Given that $\mathrm{C}+{\mathrm{O}}_2\to {\mathrm{CO}}_2;{\mathrm{\Delta H}}^0=$ $-\mathrm{xkJ}.2\mathrm{CO}+{\mathrm{O}}_2\to 2{\mathrm{CO}}_2;{\mathrm{\Delta H}}^0$ $=-\mathrm{ykJ}$. The enthalpy of formation of carbon monoxide will be (CBSE)

$\mathrm{C}+{\mathrm{O}}_2\to {\mathrm{CO}}_2;\mathrm{\Delta H}=-\mathrm{xkJ}$

$2\mathrm{CO}+{\mathrm{O}}_2\to 2{\mathrm{CO}}_2;{\mathrm{\Delta H}}_4=-\mathrm{ykJ}$

Required equation is

$\mathrm{C}+\frac{1}{2}{\mathrm{O}}_2\to \mathrm{CO},\Delta {\mathrm{H}}_1=\text{ ? }$

On reversing equation (ii), we get

$2{\mathrm{CO}}_2\to 2\mathrm{CO}+{\mathrm{O}}_2;\Delta {\mathrm{H}}_2=+\mathrm{ykJ}$

Dividing equation (iii) by 2 gives

${\mathrm{CO}}_2\to \mathrm{CO}+\frac{1}{2}{\mathrm{O}}_2;\Delta {\mathrm{H}}_3=+\frac{\mathrm{y}}{2} \mathrm{kJ}$

On adding equation (i) and (iv), we get (required equation)

$\mathrm{C}+\frac{1}{2}{\mathrm{O}}_2\to \mathrm{CO};\Delta {\mathrm{H}}_1=\Delta \mathrm{H}+\Delta {\mathrm{H}}_3$

$=\left(-\mathrm{x}+\frac{\mathrm{y}}{2}\right)\mathrm{kJ}$

$= \frac{\mathrm{y}-2\mathrm{x}}{2}\mathrm{KJ}$

Hence the correct answer is (B) $\frac{y-2x}{2}$.