From the enthalpies of reaction $$ \begin{aligned} 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}(g) & \Delta H=-221.0 \mathrm{~kJ} \\ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g)+4 \mathrm{H}_{2}(g) & \longrightarrow & 2 \mathrm{CH}_{3} \mathrm{OH}(g) & \Delta H=-402.4 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$

Our desired reaction is: $$ \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ We need to manipulate the given reactions to arrive at this final desired reaction.

First, we notice that the first given reaction has 2 CO(g) being formed on the right-hand side, while in the desired reaction we have only 1 CO(g). So, we will divide the first given reaction by 2: $$ \mathrm{C}(s) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) \qquad \Delta H_{1} = -110.5\ \mathrm{kJ} $$ Now, the second given reaction has 2 CH3OH(g) in the products, but we only need 1 CH3OH(g) for the desired reaction. Divide the second given reaction by 2: $$ \mathrm{C}(s) + \frac{1}{2} \mathrm{O}_{2}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) \qquad \Delta H_{2} = -201.2\ \mathrm{kJ} $$ At this point, we can see that the sum of these two modified reactions gives us the desired reaction. Accordingly, we will subtract the first modified reaction from the second modified reaction to obtain the desired reaction.

Now we will apply Hess's Law: $$ (\mathrm{C}(s) + \frac{1}{2} \mathrm{O}_{2}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)) - (\mathrm{C}(s) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g)) $$ The Carbon and Oxygen in the reactants cancel out, giving us the desired reaction: $$ \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ The enthalpy change of the desired reaction is the difference between the enthalpies of the two modified reactions: $$ \Delta H_{\text{desired}} = \Delta H_{2} - \Delta H_{1} = (-201.2\ \mathrm{kJ}) - (-110.5\ \mathrm{kJ}) = -90.7\ \mathrm{kJ} $$

The enthalpy change for the desired reaction is -90.7 kJ: $$ \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g); \qquad \Delta H = -90.7\ \mathrm{kJ} $$