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Thermodynamics and Thermochemistry
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Gibbs energy, spontaneity and relation to equilibrium constant
JEE Main level
Single correct option
Difficulty 2 of 3
td101
A reaction releases
57.1
kJ
of heat at constant temperature
298
K
and constant pressure. The entropy change of the surroundings is
A
+
191.6
J
K
−
1
B
+
57.1
J
K
−
1
C
+
0.192
J
K
−
1
D
−
191.6
J
K
−
1
Answer: A
The surroundings receive
q
s
u
rr
=
+
57100
J
reversibly at
298
K
, so
Δ
S
s
u
rr
=
−
Δ
H
sys
/
T
=
57100/298
=
+
191.6
J
K
−
1
.
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