Thermodynamics Questions | JEE Main Chemistry Chapter Wise Question Papers

Q 1 :
Which of the following is not correct? [2024]

$∆ G$ is zero for a reversible reaction.
$∆ G$ is negative for a spontaneous reaction.
$∆ G$ is positive for a spontaneous reaction.
$∆ G$ is positive for a non-spontaneous reaction.

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(3)

$∆ G = 0 \Rightarrow$ reaction at equilibrium

$∆ G > 0 \Rightarrow$ non spontaneous reaction

$∆ G < 0 \Rightarrow$ spontaneous reaction

Q 2 :
Consider the following reaction at 298 K. $\frac{3}{2} O_{2 (g)} ⇌ O_{3 (g)}$ . $K_{p} = 2.47 \times 10^{- 29}, Δ_{r} G^{Θ}$ for the reaction is __________ kJ.

(Given R $= 8.314$ ${JK}^{- 1}$ ${mol}^{- 1}$ ) [2024]

0%

(163)

$\log (K) = \log_{10} (2.47 \times 10^{- 29}) \approx \log_{10} (25 \times 10^{- 30})$

$= \log_{10} (5^{2} \times 10^{- 30}) = 2 \log_{10} 5 - 30$ $\log$ $10$

$= 2 \times 0.7 - 30 \times 1 = - 28.6$

$Δ G ° = - 2.303$ $RT$ $\log_{10} K$

$Δ G ° = - 2.303 \times 8.314 \frac{J}{K mol} \times 298 K \times \log_{10} (2.47 \times 10^{- 29})$

$= - 2.303 \times 8.314 \times 298 \times (- 28.6) J m o l^{- 1} = 163.2 k J m o l^{- 1}$

Q 3 :
For the reaction at 298 K, 2A + B $\to$ C. $∆ H$ = 400 kJ ${mol}^{- 1}$ and $∆ S$ = 0.2 kJ ${mol}^{- 1}$ $K^{- 1}$ . The reaction will become spontaneous above ________ K. [2024]

0%

(2000)

For reaction to be spontaneous,

$Δ G < 0$

$Δ H - T Δ S < 0$

$400 kJ$ ${mol}^{- 1} - T \times 0.2 kJ$ ${mol}^{- 1} K^{- 1} < 0$

$400 kJ$ ${mol}^{- 1} < T \times 0.2 kJ$ ${mol}^{- 1} K^{- 1}$

$T > \frac{400}{0.2} K$

$T > 2000 K$

Q 4 :
When $Δ H_{vap} = 30 kJ / mol$ and $∆ S_{vap} = 75 J$ $m o l^{- 1}$ $K^{- 1}$ , then the temperature of vapour, at one atmosphere is ___________ K. [2024]

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(400)

When external pressure is 1 atm ,then vapours are formed at normal boiling point of the liquid. So in the question, normal boiling point of the liquid is asked.

$Δ S_{vap} = \frac{Δ H_{vap}}{T_{bpt}}$

$75 J$ $m o l^{- 1} K^{- 1} = \frac{30 \times 1000 J m o l^{- 1}}{T_{b p t}}$

$T_{bpt} = 400 K$

Q 5 :
Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option. [2025]

Both $Δ H$ and $Δ S$ are (–ve)
Both $Δ H$ and $Δ S$ are (+ve)
$Δ H$ is (–ve) but $Δ S$ is (+ve)
$ΔH$ is (+ve) but $ΔS$ is (–ve)

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(2)

For reaction to be spontaneous $Δ G = Δ H - T Δ S < 0$ . As the reaction is endothermic, $Δ H > 0$ . If $Δ S > 0$ , then at low temperature $Δ G$ becomes positive i.e. reaction becomes non-spontaneous, and at high temperature $Δ G$ becomes negative i.e. the reaction becomes spontaneous.

Q 6 :
Match List-I with List-II. [2025]

List-I (Partial Derivatives)

List-II (Thermodynamic Quantity)

A. ${(\frac{\partial G}{\partial T})}_{P}$ I. $C_{p} $

B. ${(\frac{\partial H}{\partial T})}_{P}$ II. − S

C. ${(\frac{\partial G}{\partial P})}_{T}$ III. Cv

D. ${(\frac{\partial U}{\partial T})}_{V}$ IV. V

Choose the correct answer from the options given below:

A-II, B-I, C-IV, D-III
A-II, B-III, C-I, D-IV
A-II, B-I, C-III, D-IV
A-I, B-II, C-IV, D-III

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(1)

$G = H - T S$

$d G = d H - T d S - S d T$

$d G = d (U + P V) - T d S - S d T$

$d G = d U + P d V + V d P - T d S - S d T$

$d G = q_{rev} + W_{rev} + P d V + V d P - T d S - S d T$

$d G = T d S - P d V + P d V + V d P - T d S - S d T$

$d G = V d P - S d T$

At constant pressure, $d P = 0$ , so ${[\frac{\partial G}{\partial T}]}_{P} = - S$

At constant temperature, $d T = 0$ , so ${[\frac{\partial G}{\partial P}]}_{T} = V$

(B) $H = U + P V$

$d H = d U + P d V + V d P$

$d H = q_{rev} + W_{rev} + P d V + V d P$

$d H = q_{rev} - P d V + P d V + V d P$

$d H = q_{rev} + V d P$

$d H = n C d T + V d P$

At constant pressure $C = C_{p}$ and $d P = 0$

$d H = n C_{p} d T$

Taking 1 mole of substance, $(n = 1)$

${[\frac{\partial H}{\partial T}]}_{P} = C_{p}$

(D) $d U = q_{rev} + W_{rev}$

$d U = q_{rev} - P d V$

$d U = n C d T - P d V$

At constant volume, $C = C_{v}$ and $d V = 0$

$d U = n C_{v} d T$

${[\frac{\partial U}{\partial T}]}_{V} = C_{v}$

Q 7 :
The effect of temperature on spontaneity of reactions are represented as:

$Δ H$ $Δ S$ Temperature Spontaneity

(A) + – any T Non spontaneous

(B) + + low T spontaneous

(C) – – low T Non spontaneous

(D) – + any T spontaneous

The incorrect combinations are: [2025]

	$Δ H$	$Δ S$	Temperature	Spontaneity
(A)	+	–	any T	Non spontaneous
(B)	+	+	low T	spontaneous
(C)	–	–	low T	Non spontaneous
(D)	–	+	any T	spontaneous

(A) and (D) only
(B) and (D) only
(A) and (C) only
(B) and (C) only

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(4)

A reaction is spontaneous if $Δ G$ is negative.

S.no	$Δ H$	$Δ S$	T	$Δ G = Δ H - T Δ S$	Spontaneity
(A)	+	−	any	+	Non-spontaneous
(B)	+	+	low	+	Non-spontaneous
(C)	−	−	low	−	Spontaneous
(D)	−	+	any	−	Spontaneous

Q 8 :
Which of the following graphs correctly represents the variation of thermodynamic properties of Haber’s process? [2025]

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(1)

(I) $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g), Δ H ° = - 46.1 {kJ mol}^{- 1}$

As number of gaseous moles decrease, $Δ S ° < 0$ .

Both $Δ H °$ and $Δ S °$ are negative and almost remain constant with change of temperature.

(II) $Δ G ° = - R T \ln K$

$\frac{- Δ G °}{T} = R \ln K$

For exothermic reaction, value of equilibrium constant decreases with increase in temperature, so $\frac{- Δ G °}{T}$ decreases with increase in temperature.

(III) As $Δ H °$ is negative $\frac{- Δ H °}{T}$ is positive and it decreases with increase in temperature.

Q 9 :
Let us consider a reversible reaction at temperature, $T$ . In this reaction, both $Δ H$ and $Δ S$ were observed to have positive values. If the equilibrium temperature is $T_{e}$ , then the reaction becomes spontaneous at : [2025]

$T = T_{e}$
$T_{e} > T$
$T > T_{e}$
$T_{e} = 5 T$

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(3)

When reaction is at equilibrium, then:

$Δ G = 0$

$Δ H - T_{e} Δ S = 0$

$T_{e} = \frac{Δ H}{Δ S}$

For reaction to be spontaneous:

$Δ G < 0$

$Δ H - T Δ S < 0$

$T Δ S > Δ H$

$T > \frac{Δ H}{Δ S}$

$T > T_{e}$

Q 10 :
Standard entropies of $X_{2}$ , $Y_{2}$ and $X Y_{5}$ are 70, 50 and 110 ${JK}^{- 1}$ ${mol}^{- 1}$ respectively. The temperature in Kelvin at which the reaction

$\frac{1}{2} X_{2} + \frac{5}{2} Y_{2} ⇌ X Y_{5}, Δ H^{⊝} = - 35 {kJ mol}^{- 1}$

will be at equilibrium is _____ (Nearest integer). [2025]

0%

(700)

$\frac{1}{2} X_{2} + \frac{5}{2} Y_{2} ⇌ X Y_{5}$

$Δ S ° = S_{products} - S_{reactants} = S_{X Y_{5}} - [\frac{1}{2} S_{X_{2}} + \frac{5}{2} S_{Y_{2}}]$

$= 110 {JK}^{- 1} {mol}^{- 1} - [\frac{1}{2} \times 70 + \frac{5}{2} \times 50] {JK}^{- 1} {mol}^{- 1}$

$= - 50 {JK}^{- 1} {mol}^{- 1}$

$Δ G ° = Δ H ° - T Δ S ° = - 35 {kJ mol}^{- 1} - T \times (- 50 {JK}^{- 1} {mol}^{- 1})$

$= - 35000 {J mol}^{- 1} + T \times 50 {JK}^{- 1} {mol}^{- 1}$

Taking $Δ G ° = 0$ for equilibrium, $- 35000 {J mol}^{- 1} + T \times 50 {JK}^{- 1} {mol}^{- 1} = 0$

$T = \frac{35000}{50} K = 700 K$

(Note: At equilibrium $Δ G = 0$ , $Δ G °$ is not necessarily zero at equilibrium, but as per data provided in the question, the only way to solve the question is by taking $Δ G ° = 0$ ).

Topic Question Set

Q 1 :
Which of the following is not correct? [2024]

Q 2 :
Consider the following reaction at 298 K. $\frac{3}{2} O_{2 (g)} ⇌ O_{3 (g)}$ . $K_{p} = 2.47 \times 10^{- 29}, Δ_{r} G^{Θ}$ for the reaction is __________ kJ.

(Given R $= 8.314$ ${JK}^{- 1}$ ${mol}^{- 1}$ ) [2024]

0%

Q 3 :
For the reaction at 298 K, 2A + B $\to$ C. $∆ H$ = 400 kJ ${mol}^{- 1}$ and $∆ S$ = 0.2 kJ ${mol}^{- 1}$ $K^{- 1}$ . The reaction will become spontaneous above ________ K. [2024]

0%

Q 4 :
When $Δ H_{vap} = 30 kJ / mol$ and $∆ S_{vap} = 75 J$ $m o l^{- 1}$ $K^{- 1}$ , then the temperature of vapour, at one atmosphere is ___________ K. [2024]

0%

Q 5 :
Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option. [2025]

Q 7 :
The effect of temperature on spontaneity of reactions are represented as:

$Δ H$ $Δ S$ Temperature Spontaneity

(A) + – any T Non spontaneous

(B) + + low T spontaneous

(C) – – low T Non spontaneous

(D) – + any T spontaneous

The incorrect combinations are: [2025]

Q 8 :
Which of the following graphs correctly represents the variation of thermodynamic properties of Haber’s process? [2025]

Q 9 :
Let us consider a reversible reaction at temperature, $T$ . In this reaction, both $Δ H$ and $Δ S$ were observed to have positive values. If the equilibrium temperature is $T_{e}$ , then the reaction becomes spontaneous at : [2025]

0%

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	List-I (Partial Derivatives)		List-II (Thermodynamic Quantity)
A.	${(\frac{\partial G}{\partial T})}_{P}$	I.	$C_{p} $
B.	${(\frac{\partial H}{\partial T})}_{P}$	II.	− S
C.	${(\frac{\partial G}{\partial P})}_{T}$	III.	Cv
D.	${(\frac{\partial U}{\partial T})}_{V}$	IV.	V

Topic Question Set

Q 1 : Which of the following is not correct? [2024]

Q 2 : Consider the following reaction at 298 K. 32O2(g)⇌O3(g). Kp=2.47×10-29,ΔrGΘ for the reaction is __________ kJ. (Given R=8.314 JK-1 mol-1) [2024]

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Q 3 : For the reaction at 298 K, 2A + B→C. ∆H = 400 kJ mol-1 and ∆S = 0.2 kJ mol-1 K-1. The reaction will become spontaneous above ________ K. [2024]

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Q 4 : When ΔHvap=30kJ/mol and ∆Svap=75J mol-1 K-1, then the temperature of vapour, at one atmosphere is ___________ K. [2024]

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Q 5 : Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option. [2025]

Q 6 : Match List-I with List-II. [2025] List-I (Partial Derivatives) List-II (Thermodynamic Quantity) A. (∂G∂T)P I. Cp​ B. (∂H∂T)P II. − S C. (∂G∂P)T III. Cv D. (∂U∂T)V IV. V Choose the correct answer from the options given below:

Q 7 : The effect of temperature on spontaneity of reactions are represented as: ΔH ΔS Temperature Spontaneity (A) + – any T Non spontaneous (B) + + low T spontaneous (C) – – low T Non spontaneous (D) – + any T spontaneous The incorrect combinations are: [2025]

Q 8 : Which of the following graphs correctly represents the variation of thermodynamic properties of Haber’s process? [2025]

Q 9 : Let us consider a reversible reaction at temperature, T. In this reaction, both ΔH and ΔS were observed to have positive values. If the equilibrium temperature is Te, then the reaction becomes spontaneous at : [2025]

Q 10 : Standard entropies of X2, Y2 and XY5 are 70, 50 and 110 JK-1mol-1 respectively. The temperature in Kelvin at which the reaction 12X2+52Y2⇌XY5, ΔH⊝=-35 kJ mol-1 will be at equilibrium is _____ (Nearest integer). [2025]

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Q 1 :
Which of the following is not correct? [2024]

Q 2 :
Consider the following reaction at 298 K. $\frac{3}{2} O_{2 (g)} ⇌ O_{3 (g)}$ . $K_{p} = 2.47 \times 10^{- 29}, Δ_{r} G^{Θ}$ for the reaction is __________ kJ.

(Given R $= 8.314$ ${JK}^{- 1}$ ${mol}^{- 1}$ ) [2024]

Q 3 :
For the reaction at 298 K, 2A + B $\to$ C. $∆ H$ = 400 kJ ${mol}^{- 1}$ and $∆ S$ = 0.2 kJ ${mol}^{- 1}$ $K^{- 1}$ . The reaction will become spontaneous above ________ K. [2024]

Q 4 :
When $Δ H_{vap} = 30 kJ / mol$ and $∆ S_{vap} = 75 J$ $m o l^{- 1}$ $K^{- 1}$ , then the temperature of vapour, at one atmosphere is ___________ K. [2024]

Q 5 :
Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option. [2025]

Q 7 :
The effect of temperature on spontaneity of reactions are represented as:

$Δ H$ $Δ S$ Temperature Spontaneity

(A) + – any T Non spontaneous

(B) + + low T spontaneous

(C) – – low T Non spontaneous

(D) – + any T spontaneous

The incorrect combinations are: [2025]

Q 8 :
Which of the following graphs correctly represents the variation of thermodynamic properties of Haber’s process? [2025]

Q 9 :
Let us consider a reversible reaction at temperature, $T$ . In this reaction, both $Δ H$ and $Δ S$ were observed to have positive values. If the equilibrium temperature is $T_{e}$ , then the reaction becomes spontaneous at : [2025]