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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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121 Medium
121 of 329
Application +4 | -1
If the enthalpy change for the reaction is 2NH3( g)→N2( g)+3H2( g)2 \mathrm{NH}_{3(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} is 92 kJ mol−192 \mathrm{~kJ} \mathrm{~mol}^{-1} the enthalpy of formation of ammonia is
View answer and explanation
Correct option: B
−46 kJ mol−1-46 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option B (−46 kJ mol−1-46 \mathrm{~kJ} \mathrm{~mol}^{-1}) is correct because the enthalpy of formation is half the negative of the given enthalpy change for the reverse reaction.
122 Medium
122 of 329
Understanding +4 | -1
The change in internal energy equals
View answer and explanation
Correct option: D
Adiabatic work
Option D (Adiabatic work) is correct because in an adiabatic process, there is no heat exchange with the surroundings (Q=0), so the change in internal energy is equal to the work done.
123 Medium
123 of 329
Understanding +4 | -1
Select the correct statements.
View answer and explanation
Correct option: C
q=+q=+ve means heat is transferred from the surroundings to the system
Option C is correct because a positive q value indicates that heat is transferred from the surroundings to the system, meaning the system gains heat.
124 Medium
124 of 329
Understanding +4 | -1
The correct relation for KfK_f is/are:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the given expressions are valid representations of the cryoscopic constant, KfK_f.
125 Medium
125 of 329
Application +4 | -1
If at 298 K the bond energies of C−H, C−C, C=CC - H,\,C - C,\,C = C and H−HH - H are respectively, 414, 347, 615 and 435 kJ mol−1,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}, the value of enthalpy change for the reaction H2C=CH2(g)+H2(g)→H3C−CH3(g){H_2}C = C{H_2}\left( g \right) + {H_2}\left( g \right) \to {H_3}C - C{H_3}\left( g \right)
at 298 K will be:
View answer and explanation
Correct option: D
– 125 kJ
Option D (– 125 kJ) is correct because the enthalpy change is calculated by subtracting the bond energies of the products from the bond energies of the reactants.
126 Medium
126 of 329
Application +4 | -1
The temperature of 15 ml15 \mathrm{~ml} of a strong acid increases by 2∘C2^{\circ} \mathrm{C} when 15 ml15 \mathrm{~ml} of a strong base is added to it. If 5 ml5 \mathrm{~ml} of each are mixed, temperature should increase by
View answer and explanation
Correct option: C
2∘C2^{\circ} \mathrm{C}
Option C (2∘C2^{\circ} \mathrm{C}) is correct because the temperature change depends on the concentration and the heat of neutralization, which remains constant under the same conditions.
127 Medium
127 of 329
Application +4 | -1
Temperature of 5 moles of a gas is decreased
by 2K at constant pressure. Indicate the correct
statement
View answer and explanation
Correct option: B
Work done over the gas is = 10 R
Option B (Work done over the gas is = 10 R) is correct because the work done at constant pressure is calculated as -PΔV, and since the temperature decreases, the volume decreases, resulting in work being done *on* the gas.
128 Medium
128 of 329
Application +4 | -1
The latent heat of fusion of ice at 0∘C0^{\circ} \mathrm{C} being 1.44 k cal mol−11.44 \mathrm{~k~cal} \mathrm{~mol}^{-1} and difference in heat capacity of ice and water being 9.3 cal mol−19.3 \mathrm{~cal} \mathrm{~mol}^{-1}, thus latent heat of fusion of ice at −27∘C-27^{\circ} \mathrm{C} is
View answer and explanation
Correct option: B
1.254 k cal mol−11.254 \mathrm{~k~cal} \mathrm{~mol}^{-1}
The correct answer is B (1.254 k~cal mol⁻¹) because the latent heat of fusion changes with temperature according to Kirchhoff's law.
129 Medium
129 of 329
Analyzing +4 | -1
Statement I: The final temperature of an ideal
gas in adiabatic compression is less in reversible
compression than in irreversible compression
against a constant external pressure.

Statement II: The magnitude of work done on
an ideal gas in adiabatic compression is more in
irreversible process than that in reversible process.
View answer and explanation
Correct option: A
If both statements are CORRECT, and Statement II is the CORRECT explanation of Statement I
Option A (If both statements are CORRECT, and Statement II is the CORRECT explanation of Statement I) is correct because irreversible adiabatic compression involves more work done, leading to a higher final temperature compared to reversible adiabatic compression.
130 Medium
130 of 329
Application +4 | -1
H2S(g)→HS−(g)+H+(g),ΔH∘=x1,\mathrm{H}_2 \mathrm{S}(\mathrm{g}) \rightarrow \mathrm{HS}^{-}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{g}), \Delta \mathrm{H}^{\circ}=\mathrm{x}_1,
ΔHf∘(H2S(g))=x2,ΔHf∘(H+(g))=x3\Delta \mathrm{H_f^\circ}\left(\mathrm{H}_2 \mathrm{S}\right. (\mathrm{g})) =\mathrm{x}_2, \Delta \mathrm{H_f^\circ}\left(\mathrm{H}^{+}(\mathrm{g})\right)=\mathrm{x}_3
Hence, ΔHf∘(HS−)\Delta \mathrm{H_f^\circ}\left(\mathrm{HS}^{-}\right) is:
View answer and explanation
Correct option: A
x1+x2−x3\mathrm{x_1+x_2-x_3}
Option A (x1+x2−x3\mathrm{x_1+x_2-x_3}) is correct because it correctly rearranges the enthalpy equation to solve for the heat of formation of HS-.
131 Medium
131 of 329
Application +4 | -1
The heat of combustion of carbon monoxide at constant pressure and at constant volume at 27∘C27^{\circ} \mathrm{C} will differ from one another by:
View answer and explanation
Correct option: C
300 cal300 \mathrm{~cal}
Option C (300 cal) is correct because it is the difference between heat of combustion at constant pressure and constant volume, calculated using the formula ΔH - ΔU = Δn_gRT.
132 Medium
132 of 329
Analyzing +4 | -1
A: Dissolution of sugar in water proceed via increases in entropy.
R: Entropy decreases when egg is hard boiled.
View answer and explanation
Correct option: C
If A is correct but R is incorrect
Option C is correct because the dissolution of sugar in water increases entropy, but boiling an egg also increases entropy, making the reason incorrect.
133 Medium
133 of 329
Application +4 | -1
Cdiamond+O2→CO2;ΔH{\mathrm{C}_{\mathrm{diamond}}} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2};\Delta H=-395.3 kJ/mol
Cgraphite+O2→CO2;ΔH{\mathrm{C}_{\mathrm{graphite}}} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2};\Delta H=-393.4 kJ/mol
Cgraphite→Cdiamond;ΔH{\mathrm{C}_{\mathrm{graphite}}} \to {\mathrm{C}_{\mathrm{diamond}}};\Delta \mathrm{H}=?
View answer and explanation
Correct option: D
+1.9+ 1.9
Option D (+1.9+ 1.9) is correct because it is calculated by subtracting the enthalpy of formation of CO2 from diamond from that of graphite.
134 Medium
134 of 329
Analyzing +4 | -1
Given that C+O2→CO2,ΔH∘=−x kJ\mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2, \Delta \mathrm{H}^{\circ}=-\mathrm{x} \mathrm{~kJ}

2CO+O2→2CO2,ΔH∘=−y kJ2 \mathrm{CO}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2, \Delta \mathrm{H}^{\circ}=-\mathrm{y} \mathrm{~kJ}

What is heat of formation of CO\mathrm{CO}?
View answer and explanation
Correct option: A
y−2x2\frac{\mathrm{y-2 x}}{2}
Option A is correct because the heat of formation of CO can be calculated by manipulating the given equations using Hess's Law.
135 Medium
135 of 329
Application +4 | -1
For an adiabatic expansion of a perfect gas ∆P/P is equal to:
View answer and explanation
Correct option: C
  −γΔVV\; - \gamma \frac{{\Delta V}}{V}
Option C (−γΔVV- \gamma \frac{{\Delta V}}{V}) is the correct expression for an adiabatic process relating pressure and volume.
136 Medium
136 of 329
Analyzing +4 | -1
In the following reaction,

HF(aq)+NaOH(aq)⟶NaF(aq)+H2O(l);ΔH∘=−68.6 kJ3mol−1\begin{aligned} & \mathrm{HF}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaF}(a q)+\mathrm{H}_2 \mathrm{O}(l) ; \Delta H^\circ=-68.6 \mathrm{~kJ}^3 \mathrm{mol}^{-1} \\ \end{aligned}

This value is much higher than the heat of neutralisation of strong acid with strong base (−57.3 kJ mol−1)\left(-57.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\right). This is because
View answer and explanation
Correct option: B
there is a hydration of F−F^-ion and the process is
exothermic
Option B is correct because the higher heat of neutralization for HF is due to the exothermic hydration of the F- ion, which releases additional energy.
137 Medium
137 of 329
Application +4 | -1
A quantity of 1 g1 \text{ g} of water on evaporation at
atmospheric pressure forms 1671 cm31671 \, \text{cm}^3
of steam.
Heat of vaporization is 540 cal/g.540 \text{ cal/g}. The approximate
increase in internal energy is (1 L – atom = 24 cal)
View answer and explanation
Correct option: B
500 cal
Option B (500 cal) is correct because the increase in internal energy is calculated using the first law of thermodynamics, considering heat absorbed and work done.
138 Medium
138 of 329
Application +4 | -1
Calculate the number of kJ\mathrm{kJ} of heat necessary to raise the temperature of 60 g60 \mathrm{~g} of aluminium from 35∘C35^{\circ} \mathrm{C} to 55∘C55^{\circ} \mathrm{C}. Molar heat capacity of Al{Al} is 24 J mol−1 K−124 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}.
View answer and explanation
Correct option: A
1.067 kJ
The correct answer is 1.067 kJ. The heat required is calculated using the formula q = m * C * ΔT, where m is the number of moles, C is the molar heat capacity, and ΔT is the change in temperature.
139 Medium
139 of 329
Understanding +4 | -1
The enthalpy of formation of HI is 30.4 kJ. Which statement is false according to this observation?
View answer and explanation
Correct option: C
HI is a stable compound
Option C (HI is a stable compound) is correct because a positive enthalpy of formation indicates that HI is an endothermic and therefore unstable compound.
140 Medium
140 of 329
Application +4 | -1
The latent heat of vaporization of water is 540 cal g−1\mathrm{g}^{-1} at 100∘C100^{\circ} \mathrm{C}. What will be the entropy increase when one mole of water is evaporated at 100∘C100^{\circ} \mathrm{C} ?
View answer and explanation
Correct option: B
26 cal K−1 mol−126 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
The correct answer is B (26 cal K-1 mol-1) because it is calculated using the formula ΔS = ΔH/T, where ΔH is the molar heat of vaporization and T is the temperature in Kelvin.
141 Medium
141 of 329
Application +4 | -1
Find the approximate value of (ΔH−ΔU)(\Delta \mathrm{H}-\Delta \mathrm{U}) in J\mathrm{J}. mol−1\mathrm{mol}^{-1}, for the formation of CO\mathrm{CO} from its elements at 298 K.(R=8.314.J K−1⋅mol−1)298 \mathrm{~K} .\left(\mathrm{R}=8.314 . \mathrm{J}\mathrm{~K}^{-1} \cdot \mathrm{mol}^{-1}\right)
View answer and explanation
Correct option: B
1238
Option B (1238) is correct because it is the approximate value calculated using the formula ΔH - ΔU = Δn_gRT, where Δn_g is the change in the number of moles of gas.
142 Medium
142 of 329
Remembering +4 | -1
ΔG\Delta \mathrm{G} is equal to
View answer and explanation
Correct option: B
ΔG∘+RTln⁡Q\Delta G^{\circ}+R T \ln Q
Option B (ΔG∘+RTln⁡Q\Delta G^{\circ}+R T \ln Q) is correct because it represents the relationship between the Gibbs free energy change under non-standard conditions (ΔG\Delta G), the standard Gibbs free energy change (ΔG∘\Delta G^{\circ}), the gas constant (R), temperature (T), and the reaction quotient (Q).
143 Medium
143 of 329
Remembering +4 | -1
Standard enthalpy (heat) of formation of liquid water at 25∘C25^{\circ} \mathrm{C} is around

H2( g)+12O2( g)→H2O(l)\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})
View answer and explanation
Correct option: C
−286 kJ/mol-286 \mathrm{~kJ} / \mathrm{mol}
Option C is correct because the standard enthalpy of formation of liquid water at 25°C is approximately -286 kJ/mol.
144 Medium
144 of 329
Application +4 | -1
For the reaction:

A(ℓ)→2B(g)ΔU=2.1 kcal,ΔS=20 cal/K, at T=300 K.\begin{aligned} & A(\ell) \rightarrow 2 B(g) \\ & \Delta {U}=2.1 \mathrm{~kcal}, \Delta {S}=20 \mathrm{~cal} / \mathrm{K}, \text { at } {T}=300 \mathrm{~K}. \end{aligned}

Hence ΔG\Delta {G} (in kcal)\mathrm{kcal}) is −X-X. Find the value of XX.
View answer and explanation
Correct option: A
-2.7
Option C (2.7) is the correct value of X because it is calculated using the relationship between Δ G, Δ H, and TΔ S.
145 Medium
145 of 329
Remembering +4 | -1
Work done in the reversible isothermal process is given by
View answer and explanation
Correct option: A
−2.303nRTlog⁡(V2/V1)-2.303 n R T \log \left({V}_2 / {V}_1\right)
Option A (−2.303nRTlog⁡(V2/V1)-2.303 n R T \log \left({V}_2 / {V}_1\right)) is correct because it represents the formula for work done in a reversible isothermal process.
146 Medium
146 of 329
Application +4 | -1
∆G°  for the reaction X+Y⇌Z is -4.606 kcal. The value of equilibrium constant of the reaction at 227℃ is (R=2.0  cal  mol−1 K−1)\left( {R = 2.0\,\,\mathrm{cal}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}\,{\mathrm{K}^{ - 1}}} \right)
View answer and explanation
Correct option: A
100
Option A (100) is correct because the equilibrium constant is calculated using the Gibbs free energy change and the given temperature and gas constant.
147 Medium
147 of 329
Analyzing +4 | -1
Consider the reactions given below. On the basis of these reactions find out which of the algebraic relationship given in options is correct?

1. C(g)+4H(g)⟶CH4(g);ΔrH=x kJ mol−1\mathrm{C(g})+4 \mathrm{H(g}) \longrightarrow \mathrm{CH_4(g}) ; \Delta_r H=x \mathrm{~kJ~mol}^{-1} ,

2. C\mathrm{C} (graphite, s)+2H2(g)⟶CH4(g);ΔrH=y kJ mol−1\mathrm{s})+2 \mathrm{H_2(g}) \longrightarrow \mathrm{CH_4(g}); \Delta_r H=y \mathrm{~kJ~mol}^{-1}
View answer and explanation
Correct option: C
x>yx > y
The correct answer is C (x>yx > y) because breaking bonds requires energy, so the energy released when forming bonds from gaseous atoms (x) is greater than when forming bonds from graphite and hydrogen gas (y).
148 Medium
148 of 329
Understanding +4 | -1
Assertion(A): Work and internal energy are not state functions.
Reason(R): The sum of q+wq+w is path function.
View answer and explanation
Correct option: B
(A) is wrong and (R) is wrong
Option B is correct because work and internal energy are state functions, and the sum of q+w (change in internal energy) is a state function, making both the assertion and the reason incorrect.
149 Medium
149 of 329
Application +4 | -1
Calculate the difference between CpC_p and CVC_V for 10 moles of an ideal gas.
View answer and explanation
Correct option: A
83.14 J/K83.14 \mathrm{~J} / \mathrm{K}
Option A (83.14 J/K) is correct because the difference between Cp and Cv for an ideal gas is given by nR, where n is the number of moles and R is the ideal gas constant.
150 Medium
150 of 329
Analyzing +4 | -1
For which reaction(s) is the ΔH\Delta \mathrm{H} value equal to the ΔE\Delta \mathrm{E} value?
(I) CH4(g)+2O2(g)→CO(g)+2H2O(l)\mathrm{CH}_4(\mathrm{g})+2 \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O} (l)
(II) C2H4(g)+3O2(g)→2CO(g)+2H2O(g)\mathrm{C}_2 \mathrm{H}_4(\mathrm{g})+3 \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})
View answer and explanation
Correct option: B
II only
Option B (II only) is correct because ΔH=ΔE+ΔngRT\Delta H = \Delta E + \Delta n_g RT, and for reaction II, Δng=0\Delta n_g = 0, making ΔH=ΔE\Delta H = \Delta E.
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