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Byju's Answer
Standard XII
Chemistry
Gibbs Free Energy & Spontaneity
The value of ...
Question
The value of
l
o
g
10
K for a reaction
A
⇌
B
is:
(
G
i
v
e
n
:
Δ
r
H
0
298
k
=
−
54.07
k
J
m
o
l
−
1
,
Δ
r
S
0
298
k
=
10
J
K
−
1
m
o
l
−
1
a
n
d
R
=
8.314
J
K
−
1
m
o
l
−
1
;
2.303
×
8.314
×
298
=
5705
)
A
5
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B
10
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C
95
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D
100
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Solution
The correct option is
B
10
Δ
G
0
=
Δ
H
0
−
T
Δ
S
0
=
−
54.07
×
1000
−
298
×
10
=
−
54070
−
2980
=
−
57050
Δ
G
0
=
−
2.303
R
T
l
o
g
10
K
−
57050
=
−
2.303
×
298
×
8.314
l
o
g
10
K
=
−
5705
l
o
g
10
K
l
o
g
10
K
=
10
Suggest Corrections
1
Similar questions
Q.
The value of
l
o
g
10
K
for a reaction
A
⇌
B
is:
(Given
Δ
r
H
0
298
K
=
−
54.07
k
J
m
o
l
−
1
,
Δ
r
S
o
298
K
=
10
J
K
−
1
m
o
l
−
1
,
R
=
8.314
J
K
−
1
m
o
l
−
1
; and
2.303
×
8.314
×
298
=
5705
J
m
o
l
−
1
)
Q.
The value of
l
o
g
10
K for a reaction
A
⇌
B
is:
Given:
△
r
H
298
K
=
−
54.07
k
J
m
o
l
−
1
,
△
r
S
o
298
K
=
10
J
K
−
1
m
o
l
−
1
a
n
d
R
=
8.314
J
K
−
1
m
o
l
−
1
;
2.303
×
8.314
×
298
=
5705
)
.
Q.
The value of
l
o
g
10
K for a reaction
A
⇌
B
is:
Given:
△
r
H
298
K
=
−
54.07
k
J
m
o
l
−
1
,
△
r
S
o
298
K
=
10
J
K
−
1
m
o
l
−
1
a
n
d
R
=
8.314
J
K
−
1
m
o
l
−
1
;
2.303
×
8.314
×
298
=
5705
)
.
Q.
The value of
log
10
K for a reaction
A
⇌
B
is:
Given
−
△
r
H
∘
298
K
=
−
54.07
k
J
m
o
l
−
1
,
△
r
S
∘
298
K
=
10
J
K
−
1
m
o
l
−
1
and
R
−
8.314
J
K
−
1
m
o
l
−
1
Q.
The value of -
log
K
for a reaction :
A
(
g
)
⇌
B
(
g
)
[Given:
Δ
r
H
o
298
k
=
−
54.07
k
J
m
o
l
−
1
,
Δ
r
S
o
298
K
=
−
191.62
k
J
/
K
and
2.303
×
8.314
×
298
=
5705
]
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