Using the following data, calculate the standard heat of formation of ICl \((g)\) in \(\mathrm{kJ} / \mathrm{mol} :\) $$\begin{array}{ll}{\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)} & {\Delta H^{\circ}=242.3 \mathrm{kJ}} \\ {\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{I}(g)} & {\Delta H^{\circ}=151.0 \mathrm{kJ}} \\ {\mathrm{ICl}(g) \longrightarrow \mathrm{I}(g)+\mathrm{Cl}(g)} & {\Delta H^{\circ}=211.3 \mathrm{kJ}} \\ {\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)} & {\Delta H^{\circ}=62.8 \mathrm{kJ}}\end{array}$$

We want to find the standard heat of formation for ICl(g), which means we are looking for the heat change when ICl is formed from its elements in their standard states. The desired reaction is: \[ \frac{1}{2}\mathrm{I}_{2}(s) + \frac{1}{2}\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ICl}(g) \]

From the given reactions, we need to rearrange and manipulate the equations to achieve the desired reaction. 1. Multiply the first reaction by \(\frac{1}{2}\) to have one Cl(g) produced: \[\frac{1}{2}\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Cl}(g); \quad \Delta H^{\circ} = \frac{1}{2}\cdot 242.3\,\mathrm{kJ} = 121.15\,\mathrm{kJ}\] 2. Multiply the second reaction by \(\frac{1}{2}\) to have one I(g) produced: \[\frac{1}{2}\mathrm{I}_{2}(g) \longrightarrow \mathrm{I}(g); \quad \Delta H^{\circ} = \frac{1}{2}\cdot 151.0\,\mathrm{kJ} = 75.5\,\mathrm{kJ}\] 3. Reverse the third given reaction: \[ \mathrm{I}(g) + \mathrm{Cl}(g) \longrightarrow \mathrm{ICl}(g); \quad \Delta H^{\circ} = -211.3\,\mathrm{kJ}\]

Now, we will sum the manipulated reactions as well as their heats to find the heat of formation of ICl(g): Reaction 1: \[\frac{1}{2}\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Cl}(g); \quad \Delta H^{\circ} = 121.15\,\mathrm{kJ}\] Sublimation reaction: \[\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g); \quad \Delta H^{\circ} = 62.8\,\mathrm{kJ} \] Reaction 2: \[\frac{1}{2}\mathrm{I}_{2}(g) \longrightarrow \mathrm{I}(g); \quad \Delta H^{\circ} = 75.5\,\mathrm{kJ}\] Reaction 3 (reversed): \[ \mathrm{I}(g) + \mathrm{Cl}(g) \longrightarrow \mathrm{ICl}(g); \quad \Delta H^{\circ} = -211.3\,\mathrm{kJ}\] Summing these reactions, we get the desired reaction: \[\frac{1}{2}\mathrm{I}_{2}(s) + \frac{1}{2}\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ICl}(g)\]

Summing the corresponding heats of reactions to calculate the standard heat of formation of ICl(g): \[\Delta H_{f}^{\circ}(\mathrm{ICl}) = 121.15\,\mathrm{kJ} + 62.8\,\mathrm{kJ} + 75.5\,\mathrm{kJ} - 211.3\,\mathrm{kJ} = 48.15\,\mathrm{kJ/mol}\] The standard heat of formation of ICl(g) is 48.15 kJ/mol.