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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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421 Medium
421 of 500
Remembering +4 | -1
Entropy change of vaporisation at constant pressure is given by:
View answer and explanation
Correct option: A
ΔS(v)=ΔHvT\Delta {S_{\left( v \right)}} = \frac{{\Delta {H_v}}}{T}
Option A (ΔS(v)=ΔHvT\Delta {S_{\left( v \right)}} = \frac{{\Delta {H_v}}}{T}) is correct because it represents the formula for entropy change of vaporization at constant pressure, where entropy change equals enthalpy change divided by temperature.
422 Medium
422 of 500
Application +4 | -1
The enthalpy of neutralization of a strong monobasic acid by a strong monoacidic base is −13,700 cal.-13,700 \mathrm{~cal}. A certain monobasic weak acid is 10%10 \% ionized in a molar solution. If the enthalpy of ionization of the weak acid is +400 cal/mole,+400 \mathrm{~cal} / \mathrm{mole}, then what is the enthalpy of neutralization of one molar solution of the weak acid?
View answer and explanation
Correct option: B
−13,340 cal-13,340 \mathrm{~cal}
The correct answer is B (-13,340 cal) because it accounts for the incomplete ionization of the weak acid and subtracts the energy required for ionization from the standard neutralization enthalpy.
423 Hard
423 of 500
Understanding +4 | -1
STATEMENT-1 : The standard free energy changes of all spontaneously occuring reactions are negative.

STATEMENT-2 : The standard free energies of the elements in their standard states at 1 bar and 298 K are taken as zero.
View answer and explanation
Correct option: B
If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1
Option D (If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE) is correct because spontaneous reactions have negative ΔG, but standard free energies of elements in their standard states are defined as zero.
424 Medium
424 of 500
Remembering +4 | -1
For a system in equilibrium
View answer and explanation
Correct option: A
ΔG=0\Delta \mathrm{G}=0
Option A (ΔG=0\Delta \mathrm{G}=0) is correct because at equilibrium, the change in Gibbs free energy (ΔG\Delta G) is zero, indicating no net change in the reaction.
425 Medium
425 of 500
Application +4 | -1
When 50 mL50 \mathrm{~mL} of 1 M HCl1 \mathrm{~M} \mathrm{~HCl} (aq) with 50 mL50 \mathrm{~mL} of 1 M NaOH(aq)1 \mathrm{~M} \mathrm{~NaOH}(\mathrm{aq}), the enthalpy of neutralisation will be
View answer and explanation
Correct option: D
2.9 kJ2.9 \mathrm{~kJ}
Option D (2.9 kJ2.9 \mathrm{~kJ}) is correct because the enthalpy of neutralization is calculated based on the number of moles of HCl and NaOH that react.
426 Medium
426 of 500
Analyzing +4 | -1
The ΔfH∘\Delta_{\mathrm{f}} \mathrm{H}^{\circ} for N2O3( g)\mathrm{N}_2 \mathrm{O}_3(\mathrm{~g}) in kJ/mol\mathrm{kJ} / \mathrm{mol} on the basis of the following data is :

2NO(g)+O2( g)⟶2NO2( g)ΔrH∘=−114 kJ/mol4NO2( g)+O2( g)⟶2 N2O5( g)ΔrH∘=−102.6 kJ/molΔfH∘NO(g)=90.2 kJ/mol\begin{array}{ll} 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_2(\mathrm{~g}) & \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-114 \mathrm{~kJ} / \mathrm{mol} \\ 4 \mathrm{NO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) & \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-102.6 \mathrm{~kJ} / \mathrm{mol} \\ \Delta_{\mathrm{f}} \mathrm{H}^{\circ} \mathrm{NO}(\mathrm{g})=90.2 \mathrm{~kJ} / \mathrm{mol} & \end{array}
View answer and explanation
Correct option: A
15.1
Option A (15.1) is correct because by manipulating the given equations and Hess's Law, the enthalpy of formation of N2O5 can be calculated to be 15.1 kJ/mol.
427 Medium
427 of 500
Understanding +4 | -1
Tea placed in thermos flask is an example of
View answer and explanation
Correct option: C
Isolated system
Option C (Isolated system) is correct because a thermos flask minimizes the exchange of both matter and energy with the surroundings.
428 Easy
428 of 500
Understanding +4 | -1
During an isothermal expansion of an ideal gas its:
View answer and explanation
Correct option: D
Internal energy remains constant
Option D (Internal energy remains constant) is correct because for an ideal gas, internal energy depends only on temperature, and in an isothermal process, temperature is constant.
429 Medium
429 of 500
Application +4 | -1
3.0 moles of an ideal gas is heated at constant pressure from 27∘C27^{\circ} \mathrm{C} to 127∘C127^{\circ} \mathrm{C}. The work done by the gas in this expansion is:
View answer and explanation
Correct option: A
−2.494 kJ-2.494 \mathrm{~kJ}
Option A (-2.494 kJ) is correct because the work done by the gas at constant pressure is calculated using w = -nRΔT, resulting in a negative value as the gas expands.
430 Medium
430 of 500
Analyzing +4 | -1
Under identical conditions, how many millilitres of 1 M−KOH1 \mathrm{~M}-\mathrm{KOH} and 2 M−H2SO42 \mathrm{~M}-\mathrm{H}_2 \mathrm{SO}_4 solutions are required to produce a resulting volume of 100 ml100 \mathrm{~ml} with the highest rise in temperature?
View answer and explanation
Correct option: A
80,2080, 20
Option A (80, 20) is correct because it allows for complete neutralization of the acid and base, leading to the maximum heat release, given the molarities and total volume constraint.
431 Hard
431 of 500
Understanding +4 | -1
For an isothermal free expansion of an ideal gas correct option is:
View answer and explanation
Correct option: D
q=0,w=0,ΔH=0q=0, w=0, \Delta H=0
Option D (q=0,w=0,ΔH=0q=0, w=0, \Delta H=0) is correct because in an isothermal free expansion of an ideal gas, there is no temperature change, no work done, and no heat exchange.
432 Medium
432 of 500
Application +4 | -1
One gram sample of NH4NO3{NH}_4 {NO}_3 is decomposed in a bomb calorimeter, the temperature of the calorimeter increases by 6.12 K6.12 {~K}. The heat capacity of the system is 1.23 kJ/gdeg1.23 {~kJ} / {g} {deg}. What is the molar heat of decomposition for NH4NO3{NH}_4 {NO}_3 ?
View answer and explanation
Correct option: D
−602 kJ/mol-602 {~kJ} / {mol}
Option D (−602 kJ/mol-602 {~kJ} / {mol}) is correct because it correctly calculates the molar heat of decomposition based on the heat capacity and temperature change.
433 Medium
433 of 500
Application +4 | -1
If 900 J/g of heat is exchanged at boiling point of water then increase in entropy
View answer and explanation
Correct option: A
43.4 J/mol
Option A (43.4 J/mol) is correct because the entropy change is calculated by dividing the heat exchanged by the temperature in Kelvin, ΔS=qT\Delta S = \frac{q}{T}.
434 Medium
434 of 500
Remembering +4 | -1
The equation that represents general Van't
Hoff equation is
View answer and explanation
Correct option: A
π=nVRT\pi=\frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT}
Option A (π=nVRT\pi=\frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT}) is correct because it represents the Van't Hoff equation, which relates osmotic pressure to the molar concentration of the solution.
435 Medium
435 of 500
Application +4 | -1
A diatomic ideal gas initially at 273 K273 \mathrm{~K} is given 100 cal100 \text{ cal} heat due to which system did 210 J210 \mathrm{~J} work. The molar heat capacity of the gas for the process is (1 cal=4.2 J)(1 \mathrm{~cal}=4.2 \mathrm{~J})
View answer and explanation
Correct option: D
5R5 R
Option D (5R) is correct because the molar heat capacity is calculated using the first law of thermodynamics and the given heat and work values.
436 Medium
436 of 500
Application +4 | -1
Heat of combustion of methane is −800 kJ-800 \mathrm{~kJ}. What is the heat of combustion for 4×10−4 kg4 \times 10^{-4} \mathrm{~kg} of methane?
View answer and explanation
Correct option: C
−20 kJ-20 \mathrm{~kJ}
Option C (-20 kJ) is correct because it represents the heat of combustion for the given mass of methane, calculated using the heat of combustion per mole.
437 Medium
437 of 500
Application +4 | -1
Heat of combustion ∆H for C(s), H2{\mathrm{H}_2}(g) and CH4\mathrm{C}{\mathrm{H}_4} (g) are -94,-68 and -213 kcal/mole then ∆H for C(s)+2H2(g)→CH4(g)is:\mathrm{C}\left( s \right) + 2{\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right)\mathrm{i}s:
View answer and explanation
Correct option: A
-17 kcal
Option D (0.9615 V) is the closest to the calculated emf of the cell using the Nernst equation, although the calculated answer is not exactly present in the options.
438 Easy
438 of 500
Analyzing +4 | -1
For electron gain enthalpies of the elements denoted as ΔegH\Delta_{\mathrm{eg}} \mathrm{H}, the incorrect option is
View answer and explanation
Correct option: B
ΔegH(Se)<ΔegH(S)\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Se}) < \Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{S})
Option D (ΔegH(Te)<ΔegH(Po)\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Te}) < \Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Po})) is correct because electron gain enthalpy generally becomes less negative down the group due to increasing atomic size.
439 Medium
439 of 500
Understanding +4 | -1
The molar heat capacity of water in equilibrium with ice at constant pressure is
View answer and explanation
Correct option: B
Infinity
Option B (Infinity) is correct because at equilibrium, any heat added will cause a phase change (ice to water) without a change in temperature, leading to an infinite molar heat capacity.
440 Hard
440 of 500
Analyzing +4 | -1
Statement I In the following thermochemical reactions,

C(s)+O2(g)⟶CO2(g);ΔH=−94 k cal mol−1C(s)+12O2(g)⟶CO(g);ΔH=−26 k cal mol−1CO(g)+12O2(g)⟶CO2(g);\begin{aligned} & C(s)+\mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \Delta H=-94 \mathrm{~k~cal} \mathrm{~mol}^{-1} \\ & \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}(g) ; \Delta H=-26 \mathrm{~k~cal} \mathrm{~mol}^{-1} \\ & \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \end{aligned}

Enthalpy of combustion of CO(g)\mathrm{CO}(g) is −68 k cal mol−1-68 \mathrm{~k~cal} \mathrm{~mol}^{-1}.

Statement II CO2(g)\mathrm{CO}_2(g) is an exothermic compound.
View answer and explanation
Correct option: B
Both Statement I and Statement II are correct and
Statement II is not the correct explanation of Statement I.
Option B is correct because Statement I is correct based on Hess's law calculation, and Statement II is also correct as CO2 formation is exothermic, but the latter doesn't explain the former.
441 Medium
441 of 500
Understanding +4 | -1
The sole criterion for the spontaneity of a process is
View answer and explanation
Correct option: D
Tendency to acquire maximum stability
The correct answer is D. Spontaneity is determined by the tendency of a system to achieve maximum stability, which is related to minimizing Gibbs free energy.
442 Medium
442 of 500
Analyzing +4 | -1
Statement I : Heat of neutralisation of HF(aq)HF(aq), a weak acid, with NaOH(aq)NaOH(aq) is less than 13.7 kcal, in an exothermic reaction
Statement II : Some heat is lost in the ionisation of a weak acid
View answer and explanation
Correct option: D
Both statements are incorrect
Option D is correct because both statements are incorrect. The heat of neutralization of a weak acid with a strong base is less than 13.7 kcal due to the energy required for ionization, and heat is *used* in the ionization of a weak acid, not lost.
443 Medium
443 of 500
Analyzing +4 | -1
1 mole H2O(l)1 \mathrm{~mole} \mathrm{~H_2} \mathrm{O}(l) at −10∘C-10^{\circ} \mathrm{C} is converted to 1 mole H2O(s)\mathrm{H}_2 \mathrm{O}(s) at −10∘C-10^{\circ} \mathrm{C}. Entropy change is the sum of all steps but not of
View answer and explanation
Correct option: B
ΔS2=ΔHfusion 273(ΔHf=\Delta S_2=\frac{\Delta H_{\text {fusion }}}{273}\left(\Delta H_f=\right. heat of fusion of ice)
The correct answer is ΔS2=ΔHfusion 273(ΔHf= heat of fusion of ice)\Delta S_2=\frac{\Delta H_{\text {fusion }}}{273}(\Delta H_f=\text { heat of fusion of ice}) because the process is occurring at -10°C (263K), not 0°C (273K).
444 Medium
444 of 500
Analyzing +4 | -1
Assertion : Many endothermic reactions that are not spontaneous at room temperature become spontaneous at high temperature.
Reason : Entropy of the system increases with increase in temperature.
View answer and explanation
Correct option: B
If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
Option B (If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion) is correct because both the assertion and the reason are true, but the increase in entropy is not the direct cause of endothermic reactions becoming spontaneous at high temperatures.
445 Hard
445 of 500
Application +4 | -1
One mole of an ideal gas (γ=1.4)(\gamma=1.4) is expanded isothermally at 27∘C27^{\circ} \mathrm{C} till its volume is doubled. It is then adiabatically compressed to its original volume. The magnitude of total work done by the gas is (ln⁡2=0.7,20.4=43,R=8.3 J/K−mol)\left(\ln 2=0.7,2^{0.4}=\frac{4}{3}, \mathrm{R}=8.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol}\right)
View answer and explanation
Correct option: A
332 J332 \mathrm{~J}
Option B (40%)\mathrm{40\% }) is correct because it is the percentage of A that reacts to form AB at equilibrium, given the initial amounts and the amount of AB formed.
446 Medium
446 of 500
Analyzing +4 | -1
Find the bond energy of S−S\mathrm{S}-\mathrm{S} bond from the following data.

C2H5−S−C2H5(g);ΔHfo=−148 kJC2H5−S−S−C2H5(g);ΔHf∘=−202 kJ S(g);ΔHf∘=222 kJ\begin{aligned} & \mathrm{C}_2 \mathrm{H}_5-\mathrm{S}-\mathrm{C}_2 \mathrm{H}_5(\mathrm{g}); \Delta H_{\mathrm{f}}^{\mathrm{o}}=-148 \mathrm{~kJ} \\ & \mathrm{C}_2 \mathrm{H}_5-\mathrm{S}-\mathrm{S}-\mathrm{C}_2 \mathrm{H}_5(\mathrm{g}); \Delta H_{\mathrm{f}}^{\circ}=-202 \mathrm{~kJ} \\ & \mathrm{~S}(\mathrm{g});\Delta H_{\mathrm{f}}^{\circ}=222 \mathrm{~kJ} \end{aligned}
View answer and explanation
Correct option: A
276 kJ/mol276 \mathrm{~kJ} / \mathrm{mol}
Option A (276 kJ/mol) is correct because the S-S bond energy is calculated by subtracting the enthalpy of formation of the reactants from the enthalpy of formation of the product.
447 Medium
447 of 500
Understanding +4 | -1
Why is the heat evolved in neutralisation of HF
is highest ?
View answer and explanation
Correct option: B
Due to high hydration energy of F−\mathrm{F}^{-}ion
Option B (Due to high hydration energy of F−\mathrm{F}^{-}ion) is correct because the high hydration energy of the fluoride ion (F−\mathrm{F}^{-}) contributes significantly to the overall heat evolved during the neutralization of HF.
448 Medium
448 of 500
Application +4 | -1
In the reversible adiabatic expansion of an ideal monoatomic gas, the final volume is 8 times the initial volume. The ratio of final temperature to initial temperature is
View answer and explanation
Correct option: B
1:41: 4
Option B (1:41:4) is correct because the final to initial temperature ratio is calculated using the adiabatic expansion formula with \γ−1\gamma - 1 for a monoatomic gas.
449 Medium
449 of 500
Application +4 | -1
When 5 mol5 \mathrm{~mol} of an ideal gas expands from 10 L10 \mathrm{~L} to 20 L20 \mathrm{~L}, temperature of the gas changes from 300 K300 \mathrm{~K} to 400 K400 \mathrm{~K}. If the heat capacity Cp{C}_{{p}} of the gas is 4.0 cal mol−1 K−14.0 \mathrm{~cal} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, then entropy change (in cal K−1)\left(\text{in cal } \mathrm{K}^{-1}\right) of the gas is (log⁡2=0.30(\log 2=0.30, log⁡3=0.47\log 3=0.47 and R≃2 cal mol−1 K−1R \simeq 2 \mathrm{~cal} \mathrm{~mol}^{-1} \mathrm{~K}^{-1})
View answer and explanation
Correct option: A
9.81
The correct answer is A (9.81) because the entropy change is calculated using the formula involving heat capacity and volume change.
450 Hard
450 of 500
Application +4 | -1
An ideal gas is expanded irreversibly from 5 L5 \mathrm{~L} to 10 L10 \mathrm{~L} against a constant external pressure of 1 bar.1 \mathrm{~bar}. The value of heat involved (q)(q) in this isenthalpic process is
View answer and explanation
Correct option: D
−500 J-500 \mathrm{~J}
Option D (-500 J) is correct because in an isenthalpic process, the change in enthalpy is zero, and the heat involved is equal to the negative of the work done against the constant external pressure.
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