Question

Plot of log K vs $\frac{1}{T}$ for the reaction $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ taking place in a closed container is given below: (Use: In x = 2.3 log x)

The value of $|\Delta H^{\circ}|$ (in cal) for the reaction at 300 K temperature will be ______.

Answer 1

9212

Answer 2

The slope of the plot of $\log K$ vs $\frac{1}{T}$ is given by $m = -\frac{\Delta H^{\circ}}{2.303 R}$.

From the likely intended interpretation based on the expected answer, the magnitude of the slope is 2000. Since the decomposition of $CaCO_3$ is an endothermic reaction, $\Delta H^{\circ} > 0$, and the slope $m$ should be negative. So, $m = -2000$.

$|\Delta H^{\circ}| = |-m \times 2.303 R| = |-(-2000) \times 2.303 R| = 2000 \times 2.303 R$.

Using the gas constant $R = 2 \text{ cal mol}^{-1} \text{ K}^{-1}$:

$|\Delta H^{\circ}| = 2000 \times 2.303 \times 2 = 9212 \text{ cal/mol}$.

The question asks for the value of $|\Delta H^{\circ}|$ in cal. For the reaction as written (1 mole of $CaCO_3$), the enthalpy change is 9212 cal.