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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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241 Medium
241 of 500
Remembering +4 | -1
For the process to occur under adiabatic conditions, the correct condition is
View answer and explanation
Correct option: C
q=0q=0
Option C (q=0q=0) is correct because, by definition, an adiabatic process is one in which there is no heat transfer between the system and its surroundings.
242 Medium
242 of 500
Application +4 | -1
What is the entropy change (inJK−1mol−1)(in J{K^{ - 1}}mo{l^{ - 1}}) when one mole of ice is converted into water at 0℃? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol(-1) at 0℃)
View answer and explanation
Correct option: D
21.98
Option D (21.98) is correct because it's the calculated entropy change using the formula ΔS = ΔH/T.
243 Medium
243 of 500
Analyzing +4 | -1
Assertion :- Enthalpy of graphite is lower than that of diamond.

Reason :- Entropy of graphite is greater than that of diamond.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B is correct. Both the assertion and the reason are true, but the reason (entropy) does not explain why graphite has lower enthalpy than diamond.
244 Medium
244 of 500
Application +4 | -1
Heat evolved in the reaction H2+Cl2→2 HCl{H_2} + C{l_2} \to 2\,HCl is 182 KJ. Bond energies of H-H and Cl−ClCl - Cl are 430 and 242 KJ/mol respectively. The H−ClH - Cl bond energy is
View answer and explanation
Correct option: B
245 KJ/mol
Option B (245 KJ/mol) is correct because the H-Cl bond energy is calculated using the enthalpy change of the reaction and the bond energies of the reactants.
245 Medium
245 of 500
Understanding +4 | -1
For a spontaneous process, the correct statement is :
View answer and explanation
Correct option: D
Total entropy change is always positive.
Option D is correct because, for a spontaneous process, the total entropy change (system + surroundings) is always positive according to the second law of thermodynamics.
246 Hard
246 of 500
Application +4 | -1
Using the data provided, calculate the multiple bond energy (kJ mol−1)(kJ\,mo{l^{ - 1}})of a C≡C bond in C2H2.{\mathrm{C}_2}{\mathrm{H}_2}. Given that the heat of formation of C2H2{\mathrm{C}_2}{\mathrm{H}_2} is 225kJmol−1225kJmo{l^{ - 1}} (take the bond energy of C─H bond as 350  kJ  mol−1.):350\,\,kJ\,\,mo{l^{ - 1}}.):
2C(s)→2C(g)ΔH=1410  kJ  mol−12\mathrm{C}\left( s \right) \to 2\mathrm{C}\left( \mathrm{g} \right)\Delta H = 1410\,\,\mathrm{kJ\,\,mo}{\mathrm{l}^{ - 1}}
2C(s)→2C(g)ΔH=1410  kJ  mol−12\mathrm{C}\left( s \right) \to 2\mathrm{C}\left( \mathrm{g} \right)\Delta H = 1410\,\,\mathrm{kJ\,\,mo}{\mathrm{l}^{ - 1}}
H2(g)→2H(g)ΔH=330  kJ  mol−1{\mathrm{H}_2}\left( \mathrm{g} \right) \to 2\mathrm{H}\left( \mathrm{g} \right)\Delta H = 330\,\,\mathrm{kJ\,\,mo}{\mathrm{l}^{ - 1}}
View answer and explanation
Correct option: D
815
The correct answer is D (815) because it's the calculated C≡C bond energy using the given heat of formation and bond energies.
247 Easy
247 of 500
Analyzing +4 | -1
ASSERTION (A) : For the combustion of methane, ΔU>ΔH\Delta U>\Delta H

REASON (R): ΔH\Delta H is related by ΔU\Delta U by the expression ΔH=ΔU+ΔnRT\Delta H=\Delta U+\Delta n R T
View answer and explanation
Correct option: A
If both AA and RR are correct and RR is correct explanation or AA.
Option A (If both A and R are correct and R is correct explanation or A) is correct because for the combustion of methane, ΔH < ΔU, making ΔU > ΔH, and the relationship ΔH = ΔU + ΔnRT correctly explains this difference due to the change in the number of moles of gas.
248 Hard
248 of 500
Application +4 | -1
CCl2F2\mathrm{CCl}_2 \mathrm{F}_2 cooled 1.25 g1.25 \mathrm{~g} sample at constant atmospheric pressure of 1 atm1 \mathrm{~atm} from 320 K320 \mathrm{~K} to 293 K293 \mathrm{~K}. During cooling the sample volume decreased from 274274 to 248 mL248 \mathrm{~mL}. Calculate ΔH\Delta {H} for the CCl2F2\mathrm{CCl}_2 \mathrm{F}_2 for this process. (Cp=80.7 J mol−1 K−1)\left(C_p=80.7 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right).
View answer and explanation
Correct option: A
22.51 J22.51 \mathrm{~J}
Option A (22.51 J22.51 \mathrm{~J}) is correct because it is the heat evolved from the sample calculated using the given CpC_p and temperature change.
249 Medium
249 of 500
Understanding +4 | -1
For the reactions,

(i) H2( g)+Cl2( g)⟶2HCl(g)+xkJ{H}_2(\mathrm{~g})+{Cl}_2(\mathrm{~g}) \longrightarrow 2 {HCl}(\mathrm{g})+{xkJ}

(ii) H2( g)+Cl2( g)⟶2HCl(ℓ)+ykJ{H}_2(\mathrm{~g})+{Cl}_2(\mathrm{~g}) \longrightarrow 2 {HCl}(\ell)+{ykJ}

Which one of the following statement is correct :
View answer and explanation
Correct option: B
x<yx < y
Option B (x<yx < y) is correct because the heat evolved is greater when HCl is in the liquid state due to the additional energy released during condensation.
250 Medium
250 of 500
Application +4 | -1
When two moles of hydrogen expands isothermally against a constant pressure of 1 atm, at 25℃ from 15 L to 50 L, the work done (in litre atm) will be
View answer and explanation
Correct option: B
35
Option B (35) is correct because the work done during isothermal expansion against constant pressure is calculated as W = -PΔV.
251 Medium
251 of 500
Understanding +4 | -1
The change in entropy, ∆S is positive for an endothermic reaction. If enthalpy charge ∆H occurs at the same temperature T, then the reaction is feasible:
View answer and explanation
Correct option: C
When ∆H<T∆S
Option C (When ∆H<T∆S) is correct because for a reaction to be feasible (spontaneous), the Gibbs free energy change (∆G) must be negative, and ∆G = ∆H - T∆S. With positive ∆H and ∆S, ∆G is negative only when T∆S > ∆H.
252 Hard
252 of 500
Application +4 | -1
For A→B,ΔH=4 kcal mol−1,ΔS=10 cal mol−1 K−1.A \rightarrow B, \Delta H=4 \mathrm{~kcal} \mathrm{~mol}^{-1}, \Delta S=10 \mathrm{~cal} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}. Reaction is spontaneous when temperature will be
View answer and explanation
Correct option: C
500 K500 \mathrm{~K}
Option C (500 K) is correct because the reaction is spontaneous when the Gibbs free energy change is negative, which occurs at temperatures above 400 K.
253 Medium
253 of 500
Understanding +4 | -1
The enthalpy of formation of water from hydrogen and oxygen is - 286 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}}. The enthalpy of decomposition of water into hydrogen and oxygen is:
View answer and explanation
Correct option: C
+286kJmol−1+ 286\mathrm{kJmo}{\mathrm{l}^{ - 1}}
Option C (+286kJmol−1+ 286\mathrm{kJmo}{\mathrm{l}^{ - 1}}) is correct because the enthalpy of decomposition is the negative of the enthalpy of formation.
254 Medium
254 of 500
Analyzing +4 | -1
The standard molar enthalpies of formation of IF3(g)\mathrm{IF}_3(\mathrm{g}) and IF5(g)\mathrm{IF}_5(\mathrm{g}) are −470 kJ-470 \mathrm{~kJ} and −847 kJ,-847 \mathrm{~kJ}, respectively. Valence shell electron-pair repulsion theory predicts that IF5(g)\mathrm{IF}_5(\mathrm{g}) is square pyramidal in shape in which all I−F\mathrm{I}-\mathrm{F} bonds are equivalent while IF3(g)\mathrm{IF}_3(\mathrm{g}) is T-shaped (based on trigonal-bipyramidal geometry) in which I−F\mathrm{I}-\mathrm{F} bonds are of different lengths. It is observed that the axial I−F\mathrm{I}-\mathrm{F} bonds in IF3\mathrm{IF}_3 are equivalent to the I−F\mathrm{I}-\mathrm{F} bonds in IF5.\mathrm{IF}_5. Calculate the equatorial I−F\mathrm{I}-\mathrm{F} bond strength in IF3.\mathrm{IF}_3. Some other details given are as follows.

I2(s)→I2(g):ΔH=62 kJF2(g)→2 F(g):ΔH=155 kJI2(g)→2I(g):ΔH=149 kJ\begin{aligned} & \mathrm{I}_2(\mathrm{s}) \rightarrow \mathrm{I}_2(\mathrm{g}): \Delta H=62 \mathrm{~kJ} \\ & \mathrm{F}_2(\mathrm{g}) \rightarrow 2 \mathrm{~F}(\mathrm{g}): \Delta H=155 \mathrm{~kJ} \\ & \mathrm{I}_2(\mathrm{g}) \rightarrow 2 \mathrm{I}(\mathrm{g}): \Delta H=149 \mathrm{~kJ} \end{aligned}
View answer and explanation
Correct option: A
272 kJ/mol272 \mathrm{~kJ} / \mathrm{mol}
Option A (272 kJ/mol272 \mathrm{~kJ} / \mathrm{mol}) is correct because it is the calculated equatorial I-F bond strength in IF3, derived from enthalpy of formation data and bond energies.
255 Medium
255 of 500
Application +4 | -1
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are –890.3 KJ/mol, –393.5 KJ/mol and –285.8 KJ/mol respectively. Enthalpy of formation of CH4(g)C{H_4}\left( g \right) will be
View answer and explanation
Correct option: A
–74.8 KJ/mol
Option A (–74.8 KJ/mol) is correct because it is the enthalpy of formation calculated using Hess's law and the given enthalpies of combustion.
256 Hard
256 of 500
Application +4 | -1
Heat of formation of two moles of NH3( g)\mathrm{NH}_3(\mathrm{~g}) is −90KJ-90 \mathrm{KJ}. Bond energies of H−H\mathrm{H}-\mathrm{H} and N−H\mathrm{N}-\mathrm{H} bonds are 435KJ435 \mathrm{KJ} and 390KJmol−1390 \mathrm{KJ} \mathrm{mol}^{-1} respectively. The value of the bond energy of N=NN=N will be
View answer and explanation
Correct option: D
945KJ945 \mathrm{KJ}
Option D (945KJ945 \mathrm{KJ}) is correct because it is calculated using the formula relating heat of reaction to bond energies of reactants and products.
257 Medium
257 of 500
Analyzing +4 | -1
Assertion : The increase in internal energy (ΔE)(\Delta E) for the vaporization of one mole of water at 1 atm1 {~atm} and 373 K373 {~K} is zero.

Reason : For all isothermal processes, ΔE=0\Delta E=0.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
Option A (If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion) is correct because the internal energy change is zero for an isothermal process.
258 Medium
258 of 500
Analyzing +4 | -1
Statement I : In the following reaction, C(s)+O2(g)→CO2(g);ΔH=ΔE−RTC(s) + {O_2}(g) \to C{O_2}(g);\Delta H = \Delta E - RT
Statement II : ΔH\Delta H is related to ΔE\Delta E by equation ΔH=ΔE+ΔngRT\Delta H = \Delta E + \Delta {n_g}RT
View answer and explanation
Correct option: B
Only Statement II is correct
Option B is correct because Statement II correctly relates ΔH\Delta H to ΔE\Delta E using the equation ΔH=ΔE+ΔngRT\Delta H = \Delta E + \Delta n_g RT. Statement I is incorrect because Δng=0\Delta n_g = 0 for the given reaction, so ΔH=ΔE\Delta H = \Delta E.
259 Medium
259 of 500
Analyzing +4 | -1
Statement-I : At equilibrium ΔG0\Delta {G^0} always zero.
Statement-II : At STP active masses of 11.2 L of O2(g){O_2}\left( g \right) and 5.6 L of N2(g){N_2}\left( g \right) are equal.
View answer and explanation
Correct option: D
Statement I is wrong, Statement II is correct
Statement I is wrong, Statement II is correct because at equilibrium ΔG=0\Delta G = 0 and not ΔG0=0\Delta G^0 = 0. The active masses of the gases are equal at STP.
260 Easy
260 of 500
Remembering +4 | -1
When water is added to quick lime, the reaction is :
View answer and explanation
Correct option: C
Exothermic
Option C (Exothermic) is correct because the reaction of quick lime (CaO) with water is highly exothermic, releasing a significant amount of heat.
261 Medium
261 of 500
Analyzing +4 | -1
Assertion (A): Volume of water in a pond is not a state function.
Reason (R): Volume of water is independent of the route by which water is filled in the pond either by rain or by tubewell or by both.
View answer and explanation
Correct option: D
Assertion is incorrect; Reason is correct
The correct answer is D (Assertion is incorrect; Reason is correct) because the volume of water in a pond is a state function, and the reason correctly explains why.
262 Hard
262 of 500
Application +4 | -1
Determine whether or not, it is possible for sodium to reduce aluminium oxide to aluminium at 298 K298 {~K}. Also calculate equilibrium constant. Given that Gf∘G_f^{\circ} of Al2O3{Al}_2 {O}_3 at 298 K=−1582 kJ mol−1;Gf0298 {~K}=-1582 {~kJ} {~mol}^{-1} ; G_f^0 of Na2O(s){Na}_2 {O}(s) at 298 K298 {~K} =−377 kJ mol−1=-377 {~kJ} {~mol}^{-1}.
View answer and explanation
Correct option: A
9.1×10−809.1\times10^{-80}
Option A (9.1x10-80) is correct because the equilibrium constant is calculated from the Gibbs free energy change of the reaction, which is positive, indicating a non-spontaneous reaction.
263 Medium
263 of 500
Application +4 | -1
The heat of neutralisation of a strong acid and a strong alkali is 57.0 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}}. The heat released when 0.5 mole of HNO3\mathrm{HN}{\mathrm{O}_3} solution is mixed with 0.2 mole of KOH is
View answer and explanation
Correct option: B
11.4 kJ
Option A (t2g5eg2t_{2 g}{ }^5 e_g{ }^2) is correct because it corresponds to a high-spin Co(II) complex with three unpaired electrons, resulting in a magnetic moment close to 4.0 B.M.
264 Medium
264 of 500
Application +4 | -1
Heat of neutralisation of strong acid and strong base under 1 atm1 {~atm} and 25∘C25^{\circ} {C} is −13.7 kcal-13.7\, {kcal}. If standard Gibbs energy change for dissociation of water to H+{H}^{+}and OH−{OH}^{-} is −19.14 kcal-19.14\, {kcal}, the change in standard entropy for dissociation of water in cal K mol−1cal \,K \,{mol}^{-1} is :
View answer and explanation
Correct option: B
110.2
Option B (110.2) is correct because it's calculated using the Gibbs free energy equation, considering the heat of neutralization and temperature.
265 Medium
265 of 500
Application +4 | -1
Five moles of an ideal gas expand isothermally and reversibly from an initial pressure of 100 atm100 \mathrm{~atm} to a final pressure of 1 atm1 \mathrm{~atm} at 27∘C.27^{\circ} \mathrm{C}. The work done by the gas is (ln⁡100=4.6)(\ln 100=4.6)
View answer and explanation
Correct option: D
13,800 cal13,800 \mathrm{~cal}
Option D (13,800 cal) is correct because it is the work done calculated using the formula for isothermal reversible expansion of an ideal gas.
266 Easy
266 of 500
Analyzing +4 | -1
Given below are two statements one is labelled as Assertion (A) and the other is labelled as Reason (R)

Assertion (A): Heat of neutralisation of HFHF (aq.) with NaOH{NaOH} (aq.) is greater than 13.7 Kcal13.7 \mathrm{~Kcal}.

Reason (R): Generally some heat is consumed in the ionisation of weak acid.
On the basis of above statements choose the most appropriate option from the following options:
View answer and explanation
Correct option: A
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Option A is correct because both the assertion and the reason are correct, but the reason does not explain the assertion. The higher heat of neutralization for HF is due to hydration energy, not just ionization.
267 Easy
267 of 500
Understanding +4 | -1
A boiled egg show a/an …in entropy.
View answer and explanation
Correct option: A
Increase
Option A (Increase) is correct because the denaturation of proteins in a boiled egg leads to a more disordered state, thus increasing entropy.
268 Medium
268 of 500
Remembering +4 | -1
For an ideal monoatomic gas, molar heat capacity at constant volume (Cv)\left( {{C_v}} \right) is
View answer and explanation
Correct option: B
32R\frac{3}{2}R
Option B (32R\frac{3}{2}R) is correct because for an ideal monoatomic gas, the molar heat capacity at constant volume (CvC_v) is 32R\frac{3}{2}R.
269 Medium
269 of 500
Remembering +4 | -1
The van&rsquo;t Hoff reaction isotherm is:
View answer and explanation
Correct option: B
−∆G=RT log⁡eKp-∆ G = RT\,{\log _e}{K_p}
Option B (−∆G=RT log⁡eKp-∆ G = RT\,{\log _e}{K_p}) is correct because it represents the van't Hoff reaction isotherm relating Gibbs free energy change to the equilibrium constant.
270 Medium
270 of 500
Application +4 | -1
H2(g)+12O2(g)→H2O(l);ΔH=−68kcal{\mathrm{H}_2}\left( \mathrm{g} \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 68\mathrm{kcal}.The heat change, for the decomposition of 3.6 g of water is :
View answer and explanation
Correct option: B
13.6 kcal
Option B (13.6 kcal) is correct because the heat change is calculated based on the stoichiometry of the reaction and the given enthalpy change.
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