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Chemistry
Chemistry
Problem
Calculate the temperature at which the decomposition of calcium carbonate becomes spontaneous.
CaCO
3
(
s
)
→
CaO
(
s
)
+
CO
2
(
g
)
Δ
H
⊖
=
+
178
kJ mol
−
1
Δ
S
⊖
=
+
164
J K
−
1
mol
−
1
\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g}) \quad \Delta H^\ominus = +178 \text{ kJ mol}^{-1} \quad \Delta S^\ominus = +164 \text{ J K}^{-1} \text{ mol}^{-1}
CaCO
3
(
s
)
→
CaO
(
s
)
+
CO
2
(
g
)
Δ
H
⊖
=
+
178
kJ mol
−
1
Δ
S
⊖
=
+
164
J K
−
1
mol
−
1
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