0
4. ΔS₁ +ΔS₂>0

Answer: 3. ΔS₁+ΔS₂= 0

Question 89. A flask of volume 1 L contains 1 mol of an ideal gas. The flask is connected to an evacuated flask, and as a result, the volume of the gas becomes 10 L. The change in entropy (J. K^-1) of the gas in this process is

1. 9.56
2. 19.14
3. 11.37
4. 14.29

Answer: 2. 19.14

Question 90. The heats of neutralization of four acids A, B, C, and D are 13.7, 9.4, 11.2, and 12.4kcal respectively when they are neutralized against a common base. The weakest add among A, B, C, and D is

1. A
2. B
3. C
4. D

Answer: 2. B

Question 91. In a closed insulated container, a liquid is stirred with a paddle to increase the temperature. Which is true-

1. ΔU= w≠0, q=0
2. ΔU=0, w=0, q≠0
3. ΔU=0 w=0, q≠0
4. w=0 w≠0, q=0

Answer: 1. ΔU= w≠0, q=0

Question 92. How many calories are required to increase the temperature of 40g of Ar from 40-C to 100°C at a constant volume (R = 2 cal.mol^-1.K^-1)

1. 120
2. 2400
3. 1200
4. 180

Answer: 4. 180

Question 93. Water is supercooled to -4°C. The enthalpy (H) of the supercooled water is

1. Same as ice at -4°c
2. More than ice at -4°c
3. Same as ice at 0°c
4. Less than ice at -4°c

Answer: 4. Less than ice at -4°c

Question 94. The standard entropy of X₂, Y₂, and XY₃ are 60, 40, and 50 J.K-1.mol^-1 respectively. For the reaction\(\frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightarrow \mathrm{XY}_3,(\Delta H=-30 \mathrm{~kJ})\) (AH = -30 kl ) to be at equilibrium, the temperature will be

1. 1250 k
2. 750k
3. 500 k
4. 1000 k

Answer: 2. 750k

Question 95. Two moles of gas of volume 50L and pressure 1 atm are compressed adiabatically and reversibly to 10atm. What is the atomicity of the gas (T₁/T₂= 0.4)

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Question 96. Given (at 25°C): C(s, graphite)-C(g) ; ΔH⁰ = +713.64 kj. mol^-1

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g}); \Delta H^0=+218 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) C(s, graphite)+3H₂(g)→C₆H₆(g); ΔH°-+82.93kj – mol^-1At 25°C, if the bond energy of C—H and C— C bonds are 418 and 347 kj .mol^-1 respectively, then the C=C bond energy is

1. +679.81 KJ. mol^-1
2. +652.63 kj. mol^-1
3. +808.75 KJ. mol^-1
4. +763.39 kJ. mol^-1

Answer: 2. +652.63 kj. mol^-1

Question 97. Given (at 25 °C):

⇒ \(\mathrm{C}_6 \mathrm{H}_6(g)+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \( \Delta H^0=-1560 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \)

⇒ \(\mathrm{C}(s, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

At 25 °C, if the stand heat of formation of C₃H₈(g) is -103.8 kj. mol-1 , then the standard heat of reaction for the reac3h8ction; C₃H₈(g) + H₂(g)yC₂H₆(g) + CH₄(g) is

1. +98.45 kJ
2. -55.70KJ
3. 62.37 KJ
4. -47.25 KJ

Answer: 2. -55.70KJ

Question 98. At 0°C and normal pressure, the enthalpy of fusion of ice is 334.7 J . g^-1. At this temperature and pressure, if 1 mol of water is converted into 1 mol ice, then the change in entropy of the system will be

1. 16.7 J.K^-1
2. -16.7 J.K^-1
3. 22.06 J.K^-1
4. -22.06 J.K^-1

Answer: 4. -22.06 J.K^-1

Question 99. 5 mol of gas is put through a series of changes as shown graphically in a cyclic process. The process X→Y, Y→Z and Z→X respectively are

1. Isochoric, isobaric, isothermal
2. Isobaric, isochoric, isothermal
3. Isothermal, isobaric, isochoric
4. Isochoric, isothermal, isobaric

Answer: 1. Isochoric, isobaric, isothermal

Question 100. Given \(\mathrm{NH}_3(g)+3 \mathrm{Cl}_2(g) \rightarrow \mathrm{NCl}_3(g)+3 \mathrm{HCl}(g) ;-\Delta H_1\) N₂(g) + 3H₂(g)→2NH₃(g) ; ΔH₂; H₂(g) + Cl₂(g→2HCl(g);ΔH₃ Heat of formation (ΔHf) of NCl₃(g) in terms of ΔH₁ , ΔH₂ and ΔH₃ is-

1. \(-\Delta H_1+\frac{\Delta H_2}{2}-\frac{3}{2} \Delta H_3\)

2. \(\Delta H_1+\frac{\Delta H_2}{2}-\frac{3}{2} \Delta H_3\)

3.\(\Delta H_1-\frac{\Delta H_2}{2}-\frac{3}{2} \Delta H_3\)

4. None of the above

Answer: 4. None of the above

Chemical Thermodynamics MCQs Class 11 Solutions

Question 101. When lg of graphite is completely burnt in a bomb calorimeter, the temperature of the bomb and water rises from 25°C to 30.5°C. The heat capacity of the calorimeter is 5.96 kj. °C-1, then the heat of combustion per mole of graphite at constant volume is

1. -357.13 kj.mol^-1
2. -289.71 kj.mol^-1
3. -393.36 kj .mol^-1
4. -307.94 kj-mol^-1

Answer: 3. -393.36 kj .mol^-1

Question 102. The volume of a gas is reduced to half of its original volume. The specific heat will—

1. Reduce to half
2. Double
3. Remain Constant
4. Increase Four times

Answer: 3. Remain Constant

Question 103. For which molar-specific heat is temperature-independent

1. Argon
2. Hydrogen
3. Nitrogen
4. Carbon dioxide

Answer: 1. ARgn

Question 104. Which of the following quantities are state functions

1. q
2. q+w
3. w
4. U+pv

Answer: 1. q

Question 105. A monoatomic ideal gas undergoes the cyclic process, Which of the comments are true for this process

1. For the whole process q = +1.134J
2. For the whole process Δ_sys> 0
3. For the whole process Δ_sys = 0
4. For the whole process q = -2.310 J

Answer: 1. For the whole process q = +1.134J

Question 106. Which of the following comments is true

1. Only for an ideal gas, cp m > cv m
2. For any gas, cp m > cv m
3. For a solid substance, cp m – cv> m
4. For ‘ n ’ mol ofideal gas, cp m- cv m = nr

Answer: 2. For any gas, cp m > cv m

Question 107. When 3g of ethane gas is brunt at 25°C, 156 kl of heat is liberated. If the standard enthalpies of formation for CO₂(g) and H₂O(l) are -393.5 and -285.8 kj.mol^-1respectively, then for ethane gas-

1. Standard heat of combustion = -1560 kJ mol^-1
2. Standard heat of formation =-67.9 kJ. Mol^-1
3. Standard heat of combustion =-832 kJ .mol^-1
4. Standard heat of formation = -84.4 kJ. Mol^-1

Answer: 1. Standard heat of combustion = -1560 kJ. Mol-^-1

Question 108. A reaction is spontaneous at a temperature of 300K, but it is non-spontaneous at a temperature of 400 K. If ΔH and ΔS for the reaction do not depend on temperature, then

1. ΔH> 0
2. ΔH < 0
3. ΔS > 0
4. ΔS<0

Answer: 2. AH < 0

Question 109. The reaction, 3O₂(g)→2O₃(g), is non-spontaneous at any temperature. Hence

1. The reverse reaction is spontaneous at any temperature
2. ΔH < 0 and ΔS < 0 for the reverse reaction
3. ΔH > 0 , ΔS > 0 for the reverse reaction
4. ΔH < 0 and ΔS > 0 for the reverse reaction

Answer: 1. The reverse reaction is spontaneous at any temperature

Question 110. For the isothermal free expansion of ideal gas

1. ΔH =0
2. ΔS < 0
3. ΔS > 0
4. ΔH > 0

Answer: 1. ΔH =0

Question 111. The changes in which of the following quantities are for a cyclic process

1. Enthalpy
2. Work
3. Entropy
4. Internal energy

Answer: 1. Enthalpy

Question 112. Which of the following relations are true for the reaction, PCl₅(g)→PCl₃(g) + Cl₂(g)

1. ΔH< 0
2. ΔH >0
3. ΔS <0
4. ΔS>0

Answer: 2. ΔH >0

Question 113. An ideal gas performs only pressure-volume work in the given cyclic process. In the diagram, AB, BC, and CA are the reversible isothermal, isobaric, and isochoric processes respectively. Identify the correct statements regarding this cycle

1. Total Work Done In This Process ( W) = WA→B + WB→C

2. Changes In Internal Energy In The Step Ab = 0

3.ΔSA→b = ΔSB→C+ΔSC→A

4. If The Total Heat And Work Involved In The Process Are Q And W Respectively, Then q + W = 0

Answer: 2. Changes In Internal Energy In The Step Ab = 0

Question 114. Identify the correct statements

1. Standard state of bromine (25°C, 1 atm) is Br2(g)

2. C (graphite, s)-C (diamond, s); here AH=0

3. Standard enthalpy change for the reaction N₂(g) + O₂(g)→2NO(g) at 25°C and 1 atm is standard enthalpy of formation of NO(g)

4. At a particular temperature and pressure, if ΔH = xkj for the reaction A + 3.B→2C then AH \(-\frac{x}{2} \mathrm{~kJ}\) for the reaction \(C \rightarrow \frac{1}{2} A+\frac{3}{2} B\)

Answer: 2. C (graphite, s)-C (diamond, s); here ΔH=0

Question 115. Which of the following statements is correct

1. In any adiabatic process, ΔSsys = 0
2. In the isothermal expansion of ideal gas, ah = 0
3. An endothermic reaction will be spontaneous if in this reaction ΔSsys > 0
4. Heat capacity is a padi-dependent quantity

Answer: 2. In the isothermal expansion of ideal gas, ah = 0

Question 116. Correct statements are

1. A + B→D, ΔH = x kj.This reaction is performed in the following two steps: 1. A + B→C  2. C→D. If in step 1. ΔH = y kj. then 2. ΔH = (x-y)J in step
2. for a spontaneous process occurring in an isolated system ΔSsys> 0 at equilibrium
3. In a spontaneous chemical reaction at constant temperature and pressure,  ΔG = =TΔ_surr
4. In a chemical reaction ΔH > 0 and ΔS > 0. The reaction attains equilibrium at temperature, Tg. At constant pressure and constant temperature TK the reaction will be spontaneous, if T > T

Answer: 1. A + B-D, ΔH = x kj.This reaction is performed in the following two steps:

1. A + B→C→C
2. C→D.

If in step

1. ΔH = ykj. then
2. ΔH = (x-y)J in step

NCERT Class 11 Chemistry Chemical Thermodynamics MCQs

Question 117. Some reactions and their ΔH° values are given below:

⇒ C(graphite,s)\(+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) ; \Delta H^0=a \mathrm{~kJ}\)

⇒  \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=b \mathrm{~kJ}\)

⇒  \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=c \mathrm{~kJ}\)

⇒  \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=d \mathrm{~kJ}\)

⇒ \(2 \mathrm{C} \text { (graphite,s) }+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) ; \Delta H^0=m \mathrm{~kJ}\)

Which of the following statements is correct

1. Standard heat of formation of CH₄(g) = (a + b + 2c→d) kj-mol^-1
2. Standard heat of combustion of C₂Hg = (2a + 2b + 3c- m) kj.mol^-1
3. Standard heat of combustion of carbon = a kj.mol^-1
4. Standard of formation of CO₂(g)=(a+b) kj.mol^-1

Answer: 1. Standard heat of formation of CH₄(g) = (a + b + 2c→d) kj-mol^-1

Question 118. At 25°C, in which of the given reactions do standard enthalpies of reactions indicate standard enthalpies of formation of the products in the respective reactions

1. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{3} \mathrm{O}_3(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)
3. \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)
4. \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~s}) \rightarrow \mathrm{HI}(\mathrm{g})\)

Answer: 2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)

Question 119.

1. CaCO₃(s)→CaO(s)+CO₂(g)
2. \(\mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g)\)
3. PCl₅(g)⇒ Pcl₃(g) + C1₂(g)
4. N₂(g) + O₂(g)→2NO(g)

For which of these reactions, P-V work is negative

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Question 120. Which of the given reactions are endothermic

1. Combustion of methane
2. Decomposition of water
3. Dehydrogenation of ethane to ethene
4. Conversion of graphite to diamond

Answer: 2. Decomposition of water

Question 121. At constant volume and 298K, mol of gas is heated and the final temperature is 308 K. Ifheat supplied to the gas is 500 J, then for the overall process

1. w = 0
2. w = -500 J
3. AU = 500 J
4. AU = 0

Answer: 1. w = 0

Question 122. True for spontaneous dissolution of KCl in water are

1. ΔG<0
2. ΔH > 0
3. ΔS_surr < 0
4. ΔH<0

Answer: 1. ΔG<0

Question 123. When a bottle of perfume is opened, odorous molecules mix with air and diffuse gradually throughout the room. The correct facts about the process are

1. ΔS = 0
2. ΔG < 0
3. ΔS> 0
4. ΔS < 0

Answer: 2. ΔG < 0

Question 124. mol of an ideal gas undergoes a cyclic process ABC A represented by the following diagram

Which of the given statements is correct for the process —

1. Work done by the gas in the overall process is \(\frac{P_0 V_0^2}{2}\)
2. Work done by the gas in the overall process is P₀ V₀
3. Heat absorbed by the gas in path AB is 2P₀ V₀
4. Heat absorbed by the gas in path BC is \(\frac{1}{2} P_0 V_C\)

Answer: 2. Work done by the gas in the overall process is \(\frac{P_0 V_0^2}{2}\)

Question 125. At 0°C and 10 atm pressure 14g of oxygen is subjected to undergo a reversible. adiabatic expansion to a pressure of 1 atm. Hence in this process

1. The final temperature of the gas is 141.4 K
2. The final temperature of the gas is 217.3 K
3. Work done = 293.2 cal
4. Work done = -286 cal

Answer: 1. Final temperature of the gas is 141.4 K

Question 126. Choose the reactions in which the standard reaction enthalpy (at 25°C) represents the standard formation enthalpy of the product

1. \(\mathrm{H}_2(g)+\frac{1}{3} \mathrm{O}_3(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)
3. \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)
4. \(\frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(s) \rightarrow \mathrm{HI}(g)\)

Answer: 2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)

Question 127. Which of the following is (are) not a state function?

1. Enthalpy
2. Heat capacity
3. Heat
4. Work done

Answer: Heat and work

Question 128. A thermodynamic state function is a quantity

1. Used to determine heat changes
2. Whose value is independent of the path
3. Used to determine pressure-volume work
4. Whose value depends on temperature only

Answer: 2. Whose value is independent of the path

A thermodynamic state function of a system is a quantity whose value depends only on the present state of the system. Its value does not depend on the path of a process in which the system participates.

Question 129. For the process to occur under adiabatic conditions, the correct condition is 

1. ΔT=0
2. ΔP=0
3. q=0
4. w=0

Answer: 3. q=0

In an adiabatic process, no exchange of heat takes place between the system and its surroundings.

Question 130. Enthalpies of all elements in standard states are

1. Unity
2. Zero
3. <0
4. Different for each element

Answer: 2. Zero

By convention, the enthalpies of all the elements in their standard states are considered to be zero.

NCERT Class 11 Chemistry Chemical Thermodynamics MCQs

Question 131. ΔU° for combustion of methane is -XkJ .mol ^-1. The value of ΔH° is

1. ΔU°
2. >ΔU°
3. <ΔU°
4. 0

Answer: 3. <ΔU°

Question 132. An amount of work w is done by the system and a q amount of heat is supplied to the system. By which the following relations the change in internal energy of the system can be expressed-

1. Δ U = q-w
2. Δ U =q+w
3. ΔU=q
4. ΔU=w-q

Answer: 4. ΔG<0

Question 133. At 25°C which of the following has an enthalpy of formation zero

1. HCL(g)
2. O₂(g)
3. O₃(g)
4. NO(g)

Answer: 2. O₂(g)

Question 134. Which one is the correct unit of entropy

1. k-1. mol^-1
2. J.k-1. mol^-1
3. J.mol^-1
4. J-1.k-1. mol^-1

Answer: 2. J.k-1. mol^-1

Question 135. For the reversible reaction A + 2B→ C + Heat, the forward reaction will proceed at

1. Low temperature and low pressure
2. Low pressure
3. High pressure and low temperature
4. High pressure and high temperature

Answer: 3. High pressure and low temperature

For the reversible reaction A + 2 BC + A, the forward reaction will proceed at high pressure and low temperature.

Question 136. Which ofthe following is an example of a closed system

1. A hot water-filled thermos flask
2. An ice water-filled airtight metallic bottle
3. A water-filled stainless steel bowl
4. A hot water-filled glass beaker

Answer: 2. An ice water-filled airtight metallic bottle

An ice-water-filled airtight metallic bottle is an example of a closed system.

Question 137. Which is an intensive property of a system

1. Internal energy
2. Entropy
3. Mass
4. Density

Answer: 4. Density Density is an intensive property.

Question 138. If one monoatomic gas is expanded adiabatically from 2L to 10 L at atm external pressure then the value of a ΔU (in atm. L )is

1. -8
2. 0
3. -66.7
4. 58.2

Answer: 1. -8

q = 0 (since process is adiabatic.) ∴ Δu =w = -pav

=-1(10 -3) atm.L=-8 atm.L

Question 139. The equation of state for ‘ n’ mol of an ideal gas is PV – nRT. fn this equation, the respective number of intensive and extensive properties are

1. 2,3
2. 3,2
3. 1,4
4. 4, 1

Answer: 2. 3,2

NCERT Class 11 Chemistry Chemical Thermodynamics MCQs

Question 140. The enthalpy of combustion of methane, graphite, and dihydrogen at 298K are -890.3 kj.mol^-1, -393.5 kj.mol^-1, and -285.8 kj.mol^-1 respectively. Enthalpy of formation of CH₄(g) will be—

1. -74.8KJ. mol^-1
2. —52.27kl.mol^-1
3. +74.8KJ. mol^-1
4. +52.26KJ. mol^-1

Answer: 1.  -74.8KJ. mol^-1

According to the given data

⇒ \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0=-890.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{C}(s, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0=-285,8 \mathrm{~kJ} \cdot \text { mol }^{-1}\)

By 1 xeq.(2) + 2xeq.(3)-eq.(1), we get the thermochemical equation involving the formation reaction of CH₄(g)

⇒ \(\mathrm{C}(s \text {, graphite })+\mathrm{O}_2(g)+2 \mathrm{H}_2(g)+\mathrm{O}_2(g)-\mathrm{CH}_4(g)-2 \mathrm{O}_2(g)\)

⇒ \(\mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)-\mathrm{CO}_2(g)-2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\mathrm{C}(s \text {, graphite })+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_4(\mathrm{~g}) ; \Delta H^0=-74.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

This equation represents the formation reaction of CH₄(g) . Hence, the enthalpy of the formation of CH₄(g) is -74.8kJ.mol^-1

Question 141. A reaction, A + B→+C + D + q is found to have a positive entropy change. The reaction will be

1. Possible at high temperature
2. Possible only at low temperature
3. Not possible at any temperature
4. Possible at any temperature

Answer: 4. Possible at any temperature

The Thermochemical equation for the reaction indicates that the reaction is exothermic. So, for this reaction, ΔH < 0. It is given that ΔS > 0 for the reaction. So, according to the relation ΔG = ΔH- TΔS, ΔG will be <0 at any temperature. Hence, the reaction is possible at any temperature.

NCERT Solutions For Class 11 Chemistry Chapter 10 S Block Elements Short Question And Answers

Question 1. Li⁺ ion is the smallest one among the ions of group- 1 elements. It would, therefore, be expected to have much higher ionic mobility, and hence the solutions of lithium salts would be expected to have higher conductivity than the solutions of cesium salts. However, in reality, the reverse is observed. Explain
Answer: 

Due to high charge density, very small Li⁺ ion gets much more hydrated compared to the large Cs⁺ ion. Thus, the size of the hydrated lithium-ion is much larger than that of the hydrated cesium ion. For this reason, the mobility of Li⁺ ion is much lower than that of Cs⁺ ion and consequently, the solutions of lithium salts have much lower conductivity than the solutions of cesium salts.

Read and Learn More NCERT Class 11 Chemistry Short Answer Questions

Question 2. The E° value for Cl₂/Cl^– is +1.36, for I₂/I^– is + 0.53V, for Ag⁺/Ag is + 0.79 V, for Na⁺/Na is -2.71V and for Li⁺/Li is -3.04V. Arrange the following atoms and ions in decreasing order of their reducing strength: I^–, Ag, Cl^–, Li, Na
Answer:

The lower the value of standard reduction potential, the greater the tendency of the reduced form to be oxidized, i.e., the reduced form will serve as a stronger reductant. Therefore, the decreasing order of reducing strength of the given atoms and ions is

Li > Na > I^–> Ag > Cl^–

Question 3. The alkali metals are obtained not by the electrolysis of the aqueous solutions of their salts but by the electrolysis of their molten salts. Explain.
Answer:

The solutions of alkali metal salts contain metal cations, anions, H⁺ ions, and OH^– ions. The discharge potential of H⁺ ions is lower than that of metal cations. So, on electrolysis of tire solutions of alkali metal salts, hydrogen is discharged at the cathode rather than the metal. However, when the molten salts of alkali metals are electrolysed, the metal cation being the only cation present, gets discharged at the cathode.

S-Block Elements Short Answer Questions Class 11

Question 4. The alkali metals are paramagnetic but their salts are diamagnetic—why?
Answer:

Due to the presence of an unpaired valence electron the alkali metals are paramagnetic. During salt formation, this unpaired electron of the outermost shell leaves the metal atom and becomes attached to a non-metal atom. As a result, the cation and the anion thus obtained contain no unpaired electron. Hence, the alkali metal salts are diamagnetic.

Question 5. Beryllium & magnesium do not give color to flame whereas other alkaline earth metals do so. Why?
Answer: 

Due to their smaller size, valence electrons of Be and Mg are more tightly held by the nucleus. Therefore, they need a large amount of energy for the excitation of their valence electrons to higher energy levels. Since such a large amount of energy is not available from Hansen flame, these two metals do not impart any color to the flame.

Question 6. E° for M²⁺(aq) + e →M(s) (where, M = Ca, Sr or Ba) Is nearly constant. Comment.
Answer:

The value of standard electrode potential (E°) of any M²⁺/M electrode depends upon three factors.

1. Enthalpy of vaporisation
2. Ionization enthalpy, and
3. Enthalpy of hydration.

Since the combined effect of these factors is approximately the same for Ca, Sr, and Ba, their standard electrode potential (E°) values are nearly constant.

Question 7. Both alkaline earth metals and their salts are diamagnetic. Explain.
Answer:

The alkaline earth metals are diamagnetic as all the orbitals are filled with paired electrons. The ions of alkaline earth metals, M²⁺ have stable noble gas configurations in which all the orbitals are doubly occupied. Also, there are no unpaired electrons in the anions. Hence, the salts of alkaline earth metals are also diamagnetic.

Question 8. Beryllium salts can never have more than 4 molecules of water of crystallization, i.e., they can never achieve a coordination number > 4 while other metal ions tend to have a coordination number of 6, for example: [Ca(H₂O₆)²⁺. Explain.
Answer: 

Beryllium does not exhibit coordination numbers more than 4 because, in its valence shell of Be²⁺ ion, there are only four available orbitals (one s and three p) present. The remaining members of the group can have a coordination number of six by using their d -d-orbitals along with s -and p -orbitals.

Question 9. Anhydrous is used as a drying agent — why?
Answer:

Anhydrone or magnesium perchlorate, Mg(ClO₄)₂ is  used as a drying agent because, due to its smaller size and higher charge, Mg has a greater tendency to complexation with water molecules by forming Mg(ClO₄)₂ -6H₂O

Question 10. Li salts are commonly hydrated while other alkali metal salts are usually anhydrous. Explain.
Answer:

Due to the smallest size among all alkali metal ions, the Li⁺ ion can interact with water molecules more easily than the other alkali metal ions. Hence the salts of lithium are commonly hydrated. On the other hand, other alkali metal ions being larger have little tendency to get hydrated. Therefore, their salts are generally anhydrous. For example, lithium chloride crystallizes as LiCl. 2H₂O₂ but sodium chloride crystallizes as NaCl.

Question 11. Although the abundance of Na and K in the earth’s crust are comparable, sodium is nearly 30 times more abundant than potassium in seawater—why?
Answer:

These metals were leached from the aluminosilicate rocks by weathering. The potassium salts having large anions are less soluble than the sodium salts because of higher lattice energy. Moreover, potassium is preferentially absorbed by the plants. For this reason, sodium is more abundant than potassium in seawater.

Question 12. When caustic soda solution is kept in a glass bottle, the inner surface of the bottle becomes opaque. Explain
Answer:

Caustic soda (NaOH) being strongly basic reacts with acidic silica (SiO₂) present in glass to form sodium silicate (Na₂SiO₃)

SiO₂ + 2NaOH → Na₂SiO₃ + H₂O

As a result, the inner surface of the bottle becomes opaque.

Question 13. Why are alkali metals stored in kerosene?
Answer:

When the highly reactive alkali metals are exposed to air,  they readily react with oxygen, moisture, and carbon dioxide of air to form oxides, hydroxides, and carbonates respectively. To prevent these reactions, alkali metals are normally stored in kerosene, an inert liquid.

Question 14. The second ionization enthalpies of alkaline earth metals are much lower than those of the corresponding alkali metals. Explain.
Answer:

The loss of a second electron from an alkali metal cation (M⁺) causes a loss of stable noble gas configuration while the loss of a second electron from an alkaline earth metal cation leads to the attainment of a very stable noble gas configuration. This explains why the second ionization enthalpies of alkaline earth metals are much lower than those of the corresponding alkali metals.

Question 15. MgO is used as a refractory material— Explain why
Answer:

Due to greater charge on both the cation (Mg²⁺) and die anion (O^2-), MgO possesses higher lattice energy and for tills, it has a very high melting point and does not decompose on heating. For this reason, it is used as a refractory material

Question 16. BaSO₄ is insoluble in water whereas BeSO₄ is soluble in water—1-explain with reason._
Answer: 

The lattice enthalpy of BaSO₄ is much higher than its hydration enthalpy and hence it is insoluble in water. On the other hand, the hydration enthalpy of BeSO₄ is much higher than its lattice enthalpy because of the loose packing of the small Be²⁺ ion with the relatively large sulfate ion. Hence, it becomes soluble in water.

Question 17. BeCl₂ fumes in moist air but BaCl₂ does not. Explain
Answer:

BeCl₂ being a salt of a weak base, Be(OH)_2, and a strong acid, HCl undergoes hydrolysis in moist air to form HCl; which fumes in air. BaCl₂ on the other hand, being a salt of a strong base, Ba(OH)_2, and a strong acid, HCl does not undergo hydrolysis to form HCl and hence does not fume in moist air

BeCl₂ + 2H₂O→Be(OH)₂ + 2HCl↑

BaCl₂ +H₂O→ Ba(0H)₂ + 2HCl

NCERT Solutions Class 11 Chemistry Chapter 10 S-Block Elements SAQ

Question 18. Mg₃N₂ when reacts with water, gives off NH₂ but HCl is not evolved when MgCl₂ reacts with water at room temperature. Give reasons.
Answer:

Mg₃N₂ is a salt of the strong base, Mg(OH)₂ and the weak acid, NH₃

Hence it gets hydrolyzed to give NH₃.

Mg₃N₂ + 6H₂O→ 3Mg(OH)₂ + 2NH₃T↑

MgCl₂,  On the other hand, is a salt of the strong base, Mg(OH)_2, and the strong acid, HCl. Hence, it does not undergo hydrolysis.

Question 19. A piece of burning magnesium ribbon continues to bum in sulfur dioxide.
Answer:

A piece of magnesium ribbon continues to bum in SO₂ since it reacts to form MgO and S. This reaction is such exothermic that heat evolved keeps the magnesium ribbon burning.

This reaction is such exothermic that heat evolved keeps the magnesium ribbon burning.

Question 20. Ba²⁺ ions are poisonous, still, they are provided to patients before taking stomach X-rays. Explain
Answer:

A barium meal (suspension of BaSO₄ in water) is generally given to patients when X-ray photographs of the alimentary canal are required. The salt provides a coating of the alimentary canal and hence X-ray photograph can be taken since it is quite transparent to the X-ray otherwise. BaSO₄ is almost insoluble in water and hence it does not pass from the digestive system to the circulatory system and can therefore be safely used for the purpose.

Question 21. Can sodium hydride be dissolved in water? Justify.
Answer:

Sodium hydride cannot be dissolved in water because it gets hydrolyzed with brisk effervescence of hydrogen gas.

NaH + H₂ O→  H₂ + NaOH

Question 22. Why does sodium impart a yellow color in the flame?
Answer:

The ionization enthalpy of Na is relatively low. Therefore when this metal or its salt is heated in Bunsen flame, its valence shell electron is excited to higher energies by absorption of energy. When the excited electron returns to its initial position in the ground state, it liberates energy in the form of light in the yellow region of the electromagnetic spectrum. That’s why sodium imparts a yellow color to the flame.

Question 23. Wind makes lithium exhibit uncommon properties compared to the rest of the alkali metals.
Answer:

The unusual properties of lithium as compared to other alkali metals are since

• Li – atom and L⁺  ion are exceptionally small in size and
• Li⁺ ion has the highest polarising power (i.e.„ charge/size ratio).

Question 24. What is the common oxidation state exhibited by the alkali metals and why?
Answer:

The alkali metals easily lose their valence electrons (ns¹) to acquire a stable octet, (i.e., the stable electronic configuration of the nearest noble gas) and because of this, the common oxidation state exhibited by the alkali metals is +1.

Question 25. What is the difference between baking soda and baking powder?
Answer:

Both baking soda and baking powder are leavening agents. Baking soda is pure sodium bicarbonate. When baking soda is combined with moisture and an acidic ingredient (for example,  Yoghurt, buttermilk) the resulting chemical reaction produces CO₂ gas bubbles that cause baked goods to rise. Baking powder contains NaHCO₃, but it includes an acidifying agent (cream of tartar) already, and also a drying agent (usually starch).

Baking powder is available as single-acting baking powder and as double-acting baking powder. Single-acting baking powders are activated by moisture, so we must bake recipes that include this product immediately after mixing. Double-acting powder reacts in two phases and can stand for a while before baking.

Question 26. Though table salt is not deliquescent it gets wet ! in the rainy season— Explain.
Answer:

Pure NaCl is not deliquescent but table salt contains impurities like MgCl₂ and CaCl₂ These impurities being deliquescent absorb moisture from air in the rainy season. As a result, table salt gets wet.

Question 27. What precautions should be taken while handling beryllium compounds and why?
Answer:

Contact of Be compounds with skin dermatitis, and inhaling dust or smoke of Be-compounds causes a disease called berylliosis which is rather similar to success- Therefore, beryllium compounds should be handled with care.

Question 28. Explain why the elements of group 2 form M²⁺ Ions, but not M3+ ions.
Answer:

Loss of third electron from group-2 metal atoms causes loss of stable noble gas configuration and for this reason group- 2 elements form M²⁺ ions but not M³⁺ ions. fifl Arrange Be(OH)₂, Ba(OH)₂ & Ca(OH)₂in order of increasing solubility in water and explain the order. Answer: Among the alkaline earth metal hydroxides having a common anion, the cationic radius influences the lattice enthalpy. Since the lattice enthalpy decreases much more than the hydration enthalpy with increasing cationic size, the solubility increases on moving down the group.

Question 29. The reaction between marble and dilute H₂SO₄ is not used to prepare CO₂ gas—why?
Answer:

Marble (CaCO₃) reacts with dilute H₂SO₄ to form insoluble CaSO₄ which deposits on the surface of marble and prevents further reaction. So, the evolution of CO₂ ceases after some time. Thus the reaction between marble and dilute H₂SO₄ prepares CO₂ gas.

Question 30. Name the important compound of Li used in organic synthesis. How the compound is prepared?
Answer:

The compound is lithium aluminum hydride (LiAlH₄). It is a useful reducing agent and is used in organic synthesis. It is prepared by the reaction of lithium hydride and aluminum chloride in a dry ether solution.

4LiH + AlCl₃ → LiAlH₄ + 3LiCl

Question 31. What is the oxidation state of K in KO₂ and why is this compound paramagnetic?
Answer:

The superoxide ion is represented as O^–₂. It has one unit of negative charge. Since the compound is neutral, therefore, the oxidation state of K is +1. The structure of Or is 6J0 O^–₂ Since it has one unpaired electron in π∗_2p MO, therefore, the compound is paramagnetic.

Question 32. The crystalline salts of alkaline earth metals contain more water for crystallization than the corresponding alkali metals. Explain.
Answer:

In the salts of alkaline earth metals, the metal ions have a smaller size and higher charge compared to the corresponding metal ions of the alkali metals of the same period. Thus, alkaline earth metals have a greater tendency to get hydrated & form crystalline salts compared to alkali metals. Thus, NaCl is completely anhydrous whereas MgCl₂ exists as MgCl₂-6H₂O.

Question 33. Lithium salts are more stable if the anion present in the salt is small. Explain.
Answer:

Small anions have more ionic character and hence the salts of lithium containing those ions have more lattice enthalpy. Large anions, on the other hand, are highly polarisable, and hence they impart covalent character to the salt. Thus, lithium salts are more stable with small anions than that with large anions.

Question 34. Alkali metals become opaque when they are kept open in the air Why?
Answer:

As the alkali metals are highly reactive, they readily react 20; with oxygen to form oxides. These oxides undergo a reaction with the water vapor present in the air to produce hydroxides. The formed hydroxides immediately react with CO₂ of air to produce carbonate compounds. These carbonate compounds form layers on the surface of alkali metals. Consequently, they become opaque.

Question 35. BaSO₄ is insoluble in water, but BcSO₄ is soluble in water-Explain.
Answer:

The lattice enthalpy of BaSO₄ is much higher than its hydration enthalpy and hence, it is insoluble in water. On the other hand, the hydration enthalpy of BeSO₄ is much higher than its lattice enthalpy because of the loose packing of the small Be²⁺ ion with the relatively large sulfate ion. Hence, it becomes soluble in water.

Question 36. An aqueous solution of Be(NO₃)₂ is strongly acidic. Explain.
Answer:

In the hydrated ion, [Be(H₂O)₄]²⁺, water molecules are extensively polarised, ultimately leading to the weakening of the O —H bond.

Hydrolysis takes place and the solution becomes distinctly acidic:

(H₂O)₃Be₂ +—OH₂ + H₂O→(H₂O)₃ 3 Be⁺ —OH + H₃O⁺

Question 37. The hydroxides and carbonates of Na and K are readily soluble in water while the corresponding salts of Mg and Ca are sparingly soluble. Explain.
Answer:

Due to smaller size and higher ionic charge, the lattice enthalpies of alkaline earth metals are much higher than those of alkali metals and hence the solubility of alkali metal hydroxides and carbonates is much higher than those of alkaline earth metal hydroxides and carbonates;

Question 38. BeO is insoluble but BeSO₄ is soluble in water. Explain.
Answer:

The higher lattice enthalpy of BeO formed by the combination of a small cation and small anion is more than its hydration enthalpy but the lattice enthalpy of BeSO₄ formed by the combination of a small cation and large anion is less than its hydration enthalpy;

Short Answer Questions for Class 11 Chemistry S-Block Elements

Question 39. BaO is soluble but BaSO₄ is insoluble in water — why?
Answer:

The lattice enthalpy of BaO formed by the combination of a large cation and a small anion is less than its hydration enthalpy but the lattice enthalpy of BaSO₄ formed by the combination of a large cation and a large anion is more than its hydration enthalpy;

Question 40. Lil is more soluble than KIin ethanol. Explain.
Answer:

Due to the much higher polarising power of very small Li⁺ ion, Lil is predominantly covalent but due to the low polarising power of relatively large K⁺ ion, KI is predominantly ionic and for this reason, Lil is more soluble in the organic solvent ethanol;

Question 41. How can fused calcium chloride be prepared? Give two important uses of it.
Answer:

When CaCl₂ 2H₂O is heated above 533K, anhydrous CaCl₂ forms. This melts at 1046K. When the molten salt is cooled, it solidifies as white lumps of crystalline mass which is known as fused calcium chloride;

Question 42. Hydrated magnesium chloride is heated in the presence of ammonium chloride (NH₄Cl).
Answer:

Magnesium chloride when heated with NH₄Cl forms an additional compound (MgCl₂-NH₄Cl₆H₂O) which on heating forms anhydrous MgCl₂

Question 43. Why are alkali metals not found in nature?
Answer:

Alkali metals are highly electropositive and extremely reactive elements. Thus, they easily react with atmospheric oxygen and carbon dioxide. These metals have a high tendency to lose electrons to form cations because of their low ionization potential. For this reason, they readily react with highly electronegative elements or radicals to form compounds. So, alkali metals are not found in a free state in nature.

Question 44. Explain why is sodium less reactive than potassium.
Answer:

Due to its small size, the ionization enthalpy of sodium (496kL-mol^-1) is greater than potassium (419kJ. mol^-1) and the standard electrode potential value of potassium (-2.925V) is more negative than that of sodium (-2.714V). Thus, sodium is less reactive than potassium.

Question 45. Basicity of oxides
Answer:.

Basicity of oxides: The ionization enthalpy of alkali metals is less than that of the corresponding alkaline earth metals, i.e., the electropositive character of alkali metals is greater than that of alkaline earth metals. Thus, the basicity of oxides of alkali metals is more than the oxides of alkaline earth.

S-Block Elements Class 11 Short Answer Solutions

NCERT Solutions For Class 11 Chemistry Chapter 10 S Block Elements Warm-Up Exercise Question And Answers

Question 1. Name the lightest and the heaviest metal.
Answer:   Lithium is the lightest (density: 0.53g-cm^-3) and osmium is the heaviest (density: 22.6g-cm^-3 ) metal.

Question 2. Which one between water and pyrene (CCl₄), can be used to extinguish fire caused by metallic sodium?
Answer:  Pyrene (CCl₄) can be used.

Question 3. Which among the alkali metal ions has the lowest mobility In aqueous solution?
Answer: Li⁺ ions have the lowest mobility because they remain highly hydrated in aqueous solution.

Question 4. Which of the alkali metal cations has the highest polarising power?
Answer: Due to its very small size, Li⁺ ion has the highest polarising power among the alkali metal ions

NCERT Class 11 Chemistry S-Block Elements Short Answer Solutions

Question 5. The density of alkali metals is very low Why?
Answer: Due to large atomic size and weak metallic bonding, the densities of alkali metals are very low.

Question 6. Name a radioactive alkali metal and write its atomic number.
Answer:  Francium, atomic number =87

Question 7. Mention the similarity shown in the electronic configurations of the alkali metals.
Answer: All alkali metals have similar valence shell electronic configuration of ns¹

Question 8. Why alkali metals are called s -s-block elements?
Answer: Alkali metals are called s -s-block elements because the last electron enters the ns-orbital

Question 9. What is trona?
Answer:

Trona ( Na₂ CO₃ , NaHCO₃  .  2H₂O) is an important mineral of sodium

Question 10. Arrange lithium, sodium, and potassium according to their abundance in the earth’s crust.
Answer:

According to their abundance in nature, the elements are arranged as lithium < potassium < sodium.

S-Block Elements Chapter 10 Short Answer Questions Class 11

Question 11. Give an example of a double salt formed by an alkali metal and alkaline earth metal.
Answer:

The double salt formed by an alkali metal and an alkaline earth metal is carnallite (KCl-MgCl₂-6H₂O)

Question 12. Arrange LiF, NaF, KF, RbF, and CsF in increasing order of their lattice energies.
Answer:

The increasing order of lattice energies of the given compounds is CsF < RbF < KF < NaF < LiF.

Question 13. Why are alkali metals paramagnetic?
Answer: Due to the presence of unpaired electrons in the valence shell of alkali metals, they are paramagnetic.

Question 14. Which alkali metal is generally used in photoelectric cells?
Answer: Cesium is generally used in photoelectric cells

Question 15. Which alkali metals form superoxides when heated in excess air?
Answer: The alkali metals that form superoxides when they are heated in excess air are potassium (K), rubidium (Rb), and cesium (Cs).

Question 16. Differentiate between Na₂CO₃ and NaHCO₃
Answer:

When NaHCO₃ is heated, it liberates CO₂ which turns lime-water milky. On the other hand, when Na₂CO₃ is heated, it does not undergo decomposition

Question 17. Arrange MCI, MBr, MF, and MI (where M= alkali metal) according to increasing covalent character.
Answer:

The covalent character of metallic chlorides increases with an increase in the size of the anion. Therefore, the order of the metallic chlorides according to increasing covalent character is MF < MCI < MBr < MI.

Question 18. Explain why the peroxides and superoxides of the alkali metals act as strong oxidizing agents.
Answer:

In reaction with water, peroxides produce MOH along with H₂O₂ and superoxides produce MOH and O, along with H₂O₂. H₂O₂ is a strong oxidizing agent. Thus, peroxides and superoxides of the alkali metals act as strong oxidizing agents.

M₂O₂ + 2H₂O → 2MOH + H₂O₂

2MO₂ + 2H₂O→ 2MOH + H₂O₂ + O

Question 19. Give a simple test to distinguish between KNO₃ & LiNO₃.
Answer:

Colourless O₂ gas evolves on heating KNO₃. But heating, LiNO₃ dissociates into colorless gas and brown NO₂ gas.

Question 20. Explain why a solution of Na₂CO₃ is alkaline in nature whereas a solution of Na₂SO₄ is neutral.
Answer:

For Na₂CO₃, Na₂SO₄ is a salt of strong base (NaOH) and strong acid (H₂SO₄). So, the nature of the solution of Na₂SO₄ is neutral.

Question 21. Among the sulfate salts of lithium, sodium, potassium, and rubidium, which salt does not form double salt?
Answer: Lithium sulfate (Li₂, SO₄) does not form any double salt

Question 22. What happens when sodium sulfate is fused with charcoal? Give equation.
Answer:

When sodium sulfate is fused with charcoal, it reduces to sodium sulfide, and carbonÿis oxidized to carbon monoxide:

Na₂SO₄ + 4→ Na₂S + 4CO↑

Question 23. Which alkali metal bicarbonate has no existence in the solid state?
Answer:

The alkali metal bicarbonate which has no existence in a solid state is lithium bicarbonate (LiHCO₃)

Question 24. Mention the property for which lithium is used to separate N2 gas from a gas mixture.
Answer:

The alkali metal bicarbonate which has no existence in a solid state is lithium bicarbonate (LiHCO₃)

Class 11 Chemistry S-Block Elements SAQ

Question 25. Give a simple test to distinguish between Li₂CO₃ & Na₂CO₃. 
Answer:

On heating, Li₂CO₃ decomposes to CO₂ which turns lime water milky. On the other hand, Na₂ CO₃ does not decompose on heating

Question 26. Write down the name of the alkali metal compound which i

1. Effective in the treatment of manic depressive psychosis
2. Used in baking powder

Answer:

1. Li₂CO₃
2. NaHCO₃

Question 27. Why a standard solution of sodium hydroxide (NaOH) cannot be prepared?
Answer:

Sodium hydroxide, being a hygroscopic substance absorbs moisture from the atmosphere. It also absorbs CO₂ from the air and forms Na₂CO₃. Thus, sodium hydroxide cannot be accurately weighed and so a standard solution of NaOH cannot be prepared.

Question 28. How the group-2 elements are commonly known? Why are they so called?
Answer:

Group-2 elements are commonly known as ‘alkaline earth metals’. These are so-called because their oxides are basic and are found in the earth’s crust.

Question 29. Why group-2 elements are called s -s-block elements?
Answer: Group-2 elements are called s -s-block elements because the last electron enters the s -s-orbital

Question 30. Which group-2 element has a slightly different electronic configuration than the rest of the elements?
Answer:  Beryllium has 2 electrons in its penultimate shell while the rest have 8 electrons in their penultimate shells.

Question 31. Explain why the atomic and ionic radii of Mg are less than those of Na and Ca.
Answer:

The electrons of Mg having a higher nuclear charge are more strongly attracted towards the nucleus. Thus, the atomic and ionic radii of Mg are less than Na. Again on moving down the group (from Mg to Ca), the atomic and ionic radii increase due to the addition of new shells, and the increasing screening effect of joindy overcomes the effect of increasing nuclear charge. Thus, the atomic and ionic radii of Mg are less than Ca.

Question 32. Arrange Mg²⁺, Ba²⁺, Sr²⁺, Be²^+, and Cav according to decreasing order of their hydration enthalpies. Explain your answer.
Answer: The correct order is  → Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺

S-Block Elements Chapter 10 NCERT Short Answer Questions

Question 33. Alkaline earth metals predominantly form ionic compounds. However, the first member of the group forms covalent compounds—Explain why.
Answer:

Due to relatively high electronegativity, the first member of each of these groups tends to form covalent compounds.

Question 34. A white residue is obtained when metallic Mg is burnt in air. This residue when heated with water emits an ammoniacal smell. Explain these observations.
Answer:

On burning metallic magnesium in the air, magnesium oxide (MgO) and magnesium nitride (Mg₃N₂) are formed. So, the white residue obtained is a mixture of MgO & Mg₃N₂.

This residue when heated with water, results in the hydrolysis of Mg₃N₂ which emits an ammoniacal smell.

Question 35. Why calcium is better than sodium in eliminating a small amount of water from alcohol?
Answer:

Both Na and Ca react with water to form the corresponding hydroxide. However, sodium rapidly reacts with alcohol to form sodium ethoxide (NaOC₂H₅) but calcium reacts quite slowly with alcohol

2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑

Thus, for eliminating a small amount of water from alcohol, calcium is better than sodium.

Question 36. Why Mg-ribbon continues to burn in the presence of SO₂ gas
Answer:

During the combustion of Mg-ribbon, the amount of heat generated leads to the decomposition of sulfur dioxide into sulfur and oxygen. This oxygen is responsible for the continued burning of Mg-ribbon.

2Mg + SO₂→ 2MgO + S

Question 37. Except Be(OH)₂, all other alkaline earth metal hydroxides are basic and their basic strength increases down the group.
Answer:

Be has high ionization enthalpy, for which Be(OH)₂ is amphoteric.

Question 38. Why is Mg(OH)₂ less basic than NaOH?
Answer:

Due to the larger ionic size and lower ionization enthalpy of Na, the Na — OH bond in NaOH is weaker than the Mg — OH bondin Mg(OH)₂. Thus, NaOH is more basic than Mg(OH)₂

Question 39. Sparingly soluble carbonate salts of alkaline earth metals become easily soluble in water in the presence of CO₂. Why?
Answer:

Sparingly soluble carbonate salts of alkaline earth metals are converted into soluble bicarbonate salts in the presence of CO₂ So, these salts become easily soluble in water.

CaCO₃(s) + CO₂(g) + H₂O(l)→ Ca(HCO₃)₂ (aq)

Short Answer Solutions for S-Block Elements Class 11 Chemistry

Question 40. The anhydrous chloride salt of which alkaline earth metal is used as a drying agent in the laboratory?
Answer:

Anhydrous chloride salt of calcium metal, i.e., calcium chloride (CaCl₂) is used as a drying agent in the laboratory.

Question 41. Why BeCO₃ is kept in an atmosphere of CO₂?
Answer:

BeCO₃ being highly unstable easily decomposes to give off CO₂ when kept in an open atmosphere.

Question 42. Write down some important points of difference between beryllium and magnesium.
Answer:

• Beryllium is harder than magnesium,
• Compounds of Be are largely covalent, whereas most of the compounds of Mg are ionic,
• Be does not exhibit a coordination number of more than 4, while Mg exhibits a coordination number of 6

Question 43. Be usually forms covalent compounds but other elements of group ionic compounds. Why?
Answer:

Be usually forms covalent compounds due to its high ionisation enthalpy and small size. However, due to comparatively low ionization enthalpy and large size, other elements of group-2 form ionic compounds

Question 44. Which compounds of the alkaline earth metals are used as refractory substances?
Answer:

The oxides of alkaline earth metals (MO) have high melting points and so are used as refractory substances.

Question 45. Carbonaceous impurities in gypsum and any fuel are avoided during the preparation of Plaster of Paris. Explain.
Answer:

Carbonaceous impurities in gypsum and any fuel are avoided during the preparation of Plaster of Paris because carbon will reduce CaSO₄ to CaS.

CaSO₄ + 4C →  CaS + 4CO ↑

Question 46. The fire caused by sodium in the laboratory cannot be extinguished by spraying water Why?
Answer:

Sodium reacts vigorously with water producing H₂ gas which catches fire by the heat evolved in the reaction. So water cannot be used for extinguishing sodium-fire

Question 47. Why does Li not exist with Na or K in their minerals?
Answer:

Lithium forms independent minerals and does not exist with Na or K because the Li⁺ ion is too small to replace the more abundant Na+ or K⁺ ions in their minerals.

Question 48. How can you prepare propyne from magnesium carbide?
Answer:

When magnesium carbide (MgC₂) is heated, Mg₂C₃ (magnesium allylide) is obtained. This on hydrolysis, produces propyne (CH₃C=CH)

Question 49. Why do halides of Be dissolve in organic solvents while those of Ba do not?
Answer:

Halides of Be are covalent. Hence, they dissolve in organic solvents while Ba is ionic. Hence, they do not dissolve in organic solvents.

Question 50. Write with a balanced equation the reaction for the manufacture of sodium bicarbonate from sodium carbonate.
Answer:

Sodium bicarbonate (NaHCO₃) Is manufactured by passing CO2 through a saturated solution of sodium carbonate (Na₂CO₃).

Na₂CO₃ + CO₂ + H₂O ⇌ 2NaHCO₃↓

NCERT Class 11 Chemistry Chapter 10 S-Block Elements Short Answer Questions

Question 51. Write the balanced equation(s) for the reaction when excess carbon dioxide is passed through brine saturated with ammonia
Answer:

When carbon dioxide is passed through an aqueous solution of NaCI (brine, 28% NaCl solution) saturated with ammonia, sodium bicarbonate is formed.

Question 52. Why is LiCl soluble in organic solvents?
Answer:

As the polarising power of Li⁺ is very high, LiCl is covalent. Hence, it is soluble in an organic solvent.

Question 53. Explain why can alkali and alkaline earth metals not be obtained by chemical reduction methods.
Answer:

Alkali and alkaline earth metals are strong, reducing agents and it is difficult to reduce their oxides or chlorides by any other reducing agent.

Question 54. With the help of a drop of an indicator solution, how would you know whether a solution consists of Na₂CO₃ or NaHCO₃?
Answer:

A drop of phenolphthalein will change the colorless solution of Na₂CO₃ to purple but the colorless solution of NaHCO₃ will remain unchanged;

Question 55. The crystalline salts of alkaline earth metals contain more water of crystallization than the corresponding alkali metal salts—why?
Answer:

Due to their smaller size and higher charge, the alkaline earth metal ions have a greater tendency to coordinate with water molecules as compared to alkali metal ions.

Question 56. Beryllium chloride hydrate loses no water over P₄O₁₀ — why?
Answer:

Due to very small size and stronger hydrating tendency of Be²⁺ ion, it is not possible for P₄O₁₀ to abstract water molecules from beryllium chloride hydrate, [Be(H₂O)4]Cl₂;

Question 57. Why are fumes seen when barium halides are kepi In open air?
Answer:

In the open air, barium halide undergoes hydrolysis by water vapor (moisture), and fumes of halogen acid (except HF) evolve.

Question 58. Explain why the compounds of beryllium are much more covalent than the other Gr-2 metal compounds.
Answer:

Due to the very small size and high charge of Be²⁺ ion, it has much higher polarising power and because of this, the compounds of Be are much more covalent than the other group-2 metal compounds.

Question 59. Beryllium compounds are extremely toxic why?
Answer:

Beryllium compounds are extremely toxic because of their very high solubility and their ability to form complexes with enzymes in the body.  Also, beryllium displaces magnesium from some enzymes.

NCERT Class 11 Chemistry Chapter 10 S-Block Elements Short Answer Questions

Question 60. Which calcium salt causes the formation of the kidney? Stones?
Answer:

Calcium oxalate, \(\mathrm{Ca}^2+\stackrel{\ominus}{\mathrm{O}}_2 \mathrm{C}-\mathrm{C} \stackrel{\ominus}{\mathrm{O}}_2\) causes formation of kidney stones

Question 61. Alkali metals are good reducing agents—Why?
Answer: 

The smaller the ionization enthalpy, the greater the reducing strength. Since alkali metals have lower ionization enthalpies, they are good reducing agents.

Question 62. Explain why the alkali metals cannot be obtained by the reduction method.
Answer: Alkali metals are strong reducing agents and it is difficult to reduce their oxide with any other reducing agent

Question 63. Which alkali metal ion has the maximum polarising j power and why?
Answer:  Li⁺ ion has the highest polarising power among all the alkali metal ions as the value of charge to size ratio of the smallest Li+ ion is the highest.

NCERT Solutions For Class 11 Chemistry Chapter 10 S Block Elements Multiple Choice Questions

Question 1. NO₂ is not obtained on heating—

1. AgNO₃
2. KNO₃
3. Cu(NO₃)₂
4. Pb(NO₃)₂

Answer: 2. KNO₃

KNO₃ on heating decomposes to form potassium nitrite (KNO₂) and oxygen gas

Question 2. Which one of the following has the lowest ionization

1. ls²2s²2p⁶
2. ls²s²2p⁶3s¹
3. ls²2s²2p⁵
4. ls²2s²2p³