The enthalpy changes for the following reacti | vedclass

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(B) The formation reaction for $ICl_{(g)}$ is: $\frac{1}{2} I_{2(s)} + \frac{1}{2} Cl_{2(g)} \rightarrow ICl_{(g)}$Step $1$: Sublimation of $I_{2(s)}$

Vedclass · Accepted Answer

$+16.8$

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(B) The formation reaction for $ICl_{(g)}$ is: $\frac{1}{2} I_{2(s)} + \frac{1}{2} Cl_{2(g)} ightarrow ICl_{(g)}$Step $1$: Sublimation of $I_{2(s)}$: $\frac{1}{2} I_{2(s)} ightarrow \frac{1}{2} I_{2(g)}$,$\Delta H_1 = \frac{1}{2} 	imes 62.76 = 31.38 \, kJ \, mol^{-1}$Step $2$: Dissociation of $I_{2(g)}$: $\frac{1}{2} I_{2(g)} ightarrow I_{(g)}$,$\Delta H_2 = \frac{1}{2} 	imes 151.0 = 75.5 \, kJ \, mol^{-1}$Step $3$: Dissociation of $Cl_{2(g)}$: $\frac{1}{2} Cl_{2(g)} ightarrow Cl_{(g)}$,$\Delta H_3 = \frac{1}{2} 	imes 242.3 = 121.15 \, kJ \, mol^{-1}$Step $4$: Formation of $ICl_{(g)}$ from atoms: $I_{(g)} + Cl_{(g)} ightarrow ICl_{(g)}$,$\Delta H_4 = -211.3 \, kJ \, mol^{-1}$Total enthalpy of formation $\Delta H_f = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 = 31.38 + 75.5 + 121.15 - 211.3 = 16.73 \, kJ \, mol^{-1}$.

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