Thermodynamics And Thermochemistry Result Question 3
3. Given, $\mathrm{C}_{\text {(graphite) }}+\mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \quad \Delta_r H^{\circ}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$ \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; $ $\quad \Delta_r H^{\circ}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} $
$ \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{CH}_4(g)+2 \mathrm{O}_2(g) ; $ $\quad \Delta_r H^{\circ}=+890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Based on the above thermochemical equations, the value of $\Delta_r H^{\circ}$ at 298 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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Answer:
Correct Answer: 3. ( c )
Solution:
- Based on given $\Delta_r H^{\circ}$
$ \Delta_f H^{\circ}=H_{\mathrm{CO}_2}^{\circ}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} $ …..(i)
$ \Delta_f H^{\circ}=H_{\mathrm{H}_2 \mathrm{O}}^{\circ}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} $ …..(ii)
$ \Delta_f H^{\circ}=H_{\mathrm{O}_2}^{\circ}=0.00 \text { (elements) }$ …..(iii)
Required thermal reaction is for $\Delta_f H^{\circ}$ of $\mathrm{CH}_4$
Thus, from III
$890.3=[\Delta_f H^{\circ}(\mathrm{CH}_4)+2 \Delta_f H^{\circ}(\mathrm{O}_2)] -[\Delta_f H^{\circ}(\mathrm{CO}_2)+2 \Delta_f H^{\circ}(\mathrm{H}_2 \mathrm{O})] =\Delta_f H^{\circ}(\mathrm{CH}_4)+0]-[-393.5-2 \times 285.5] $
$\therefore \quad \Delta_f H^{\circ}(\mathrm{CH}_4)=-74.8 \mathrm{~kJ} / \mathrm{mol}$
BETA
