NCERT Class 11 Chemistry Chemical Thermodynamics Multiple Choice Questions

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.

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, \(T>\frac{400}{} \mathrm{~K}_1 \text { or, } T>20 \mathrm{~K}\)

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

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

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.

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

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

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

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

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

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.

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

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.