Thermodynamics and Thermochemistry 2 Question 3
3. Given, $\mathrm{C}{\text {(graphite) }}+\mathrm{O}{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})$;
$$ \begin{aligned} & \Delta_{r} H^{\circ}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \ & \mathrm{H}{2}(g)+\frac{1}{2} \mathrm{O}{2}(g) \longrightarrow \mathrm{H}{2} \mathrm{O}(l) ; \ & \Delta{r} H^{\circ}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$
Based on the above thermochemical equations, the value of $\Delta_{r} H^{\circ}$ at $298 \mathrm{~K}$ for the reaction,
(2017 Main) $\mathrm{C}{\text {(graphite) }}+2 \mathrm{H}{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{~g})$ will be
(a) $+78.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(b) $+144.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(c) $-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(d) $-144.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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Solution:
- Based on given $\Delta_{r} H^{\circ}$
$\Delta_{f} H^{\circ}=H_{\mathrm{CO}_{2}}^{\circ}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\Delta_{f} H^{\circ}=H_{\mathrm{H}_{2} \mathrm{O}}^{\circ}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\Delta_{f} H^{\circ}=H_{\mathrm{O}_{2}}^{\circ}=0.00$ (elements)
Required thermal reaction is for $\Delta_{f} H^{\circ}$ of $\mathrm{CH}_{4}$
Thus, from III
$890.3=\left[\Delta_{f} H^{\circ}\left(\mathrm{CH}{4}\right)+2 \Delta{f} H^{\circ}\left(\mathrm{O}_{2}\right)\right]$
$$ -\left[\Delta_{f} H^{\circ}\left(\mathrm{CO}{2}\right)+2 \Delta{f} H^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right)\right] $$
$\left.=\Delta_{f} H^{\circ}\left(\mathrm{CH}_{4}\right)+0\right]-[-393.5-2 \times 285.5]$
$\therefore \quad \Delta_{f} H^{\circ}\left(\mathrm{CH}_{4}\right)=-74.8 \mathrm{~kJ} / \mathrm{mol}$
