You are given the following data: $$\begin{aligned} \mathrm{H}_{2}(g) & \longrightarrow 2 \mathrm{H}(g) & & \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\\\\mathrm{Br}_{2}(g) & \longrightarrow 2 \mathrm{Br}(g) & & \Delta H^{\circ}=192.5 \mathrm{~kJ} /\mathrm{mol} \\\\\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) & \longrightarrow 2 \mathrm{HBr}(g) & & \Delta H^{\circ}=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$Calculate \(\Delta H^{\circ}\) for the reaction$$\mathrm{H}(g)+\operatorname{Br}(g) \longrightarrow \operatorname{HBr}(g)$$

First, you need to observe the reactions carefully and determine how you can combine or modify them to yield the desired reaction. In this case, the target is to form the equation \( \mathrm{H}(g) + \mathrm{Br}(g) \longrightarrow \mathrm{HBr}(g) \).

Next, you'll notice that the reaction \( \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) \) needs to be divided by 2 to give \( \mathrm{H}_{2}(g)/2 \longrightarrow \mathrm{H}(g) \) with enthalpy \( \Delta H^{\circ}=218.2 \mathrm{~kJ} / \mathrm{mol} \). Similarly, the reaction \( \mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{Br}(g) \) should be divided by 2 to get \( \mathrm{Br}_{2}(g)/2 \longrightarrow \mathrm{Br}(g) \) with enthalpy \( \Delta H^{\circ}=96.25 \mathrm{~kJ} / \mathrm{mol} \). The reaction \( \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g) \) should be divided by 2 to get \( \mathrm{H}_{2}(g)/2+\mathrm{Br}_{2}(g)/2 \longrightarrow \mathrm{HBr}(g) \) with enthalpy \( \Delta H^{\circ}=-36.2 \mathrm{~kJ} / \mathrm{mol} \).

The last step is to subtract the third reaction from the sum of the first two reactions. Algebraically, we calculate \( (218.2 kJ/mol + 96.25 kJ/mol) - (-36.2 kJ/mol) \) to yield the ∆H for the desired reaction.