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 target reaction is: \(CO(g) + 2 H_{2}(g) \longrightarrow CH_{3}OH(g)\)

We are given two reactions: Reaction 1: \(2 C(s) + O_{2}(g) \longrightarrow 2 CO(g) \qquad \Delta H_1 = -221.0 kJ\) Reaction 2: \(2 C(s) + O_{2}(g) + 4 H_{2}(g) \longrightarrow 2 CH_{3}OH(g) \qquad \Delta H_2 = -402.4 kJ\)

To obtain the target reaction from the given reactions, we can perform the following operations: 1. Divide Reaction 1 by 2. 2. Reverse the modified Reaction 1. 3. Subtract the modified Reaction 1 from Reaction 2. Reaction 1 (modified): \(\frac{1}{2}(2 C(s) + O_{2}(g) \longrightarrow 2 CO(g))\) \(C(s) + \frac{1}{2}O_{2}(g) \longrightarrow CO(g) \qquad \frac{\Delta H_1}{2} = -110.5 kJ\) Reverse Reaction 1 (modified): \(CO(g) \longrightarrow C(s) + \frac{1}{2}O_{2}(g) \qquad \Delta H'_1 = 110.5 kJ\) Finally, \(Reaction\, 2 - Reverse\, Reaction\, 1 (modified)\) \(2 C(s) + O_{2}(g) + 4 H_{2}(g) - [CO(g) - (C(s) + \frac{1}{2}O_{2}(g))] \longrightarrow 2 CH_{3}OH(g) - C(s) - \frac{1}{2}O_{2}(g)\) \(CO(g) + 2 H_{2}(g) \longrightarrow CH_{3}OH(g)\)