Q6 / 170 · Manipal (MET) 2015 · Chemistry
The enthalpy of combustion of carbon and carbon monoxide are -393.5 and $-283 \mathrm{~kJ} / \mathrm{mol}$ respectively. The enthalpy of formation of carbon monoxide per mole is
- A110.5 kJ
- B676.5 kJ
- C$-676.5 \mathrm{~kJ}$
- Correct answerD$-110.5 \mathrm{~kJ}$
Solution
$\begin{aligned}
& \text { I: } \mathrm{C}(\mathrm{~s})+\mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2 ; \Delta H=-393.5 \mathrm{~kJ} / \mathrm{mol} \\
& \text { II: } \mathrm{CO}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \\
& \Delta H=-283.0 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}$
On substracting II from I III: $\mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})$; $\Delta H=-110.5 \mathrm{~kJ} / \mathrm{mol}$ This equation III also represents formation of 1 mole of CO and thus enthalpy of formation of $\mathrm{CO}(g)$ is $-110.5 \mathrm{~kJ} / \mathrm{mol}$
On substracting II from I III: $\mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})$; $\Delta H=-110.5 \mathrm{~kJ} / \mathrm{mol}$ This equation III also represents formation of 1 mole of CO and thus enthalpy of formation of $\mathrm{CO}(g)$ is $-110.5 \mathrm{~kJ} / \mathrm{mol}$
Hence, the correct answer is option: 'D'.