Question

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.

Short answer

a) \(K_p = 1.26 \times 10^{-3}\) b) \(K'_c = 793.65\) c) \(\Delta H_{\text{rxn}}^{\circ} = -53 \text{kJ/mol}\) d) Use van't Hoff equation to find \(\Delta H_{\text{rxn}}\).

Step-by-step solution

Calculate Kp* from Kc

Use the relationship between the equilibrium constants in terms of concentration and pressure: \(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n\) is the change in moles of gas. For the given reaction, \(\Delta n = (1 + 1 - 2) = 0\). Thus, \(K_p = K_c\). Given that \(K_c = 1.26 \times 10^{-3}\) at \(298 \mathrm{~K}\), and since \(\Delta n = 0\), \(K_p = 1.26 \times 10^{-3}\).

Calculate Kc for the formation of HI

For the reverse reaction, the equilibrium constant \(K'_c\) for the formation of HI is the reciprocal of \(K_c\) for the decomposition: \(K'_c = \frac{1}{K_c} = \frac{1}{1.26 \times 10^{-3}} = 793.65\)

Calculate \(\Delta H_{\text{rxn}}^{\circ}\) for HI decomposition

Use the standard enthalpies of formation \(\Delta H_f^{\circ}\) for the substances. Suppose the standard enthalpies of formation are given as \[\Delta H_f^{\circ}(HI) = -26.5 \text{kJ/mol}\]\[\Delta H_f^{\circ}(H_2) = 0 \text{kJ/mol}\]\[\Delta H_f^{\circ}(I_2) = 0 \text{kJ/mol}\]The reaction enthalpy \(\Delta H_{\text{rxn}}^{\circ}\) is: \[\Delta H_{\text{rxn}}^{\circ} = 2 \times \Delta H_f^{\circ}(HI) - \left( \Delta H_f^{\circ}(H_2) + \Delta H_f^{\circ}(I_2) \right)\]Substitute the values: \[\Delta H_{\text{rxn}}^{\circ} = 2 \times (-26.5) - (0 + 0) = -53 \text{kJ/mol} \]

Calculate \(\Delta H_{\text{rxn}}\) from the van't Hoff equation

The van't Hoff equation relates the change in the equilibrium constant with temperature: \[\ln\left(\frac{K_{c1}}{K_{c2}}\right) = \frac{\Delta H_{\text{rxn}}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]Given data are: \(K_{c1} = 1.26 \times 10^{-3}, T_1 = 298 \mathrm{K}, K_{c2} = 2.0 \times 10^{-2}, T_2 = 729 \mathrm{K}\)Rearrange to solve for \(\Delta H_{\text{rxn}}\): \[\Delta H_{\text{rxn}} = R \times \frac{\ln(K_{c2}/K_{c1})}{(1/T_2 - 1/T_1)}\]Substitute the constants and values: \[R = 8.314 \text{J/mol K}\]\[\Delta H_{\text{rxn}} = 8.314 \times \frac{\ln(2.0 \times 10^{-2} / 1.26 \times 10^{-3})}{(1/729 - 1/298)}\] Calculate to find \(\Delta H_{\text{rxn}}\).

Key concepts

Equilibrium Constant (Kc)

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.

Enthalpy Change (ΔH)

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.

Van't Hoff Equation

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.