Q10 / 180 · COMEDK 2014 · Chemistry

Standard molar enthalpies of formation of $\mathrm{CaCO}_{3}(s), \mathrm{CaO}(s)$ and $\mathrm{CO}_{2}(g)$ are $-1206.92 \mathrm{~kJ} \mathrm{~mol}^{-1},-635.09 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-393.51 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The $\Delta_{r} H$, for decomposition of $\mathrm{CaCO}_{3}(s)$ is

  • Correct answer
    A
    $178.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \quad$
  • B
    $-178.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • C
    $1448.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \quad$
  • D
    $-1448.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

$\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$ $$ \begin{aligned} \Delta_{r} H^{\circ} &=\Delta_{f} H^{\circ}(\mathrm{CaO})+\Delta_{f} H^{\circ}\left(\mathrm{CO}_{2}\right)-\Delta_{f} H^{\circ}\left(\mathrm{CaCO}_{3}\right) \\ &=-635.09+(-393.51)-(-1206.92) \\ &=178.3 \mathrm{~kJ} \mathrm{~mol}^{-1} . \end{aligned} $$

Hence, the correct answer is option: 'A'.

Relevant Questions