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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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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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331 Easy
331 of 500
Remembering +4 | -1
Which is the correct Gibb's Helmholtz
equation?
View answer and explanation
Correct option: C
ΔS=1 T[ΔH−ΔG]\Delta \mathrm{S}=\frac{1}{\mathrm{~T}}[\Delta \mathrm{H}-\Delta \mathrm{G}]
Option C, ΔS=1 T[ΔH−ΔG]\Delta \mathrm{S}=\frac{1}{\mathrm{~T}}[\Delta \mathrm{H}-\Delta \mathrm{G}], is the correct Gibb's Helmholtz equation, derived from the definition of Gibb's free energy.
332 Medium
332 of 500
Application +4 | -1
A gas is allowed to expand in an insulated container against a constant external pressure of 4 atm from the initial volume of 1.8 L to a final volume of 5.2 L. The change in internal energy of the gas is
View answer and explanation
Correct option: C
–1.38 kJ
Option C (–1.38 kJ) is correct because the change in internal energy is equal to the work done in an adiabatic process, calculated as -Pext * ΔV.
333 Hard
333 of 500
Application +4 | -1
Geological conditions are sometimes so extreme that quantities neglected in normal laboratory experiments take on an overriding importance. For example, consider the formation of diamond under geophysically typical conditions. The density of graphite is 2.4 g/cm32.4 \mathrm{~g} / \mathrm{cm}^3 and that of diamond is 3.6 g/cm33.6 \, \mathrm{g} / \mathrm{cm}^3 at a certain temperature and 500 kbar.500 \mathrm{~kbar}. By how much does ΔUtrans\Delta U_{\text {trans}} differs from ΔHtrans\Delta H_{\text {trans}} for the graphite to diamond transition?
View answer and explanation
Correct option: A
83.33 kJ/mol83.33 \mathrm{~kJ} / \mathrm{mol}
Option A (83.33 kJ/mol) is correct because it calculates the difference between \ΔH\Delta H and \ΔU\Delta U using the given pressure and volume change during the graphite to diamond transition.
334 Medium
334 of 500
Application +4 | -1
16 gm O216 \mathrm{~gm} \, {O}_2 is expanded reversibly at 300 K300 \mathrm{~K} temperature from 5 dm35 \mathrm{~dm}^3 to 25 dm325 \mathrm{~dm}^3 then find out W{W} and q{q} (in cal)
View answer and explanation
Correct option: A
−483.63 cal,483.63 cal-483.63 \text{ cal}, 483.63 \text{ cal}
Option A (-483.63 cal, 483.63 cal) is correct because the process is isothermal and reversible, allowing for the calculation of work and heat using the appropriate formulas.
335 Easy
335 of 500
Application +4 | -1
Consider the following thermochemical equation 2H2(g)+O2(g)→2H2O(g) ΔH∘=−470 kJ2 \mathrm{H}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \, \Delta \mathrm{H}^{\circ}=-470 \mathrm{~kJ} If standard enthalpy of vaporisation of H2O(ℓ)\mathrm{H}_2 \mathrm{O}(\ell) is 50 kJ/mole50 \mathrm{~kJ} / \mathrm{mole}, then find calorific value of H2(g)\mathrm{H}_2(\mathrm{g}) in standard conditions:
View answer and explanation
Correct option: D
142.5 kJ/g142.5 \mathrm{~kJ} / \mathrm{g}
The correct answer is D (142.5 kJ/g) because it involves calculating the heat of combustion for H2(g) to H2O(l) and then normalizing by the molar mass of H2.
336 Medium
336 of 500
Application +4 | -1
Two moles of an ideal gas undergo reversible isothermal expansion from 2.303 L2.303 \mathrm{~L} to 23.03 L23.03 \mathrm{~L} at 27∘C27^{\circ} \mathrm{C}. The change in entropy is: [R=8.314 J K−1\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1}\right. mol−1\mathrm{mol}^{-1}]
View answer and explanation
Correct option: D
38.3 J K−138.3 \mathrm{~J} \mathrm{~K}^{-1}
Option D (38.3 J K−138.3 \mathrm{~J} \mathrm{~K}^{-1}) is correct because it is calculated using the formula for entropy change in a reversible isothermal expansion: ΔS=nRln⁡V2V1\Delta S = nR \ln{\frac{V_2}{V_1}}.
337 Easy
337 of 500
Analyzing +4 | -1
STATEMENT-1 : There cannot be chemical equilibrium in an open system.

STATEMENT-2 : There is no fixed mass in an open system.
View answer and explanation
Correct option: A
If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1
Option A (If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1) is correct because chemical equilibrium requires a closed system with a fixed mass, which is not present in an open system.
338 Hard
338 of 500
Analyzing +4 | -1
Statement I Based on the following thermochemical
reactions;
HF(aq)+OH−(aq)⟶H2O(l)+F−(aq);ΔH∘=−16.4 k cal\mathrm{HF}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_2 \mathrm{O}(l)+\mathrm{F}^{-}(a q) ; \Delta H^{\circ}=-16.4 \mathrm{~k~cal}

H+(aq)+OH−(aq)⟶H2O(l),ΔH∘=−13.7 k cal\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_2 \mathrm{O}(l), \Delta H^{\circ}=-13.7 \mathrm{~k~cal}

Ionisation of HF(aq)\mathrm{HF}(\mathrm{aq}) is an exothermic reaction.

Statement II Enthalpy decrease in hydration of H+(aq)\mathrm{H}^{+}(a q) and F−(aq)\mathrm{F}^{-}(a q) exceeds the ionisation enthalpy of HF(aq)\mathrm{HF}(a q).
View answer and explanation
Correct option: A
Both Statement I and Statement II are correct and
Statement II is the correct explanation of Statement I
Option A (Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because the ionization of HF is exothermic due to the high hydration enthalpy of H+ and F- exceeding the ionization enthalpy.
339 Medium
339 of 500
Understanding +4 | -1
The enthalpy change is negative for :
View answer and explanation
Correct option: A
Cl−(g)+aq→Cl−(aq)\mathrm{C}{\mathrm{l}^ - }\left( \mathrm{g} \right) + \mathrm{aq} \to \mathrm{C}{\mathrm{l}^ - }\left( {aq} \right)
Option A is correct because the hydration of ions is an exothermic process, releasing energy and resulting in a negative enthalpy change.
340 Medium
340 of 500
Analyzing +4 | -1
Assertion :- Catalyst changes Gibbs free energy of system.

Reason :- Catalyst changes preexpotential factor of a chemical reaction.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
Option D (If both Assertion & Reason are False) is correct because a catalyst does not change the Gibbs free energy of the system, and it changes the activation energy, not the pre-exponential factor.
341 Medium
341 of 500
Remembering +4 | -1
In the exothermic reaction, the enthalpy of
reaction is always:
View answer and explanation
Correct option: C
negative
Option C (negative) is correct because exothermic reactions release heat, meaning the enthalpy change (ΔH) is negative.
342 Easy
342 of 500
Analyzing +4 | -1
STATEMENT-1: The expansion of a gas into an evacuated space takes place non-spontaneously.

STATEMENT-2 : A process in which all steps cannot be retraced by themselves is called a spontaneous process.
View answer and explanation
Correct option: D
If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE
Option D (If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE) is correct because gas expansion into a vacuum is spontaneous, and a spontaneous process is one that cannot be retraced by itself.
343 Medium
343 of 500
Application +4 | -1
10 moles of an ideal gas compressed from 100  L100\, \mathrm{~L} to 10  L10 \,\mathrm{~L} reversibly at constant temperature of 27∘C27^{\circ} \mathrm{C}. The entropy change during the process is ( R=R= Universal gas constant)
View answer and explanation
Correct option: D
−23.03 R-23.03\, {R}
Option D (−23.03 R-23.03\, {R}) is correct because the entropy change for an isothermal reversible compression is calculated using ΔS = nRln(V2/V1).
344 Medium
344 of 500
Understanding +4 | -1
Identify the correct statement for change of Gibb’s energy for a system (ΔGsystem\Delta {G_{\mathrm{system}}}) at constant temperature and pressure:
View answer and explanation
Correct option: D
If ΔGsystem\Delta {G_{\mathrm{system}}}=0, the system has attained equilibrium
Option D is correct because at equilibrium, the change in Gibb's free energy is zero, indicating no further net change in the system.
345 Medium
345 of 500
Analyzing +4 | -1
When 1 g1 \mathrm{~g} equivalent of strong acid reacts with strong base, the heat released is 13.5 kcal.13.5 \mathrm{~kcal}. When 1 g1 \mathrm{~g} equivalent H2A\mathrm{H}_2 \mathrm{A} is completely neutralized against strong base, 13 kcal13 \mathrm{~kcal} is released. When 1 g1 \mathrm{~g} equivalent B(OH)2\mathrm{B}(\mathrm{OH})_2 is completely neutralized against strong acid, 10 kcal10 \mathrm{~kcal} heat is released. What is the enthalpy change when 1 mole of H2A\mathrm{H}_2 \mathrm{A} is completely neutralized by B(OH)2?\mathrm{B}(\mathrm{OH})_2?
View answer and explanation
Correct option: D
−19 kcal-19 \mathrm{~kcal}
The correct answer is D (-19 kcal) because the enthalpy change is calculated by considering the enthalpy of ionization for both the acid and the base and subtracting them from the total heat released during neutralization.
346 Medium
346 of 500
Analyzing +4 | -1
If H2+1/2O2⟶H2O,ΔH=−68 kcalK+H2O+water⟶KOH(aq)+1/2H2,ΔH=−48 kcalKOH+water⟶KOH(aq),ΔH=−14 kcal\begin{aligned} & \text {If } \mathrm{H}_2+1 / 2 \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}, \Delta \mathrm{H}=-68 \mathrm{~kcal} \\ & \mathrm{K}+\mathrm{H}_2 \mathrm{O}+\text {water} \longrightarrow \mathrm{KOH}(\mathrm{aq})+1 / 2 \mathrm{H}_2, \Delta \mathrm{H}=-48 \mathrm{~kcal} \\ & \mathrm{KOH}+\text {water} \longrightarrow \mathrm{KOH}(\mathrm{aq}), \Delta \mathrm{H}=-14 \mathrm{~kcal} \end{aligned}

Find the heat of formation of KOH\mathrm{KOH}.
View answer and explanation
Correct option: A
-102 k cal
Option A (-102 k cal) is correct because it is calculated by manipulating the given equations to arrive at the formation of KOH.
347 Medium
347 of 500
Application +4 | -1
The bond dissociation energies for Cl2,I2Cl_2, I_2 and
IClICl are 242.3, 151 and 211.3 kJ/mol
respectively. The enthalpy of sublimation of
iodine is 62.8 kJ/mol. What is the standard
enthalpy of formation of ICl(g)?ICl \text{(g)}?
View answer and explanation
Correct option: C
-16.8 kJ/mol
Option C (-16.8 kJ/mol) is correct because it is the standard enthalpy of formation calculated using the given bond dissociation energies and enthalpy of sublimation.
348 Medium
348 of 500
Understanding +4 | -1
Which of the following statements is incorrect?
View answer and explanation
Correct option: C
The standard state of solid iron at 298 K298 \mathrm{~K} is equal to pure iron at 1 bar
Option D (Standard conditions are denoted by adding the superscript 'o' to the symbol ΔH\Delta H) is incorrect because standard conditions are denoted by adding the superscript 'θ\theta' (nought) to the symbol ΔH\Delta H, not 'o'.
349 Easy
349 of 500
Understanding +4 | -1
Taking each reactant in 1:1 molar ratio, enthalpy changes are ΔH1∘\Delta H_1^\circ and ΔH2∘\Delta H_2^\circ in the following reactions.

Ca(OH)2+HCl⟶Ca(OH)Cl+H2O;ΔH1∘Ca(OH)Cl+HCl⟶CaCl2+H2O;ΔH2∘\begin{aligned} & \mathrm{Ca}(\mathrm{OH})_2+\mathrm{HCl} \longrightarrow \mathrm{Ca}(\mathrm{OH}) \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} ; \Delta \mathrm{H}_1{ }^{\circ} \\ & \mathrm{Ca}(\mathrm{OH}) \mathrm{Cl}+\mathrm{HCl} \longrightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O} ; \Delta \mathrm{H}_2{ }^{\circ} \end{aligned}

ΔH1∘\Delta H_{1^{\circ}} and ΔH2∘\Delta H_{2^{\circ}} are (in kcal)
View answer and explanation
Correct option: B
ΔH1∘ΔH2∘−13.70−13.70\begin{array}{cccccccccccccccccccc} {\Delta H_1^ \circ }&{\Delta H_2^ \circ } \\ { - 13.70}&{ - 13.70} \end{array}
Option D is correct because the enthalpy change for neutralizing a strong acid with a strong base is constant, and in the second reaction, the enthalpy change is twice that of the first reaction since two moles of HCl are neutralized.
350 Medium
350 of 500
Understanding +4 | -1
Which one among the following pairs does not
represent example for intensive property?
View answer and explanation
Correct option: D
Heat capacity and enthalpy
Option D (Heat capacity and enthalpy) is correct because heat capacity and enthalpy are extensive properties, while the others are intensive.
351 Easy
351 of 500
Understanding +4 | -1
Which among the following is an extensive property?
View answer and explanation
Correct option: D
Entropy
Option D is correct because entropy is an extensive property, meaning its value depends on the amount of substance, while boiling point, viscosity, and surface tension are intensive properties.
352 Medium
352 of 500
Remembering +4 | -1
The change of enthalpy of a chemical reaction is the same whether the reaction takes place in one step or in several steps". This law was presented by
View answer and explanation
Correct option: A
Hess
Option A (Hess) is correct because Hess's Law states that the enthalpy change of a reaction is independent of the pathway.
353 Easy
353 of 500
Understanding +4 | -1
In which one of the following sets, all the properties belong to same category (all extensive or all intensive)?
View answer and explanation
Correct option: D
Enthalpy, internal energy, volume
Option D (Enthalpy, internal energy, volume) is correct because all three properties are extensive, meaning they depend on the amount of substance.
354 Medium
354 of 500
Remembering +4 | -1
Which one among the following is the correct option for right relationship between CpandCv{C_p}\mathrm{ and }{C_v} for one mole of ideal gas?
View answer and explanation
Correct option: D
Cp−Cv=R{C_p} - {C_v} = R
Option D (Cp−Cv=R{C_p} - {C_v} = R) is correct because the relationship between Cp and Cv for one mole of an ideal gas is given by Mayer's formula: Cp - Cv = R.
355 Hard
355 of 500
Application +4 | -1
The heats of combustion of carbon and carbon moxoide are – 393.5 and −283.5 kJ mol−1- 283.5\,kJ\,mo{l^{ - 1}}, respectively. The heat of formation (in kJ) of carbon monoxide per mole is
View answer and explanation
Correct option: C
-110.5
Option C (-110.5) is correct because it is calculated by subtracting the heat of combustion of CO from the heat of combustion of C.
356 Medium
356 of 500
Application +4 | -1
Given that ΔHf(H)=218 kJ/mol\Delta H_f(H)=218 \mathrm{~kJ} / \mathrm{mol}, express the H−H\mathrm{H}-\mathrm{H} bond energy in k cal/mol\mathrm{k ~cal} / \mathrm{mol}.
View answer and explanation
Correct option: C
104
Option C (104) is correct because the H-H bond energy is equivalent to the enthalpy of formation of 2 moles of H atoms from H2, and 436 kJ/mol is approximately 104 kcal/mol.
357 Hard
357 of 500
Application +4 | -1
The reaction, MgO(s)+C(s)→Mg(s)+CO(g)MgO(s) + C(s) \to Mg(s) + CO(g), for which ΔrH∘=+491.1kJmol−1{\Delta _r}{H^ \circ } = + 491.1kJ\mathrm{ }mo{l^{ - 1}} and ΔrS∘=198.0JK−1mol−1{\Delta _r}{S^ \circ } = 198.0J{K^{ - 1}}mo{l^{ - 1}} , is not feasible at 298 K. Temperature above which reaction will be feasible is
View answer and explanation
Correct option: D
2480.3 K
Option D (2480.3 K) is correct because the reaction becomes feasible when ΔG becomes negative, which occurs at temperatures above 2480.3 K.
358 Hard
358 of 500
Application +4 | -1
A system undergoes a two-step process. In step 1, 15 J15 \mathrm{~J} of work is done on the system as its internal energy increases by a total of 30 J.30 \mathrm{~J}. In step 2, at constant volume, the system decreases its internal energy by 20 J.20 \mathrm{~J}. Which statement is true about these changes?
View answer and explanation
Correct option: A
The system loses 5 J5 \mathrm{~J} as a result of the net heat flow in steps 1 and 2
Option A (The system loses 5 J5 \mathrm{~J} as a result of the net heat flow in steps 1 and 2) is correct because the net heat flow is -5J, indicating a loss of 5J.
359 Easy
359 of 500
Application +4 | -1
A system absorbs 10 kJ of heat and does 4 kJ of work. The internal energy of the system
View answer and explanation
Correct option: B
Increases by 6 kJ
Option B (Increases by 6 kJ) is correct because, according to the first law of thermodynamics, the change in internal energy is the sum of heat absorbed and work done by the system.
360 Medium
360 of 500
Analyzing +4 | -1
For which among the following reactions,
change in entropy is less than zero
View answer and explanation
Correct option: B
Dissociation of hydrogen
Option C (Formation of water) is correct because the formation of water from its gaseous elements results in a decrease in entropy.
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