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) ?Hm = ?H1 (2)
Since this is the reverse of formation of 2 mol CO from its elements, the enthalpy change is
?H1 = 2 × {–?Hf [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) ?Hm = ?H2 (3)
In this case
?H2 = 2 × ?Hf [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) ?Hm = ?Hnet
Therefore
?Hnet = ?H1 + ?H2 = 221.0 kJ mol–1 – 787.0 kJ mol–1 = – 566.0 kJ mol–1
