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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Medium

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Practice format and chapter quiz

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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271 Medium
271 of 329
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.
272 Medium
272 of 329
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.
273 Medium
273 of 329
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.
274 Medium
274 of 329
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.
275 Medium
275 of 329
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.
276 Medium
276 of 329
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.
277 Medium
277 of 329
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.
278 Medium
278 of 329
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.
279 Medium
279 of 329
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.
280 Medium
280 of 329
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}.
281 Medium
281 of 329
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.
282 Medium
282 of 329
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.
283 Medium
283 of 329
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.
284 Medium
284 of 329
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.
285 Medium
285 of 329
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.
286 Medium
286 of 329
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.
287 Medium
287 of 329
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.
288 Medium
288 of 329
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).
289 Medium
289 of 329
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.
290 Medium
290 of 329
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.
291 Medium
291 of 329
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.
292 Medium
292 of 329
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.
293 Medium
293 of 329
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.
294 Medium
294 of 329
Application +4 | -1
ΔfH∘\Delta_{\mathrm{f}} H^{\circ} for NF3(g)\mathrm{NF}_3(\mathrm{g}) is −113 kJ/mol.-113 \mathrm{~kJ} / \mathrm{mol}. Bond energy for the N−F\mathrm{N}-\mathrm{F} bond is 273.5 kJ/mol.273.5 \mathrm{~kJ} / \mathrm{mol}. The bond energies of N2\mathrm{N}_2 and F2,\mathrm{F}_2, if their magnitudes are in the ratio 6:1,6: 1, are respectively.
View answer and explanation
Correct option: C
943.33,157.22 kJ/mol943.33,157.22 \mathrm{~kJ} / \mathrm{mol}
Option C (943.33, 157.22 kJ/mol) is correct because it accurately calculates the bond energies of N2 and F2 using the given enthalpy of formation and bond energy of N-F.
295 Medium
295 of 329
Remembering +4 | -1
The term −TΔStotal-T \Delta S_{\text {total}} at constant temperature and pressure is known as
View answer and explanation
Correct option: A
Gibb's function
Option A (Gibb's function) is correct because the term −TΔStotal-T \Delta S_{\text {total}} at constant temperature and pressure defines the change in Gibb's free energy.
296 Medium
296 of 329
Analyzing +4 | -1
Assertion: The enthalpy of atomization increases with an increase in atomic number in a period.

Reason: The increase in atomic number results in an increase in the number of protons and electrons in the atom.
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation for Assertion
Option A is correct. The enthalpy of atomization generally increases across a period due to increasing nuclear charge and decreasing atomic size, which strengthens the metallic bond. The reason correctly explains this trend.
297 Medium
297 of 329
Understanding +4 | -1
Internal energy and pressure of a gas of unit volume are related as:
View answer and explanation
Correct option: A
P=23UP = \frac{2}{3}U
Option A (P=23UP = \frac{2}{3}U) is correct because the pressure of a gas is related to its internal energy by the equation P = (2/3)U for a monoatomic gas.
298 Medium
298 of 329
Understanding +4 | -1
Which compound will absorb the maximum amount of heat when dissolved in the same amount of water? (Integral heats of solution at 250c{25^0}c in kcal/mol of each solute are given in brackets):
View answer and explanation
Correct option: C
NH4NO3(ΔH=+16.08)\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}\left( {\Delta H = + 16.08} \right)
Option C (NH4NO3(ΔH=+16.08)\mathrm{N}{{\mathrm{H}}_4}{\mathrm{N}}{{\mathrm{O}}_3}(\Delta H = + 16.08)) is correct because the compound with the largest positive enthalpy of solution will absorb the maximum amount of heat when dissolved.
299 Medium
299 of 329
Application +4 | -1
If the bond energies of C=C\mathrm{C}=\mathrm{C} and C−C\mathrm{C}-\mathrm{C} at 298 K298 \mathrm{~K} are 590590 and 331 kJ mol−1331 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively then enthalpy of polymerization per mole of ethylene is:
View answer and explanation
Correct option: A
−72 kJ-72 \mathrm{~kJ}
Option A (-72 kJ) is correct because the enthalpy of polymerization is calculated by subtracting the energy of two C-C bonds from the energy of one C=C bond.
300 Medium
300 of 329
Analyzing +4 | -1
Enthalpies of combustion of CH4,C2H6\mathrm{CH}_4, \mathrm{C}_2 \mathrm{H}_6 and C3H8\mathrm{C}_3 \mathrm{H}_8 are −210,−370-210,-370 and −526 kcal/mol,-526 \mathrm{~kcal} / \mathrm{mol}, respectively. Enthalpy of combustion of hexane can be predicted as
View answer and explanation
Correct option: C
−1000 kcal mol−1-1000 \mathrm{~kcal} \mathrm{~mol}^{-1}
Option C (-1000 kcal mol⁻¹) is correct because the enthalpy of combustion increases by approximately 158 kcal/mol for each additional CH₂ group, allowing for prediction of hexane's enthalpy of combustion.
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