Question
Following data is given for the reaction: $\text{CaCO}_3\text{(s)}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }\text{CaO(s)}+\text{CO}_2\text{(g)}$
$\Delta_\text{f}\text{H}^\ominus[\text{CaO}\text{(s)}]=-635.1\text{KJ mol}^{-1}$
$\Delta_\text{f}\text{H}^\ominus[\text{CO}_2\text{(g)}]=-393.5\text{KJ mol}^{-1}$
$\Delta_\text{f}\text{H}^\ominus[\text{CaCO}_3\text{(s)}]=-1206.9\text{KJ mol}^{-1}$
Predict the effect of temperature on the equilibrium constant of the above reaction.

Answer

$\Delta_\text{f}\text{H}^\circ=\Delta_\text{r}\text{H}^\circ[\text{CaO}]+\Delta_\text{f}\text{H}^\circ[\text{CO}_2]-\Delta_\text{f}\text{H}^\circ[\text{CaCO}_3]$
$=[-635.1]+[-393.5]-[-1206.9]=178.3\text{KJ mol}^{-1}$
Thus, the reaction is endothermic.
Hence, according to Le Chatelier’s principle, on increasing the temperature, the equilibrium will proceed in the forward direction.

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