A certain reaction is non-spontaneous at 300 K. The entropy change during the reaction is 120 JK -1. Then the minimum value of Δ H for the reaction and the nature of the reaction will be:

Question

A certain reaction is non-spontaneous at 300 K. The entropy change during the reaction is 120 JK -1. Then the minimum value of Δ H for the reaction and the nature of the reaction will be: (A) 36.0 kJ; endothermic (B) 36.0 kJ; exothermic (C) 2.8 kJ; endothermic (D) -2.8 kJ; exothermic

Tardigrade Admin · Accepted Answer

For non-spontaneous reactions, Δ G is +ve. Δ G =Δ H - T Δ S ; Δ S text is +120 JK -1 To make Δ G as tve, Δ H should be +ve, i.e., endothermic For minimum value of entropy, Δ H = T Δ S =300 × 120=36000 J =36.0 kJ

Q. A certain reaction is non-spontaneous at $300 K$ . The entropy change during the reaction is $120 J K^{- 1}$ . Then the minimum value of $Δ H$ for the reaction and the nature of the reaction will be:

$36.0 k J$ ; endothermic

$36.0 k J$ ; exothermic

$2.8 k J$ ; endothermic

$- 2.8 k J$ ; exothermic

For non-spontaneous reactions, $Δ G$ is $+ v e$ .
$Δ G = Δ H - T Δ S; Δ S is + 120 J K^{- 1}$
To make $Δ G$ as tve, $Δ H$ should be $+ v e,$
i.e., endothermic For minimum value of entropy,
$Δ H = T Δ S$
$= 300 \times 120 = 36000 J$
$= 36.0 k J$

Q. A certain reaction is non-spontaneous at 300K. The entropy change during the reaction is 120JK−1. Then the minimum value of ΔH for the reaction and the nature of the reaction will be:

36.0kJ; endothermic

36.0kJ; exothermic

2.8kJ; endothermic

−2.8kJ; exothermic