Tardigrade
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Tardigrade
Question
Chemistry
The entropy change for a non-spontaneous reaction is 140 J K -1 mol -1 at 298 K. The reaction is
Q. The entropy change for a non-spontaneous reaction is
140
J
K
−
1
m
o
l
−
1
at
298
K
. The reaction is
1731
206
Thermodynamics
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A
reversible
17%
B
irreversible
21%
C
exothermic
25%
D
endothermic
38%
Solution:
Since the reaction is non - spontaneous.
∴
Δ
G
is positive.
As
Δ
S
is positive (given) hence
Δ
H
must be positive also. It means reaction is endothermic.