See the calculation on this YouTube video

Solution We can imagine that the reaction takes place in two steps, each of which involves only a standard enthalpy of formation. In the first step CO (carbon monoxide) is decomposed to its elements:

2CO(g) ? 2C(s) + O₂(g)      ?H_m = ?H₁      (2)

Since this is the reverse of formation of 2 mol CO from its elements, the enthalpy change is

?H₁ = 2 × {–?H_f [CO(g)]} = 2 × [– (–110.5 kJ mol^–1)] = +221.0 kJ mol^–1

In the second step the elements are combined to give 2 mol CO₂(carbon dioxide):

2C(s) + 2O₂(g) ? 2CO₂(g)      ?H_m = ?H₂      (3)

In this case

?H₂ = 2 × ?H_f [CO₂(g)] = 2 × (– 393.5 kJ mol^–1) = – 787.0 kJ mol^–1

You can easily verify that the sum of Eqs. (2) and (3) is

2CO(g) + 2 O₂(g) ? 2CO₂(g)      ?H_m = ?H_net

Therefore

?H_net = ?H₁ + ?H₂ = 221.0 kJ mol^–1 – 787.0 kJ mol^–1 = – 566.0 kJ mol^–1