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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Medium

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Showing 1-30 of 329 questions
1 Medium
1 of 329
Analyzing +4 | -1
Statement-1: CsI{CsI} is less soluble in water compared to CsCl{CsCl}
Satement-2: More amount of heat is liberated during the formation of CsCl{CsCl} than the formation CsI{Cs} I from their elements
View answer and explanation
Correct option: A
Both statements 1 and 2 correct
Both statements 1 and 2 are correct. CsI is less soluble than CsCl due to lower hydration enthalpy, and more heat is released during CsCl formation.
2 Medium
2 of 329
Application +4 | -1
Consider the following processes :

ΔH(KJ/mol)1/2A→B+1503B→2C+D−125E+A→2D+350\begin{array}{cccccccccccccccccccc} {}&{\Delta H(KJ/mol)} \\ {1/2A \to B}&{ + 150} \\ {3B \to 2C + D}&{ - 125} \\ {E + A \to 2D}&{ + 350} \end{array}

For B+D→E+2C,ΔH{B}+{D} \rightarrow {E}+2 {C}, \Delta {H} will be :
View answer and explanation
Correct option: B
−175 kJ/mol-175 {~kJ} / {mol}
Option B (−175 kJ/mol-175 {~kJ} / {mol}) is correct because it is calculated by manipulating the given equations and their enthalpy changes according to Hess's Law.
3 Medium
3 of 329
Understanding +4 | -1
The difference b/w ΔH\Delta H and Δu\Delta u at constant volume is equal to
View answer and explanation
Correct option: C
VΔPV\Delta P
Option C (VΔPV\Delta P) is correct because the difference between ΔH\Delta H and Δu\Delta u at constant volume is equal to VΔPV\Delta P based on the enthalpy equation.
4 Medium
4 of 329
Application +4 | -1
The heat evolved in the combustion of methane is given by the following equations CH4(g)+2O2(g)→CO2(g)+H2O(l);ΔH=−890.3kJ\mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right) + 2{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 890.3\mathrm{kJ} How many grams of methane would be required to produce 44.15 kJ of heat of combustion?
View answer and explanation
Correct option: B
8 g
Option B (8 g) is correct because it is calculated based on the heat evolved in the combustion of methane and the given enthalpy change.
5 Medium
5 of 329
Understanding +4 | -1
A reaction accompanied with the absorption of energy is :
View answer and explanation
Correct option: C
Electrolysis of water
Option C (Electrolysis of water) is correct because electrolysis requires the absorption of electrical energy to break chemical bonds.
6 Medium
6 of 329
Understanding +4 | -1
ΔU\Delta {U} of combustion of methane is −x kJ mol−1-{x} \, \mathrm{kJ} \mathrm{~mol}^{-1}. The value of ΔH\Delta {H} is
View answer and explanation
Correct option: C
<ΔU< \Delta {U}
Option C (< ΔU) is correct because the change in the number of moles of gas is negative, making ΔH less than ΔU.
7 Medium
7 of 329
Application +4 | -1
A 3 mole sample of a triatomic ideal gas at 300 K300 \mathrm{~K} is allowed to expand under adiabatic reversible condition from 5 L5 \mathrm{~L} to 40 L40 \mathrm{~L}. The value of ΔH\Delta {H} is
View answer and explanation
Correct option: B
−14.965 kJ-14.965 \mathrm{~kJ}
Option B (-14.965 kJ) is correct because the enthalpy change is calculated using adiabatic reversible expansion and the heat capacity at constant pressure.
8 Medium
8 of 329
Analyzing +4 | -1
48 g of C (diamond) on complete combustion evolves 1584 kJ of heat. The standard heat of formation of gaseous carbon is 725 kJ/mol. The energies required for the process
(i)C (graphite) → C(gas)
(ii)C (diamond) → C(gas)are:
View answer and explanation
Correct option: C
725, 723
Option C (725, 723) is correct because the heat of formation of gaseous carbon from graphite is 725 kJ/mol, and the heat of formation from diamond is slightly less due to diamond's higher energy.
9 Medium
9 of 329
Understanding +4 | -1
When 1 mole of gas is heated at constant volume. Temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct?
View answer and explanation
Correct option: B
q=∆U=500 J,W=0
Option B (q=∆U=500 J,W=0) is correct because at constant volume, no work is done, and all the heat supplied goes into changing the internal energy.
10 Medium
10 of 329
Analyzing +4 | -1
Heat of combustion of CH4,C2H4,C2H6{CH}_4, {C}_2 {H}_4, {C}_2 {H}_6 are −890,−1411-890,-1411 and −1560 KJ/mol-1560 \mathrm{~KJ} / \mathrm{mol} respectively. Which has the lowest fuel value is:
View answer and explanation
Correct option: B
C2H4{C}_2{H}_4
Option B (C2H4{C}_2{H}_4) has the lowest fuel value because, when the heat of combustion is normalized per gram, it yields the smallest magnitude.
11 Medium
11 of 329
Application +4 | -1
If equilibrium constant KcK_c for a reaction is 101010^{10} at 27∘C27^{\circ} \mathrm{C}, then ΔG0\Delta G^0 of the reaction at 27∘C27^{\circ} \mathrm{C} will be
View answer and explanation
Correct option: B
−6909R-6909 R
Option B (−6909R-6909 R) is correct because it accurately calculates the standard Gibbs free energy change using the given equilibrium constant and temperature.
12 Medium
12 of 329
Analyzing +4 | -1
The heats of neutralization of CH3COOH,   HCOOH,  HCNC{H_3}COOH,\,\,\,HCOOH,\,\,HCN and H2S{H_2}S are −13.2, − 13.4, −2.9- 13.2,\, - \,13.4,\, - 2.9 and - 3.8 kcal per equivalent respectively. Arrange the increasing order of strength.
View answer and explanation
Correct option: A
HCCOH>CH3COOH>H2S>HCNHCCOH > C{H_3}COOH > {H_2}S > HCN
Option A is correct because the strength of an acid is inversely proportional to its heat of neutralization (more negative = stronger acid).
13 Medium
13 of 329
Understanding +4 | -1
The absolute enthalpy of neutralisation of the reaction

MgO(s)+2HCl(aq)⟶MgCl2(aq)+H2O(l) will be\mathrm{MgO}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_2(a q)+\mathrm{H}_2 \mathrm{O}(l) \text { will be}
View answer and explanation
Correct option: C
less than −57.33 kJ mol−1-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option A (greater than −57.33 kJ mol−1-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}) is incorrect. The correct answer is less than −57.33 kJ mol−1-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1} because the reaction involves the formation of water and the neutralization of a strong acid with a metal oxide, which releases more heat than the neutralization of a strong acid with a strong base.
14 Medium
14 of 329
Understanding +4 | -1
Consider the following statements about the Hess's law of constant heat summation.
I. If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.
II. The enthalpy of an overall reaction A→BA \rightarrow B along one route is ΔrH\Delta_r H and ΔrH1,ΔrH2,ΔrH3…\Delta_r H_1, \Delta_r H_2, \Delta_r H_3 \ldots representing enthalpies of reactions leading to the same product, BB along another route, then we have ΔrH=ΔrH1+ΔrH2+ΔrH3⋯\Delta_r H=\Delta_r H_1+\Delta_r H_2+\Delta_r H_3 \cdots
Select the correct statement and choose the correct option.
View answer and explanation
Correct option: C
Both I and II
Option C (Both I and II) is correct because both statements accurately describe Hess's Law, which states that the enthalpy change of a reaction is independent of the pathway taken.
15 Medium
15 of 329
Application +4 | -1
One mole of a non-ideal gas undergoes a change of state (2.0 atm,3.0 L,95 K)&rarr;(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 it is the change in enthalpy calculated using the formula ∆H = ∆U + (P₂V₂ - P₁V₁).

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16 Medium
16 of 329
Understanding +4 | -1
Which of the following statements is true?
View answer and explanation
Correct option: C
The total energy of the universe remains
constant
Option C (The total energy of the universe remains constant) is correct because it is a statement of the first law of thermodynamics.
17 Medium
17 of 329
Understanding +4 | -1
Which of the following conditions lead to a
spontaneous process?
View answer and explanation
Correct option: D
Both ΔH\Delta \mathrm{H} and TΔS\mathrm{T} \Delta \mathrm{S} are + ve but TΔS>ΔH\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}
Option D (Both ΔH and TΔS are + ve but TΔS>ΔH) is correct because for a spontaneous process, ΔG must be negative, which occurs when TΔS > ΔH and both are positive.
18 Medium
18 of 329
Application +4 | -1
For vaporization of water at 1 atmospheric pressure, the values of ∆H and ∆S are 40.63 kJ mol−1mo{l^{ - 1}} and 108.8 JK−1mol−1,J{K^{ - 1}}mo{l^{ - 1}}, respectively. The temperature when Gibbs energy change (∆G) for this transformation will be zero, is:
View answer and explanation
Correct option: C
373.4 K
Option C (373.4 K) is correct because at this temperature, the Gibbs free energy change (∆G) is zero, indicating equilibrium between liquid and gas phases.
19 Medium
19 of 329
Understanding +4 | -1
Assertion (A): ΔG\Delta G has the unit of energy.
Reason (R): ΔH\Delta H and TΔST \Delta S are the energy dependent terms.
View answer and explanation
Correct option: A
Both Assertion and Reason are correct; Reason is the correct explanation of Assertion
Option A (Both Assertion and Reason are correct; Reason is the correct explanation of Assertion) is correct because Gibbs free energy change has units of energy, and both enthalpy and entropy terms contribute to the energy change.
20 Medium
20 of 329
Application +4 | -1
For a reaction at 25℃ enthalpy change (∆H) and entropy change (∆S) are -11.7×103{10^3}JmolJmo{l} - 105Jmo{l - 1}{K - 1}−1^{ - 1}Jmol−1Jmo{l^{ - 1}} respectively. The reaction is:
View answer and explanation
Correct option: B
Non-spontaneous
The correct answer is B (Non-spontaneous) because the calculated Gibbs free energy change (∆G) is positive, indicating a non-spontaneous reaction.
21 Medium
21 of 329
Application +4 | -1
The internal energy change (in J) when 90 g of water undergoes complete evaporation at 100∘C100^\circ C is &hellip;..,
(Given : ΔHvap\Delta {H_{\mathrm{vap}}} for water at 373  K=41kJ/mol, R=8.314   JK−1mol−1)373\,\,K = 41kJ/mol,\,R = 8.314\,\,\,J{K^{ - 1}}mo{l^{ - 1}})
View answer and explanation
Correct option: A
189494
Option A (189494) is correct because it is the internal energy change calculated using the enthalpy of vaporization and the ideal gas constant.
22 Medium
22 of 329
Understanding +4 | -1
Net work done by the system in a cyclic process is equal to:
View answer and explanation
Correct option: D
q
Option D (q) is correct because in a cyclic process, the change in internal energy (∆U) is zero, so the net work done (W) is equal to the heat absorbed (q).
23 Medium
23 of 329
Application +4 | -1
Cdiamond+O2(g)→CO2(g);ΔH=−395kJ…….(i){C_{\mathrm{diamond}}} + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\Delta H = - 395\mathrm{kJ} \ldots \ldots \mathrm{.}\left( \mathrm{i} \right)
Cgraphite+O2(g)→CO2(g);ΔH=−393.5kJ…….(ii){C_{\mathrm{graphite}}} + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\Delta H = - 393.5\mathrm{kJ} \ldots \ldots \mathrm{.}\left( {\mathrm{ii}} \right) The ∆H, when diamond is formed from graphite, is :
View answer and explanation
Correct option: B
+1.5kJ+ 1.5\mathrm{kJ}
Option B (+1.5 kJ) is correct because the enthalpy change for the conversion of graphite to diamond is calculated by subtracting the enthalpy of formation of CO2 from diamond from that of graphite.
24 Medium
24 of 329
Application +4 | -1
The difference between the heats of reaction at constant pressure and constant volume for the reaction, at a constant temperature T\mathrm{T} is

2C6H6(l)+15O2( g)→12CO2( g)+6H2O(l)2 \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+15 \mathrm{O}_2(\mathrm{~g}) \rightarrow 12 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l})
View answer and explanation
Correct option: A
−3RT-3 R T
Option A (-3 RT) is correct because the difference between heats of reaction at constant pressure and volume is given by ΔH−ΔU=ΔngRT\Delta H - \Delta U = \Delta n_g RT, where Δng\Delta n_g is the change in the number of gaseous moles.
25 Medium
25 of 329
Understanding +4 | -1
Heat liberated by a given amount of an ideal gas undergoing reversible isothermal process is 1200 cal1200 \, \mathrm{cal} at 300 K.300 \mathrm{~K}. What is the Gibbs free energy change of the gas in this process?
View answer and explanation
Correct option: B
+1200 cal+1200 \mathrm{~cal}
Option C (+1200 cal) is correct because the Gibbs free energy change is the negative of the heat absorbed (or positive of the heat liberated) in a reversible isothermal process.
26 Medium
26 of 329
Application +4 | -1
On the basis of the following thermochemical data [ΔfG∘H+(aq)=0]\left[\Delta_{\mathrm{f}}{G}^{\circ}\mathrm{H}^{+}(\mathrm{aq})=0\right]
H2O(l)→H+(aq)+OH−(aq);\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) ; ΔH=57.32 kJ\Delta \mathrm{H}=57.32 \mathrm{~kJ}
H2( g)+12O2( g)→H2O(l);ΔH=−286.02 kJ\begin{aligned} & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta \mathrm{H}=-286.02 \mathrm{~kJ}\end{aligned}
The value of enthalpy of formation of OH−\mathrm{OH}^{-}ion at 25∘C25^{\circ} \mathrm{C} is
View answer and explanation
Correct option: B
−228.88 kJ-228.88 \mathrm{~kJ}
Option B (−228.88 kJ-228.88 \mathrm{~kJ}) is correct because it is calculated using Hess's law and the given thermochemical data.
27 Medium
27 of 329
Understanding +4 | -1
Changes in a system from an initial state to the final state were made by a different manner that remains ΔH\Delta H remains same but qq changes because
View answer and explanation
Correct option: B
ΔH\Delta \mathrm{H} is a state function and q\mathrm{q} is a path function
Option B (ΔH\Delta \mathrm{H} is a state function and q\mathrm{q} is a path function) is correct because enthalpy change depends only on initial and final states, while heat depends on the path taken.
28 Medium
28 of 329
Understanding +4 | -1
The temperature of one mole of an ideal gas increases from 298 K298 \mathrm{~K} when it absorbs 200 J200 \mathrm{~J} of heat at constant volume. The change in the internal energy of the gas is
View answer and explanation
Correct option: A
200 J200 \mathrm{~J}
Option A (200 J) is correct because at constant volume, the heat absorbed is equal to the change in internal energy.
29 Medium
29 of 329
Analyzing +4 | -1
Statement - I: In an adiabatic expansion of an ideal gas, the entropy of the surrounding
remains unchanged independent whether the expansion is reversible or irreversible.

Statement - II: In an adiabatic process, the surrounding does not exchange heat with the
system and the process may be reversible or irreversible for system but it always remain
reversible for the sorrounding.
View answer and explanation
Correct option: A
If both Statement-I and Statement-II are true, and Statement-II is the correct
explanation of Statement-I
Option A is correct because both statements are true, and Statement-II correctly explains why the entropy of the surrounding remains unchanged in an adiabatic expansion.
30 Medium
30 of 329
Analyzing +4 | -1
For equilibrium CO(g)+12O2(g)⇋CO2(g),CO(g) + \frac{1}{2}{O_2}(g)\underset{{}}{\overset{{}}{\leftrightharpoons}}C{O_2}(g), when small amount of caustic soda (NaOH)(NaOH) is put in container then following change will occur
View answer and explanation
Correct option: A
Concentration of COCO decreases
Option A (Concentration of COCO decreases) is correct because NaOHNaOH will react with CO2CO_2, shifting the equilibrium to the right and consuming COCO.
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