NCERT Class 11 Chemistry Chapter 7 Equilibrium Short Question And Answers

NCERT Class 11 Chemistry Chapter 7 Equilibrium Short Question And Answers

Question 1. The unit of the equilibrium constant of the reaction, A + 3B ⇌ nC is L².mol^-2. What is the value of
Answer:

For the given \(K_c=\frac{[C]^n}{[A][B]^3}\)

Thus, the unit of

K_c = \(\frac{\left(\mathrm{mol} \cdot \mathrm{L}^{-1}\right)^n}{\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right) \times\left(\mathrm{mol} \cdot \mathrm{L}^{-1}\right)^3}\)

= (mol .L^-14)n^-4

Hence, L₂.mol^-2 = (mol . L^-1)n^-4

Or, n = 2

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Question 2. Find out the value of K_p/K_c for the reaction \(\mathrm{PCl}_5(g) \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)\) at 298K, consider the unit of concentration is mol L-1 and the unit of pressure is atm.
Answer:

We know. K_p = K_c(RT)An For the given reaction, Δn = (1 +1-1) = 1

Thus, \(K_p=K_c \times R T \quad \text { or, } \frac{K_p}{K_c}=R T\)

⇒ \(0.0821 \mathrm{~L} \cdot {atm} \cdot \mathrm{mol}^{-1} \mathrm{~K}^{-1} \times 298 \mathrm{~K}\)

= \(24.465 \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1}\)

Question 3. How will the following reaction equilibrium be affected if the volume of each reaction system is increased at a constant temperature?
Answer:

At constant temperature, if the volume of the reaction system is increased, then the total pressure at equilibrium will be decreased. Thus, the equilibrium of the system will be disturbed. According to Le Chatelier’s principle, equilibrium will be shifted, in a direction that increases the total number of molecules.

In the ofthe first reaction, the equilibrium will shift to the left side. As a result, the product yield, [SO₃] will decrease. On the other hand, in the case ofthe second reaction, the equilibrium will shift to the right side “thereby increasing the yield of the product [CO(g)].

Question 4. Any reversible reaction’s equilibrium may be shifted to the left or reality changing the conditions. Will this change cause any alteration in the value of the equilibrium constant?
Answer:

At ascertain temperature The equilibrium constant of a reversible reaction has a definite value.

If temperature remains fixed, then the equilibrium can be shifted to the left or right by changing the conditions of pressure, temperature, etc: on which the equilibrium of a reversible reaction depends. Consequently, the respective amount of both reactants and products will change, but the value of the equilibrium constant remains unchanged since the temperature remains fixed.

If the equilibrium is shifted due to temperature change, then the amounts of both reactants and products as well as the value of the equilibrium constant will change. With the increase in temperature, the value of the equilibrium constant will increase for an endothermic reaction and decrease for an exothermic reaction.

Question 5. At constant temperature, if the pressure is changed at the equilibrium of a gaseous reaction, then will the values of K_p, K_c, and K_x change?
Answer:

We known \(\Delta G^0=-R T \ln K_p \quad \text { or, } K_p=e^{-\frac{\Delta G^0}{R T}}\) where G° = standard free energy change ofthe reaction.

The value of ΔG° depends only on temperature. Its value is independent of pressure. So, the value of Kp is independent of pressures.

We know, K_p = K_c(RT)^Δn. Since the value of Kp does not depend on pressure, according to this relationship, the value of Kc is also independent of pressure.

Again, we know, K_p= K_x(p)n or \(K_x=\frac{K_p}{(P)^{\Delta n}}\)

As K_p does not depend on pressure, according to this relation, the value of Kx depends on pressure. However if Δn = 0, then pressure will not affect K_x.

Question 6. How can the yield of the products be increased by changing the volume of the reaction system in the given reactions at constant temperature?

C(s) + H₂O(g)⇌ CO(g) + H₂(g)

2H₂(g) + O₂ (g)⇌ 2H₂O(l)

Answer:

In this case, the volume increases in the reaction as written (since, Δn = +1 ). Thus, if the volume of the reaction system is increased at a constant temperature, then equilibrium will shift to the right, and consequently, the yields ofthe products will increase.

In this case, the volume decreases in the reaction as written (since, Δn = -3 ). Thus, if the volume of the reaction system is decreased at constant temperature then equilibrium will shift to the right. As a result, the yield of products will increase.

Question 7. Mention two factors for which die yields of the products in the given reaction increase.

⇒ \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})  \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \text {-heat }\)

Answer:

Keeping both temperature and volume fixed, if we add some reactants [CO(g) or H₂O(g)] to the reaction system or remove some products [CO₂(g) or H₂(g) ] from the reaction system, equilibrium will shift to the right, which will result in higher yields ofthe products.

The reaction is endothermic. Thus, on increasing the temperature at equilibrium, the equilibrium of the reaction will shift to the right. This will cause higher yields of the products.

Question 8. What will be the change in concentrations of H₃O⁺ & OH^– and the ionic product of water (K_w) if NaOH is added to pure water at a certain temperature?
Answer:

Since Kw is fixed at a certain temperature, it will not undergo any change due to the addition of NaOH in pure water.

However, the concentration of OH- ions increases due to the addition of NaOH, causing the dissociation equilibrium of H₂O to shift to the left

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

As a result, the concentration of H₃O+ ions in the solution is reduced

Question 9. 20 mL of 0.15(M) HCl solution is mixed with 50 mL of 0.1(M) CH₃COONa solution. State whether the mixed solution will act as a buffer or not.
Answer:

Number of millimoles of CH₃COONa₃ in 50 mL 0.1(M) CH₆COONa₃ =0.1 × 50 = 5 and that of HCl in 20mL 0.15(M)HCl= 0.15 × 20 = 3.

The reaction between CH₆COONa₂ and HCl is:

CH₃COONa(O + HCl(aq)→CH₃COOH(aq) + NaCl(aq)

Hence, 3 millimol of HCl + 3 millimol of CH₆COONa → 3 millimol of CH₃COOH + 3 millimol of NaCl .

Therefore, at the end of the reaction, there remains 3 millimol of CH₃COOH and (5-3) = 2 millimol of CH₃COONa.

∴ The resulting solution consists of weak acid (CH₃COOH) and its salt (CH₃COONa). So, it acts as a buffer.

Question 10. What would the effect on the yield of products be If the temperature of the following reaction systems Is changed at equilibrium?

N₂(g) + O₂(g) ⇌ 2NO(g); ΔH > 0

2SO₂(g) + O₂ (g) ⇌ 2SO₃(g); ΔH < 0

Answer:

The reaction is endothermic (as ΔH > 0 ). If the temperature is increased at the equilibrium of this reaction, then the equilibrium will shift to the right, and the yield of products will increase. On the other hand, if the temperature is reduced at the equilibrium of the reaction, then the equilibrium will shift to the left, and the die yield of products will be reduced.

The reaction is exothermic (as ΔH < 0 ). If the temperature is decreased at the equilibrium of this reaction, then the equilibrium will shift to the right, and the yield of products will increase. On die other hand, if the temperature is increased at the equilibrium of the reaction, then equilibrium will shift to the left, and the yield of products will be reduced.

Question 11. Identify Lewis acids and Lewis bases in the following reactions and give reasons:

1. SiF₄ + 2F→  SiF₆^2-

2. \(R M g X+2\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \ddot{\mathrm{O}}: {RMg}\left[\mathrm{O}\left(\mathrm{C}_2 \mathrm{H}_5\right)_2\right]_2 \mathrm{X}\)

3. \(\mathrm{Ag}^{+}+2 \ddot{\mathrm{N}} \mathrm{H}_3 \rightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}\)

4. \(\ddot{\mathrm{N}} \mathrm{H}_3+\mathrm{H}^{+} \rightarrow \stackrel{+}{\mathrm{N}} \mathrm{H}_4\)

Answer:

According to Lewis’s concept, an acid is a substance that can accept one or more electron pair(s). Generally, cations (such as Ag⁺, H⁺, K⁺), compounds having a central atom with an incomplete octet (such as SiF₄, AlF₃, RMgX, BF₃), and compounds whose central atom is linked to an electronegative atom by a double bond (such as GO₂ ) can act as Lewis acids. In the given reactions, Lewis acids are SiF₄, RMgX, Ag⁺, and H⁺.

According to Lewis’s concept, a base is a substance that can donate one or more electron pair(s). Anions (such as F^–, OH₂Si₂) and compounds with lone pairs of electrons can act as Lewis bases. Therefore, in the given reactions, Lewis bases are F-, NH₃, (C₂H₅)₂O, and NH₃.

Question 12. What will happen when a solution of potassium chloride is added to a saturated solution of lead chloride? Give reason.
Answer:

When potassium chloride solution is added to a saturated lead chloride solution, then the solubility of lead chloride decreases due to the common ion (Cl^–) effect.

Explanation:

The following equilibrium is established in an aqueous PbCl₂ solution:

PbCl₂(s) Pb²⁺(ag) + 2Cl^– (ag)

The addition of KCl to the saturated solution of PbCl₂ increases the concentration of common ion Cl^– the above equilibrium to get disturbed. To re-establish the equilibrium, some of the Cl^– ions will combine with an equivalent amount of Pb²⁺ ions to form solid PbCl₂. Therefore, as an overall effect, the equilibrium is shifted to the left. Hence, the solubility of PbCl₂ decreases.

Question 13. Why does not MgS04 form any precipitate when it reacts with NH₃ in the presence of NH₄Cl?
Answer:

NH₃ is a weak base. In an aqueous solution, it ionizes partially to produce NH+ and OH- ions.

⇒ \(\mathrm{NH}_3(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q)\)

In the presence of NH₄Cl, owing to the common ion effect of NH₄, the degree of ionization of NH₃ is further suppressed. Thus, the concentration of OH^– ions decreases to a large extent.

At this low concentration of OH^– ions, the product of the concentration of Mg²⁺ ions and square of the concentration of OH^– ions (as K_sp[Mg(OH)₂] =[Mg²⁺] × [OH^–]² cannot exceed the solubility product of Mg(OH)₂, i.e., [Mg²⁺] x [OH^–]² < K_sp (solubility product). As a result, Mg(OH)₂ does not get precipitated.

Question 14. Will the pH of pure water at 20°C be lower or higher than that at 50°C?
Answer:

The ionic product of water (Kw) increases with, a temperature rise Hence

⇒ \(K_w\left(50^{\circ} \mathrm{C}\right)>K_w\left(20^{\circ} \mathrm{C}\right)\) or, \(p K_w\left(50^{\circ} \mathrm{C}\right)<p K_w\left(20^{\circ} \mathrm{C}\right).\)

Since pK_w=-log₂₀K_w

Now, for pure water \(p H=\frac{1}{2} p K_w\)

Question 15. Both CuS and ZnS are precipitated if H₂S gas is passed through an alkaline solution of Cu²⁺ and Zn²⁺. Explain.
Answer:

In an aqueous solution, H₂S ionizes to establish the following equilibrium,

H₂S(aq) + 2HzO(l) ⇌ 2H₃O⁺(aq) + S^2-(aq)

The degree of ionization of H₂S increases in alkaline solution because OH^– ions present in the solution react with H₃O⁺ ignite) form unionized water molecules. This shifts the equilibrium to the right, thereby increasing the concentration of S^2- ions? ‘in the presence of a high concentration of S^2- ions, [Cu²⁺] (S^2-] > and [Zn²⁺][S^2-] > Ksp(ZnS). As a result, both CuS and ZnS are precipitous.

Question 16. Why is an aqueous solution of NaNO₃ neutral?
Answer:

NaNO₃ is a salt of strong acid HNO₃ and strong base NaOH. In its aqueous solution, NaNO₃ dissociates completely, forming Na+ and NO₃ ions. In an aqueous solution, Na⁺(aq) is a weaker acid than H₂O and NO₃ is a weaker base than H₂O.

So, neither Na+(a₂) nor NO₂(aq) reacts with water. As a result, there is no change in the concentration of either H₃O⁺ ions or OH^– ions. Due to this, an aqueous solution of NaNO₃ is neutral.

Question 17. An aqueous solution o/(NH4)₂SO₄ is acidic. Explain
Answer:

(NH₄)₂SO₄ is a salt of a weak base (NH₃) and a strong acid (H₂SO₄). In its aqueous solution, (NH₄)₂SO₄ dissociates almost completely forming NH₄⁺ and SO₄^2- ions. SO₄^2- ion is a conjugate base of strong acid H₂SO₄ and hence in aqueous solution, it is a very weak base in comparison to H₂O.

As a result, the SO₄^2- ion does not react with water in aqueous solution. On the other hand, NH₄ is a conjugate acid of weak base NH₃. In an aqueous solution, the NH₄⁺ ion shows a higher acidic character than H₂O.

As a result, NH₄ ions react with water.

[NH₄(O^-7) + H₂O(l)⇌NH₃(aq) + H₃O+(aq)]

Causing an increase in the concentration of H₃O⁺(aq) ions in the solution. This makes the solution of (NH₄)₂SO₄ acidic.

Question 18. At 25°C what is the concentration of H₃O⁺ ions in an aqueous solution in which the concentration of OH^– ions is 2 × 10^-5(M)?
Answer:

At 25 °C, K_w = 10-14. Now, for an aqueous solution, [H₃O+] × [OH^–]

= K_w At 25 °C, [H₃O⁺][OH^–] = 10^-14

Given: [OH^–] = 2 × 10^-5(M)

Therefore, [H₃O⁺] \(=\frac{10^{-14}}{2 \times 10^{-5}}=5 \times 10^{-10}(\mathrm{M})\)

Question 19. At a certain temperature, the ratio of ionization constants of weak acids HA and HB is 100: 1. The molarity of the solution is the same as that of HB, and the degrees of ionization of HA and HB in their respective solutions are α₁ and α₂ respectively, then show that α₁ = 10α₂
Answer:

Suppose, the ionization constants of HA and MB arc K₁ and K₂, respectively. If the concentration of each of the solutions of HA and HB is c mol.L^-1, then

⇒ \(\alpha_1=\sqrt{\frac{K_1}{c}} \text { and } \alpha_2=\sqrt{\frac{K_2}{c}} \text {; }\)

Where α₁ and α₂ are the degrees of ionization of HA and HB in their respective solutions.

It is given that K₁ : K₂ = 100: 1

⇒ \(\text { So, } \frac{\alpha_1}{\alpha_2}=\sqrt{\frac{K_1}{K_2}}=\sqrt{100}=10 \text {, i.e., } \alpha_1=10 \alpha_2\)

Question 20. Find [OH^–] in pure water if [H₃O⁺] in it is x mol L^-1. Also, find the relation between x K_w.
Answer:

In pure water,

⇒  \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]\)

Therefore, [OH^– ] in pure water \(=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=x \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

We know, Kw = [H₃O⁺] × [OH^–]

∴ K_w = x × x

or, K_w = x²

∴ \(x=\sqrt{K_w}\)

Question 21. Identify the Lewis acids and Lewis bases in the following reactions

1. H⁺ + OH^–→ H₂O
2. Co³⁺ + 6NH₃ ⇌  [CO(NH₃)6)³⁺
3. BF₃ + :NH₃→ [H₃N→ BF₃]
4. CO₂ + OH^–→HCO₃^–
5. AlF₃ + 3F^–→AlFl₆^3-

Answer:

According to Lewis’s concept, an acid is a substance that can accept a pair of electrons. Generally, cations 

Example: Ag⁺, H⁺, K⁺, etc.

Molecules with the central atom having incomplete octet

Example: SiF₄, AlF₃, RMgX, BF₂,

And molecules in which the central atom is linked to an electronegative atom through double bonds

Example: CO₂  can act as Lewis acid. In the given reactions, Lewis acids are H⁺, CO³⁺, BF₃, CO₂, and AlF₃.

According to Lewis’s concept, a base is a substance that can donate a pair of electrons. Anions

Example; F^–, OH^–, etc.) and molecules having unshared electron pairs act as Lewis bases. In the given reactions, Lewis bases are OH^–,: NH₃, F^–

Question 22. We know, ΔG° = -RTInKc and ΔG° = -RTlnKp. Therefore in case of a reaction occurring in the gaseous phase at a given temperature, ΔG° is the same even if the values of K_p and K_c are different. Is the statement true? Give reasons.
Answer:

The statement is not true. In the equation, ΔG° = -RTnKc, the concentration of each of the reactants and products at standard state is taken as l(M).

On the other hand, in the equation, ΔG° = -RTnKp, the partial pressure of each of the reactants and products at standard state is taken as 1 atm. Therefore, the different values of AG° will be obtained from these two equations.

The values of the equilibrium constant (K) of a reaction at 25°C and 50°C are 2 × 10^-1 and 2 ×10^-2 respectively. Is the reaction an exothermic or endothermic?

Question 23. Consider the reaction, \(2 \times a=c \text { or, } a=\frac{c}{2} \text {. }\) Heat and answer the following questions:

1. Find the relation among a, b, and c.
2. State whether the equilibrium will be shifted towards right or left if the temperature is Increased.

Answer:

According to the equation ofthe reaction, c mol of XY forms when a mol of X₂ reacts with b mol of Y₂. So, the number of X atoms in a mol of X₂ = The number of X atoms in c mol of XY.

since \(2 \times a=c \text { or, } a=\frac{c}{2} \text {. }\)

Similarly, the number of Y atoms in b mol of Y₂ = the number of Y atoms in c mol of XY.

∴ \(2 \times b=c \text { or, } b=\frac{c}{2}\)

∴ \(a=b=\frac{c}{2} \text {. }\)

The reaction is exothermic. So, according to Le Chatelier’s principle, a temperature rise will cause the equilibrium ofthe reaction to shift to the left.

Question 24. State Le Chatelier’s principle, explain the effect of (a) pressure and (b) continuous removal at the constant temperature on the position of equilibrium of the following reaction:

H₂(g) + I₂(g) ⇌ 2HI(g)

Answer:

Pressure does not have any effect on the position of the equilibrium because the reaction is not associated with any volume change [Total no. of molecules of HI(g) = Total no. of molecules of H₂(g) and I₂(g)].

If HI is removed continuously from the reaction system, then the equilibrium goes on shifting towards the right, and finally, the reaction moves towards completion.

Question 25. Consider the following reaction:

2A(g) + B₂(g) ⇌ 2AB(g); ΔH < 0. How can the yield of AB(g) be Increased?
Answer:

For the reaction ΔH < 0, it is an exothermic reaction. According to Le Chatelier’s principle, if the temperature of an exothermic reaction at equilibrium is decreased, the equilibrium of the reaction shifts to the right. So, the decrease in temperature will result in a higher yield of AB(g).

The given reversible reaction is associated with the decrease in number of moles [An = 2- (2 + 1) =-l ] in the forward direction. So, according to Le Chatelier’s principle, if the pressure is increased at the equilibrium of the reaction, the equilibrium shifts to the right, thereby increasing the yield of AB(g)

At constant temperature and volume, if the reactant A₂(g) or B₂(g) is added to the reaction system at equilibrium, then, according to Le Chatelier’s principle, the equilibrium of the reaction will shift to the right. As a result, the yield of AB(g) will increase.

Question 26. At a particular temperature, the following reaction is carried out with 1 mol of A(g) and 1 mol of B(g) in a closed vessel:

A(g) + 4B(g) ⇌ AB₄(g). Will the equilibrium concentration of AB₄(g) be higher than that of A(g)?
Answer:

According to the given equation of the reaction, 1 mol of AB₄ forms due to the reaction between 1 mol of A and 4 mol of B. Suppose, the concentration of AB₄ at the equilibrium of the reaction is x mol- L^-1

So, according to the given equation, the concentration of A and B at equilibrium will be (1 – x) mol- L^-1 and (1-4) mol. L^-1 respectively. At equilibrium, if the concentration of AB, was greater than that of A, then x would be greater than (1 – x), i.e., x > 1 – x or, x > 0.5.

If x was greater than 0.5, then the concentration of B would be negative. This is impossible. Therefore, the concentration of AB₄ can never be greater than that of A.

Question 27. For the reaction 2H₂(g) + O₂(g) ⇌  2H₂O(g) — K_p = K_c(RT)^x. Find the value of.
Answer:

For the given reaction, Δn = 2-(2+1) = -1

Δn = 2-3 = -1

We know, K_p = K_c(RT)^Δn

As Δn = -1, K_p = K_c(RT)^-1 …………………(1)

Given, K_p = K_c(RT)^x…………………(2)

Comparing equations (1) and (2), we have x = -1

Question 28. At 200°C, the equilibrium N₂O₄(g) 2NO₂(g) is achieved through the following two pathways: 0.1 mol N₂O₄ is heated in a closed vessel L volume, A mixture of 0.05 mol N₂O₄(g) and 0.05 mol NO₂(g) is heated at 200°C in a closed vessel of 1 L volume. In these two cases, will the equilibrium concentrations of N₂O₄(g) and NO₂(g) and the values of equilibrium constants be the same?
Answer:

The value of the equilibrium constant of a reaction depends only on temperature. It does not depend on the initial concentrations of the reactants. Since the temperature is the same for both experiments, the value of the equilibrium constant will be the same in both cases.

The initial concentrations of the reactant(s) in the two experiments are not the same. As a result, the molar concentrations of N₂O₄(g) and NO₂(g) at equilibrium will be different in the two experiments.

Question 29. What will be the relation between Kp and Kc for the given equilibrium?CO(g) + H₂O(g )⇒ CO₂(g) + H₂(g)
Answer:

We know, \(K_p=K_c(R T)^{\Delta n}\)

For the given reaction, An = (1 + 1) – (1 +1) = 0

Therefore, for the given reaction K_p = K_c(RT)° – K_c

Question 30. At a given temperature, for reaction A B, the rate constant (k) of the forward reaction is greater than that of the backward reaction (kb). Is the value of the equilibrium constant (K) for this reaction greater than, less than, or equal to 1?
Answer:

The equilibrium constant (K) of a reaction

⇒ \(=\frac{\text { Rate constant of the forward reaction }}{\text { Rate constant of the backward reaction }}=\frac{K_f}{K_b}\)

Given: Kj→ Kb. Therefore K > 1.