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Byju's Answer
Standard XII
Chemistry
Gibb's Energy and Nernst Equation
The Gibbs fre...
Question
The Gibbs free energy change for the reaction for which
Δ
H
=
−
393.4
k
J
and
Δ
S
=
−
2.9
J
K
−
1
m
o
l
−
1
at 298K is:
A
−
352.5
k
J
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B
−
392.5
k
J
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C
+
352.5
k
J
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D
+
392.5
k
J
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Solution
The correct option is
A
−
392.5
k
J
Δ
G
=
Δ
H
−
T
Δ
S
Given,
Δ
H
=
−
393.4
k
J
/
m
o
l
and
Δ
S
=
−
2.9
J
K
−
1
m
o
l
−
1
Δ
G
=
−
393400
−
298
×
2.9
=
−
392535.8
J
≈
−
392
k
J
Hence, option B is correct.
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Similar questions
Q.
Standard enthalpy and standard entropy change for the oxidation of
N
H
3
at
298
K
are
−
382.64
k
J
m
o
l
−
1
and
−
145.6
k
J
m
o
l
−
1
respectively. Standard Gibbs energy change for the same reaction at
298
K
is:
Q.
The free energy for a reaction having
Δ
H
=
31400
c
a
l
,
Δ
S
=
32
c
a
l
K
−
1
m
o
l
−
1
at
1000
o
C
is:
Q.
For reaction,
A
g
2
O
(
s
)
→
2
A
g
(
s
)
+
(
1
/
2
)
O
2
(
g
)
, the value of
Δ
H
=
30.56
k
J
m
o
l
−
1
and
Δ
S
=
66
J
K
−
1
m
o
l
−
1
. Temperature at which free energy change for reaction will be zero, is:
Q.
For the reaction,
A
g
2
O
(
s
)
⇌
2
A
g
(
s
)
+
1
2
O
2
(
g
)
Δ
H
,
Δ
S
,
Δ
T
are
40.63
k
J
m
o
l
−
1
,
1.808
J
K
−
1
m
o
l
−
1
and
373.4
K
respectively. Free energy change
Δ
G
of the reaction will be:
Q.
Δ
H
and
Δ
S
for the reaction,
A
g
2
O
(
s
)
→
2
A
g
(
s
)
+
1
2
O
2
(
g
)
, are
30.56
k
J
m
o
l
−
1
and
66.00
J
K
−
1
m
o
l
−
1
respectively. The temperature at which the free energy change for the reaction will be zero is :
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