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Byju's Answer
Standard XII
Chemistry
Gibbs Free Energy & Spontaneity
The following...
Question
The following reaction is possible at 300 K.
2
C
u
O
(
s
)
⟶
C
u
2
O
(
s
)
+
1
2
O
2
(
g
)
Given,
Δ
H
=
144.6
k
J
m
o
l
−
1
and
Δ
S
=
0.116
k
J
m
o
l
−
1
K
−
1
.
The value of
Δ
G
is _______.
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Solution
Δ
G
=
Δ
H
−
T
Δ
S
At
300
K
,
Δ
G
=
144.6
−
300
×
0.116
Δ
G
=
144.6
−
34.8
=
109.8
k
J
m
o
l
−
1
Δ
G
>
0
Hence, the value of
Δ
G
is positive.
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Similar questions
Q.
Following reaction is spontaneous at
300
K
.
2
C
u
O
(
s
)
⟶
C
u
2
O
(
s
)
+
I
/
2
O
2
(
g
)
Δ
H
=
−
144.6
k
J
m
o
l
−
1
;
Δ
S
=
0.116
k
J
m
o
l
−
1
if true enter 1 else 0
Q.
For the reaction at
300
K
A
(
g
)
+
B
(
g
)
→
C
(
g
)
Δ
H
=
−
3.0
k
c
a
l
;
Δ
S
=
−
10.0
c
a
l
/
K
value of
Δ
G
is:
Q.
Compute the value of
Δ
S
at
298
K
for the reaction,
H
2
(
g
)
+
1
2
O
2
(
g
)
⟶
H
2
O
(
g
)
Given that,
Δ
G
=
−
228.6
k
J
;
Δ
H
=
−
241.8
k
J
Q.
The temperature in K at which
Δ
G
=
0
, for a given reaction with
Δ
H
=
−
20.5
k
J
m
o
l
−
1
and
Δ
S
=
−
50.0
J
K
−
1
m
o
l
−
1
is:
Q.
For the reaction,
X
2
O
4
(
i
)
→
2
X
O
2
(
g
)
,
Δ
U
=
2.1
k
c
a
l
,
Δ
s
=
20
c
a
l
K
−
1
at 300 K.
Hence,
Δ
G
is:
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