Chemistry - Thermochemistry Question with Solution | TestHub

The given reaction is as follows, $\frac{1}{2}{\mathrm{I}}_2(g)+\frac{1}{2}{\mathrm{Cl}}_2(g)⟶\mathrm{ICl}(g)$ We are given, (i) ${\mathrm{Cl}}_2(g)⟶2\mathrm{Cl}(\mathrm{g}),\Delta \mathrm{H}=242.3\mathrm{kJ}$ (ii) ${\mathrm{I}}_2(\mathrm{g})⟶2\mathrm{I}(\mathrm{g}),\Delta H=151\mathrm{kJ}$ (iii) $\mathrm{ICl}(\mathrm{g})⟶\mathrm{I}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}),\Delta H=211.3\mathrm{kJ}$ (iv) ${\mathrm{I}}_2(g)⟶{\mathrm{I}}_2(g),\Delta H=62.8\mathrm{kJ}$ $\frac{1}{2}$ (ii) $+\frac{1}{2}$ (i) - (iii) $+\frac{1}{2}$ (iv) gives the required equation. $\begin{array}{rl}\therefore \Delta H& =\frac{1}{2}(151)+\frac{1}{2}(242.3)-211.3+\frac{1}{2}(62.8)\\ \Delta H& =16.8\mathrm{kJ}\end{array}$