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Entropy and second law, entropy changes in processes
JEE Main level
Single correct option
Difficulty 2 of 3
td139
A reaction has
Δ
H
=
+
42.6
kJ
mo
l
−
1
and
Δ
S
=
+
120
J
mo
l
−
1
K
−
1
, both assumed independent of temperature. The reaction is spontaneous
A
only below
355
K
B
at all temperatures
C
only above
355
K
D
at no temperature
Answer: C
Δ
G
=
Δ
H
−
T
Δ
S
<
0
requires
T
>
Δ
H
/Δ
S
=
42600/120
=
355
K
. Since both
Δ
H
and
Δ
S
are positive, the entropy term wins only at high temperature.
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