LIST-I Process LIST-II Sign of Thermodynamic Parameters (P) For the process A ( l ) ⇌ A ( s ) , Δ H & Δ V may be (1) -ve ; +ve (Q) A 2 ( s ) + B 2 ( g ) ⇌ C 2 ( s ) + D 2 ( s ) , Δ H & Δ G may

Question

LIST-I

Process

LIST-II

Sign of Thermodynamic Parameters

(P)

For the process

$A (l)$ ⇌ $A (s)$

, $Δ H & Δ V$ may be

(1) -ve ; +ve

(Q)

$A_{2} (s) + B_{2} (g)$ ⇌ $C_{2} (s) + D_{2} (s)$ , $Δ H & Δ G$ may be

(2) +ve ; -ve

(R)

For the reaction

$A_{2} (g)$ ⇌ $B (g) + C (g)$

$E_{a} (f or w a r d)$ = 50kJmol^-1 ; $E_{a} (ba c k w a r d)$ = 40kJmol^-1, then at very high temperature $Δ H & Δ G$ are

(3) +ve ; +ve

(S)

For the reaction

$A (g)$ ⇌ $B (g)$ , at very low temperature $Δ H & Δ G$ are

(4) -ve ; -ve

Choose the correct MATCH

TestHub · Accepted Answer

P: Freezing is exothermic ( Δ H < 0 ). Volume change can be positive (e.g., water) or negative (most other substances). So, Δ H is -ve, Δ V can be +ve or -ve. This corresponds to (1) and (4). Q: A 2 ( s ) + B 2 ( g ) ⇌ C 2 ( s ) + D 2 ( s ) . Moles of gas decrease, so Δ S < 0 . Δ G = Δ H − T Δ S . Since − T Δ S > 0 , Δ G can be positive or negative depending on Δ H and temperature. Δ H can also be positive or negative. Thus, combinations like (-ve, +ve), (+ve, +ve), (-ve, -ve) are all possible. This corresponds to (1), (3), and (4). R: A 2 ( g ) ⇌ B ( g ) + C ( g ) . Δ H = E a ( f or w a r d ) − E a ( ba c k w a r d ) = 50 − 40 = + 10 k J m o l − 1 (positive). Moles of gas increase, so Δ S > 0 . At very high temperature, T Δ S is large and positive, making Δ G = Δ H − T Δ S negative. So, Δ H is +ve, Δ G is -ve. This corresponds to (2). S: A ( g ) ⇌ B ( g ) . At very low temperature, T Δ S ≈ 0 . So, Δ G ≈ Δ H . If Δ H is positive, Δ G is positive. If Δ H is negative, Δ G is negative. This corresponds to (3) and (4). The final answer is C

Chemistry - Thermodynamics - II Question with Solution | TestHub

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JEE Advanced 2026