Given

(A)	$\mathrm{2CO(g)+O_2(g)\to 2CO_2(g)}$		$\mathrm{\Delta H_1^0=-x~kJ~mol^{-1}}$
(B)	$\mathrm{C(graphite)+O_2(g)\to CO_2(g)}$		$\mathrm{\Delta H_2^0=-y~kJ~mol^{-1}}$

$\mathrm{\Delta H^0}$ for the reaction

$\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}$ is :

<p>The standard enthalpy change for the reaction</p> <p>$\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}$</p> <p>can be calculated using Hess&#39;s Law, which states that the total enthalpy change for a reaction is the same whether it occurs in one step or in many steps. </p> <p>If we modify the given reactions to reflect the formation of CO, we can add the modified reactions together to get the reaction of interest. </p> <p>Consider the reactions:</p> <p>(A) $\mathrm{2CO(g)+O_2(g)\to 2CO_2(g)} \quad \Delta H_1^0=-x~kJ~mol^{-1}$</p> <p>(B) $\mathrm{C(graphite)+O_2(g)\to CO_2(g)} \quad \Delta H_2^0=-y~kJ~mol^{-1}$</p> <p>First, divide reaction (A) by 2 to get the reaction for 1 mol of CO:</p> <p>(A/2) $$\mathrm{CO(g)+\frac{1}{2}O_2(g)\to CO_2(g)} \quad \Delta H_1^0=-\frac{x}{2}~kJ~mol^{-1}$$</p> <p>Then, subtract the divided reaction (A) from reaction (B) to get the reaction for the formation of CO:</p> <p>$\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}$<br/><br/>$$\Delta H^0=\Delta H_2^0-\Delta H_1^0=-y-(-\frac{x}{2})= \frac{x}{2} - y~kJ~mol^{-1}$$</p> <p>Therefore, the correct answer is Option A: $\frac{x-2y}{2}$</p>