Question

The enthalpy changes for the following processes are listed below $ C{{l}_{2}}(g)2Cl(g),242.3kJmo{{l}^{-1}} $ $ {{I}_{2}}(g)2I(g),151.0\,kJmo{{l}^{-1}} $ $ ICl(g)I(g)+Cl(g),211.3\,kJmo{{l}^{-1}} $ $ {{I}_{2}}(s){{I}_{2}}(g),62.76\,kJmo{{l}^{-1}} $ Given that the standard states for iodine and chlorine are $ {{I}_{2}}(s) $ and $ C{{l}_{2}}(g) $ the standard enthalpy of formation of $ ICI(g) $ is

• A. $ -14.6kJmo{{l}^{-1}} $
• B. $ -16.8\text{ }kJ\text{ }mo{{l}^{-1}} $
• C. $ +\text{ }16.8\text{ }kJ\text{ }mo{{l}^{-1}} $
• D. $ +\text{ }244.8\text{ }kJ\text{ }mo{{l}^{-1}} $

Answer 1

Answer: (C)

$ \frac{1}{2}{{I}_{2}}(s)+\frac{1}{2}C{{l}_{2}}(g)ICl(g) $
$ \Delta H=\left[ \frac{1}{2}\Delta {{H}_{s\to g}}{{I}_{2}}+\frac{1}{2}\Delta {{H}_{diss}}C{{l}_{2}}+\frac{1}{2}\Delta {{H}_{diss}}{{I}_{2}} \right] $
$ -\Delta {{H}_{ICl}} $
$ =\left[ \frac{1}{2}\times 62.76+\frac{1}{2}\times 242.3+\frac{1}{2}\times 151.0 \right]-211.3 $
$ =228.03-211.3 $
$ \Delta H=16.73\text{ }kJ\,mo{{l}^{-1}} $