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Chemistry
Chemistry
Problem
Δ
H
\Delta H
Δ
H
for
CaCO
3
(
s
)
→
CaO
(
s
)
+
CO
2
(
g
)
\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})
CaCO
3
(
s
)
→
CaO
(
s
)
+
CO
2
(
g
)
, is 176
kJ mol
−
1
\text{kJ mol}^{-1}
kJ mol
−
1
at 1240 K. The
Δ
U
\Delta U
Δ
U
for the change is equal to : (A) 160 kJ (B) 165.6 kJ (C) 186.3 kJ (D) 180.0 kJ
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