Mark out the enthalpy of formation of carbon monoxide ( CO ) . Given, C s + O 2 g → CO 2 g ,  …

To find:$\begin{array}{ccc}\mathrm{C}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right) \to \mathrm{CO}\left(\mathrm{g}\right)=0& \mathrm{;} & \Delta {\mathrm{H}}_{\mathrm{f}}^{\circ }=??\end{array} \left(\mathrm{I}\right)$

Given:

$$\begin{gathered} \begin{array}{ccc}\mathrm{C}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)& \mathrm{;}& {\left(\Delta {\mathrm{H}}_{\mathrm{f}}^{\circ }\right)}_1=-393.3 \mathrm{kJ}/\mathrm{mol}\end{array} \left(\mathrm{II}\right) \\ \mathrm{CO}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right) ; {\left(\Delta {\mathrm{H}}_{\mathrm{f}}^{\circ }\right)}_2=-282.2 \mathrm{kJ}/\mathrm{mol} \left(\mathrm{III}\right) \end{gathered}$$

Equation (I) can be founded by subtracting equation (III) from (II),

So,  $$\begin{gathered} \Delta {\mathrm{H}}_{\mathrm{f}}^{\circ } = {\left(\Delta {\mathrm{H}}_{\mathrm{f}}^{\circ }\right)}_1 - {\left(\Delta {\mathrm{H}}_{\mathrm{f}}^{\circ }\right)}_2 \\ \Delta {\mathrm{H}}_{\mathrm{f}}^{\circ }=\left(-393.3 \right)- \left(-282.2\right) \\ \Delta {\mathrm{H}}_{\mathrm{f}}^{\circ } = -111.1 \mathrm{KJ}/{\mathrm{mol}}^{-1} \end{gathered}$$