The enthalpy of the partial oxidation reaction $\mathrm{C}_{( ext{graphite})} + \frac{1}{2}\mathrm{

Question

The enthalpy of the partial oxidation reaction $\mathrm{C}_{(	ext{graphite})} + \frac{1}{2}\mathrm{O}_{2(g)} ightarrow \mathrm{CO}_{(g)}$ is strictly negative. If the absolute magnitudes of the enthalpy of complete combustion for Graphite and Carbon Monoxide are respectively denoted as $|X|$ and $|Y|$, then

Professor PC Thomas Classes · Answer

(a) : Let us construct the relevant thermochemical equations. (i) Complete combustion of Graphite: $\mathrm{C}_{(s)} + \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}$ with $\Delta H_1 = -|X|$. (ii) Complete combustion of CO: $\mathrm{CO}_{(g)} + \frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}$ with $\Delta H_2 = -|Y|$. According to Hess's Law, subtracting eq (ii) from eq (i) yields the target reaction: $\mathrm{C}_{(s)} + \frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}$, where $\Delta H_3 = \Delta H_1 - \Delta H_2 = -|X| - (-|Y|) = -|X| + |Y|$. The problem statement confirms that this target reaction is exothermic ($\Delta H_3 < 0$). Therefore, $-|X| + |Y| < 0$, which unequivocally means $|Y| < |X|$, or $|X| > |Y|$.