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Tardigrade
Question
Chemistry
Δ H for CaCO3(s) → CaO(s) + CO2(g) is 176 kJ mol-1 at 1240 K. The Δ E for the change is equal to
Q.
Δ
H
for
C
a
C
O
3
(
s
)
→
C
a
O
(
s
)
+
C
O
2
(
g
)
is
176
k
J
m
o
l
−
1
at
1240
K
. The
Δ
E
for the change is equal to
2615
220
Thermodynamics
Report Error
A
160
k
J
9%
B
165.6
k
J
64%
C
186.3
k
J
18%
D
180.0
k
J
9%
Solution:
Δ
H
=
Δ
E
+
Δ
n
RT
∴
Δ
E
=
176
−
1
×
8.314
×
8.314
×
1240
×
1
0
−
3
=
165.6
k
J