The kinetics and equilibrium of the decomposition of hydrogen iodide have been studied extensively: $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ (a) At \(298 \mathrm{~K}, K_{\mathrm{c}}=1.26 \times 10^{-3}\) for this reaction. Calculate \(K_{\mathrm{p}^{*}}\) (b) Calculate \(K_{\mathrm{c}}\) for the formation of \(\mathrm{HI}\) at \(298 \mathrm{~K}\). (c) Calculate \(\Delta H_{\mathrm{ran}}^{\circ}\) for \(\mathrm{HI}\) decomposition from \(\Delta H_{\mathrm{f}}^{\circ}\) values. (d) At \(729 \mathrm{~K}, K_{\mathrm{c}}=2.0 \times 10^{-2}\) for HI decomposition. Calculate \(\Delta H_{\mathrm{rxn}}\) for this reaction from the van't Hoff equation.

In chemical reactions, the equilibrium constant (\text{Kc}) is a value that expresses the ratio of concentrations of products to reactants at equilibrium. For the decomposition of hydrogen iodide given by: \[2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)\] at 298 K, the equilibrium constant \[K_c = 1.26 \times 10^{-3}.\] This low \text{Kc} indicates that at equilibrium, the concentration of products (H\(_2\) and I\(_2\)) remains low relative to the reactants (HI). \ To find \text{Kc} for the reverse formation of HI, we simply take the reciprocal of the original \text{Kc}: \[K'_c = \frac{1}{K_c} = \frac{1}{1.26 \times 10^{-3}} = 793.65\] This shows that HI forms readily from H\(_2\) and I\(_2\) under the given conditions.

The enthalpy change (\text{ΔH}) of a reaction indicates the total heat content change when reactants convert into products. In our context, it's the heat change associated with the decomposition of hydrogen iodide. Given the standard enthalpies of formation (\text{ΔH_f^\text{°}}) of involved substances: \ \ \[ \text{ΔH_f^\text{°}}(\text{HI}) = -26.5 \text{kJ/mol}, \ \text{ΔH_f^\text{°}}(\text{H}_2) = 0 \text{kJ/mol}, \ \text{ΔH_f^\text{°}}(\text{I}_2) = 0 \text{kJ/mol}, \] we calculate the \text{ΔH_{\text{rxn}}^\text{°}} (standard reaction enthalpy) for the decomposition of 2 moles of HI: \[ \text{ΔH_{\text{rxn}}^\text{°}} = 2 \times \text{ΔH_f^\text{°}}(\text{HI}) - \left( \text{ΔH_f^\text{°}}(\text{H}_2) + \text{ΔH_f^\text{°}}(\text{I}_2) \right) = 2 \times (-26.5) - (0 + 0) = -53 \text{kJ/mol} \] This indicates the reaction is exothermic, releasing 53 kJ of energy per mole of HI decomposed.

The van't Hoff equation provides a way to understand how the equilibrium constant of a reaction changes with temperature. The equation is given by: \[ \text{ln}\left( \frac{K_{c1}}{K_{c2}} \right) = \frac{\Delta H_{\text{rxn}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] where \text{K_{c1}} and \text{K_{c2}} are equilibrium constants at temperatures \text{T_1} and \text{T_2} respectively, \text{R} is the universal gas constant, and \text{ΔH_{\text{rxn}}} is the reaction enthalpy. \ For the decomposition of HI, given \text{K_{c1} = 1.26 \times 10^{-3}} at \text{T_1 = 298 K} and \text{K_{c2} = 2.0 \times 10^{-2}} at \text{T_2 = 729 K}, substitute the values into the van't Hoff equation to solve for \text{ΔH_{\text{rxn}}}: \[ \Delta H_{\text{rxn}} = R \times \frac{\ln\left( \frac{2.0 \times 10^{-2}}{1.26 \times 10^{-3}} \right)}{\left( \frac{1}{729} - \frac{1}{298} \right)}\] Using \text{R = 8.314 \text{J/mol K}}, the reaction enthalpy \text{ΔH_{\text{rxn}}} is calculated. This helps understand how heat transfer influences chemical equilibrium.