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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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391 Hard
391 of 500
Application +4 | -1
Enthalpy of vaporisation for water is 186.5 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}} . The entropy change during vaorisation is …kJ K−1mol−1{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}} .
View answer and explanation
Correct option: A
0.5
Option A (0.5) is correct because the entropy change is calculated by dividing the enthalpy of vaporization by the temperature.
392 Easy
392 of 500
Understanding +4 | -1
Which of the following is/are state function?
View answer and explanation
Correct option: C
Energy
Option C (Energy) is the state function because its value depends only on the initial and final states of the system, not on the path taken.
393 Easy
393 of 500
Understanding +4 | -1
Assertion (A): Work is a path function which is expressed in joule.

Reason (R): Work appears only at the boundary of the system.
View answer and explanation
Correct option: B
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
Option B (Both (A) & (R) are true but the (R) is not the correct explanation of the (A)) is correct because both statements are true, but the reason doesn't explain why work is a path function.
394 Medium
394 of 500
Analyzing +4 | -1
Statement I: Heat evolved in the neutralization of either 1 mole of HCl{HCl} or 1 mole of H2SO4{H}_2 {SO}_4 with NaOH{NaOH} is same.

Statement II: Both HCl{HCl} and H2SO4{H}_2 {SO}_4 are strong acids.
View answer and explanation
Correct option: D
If Statement I is INCORRECT, but Statement II is CORRECT
Option D is correct because the heat evolved in the neutralization of 1 mole of H2SO4{H}_2 {SO}_4 with NaOH{NaOH} is different from that of 1 mole of HCl{HCl} because H2SO4{H}_2 {SO}_4 is a dibasic acid, while HCl{HCl} is a monobasic acid.
395 Medium
395 of 500
Understanding +4 | -1
Among the following the set of parameters that represents path functions, is
(A) q+W          q + W\,\,\,\,\,\,\,\,\,\, (B) q          q\,\,\,\,\,\,\,\,\,\, (C) W          W\,\,\,\,\,\,\,\,\,\, (D) H−TSH - TS
View answer and explanation
Correct option: D
(B) and (C)
Option D ((B) and (C)) is correct because heat (qq) and work (WW) are path functions, meaning their values depend on the path taken during a process.
396 Easy
396 of 500
Analyzing +4 | -1
The difference between electron gain enthalpies will be maximum between:
View answer and explanation
Correct option: C
Ne{Ne} and Cl{Cl}
Option C (Ne{Ne} and Cl{Cl}) is correct because the electron gain enthalpy of Ne is positive (noble gas) and that of Cl is highly negative, resulting in the maximum difference.
397 Medium
397 of 500
Remembering +4 | -1
Identify the correct statement regarding entropy
View answer and explanation
Correct option: C
At absolute zero of temperature, entropy of a perfectly crystalline substance is taken to be zero
Option C (At absolute zero of temperature, entropy of a perfectly crystalline substance is taken to be zero) is correct because the Third Law of Thermodynamics states that the entropy of a perfectly crystalline substance at absolute zero (0 K) is zero.
398 Medium
398 of 500
Application +4 | -1
The heat of combustion of CH4(g),C(graphite){CH}_{4({g})}, {C}_{(\mathrm{graphite})} and H2(g){H}_{2({g})} are 20 kcal,−40 kcal20 \, \mathrm{kcal},-40 \, \mathrm{kcal} and −10 kcal-10 \, \mathrm{kcal} respectively. The heat of formation of methane is
View answer and explanation
Correct option: C
−80 kcal-80 \, \mathrm{kcal}
Option C (-80 kcal) is correct because it correctly applies Hess's Law to calculate the heat of formation.
399 Medium
399 of 500
Application +4 | -1
The entropy change for vaporisation of liquid water to steam at 100℃ is…JK−1mol−1\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}Given that heat of vaporisation is 40.8kJmol−1.40.8kJmo{l^{ - 1}}.
View answer and explanation
Correct option: A
109.38
Option A (109.38) is correct because the entropy change is calculated by dividing the heat of vaporization by the temperature in Kelvin.
400 Medium
400 of 500
Remembering +4 | -1
Specific heat is defined as:
View answer and explanation
Correct option: A
Heat capacity/g
Option A (Heat capacity/g) is correct because specific heat is defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
401 Hard
401 of 500
Analyzing +4 | -1
Calculate proton affinity of NH3(g){NH}_3(\mathrm{g}) from the following data.

ΔHdissociation H2=218 kJ mole−1ΔHdissociation Cl2=124 kJ mole−1ΔHf∘ of NH3(g)=−46 kJ mole−1ΔHf∘ of NH4Cl(s)=−314 kJ mole−1\begin{aligned} & \Delta H_{\text {dissociation }} {H}_2=218 \mathrm{~kJ} \mathrm{~mole}^{-1} \\ & \Delta H_{\text {dissociation }} {Cl}_2=124 \mathrm{~kJ} \mathrm{~mole}^{-1} \\ & \Delta H_{\mathrm{f}}^\circ \text { of } {NH}_3(\mathrm{g})=-46 \mathrm{~kJ} \mathrm{~mole}^{-1} \\ & \Delta H_{\mathrm{f}}^\circ \text { of } {NH}_4 {Cl}(\mathrm{s})=-314 \mathrm{~kJ} \mathrm{~mole}^{-1} \end{aligned}

Ionization energy of H=1310 kJ mole−1H=1310 \mathrm{~kJ} \mathrm{~mole}^{-1}

E.A. of Cl(g)=−348 kJ mole−1{Cl}(\mathrm{g})=-348 \mathrm{~kJ} \mathrm{~mole}^{-1}

Lattice energy of NH4Cl(s)=−683 kJ mole−1{NH}_4 {Cl}(\mathrm{s})=-683 \mathrm{~kJ} \mathrm{~mole}^{-1}
View answer and explanation
Correct option: B
−718 kJ mole−1-718 \mathrm{~kJ} \mathrm{~mole}^{-1}
The correct answer is B (-718 kJ/mol) because it is the calculated proton affinity of NH3(g) using the given thermochemical data and Hess's Law.
402 Medium
402 of 500
Remembering +4 | -1
The enthalpy of a monoatomic gas at TT Kelvin is:
View answer and explanation
Correct option: D
52RT\frac{5}{2} \mathrm{RT}
Option D (52RT\frac{5}{2} \mathrm{RT}) is correct because the enthalpy of a monoatomic gas is given by H = E + PV = 32RT+RT=52RT\frac{3}{2}RT + RT = \frac{5}{2}RT.
403 Medium
403 of 500
Application +4 | -1
At 25∘C25^{\circ} \mathrm{C} 1 mole of butane upon heating forms CO2\mathrm{CO}_2 and H2O\mathrm{H}_2 \mathrm{O}. The work done is
View answer and explanation
Correct option: B
85.6 L85.6 \mathrm{~L} atm
Option B (85.6 L atm85.6 \mathrm{~L} \mathrm{~atm}) is correct because it involves calculating the work done using the change in the number of moles of gas and the ideal gas constant.
404 Medium
404 of 500
Application +4 | -1
The enthalpy of neutralisation of NH4OHNH_4 OH and CH3COOHCH_3 COOH is −10.5k cal mol−1-10.5 \mathrm{k\,cal\,} \mathrm{mol}^{-1} and enthalpy of neutralisation of CH3COOHCH_3COOH with strong base is −12.5k cal mol−1-12.5 \mathrm{k\,cal\,} \mathrm{mol}^{-1}. The enthalpy of ionisation of NH4OHNH_4OH will be
View answer and explanation
Correct option: C
2.0k cal mol−12.0 \mathrm{k\,cal}\, \mathrm{mol}^{-1}
Option C (2.0k cal}mol−12.0 \mathrm{k\,cal\} \mathrm{mol}^{-1}}) is correct because it is calculated by considering the enthalpy of neutralization of a weak acid with a weak base and a weak acid with a strong base.
405 Hard
405 of 500
Application +4 | -1
For the complete combustion of ethanol, C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l),{C_2}{H_5}OH(l) + 3{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l), the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ−1 mol−11364.47\,k{J^{ - 1}}\,mo{l^{ - 1}} at 25∘C25^\circ C. Assuming ideality the enthalpy of combustion, ΔCH,{\Delta _C}H, for the reaction will be
(R=8.314 JK−1 mol−1R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}})
View answer and explanation
Correct option: A
−1366.95 kJ−1 mol−1- 1366.95\,k{J^{ - 1}}\,mo{l^{ - 1}}
Option A (−1366.95 kJ−1 mol−1- 1366.95\,k{J^{ - 1}}\,mo{l^{ - 1}}) is correct because it accounts for the change in the number of moles of gas using the relationship between \ΔH\Delta H and \ΔU\Delta U.
406 Medium
406 of 500
Application +4 | -1
Given that N2( g)+3H2( g)→2NH3( g)\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g});
ΔrH∘=−92 kJ\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-92 \mathrm{~kJ}, the standard molar enthalpy of formation in kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} of NH3( g)\mathrm{NH}_3(\mathrm{~g}) is
View answer and explanation
Correct option: D
-46
Option D (-46) is correct because the standard molar enthalpy of formation is half of the given reaction enthalpy since 2 moles of NH3 are formed.
407 Medium
407 of 500
Analyzing +4 | -1
Butane exists in various conformations in nature. At any given instant, the probability that a given butane molecule is in anti, gauche, eclipsed and fully eclipsed conformation is 0.7,0.2,0.060.7,0.2,0.06 and 0.04,0.04 , respectively. If the molar enthalpy of combustion of natural butane is −690 kcal/mol-690 \mathrm{~kcal} / \mathrm{mol} at 25∘C,25^{\circ} \mathrm{C}, then calculate the enthalpy of combustion of butane if all the butane molecules are in gauche conformation. The potential energy of gauche form is 2 kcal/mol2 \mathrm{~kcal} / \mathrm{mol} higher than anti form but it is 3 kcal/mol3 \mathrm{~kcal} / \mathrm{mol} less than eclipsed form and 5.5 kcal/mol5.5 \mathrm{~kcal} / \mathrm{mol} less than fully eclipsed form.
View answer and explanation
Correct option: C
−691 kcal/mol-691 \mathrm{~kcal} / \mathrm{mol}
Option C (-691 kcal/mol) is correct because it calculates the enthalpy of combustion for the gauche conformation based on the given probabilities and energy differences.
408 Medium
408 of 500
Understanding +4 | -1
According to the first law of thermodynamics which of the following quantities represents change in a state function?
View answer and explanation
Correct option: D
qrev+Wrev{q}_{{rev}}+{W}_{{rev}}
Option D (qrev+Wrev{q}_{{rev}}+{W}_{{rev}}) is correct because, according to the first law of thermodynamics, the change in internal energy, a state function, is equal to the sum of heat and work.
409 Medium
409 of 500
Application +4 | -1
1 mol of an ideal gas expanded reversibly from 10 L to 100 L at 27∘C.27^{\circ} \mathrm{C}. Work done involved in the process is
View answer and explanation
Correct option: B
-5.74 kJ
Option B (-5.74 kJ) is correct because the work done in a reversible isothermal expansion is calculated using the formula W = -2.303 nRT log(V2/V1).
410 Medium
410 of 500
Remembering +4 | -1
A process in which no heat change takes place is called.
View answer and explanation
Correct option: B
An adiabatic process
Option B (An adiabatic process) is correct because an adiabatic process is defined as a thermodynamic process in which there is no heat exchange between the system and its surroundings (ΔQ = 0).
411 Medium
411 of 500
Application +4 | -1
The enthalpy changes for the following processes are listed below :

Cl2(g)→2Cl(g),242.3 kJ mol−1I2(g)→2I(g),151.0 kJ mol−1ICl(g)→I(g)+Cl(g),211.3 kJ mol−1I2(s)→I2(g),62.76 kJ mol−1\begin{array}{ll}\mathrm{Cl}_2(g) \rightarrow 2 \mathrm{Cl}(g), & 242.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{I}_2(g) \rightarrow 2 \mathrm{I}(g), & 151.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{ICl}(g) \rightarrow \mathrm{I}(g)+\mathrm{Cl}(g), & 211.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{I}_2(s) \rightarrow \mathrm{I}_2(g), & 62.76 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{array}

Given that the standard states for iodine and chlorine are I2(s)\mathrm{I}_2(s) and Cl2(g)\mathrm{Cl}_2(g), the standard enthalpy of formation for ICl⁡(g)\operatorname{ICl}(g) is :
View answer and explanation
Correct option: A
+16.8 kJ mol−1+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option A (+16.8 kJ mol⁻¹) is correct because it is calculated using Hess's Law and the given enthalpy changes.
412 Easy
412 of 500
Understanding +4 | -1
The standard heat of combustion of solid boron is :
View answer and explanation
Correct option: D
−12ΔfΔH∘(B2O3)- \frac{1}{2}{\Delta _f}\Delta H^\circ \left( {{\mathrm{B}_2}{\mathrm{O}_3}} \right)
Option B (12ΔfΔH∘(B2O3)\frac{1}{2}{\Delta _f}\Delta H^\circ \left( {{\mathrm{B}_2}{\mathrm{O}_3}} \right)) is correct because the standard heat of combustion of solid boron is half the standard heat of formation of B2O3.
413 Medium
413 of 500
Application +4 | -1
The heat of formations of CO(g)\mathrm{CO(g}) and CO2(g)\mathrm{CO_2(g}) are −26.4 K cal-26.4 \mathrm{~K} \mathrm{~cal} and −94.0 kcal-94.0 \mathrm{~kcal} respectively. The heat of combustion of carbon monoxide will be
View answer and explanation
Correct option: D
−67.6 kcal-67.6 \mathrm{~kcal}
Option D (-67.6 kcal) is correct because it is the result of subtracting the heat of formation of CO from that of CO2.
414 Medium
414 of 500
Application +4 | -1
Consider a reversible isontropic expansion of 1.0 mole of an ideal monoatomic gas from 25∘C25^{\circ} \mathrm{C} to 75∘C75^{\circ} \mathrm{C}. If the initial pressure was 1.0 bar1.0 \mathrm{~bar}, final pressure is
View answer and explanation
Correct option: A
1.474 bar
Option A (1.474 bar) is correct because it is calculated using the isentropic process formula relating pressure and temperature.
415 Medium
415 of 500
Application +4 | -1
If the bond energies of H2,Br2\mathrm{H}_2, \mathrm{Br}_2 and H−Br\mathrm{H}-\mathrm{Br} are 433,192 and 364 kJ/364 \mathrm{~kJ} / mol respectively the ΔH∘\Delta \mathrm{H}^{\circ} for the reaction H2(g)+Br2(g)⟶2HBr(g)\mathrm{H}_2(\mathrm{g})+\mathrm{Br}_2(\mathrm{g}) \longrightarrow 2 \mathrm{HBr}(\mathrm{g}) is:
View answer and explanation
Correct option: C
−103 kJ-103 \mathrm{~kJ}
Option C (-103 kJ) is correct because it calculates the enthalpy change of the reaction using bond energies of reactants and products.
416 Medium
416 of 500
Understanding +4 | -1
Which of the following on combustion gives
maximum energy?
View answer and explanation
Correct option: D
Butane
Option D (Butane) is correct because, for alkanes, the heat of combustion increases with the number of carbon atoms.
417 Medium
417 of 500
Understanding +4 | -1
A process which is unfavourable with respect to enthalpy,
but favourable with respect to entropy
View answer and explanation
Correct option: B
could occur at high temperatures, but not at low
temperatures
Option B is correct because a process unfavorable in enthalpy but favorable in entropy can occur at high temperatures where the entropy term dominates the Gibbs free energy equation.
418 Medium
418 of 500
Understanding +4 | -1
Heat absorbed at constant volume is measured in a calorimeter in which steel vessel is immersed to ensure that
View answer and explanation
Correct option: C
No heat is lost to the surrounding
Option C is correct because a calorimeter is designed to prevent heat exchange with the surroundings, ensuring accurate measurement of heat absorbed at constant volume.
419 Medium
419 of 500
Analyzing +4 | -1
Assertion (A): The dissolution of NH4Cl\mathrm{NH}_4 \mathrm{Cl} in water is endothermic but it dissolves in water.
Reason (R): On dissolution of NH4Cl\mathrm{NH}_4 \mathrm{Cl}, entropy increases.
View answer and explanation
Correct option: B
Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion
Option B (Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion) is correct because while NH4Cl dissolution is endothermic and dissolves in water, the driving force is entropy increase, not a direct consequence of endothermicity.
420 Medium
420 of 500
Remembering +4 | -1
The molar heat capacity, Cv\mathrm{C}_{\mathrm{v}} of an ideal gas whose energy is that of translational motion only is
View answer and explanation
Correct option: B
12.47 J deg−1 mol−112.47 \mathrm{~J} \mathrm{~deg}^{-1} \mathrm{~mol}^{-1}
Option B (12.47 J deg−1 mol−112.47 \mathrm{~J} \mathrm{~deg}^{-1} \mathrm{~mol}^{-1}) is correct because for an ideal gas with only translational motion, Cv = (3/2)R, which equals 12.47 J/mol·K.
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