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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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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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Showing 481-500 of 500 questions
481 Medium
481 of 500
Application +4 | -1
Assign the sign of work done (based on SI convention) in
the following chemical changes taking place against
external atmospheric pressure
I. N2(g)+3H2(g)⟶2NH3(g)\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \longrightarrow 2 \mathrm{NH}_3(g)

II. NH4HS(s)⟶NH3(g)+H2 S(g)\mathrm{NH}_4 \mathrm{HS}(s) \longrightarrow \mathrm{NH}_3(g)+\mathrm{H}_2 \mathrm{~S}(g)

III. C(s)+CO2(g)⟶2CO(g)\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(g) \longrightarrow 2 \mathrm{CO}(g)

IV. C2H6(g)+3.5O2(g)⟶2CO2(g)+3H2O(l)\mathrm{C}_2 \mathrm{H}_6(g)+3.5 \mathrm{O}_2(g) \longrightarrow 2 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)
View answer and explanation
Correct option: A
IIIIIIIV+−−+\begin{array}{cccccccccccccccccccc} {\text{I}}&{{\text{II}}}&{{\text{III}}}&{{\text{IV}}} \\ + & - & - & + \end{array}
Option A (+, -, -, +) is correct because it correctly assigns the sign of work based on the change in the number of moles of gas (Δng\Delta n_g).
482 Medium
482 of 500
Application +4 | -1
At the given condition of CH4−O2CH_4 - O_2 fuel
cell the cell emf is 0.8 V and the
enthalpy of combustion of CH4 (g)C{H_4} \text{ (g)} is
-772 kJ/mol. The maximum efficiency of
the given fuel cell in the given condition
is :
View answer and explanation
Correct option: C
0.8
Option C (80%) is correct because the maximum efficiency is calculated by dividing the Gibbs free energy change by the enthalpy change of combustion.
483 Medium
483 of 500
Remembering +4 | -1
The state of a gas can be described by quoting the relationship between
View answer and explanation
Correct option: D
Pressure, volume, temperature and amount
Option D (Pressure, volume, temperature and amount) is correct because the state of a gas is defined by the relationship between these four variables, as described by the ideal gas law.
484 Medium
484 of 500
Application +4 | -1
One mole of a non-ideal gas undergoes a change of state (2.0 atm,3.0 L,95 K)→(4.0 atm,5.0 L,245 K) with a change in internal energy, ∆U=30.0 L atm. The change in enthalpy (∆H) of the process in L atm is:
View answer and explanation
Correct option: C
44
Option C (44) is correct because the change in enthalpy is calculated using the formula ∆H = ∆U + (P₂V₂ - P₁V₁), which gives 44 L atm.
485 Medium
485 of 500
Remembering +4 | -1
If water is formed from H+{H^ + }ions andOH−,O{H^ - },the heat of formation of water is :
View answer and explanation
Correct option: A
−13.7kcal- 13.7\mathrm{kcal}
Option A (−13.7kcal- 13.7\mathrm{kcal}) is correct because the heat of formation of water from H+ and OH- is the heat of neutralization, which is -13.7 kcal.
486 Medium
486 of 500
Application +4 | -1
The volume of O2\mathrm{O}_2 required to completely burn 52 g52 \mathrm{~g} at Ethyne (C2H2)\left(\mathrm{C}_2 \mathrm{H}_2\right) of STP is:
View answer and explanation
Correct option: C
112 L112 \mathrm{~L}
Option C (112 L112 \mathrm{~L}) is correct because the balanced equation shows that 1 mole of ethyne requires 2.5 moles of oxygen for complete combustion, and 52g of ethyne is 2 moles, thus requiring 5 moles of oxygen which equals 112L at STP.
487 Medium
487 of 500
Application +4 | -1
The entropy change in the conversion of one mole of liquid water at 373 K373 \mathrm{~K} to vapour at the same temperature is: (Latent heat of vaporization of water, ΔHvap =2.257 kJ/g\Delta H_{\text {vap }}=2.257 \mathrm{~kJ} / \mathrm{g} )
View answer and explanation
Correct option: C
108.9JK−1108.9 \mathrm{JK}^{-1}
Option C (108.9JK−1108.9 \mathrm{JK}^{-1}) is correct because it is calculated by dividing the molar heat of vaporization by the temperature.
488 Medium
488 of 500
Remembering +4 | -1
Which of the following is a highly corrosive salt?
View answer and explanation
Correct option: D
HgCl2{HgCl}_2.
Option D (HgCl2{HgCl}_2) is correct because Mercuric chloride (HgCl2HgCl_2) is known as corrosive sublimate due to its highly corrosive and poisonous nature.
489 Medium
489 of 500
Application +4 | -1
The standard free energy change for a gaseous reaction at 27∘C  is  x k.cal.27^\circ C\,\,\mathrm{ is }\,\,x\,k.cal. If equilibrium constant for a reaction is 100 and R is 2 cal.k−1.mol−1.2\,cal.{k^{ - 1}}.mo{l^{ - 1}}. Then xx is
View answer and explanation
Correct option: A
– 2.7636
Option A (– 2.7636) is correct because it is the standard free energy change calculated using the given equilibrium constant and temperature.
490 Medium
490 of 500
Understanding +4 | -1
To calculate the amount of work done in joules during reversible isothermal expansion of an ideal gas, the volume must be expressed in
View answer and explanation
Correct option: D
Any of these
Option D (Any of these) is correct because the work done in a reversible isothermal expansion involves a ratio of volumes, so the units cancel out, making any volume unit acceptable as long as both volumes are in the same unit.
491 Hard
491 of 500
Application +4 | -1
Enthalpy of the reaction Ag+(aq)+Br−(aq)→\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow AgBr(s)\mathrm{AgBr}(\mathrm{s}) is −84.54 kJ.-84.54 \mathrm{~kJ}. Magnitude of enthalpies of formation of Ag+(aq)\mathrm{Ag}^{+}(\mathrm{aq}) and Br−(aq)\mathrm{Br}^{-}(\mathrm{aq}) is in 8:98:9 ratio but their signs are opposite. Enthalpy of formation of AgBr\mathrm{AgBr} is −99.54 kJ/mol.-99.54 \mathrm{~kJ} / \mathrm{mol}. The magnitude of enthalpy of formation of Ag+(aq)\mathrm{Ag}^{+}(\mathrm{aq}) is
View answer and explanation
Correct option: A
120 kJ/mol120 \mathrm{~kJ} / \mathrm{mol}
The correct answer is A (120 kJ/mol120 \mathrm{~kJ} / \mathrm{mol}) because it is derived from Hess's Law and the given enthalpy changes for the reactions, considering the ratio of enthalpies of formation of Ag+(aq) and Br-(aq).
492 Medium
492 of 500
Understanding +4 | -1
For which reaction from the following, ΔS\Delta {S} will be maximum ?
View answer and explanation
Correct option: B
CaCO3( s)⟶CaO(s)+CO2( g){CaCO}_3(\mathrm{~s}) \longrightarrow {CaO}(\mathrm{s})+{CO}_2(\mathrm{~g})
Option B is correct because the reaction CaCO3( s)⟶CaO( s)+CO2( g){CaCO}_3(\mathrm{~s}) \longrightarrow {CaO}(\mathrm{~s})+{CO}_2(\mathrm{~g}) has the largest increase in the number of gaseous moles (Δng=1\Delta n_g = 1), leading to the maximum increase in entropy (ΔS\Delta S).
493 Medium
493 of 500
Understanding +4 | -1
Which of the following is/are intensive properties?

(I) Temperature

(II) Heat

(III) Density

(IV) Refractive index
View answer and explanation
Correct option: C
(I), (III) and (IV) only
Option C ((I), (III) and (IV) only) is correct because temperature, density, and refractive index are intensive properties, while heat is an extensive property.
494 Medium
494 of 500
Understanding +4 | -1
For the reaction: PCl5(g)→PCl3(g)+Cl2(g)\mathrm{PCl}_{5(g)} \rightarrow \mathrm{PCl}_{3(g)}+\mathrm{Cl}_{2(g)}, which one is true?
View answer and explanation
Correct option: B
ΔH=+ve;ΔS=+ve\Delta H=+\mathrm{ve} ; \Delta S=+\mathrm{ve}
Option B is correct because the decomposition of PCl5(g)\mathrm{PCl}_{5(g)} is endothermic (ΔH>0\Delta H > 0) and increases the number of gas molecules, leading to an increase in entropy (ΔS>0\Delta S > 0).
495 Easy
495 of 500
Analyzing +4 | -1
Enthalpy of neutralization of the reaction between CH3COOH(aq)\mathrm{CH}_3 \mathrm{COOH} \text{(aq)} and NaOH(aq)\mathrm{NaOH} \text{(aq)} is −13.2 kcal/eq-13.2 \mathrm{~kcal/eq} and that of the reaction between H2SO4(aq)\mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq}) and KOH(aq)\mathrm{KOH}(\mathrm{aq}) is −13.7 kcal/eq.-13.7 \mathrm{~kcal} / \mathrm{eq}. The enthalpy of dissociation of CH3COOH(aq)\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq}) is
View answer and explanation
Correct option: B
+0.5 kcal eq−1+0.5 \mathrm{~kcal} \mathrm{~eq}^{-1}
Option C (O−H{O}-{H} bond will be easily broken is case of hydroxide of ' X{X} ') is correct because if X(OH) is more acidic than Y(OH), the O-H bond in X(OH) is more easily broken, indicating a stronger tendency to donate a proton.
496 Medium
496 of 500
Analyzing +4 | -1
One mole of CH3COOH\mathrm{CH}_3 \mathrm{COOH} undergoes dimerization in vapour phase at 127∘C127^{\circ} \mathrm{C} as:

2CH3COOH(g)⇌(CH3COOH)2( g)2 \mathrm{CH}_3 \mathrm{COOH}(\mathrm{g}) \rightleftharpoons\left(\mathrm{CH}_3 \mathrm{COOH}\right)_2(\mathrm{~g})

if dimer formation is due to two H\mathrm{H}-bonds involved in dimer, each of 33 kJ33 \mathrm{~kJ} stength and the degree of dimerisation of acetic acid 98.2%98.2 \%. The change in standard entropy during dimerization is
View answer and explanation
Correct option: A
−104JK−1 mol−1-104 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
Option A is correct because the change in standard entropy during dimerization is calculated using the equilibrium constant and enthalpy change, resulting in approximately -104 J/K·mol.
497 Hard
497 of 500
Application +4 | -1
2 moles of an ideal gas at 270C{27^0}C temperature is expanded reversibly from 2L to 20L. Find entropy change (R=2cal/mol K).
View answer and explanation
Correct option: D
9.2
Option D (9.2) is correct because the entropy change is calculated using the formula for reversible expansion of an ideal gas.
498 Medium
498 of 500
Application +4 | -1
Given, 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∘=−92.4 kJ mol−1\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}

What is the magnitude of standard enthalpy of formation of NH3\mathrm{NH}_3 gas?
View answer and explanation
Correct option: A
46.2
Option A (46.2) is correct because the standard enthalpy of formation is half of the given reaction enthalpy since 2 moles of NH3 are formed in the reaction.
499 Medium
499 of 500
Understanding +4 | -1
Which of the following statements is correct?
View answer and explanation
Correct option: B
During complete combustion of 1 mole of butane, 2658 kJ2658 \mathrm{~kJ} of heat is released
Option B (During complete combustion of 1 mole of butane, 2658 kJ2658 \mathrm{~kJ} of heat is released) is correct because the combustion of butane is an exothermic reaction that releases a specific amount of heat.
500 Medium
500 of 500
Understanding +4 | -1
Two mole of an ideal gas is expanded isothermally and reversibly from 1 L to 10 L at 300 K. The enthalpy change (in kJ) for the process is
View answer and explanation
Correct option: C
0
Option C (0) is correct because enthalpy change for an isothermal process involving an ideal gas is zero, as enthalpy depends only on temperature.
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