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Byju's Answer
Standard XII
Chemistry
Second Law of Thermodynamics
Δ Ssurroundin...
Question
Δ
S
s
u
r
r
o
u
n
d
i
n
g
s
=
+
959.1
J
K
−
1
m
o
l
−
1
and
Δ
S
s
y
s
t
e
m
=
163.1
J
k
−
1
m
o
l
−
1
Then the
process is
A
Spontaneous
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B
Non spontaneous
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C
At equilibrium
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D
Cannot be predicted from the information
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Solution
The correct option is
A
Spontaneous
Δ
S
t
o
t
a
l
=
Δ
S
s
u
r
r
o
u
n
d
i
n
g
s
+
Δ
S
s
y
s
t
e
m
=
959.1 + 163.1
=
1122.2 J
Δ
S
t
o
t
a
l
>
O
.
Therefore the reaction is spontaneous
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0
Similar questions
Q.
For the process,
H
2
O
(
l
)
→
H
2
O
(
g
)
a
t
T
=
100
∘
C
and 1 atmosphere pressure, the correct choice
Q.
Entropy changes for the process,
H
2
O
(
l
)
→
H
2
O
(
s
)
at normal pressure and 274 K are given below.
Δ
S
s
y
s
t
e
m
=
−
22.13
,
Δ
S
s
u
r
r
o
u
n
d
i
n
g
=
+
22.05
J
/
m
o
l
the process is non-spontaneous because :-
Q.
Enthalpy change for the process,
H
2
O
(
s
)
⇌
H
2
O
(
l
)
is
6.01
k
J
m
o
l
−
1
.
The entropy change for 1 mole of ice at its melting point will be
Q.
At
0
0
C
, ice and water are in equilibrium and
Δ
H
0
=
6.00
k
J
/
m
o
l
for the process
H
2
O
(
s
)
⇋
H
2
O
(
l
)
. The value of
Δ
S
0
for the conversion of ice to liquid water is:
Q.
Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statements are:
(i)
Δ
G
is positive
(ii)
Δ
S
s
y
s
t
e
m
is positive
(iii)
Δ
S
s
u
r
r
o
u
n
d
i
n
g
s
=
0
(iv)
Δ
H
=
0
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