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Gibbs energy, spontaneity and relation to equilibrium constant
JEE Main level
Numerical answer
Difficulty 2 of 3
td098
For a reaction at
298
K
,
Δ
G
∘
=
−
25.0
kJ
mo
l
−
1
. Calculate
lo
g
10
K
(two decimals). Use
R
=
8.314
J
mo
l
−
1
K
−
1
and
2.303
for
ln
10
.
Numerical answer type, tolerance 0.01.
Answer: 4.3815
Δ
G
∘
=
−
2.303
RT
lo
g
10
K
, so
lo
g
10
K
=
25000/
(
2.303
×
8.314
×
298
)
=
25000/5705.8
=
4.381
.
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