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Standard XII
Chemistry
Question
Calculate the equilibrium constant at 25 degrees celsius given the Standard Free Energy value of - 107.2 kJ
- 43.2
43.2
6.18 x
10
8
1.04
6.18 x
10
9
A
1.04
B
6.18 x
10
8
C
43.2
D
- 43.2
E
6.18 x
10
9
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Solution
Verified by Toppr
The temperature is 25 deg C or 298 K
Δ
G
0
=
−
R
T
l
n
K
Δ
G
0
=
−
107.2
k
J
=
−
107200
J
−
107200
=
−
8.314
×
298
×
l
n
K
l
n
K
=
43.268
K
=
6.18
×
10
18
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