Given that $C + O_{2} \to {CO}_{2} : {ΔH}^{\circ} = - xkJ$ and $2 CO + O_{2} \to 2 {CO}_{2}; {ΔH}^{\circ} = - ykJ$ The enthalpy of formation of carbon monoxide will be

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$\frac{2 x - y}{2}$

y=2x

2x-y

$\frac{y - 2 x}{2}$

answer is B.

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Detailed Solution

$C + \frac{1}{2} O_{2} \to C O$ .......(iii)
$C + O_{2} \to {CO}_{2}; {ΔH}^{\circ} = - xkJ$ ....(i)
${CO}_{2} \to CO + \frac{1}{2} O_{2}, {ΔH}^{\circ} = \frac{y}{2} kJ$ ....(ii)
(i)+(ii) = (iii)

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