Tardigrade
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Tardigrade
Question
Chemistry
The entropy (So) of the following substances are: CH4 (g) 186.2 J K-1 mol-1 O2 (g) 205.0 J K-1 mol-1 CO2 (g) 213.6 J K-1 mol-1 H2O (g) 69.9 J K-1 mol-1 The entropy change (Δ So) for the reaction CH4(g) + 2O2(g) arrow CO2(g) + 2H2O(l) is :
Q. The entropy (S
o
) of the following substances are :
C
H
4
(
g
)
186.2
J
K
−
1
m
o
l
−
1
O
2
(
g
)
205.0
J
K
−
1
m
o
l
−
1
C
O
2
(
g
)
213.6
J
K
−
1
m
o
l
−
1
H
2
O
(
g
)
69.9
J
K
−
1
m
o
l
−
1
The entropy change
(
Δ
S
o
)
for the reaction
C
H
4
(
g
)
+
2
O
2
(
g
)
→
C
O
2
(
g
)
+
2
H
2
O
(
l
)
is :
5410
226
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JEE Main 2014
Thermodynamics
Report Error
A
−
312.5
J
K
−
1
m
o
l
−
1
11%
B
−
242.8
J
K
−
1
m
o
l
−
1
67%
C
−
108.1
J
K
−
1
m
o
l
−
1
14%
D
−
37.6
J
K
−
1
m
o
l
−
1
8%
Solution:
186.2
C
H
4
(
g
)
+
205
2
O
2
(
g
)
→
213.6
C
O
2
(
g
)
+
69.9
2
H
2
O
(
ℓ
)
Δ
S
∘
=
∑
(
s
P
∘
−
∑
(
s
∘
)
R
=
[
213.6
+
(
2
×
69.9
)]
−
[
186.2
+
2
×
205
]
=
353.4
−
596.2
=
−
242.8
J
/
m
o
l
K