Mark out the enthalpy of formation of carbon monoxide (CO).
Given,
\(C (s)+ O _2(g) \rightarrow CO _2(g), \Delta H =\)\(-393.3 kJ / mol\) and \(CO _{( g )}+\)\(\frac{1}{2} O _{2(g)} \rightarrow CO _{2(g)} ; \Delta H\)
\(=-282.2 kJ / mol\).

To find: \(C ( s )+\frac{1}{2} O _2(g) \rightarrow CO ( g )=0 ; \Delta H _{ f }^{\circ}=? ?( I )\)
Given:
\(C ( s )+ O _2(g) \rightarrow CO _2(g) \quad ; \quad\left(\Delta H _{ f }^{\circ}\right)_1\)\(=-393.3 kJ / mol ( II )\)
\(CO ( s )+\frac{1}{2} O _2(g) \rightarrow CO _2(g) ;\left(\Delta H _{ f }^{\circ}\right)_2\)\(=-282.2 kJ / mol ( III )\)
Equation (I) can be founded by subtracting equation (III) from (II),So,
\(\begin{aligned} \Delta H _{ f }^{\circ} & =\left(\Delta H _{ f }^{\circ}\right)_1-\left(\Delta H _{ f }^{\circ}\right)_2 \\ \Delta H _{ f }^{\circ} & =(-393.3)-(-282.2) \\ \Delta H _{ f }^{\circ} & =-111.1 KJ / mol ^{-1}\end{aligned}\)