The dissociation constant of at from the following data:<br><br>; ; <br>; ; <br><br>Given: R = 8.3 J — Ionic Equilibrium Chemistry Question
Question
The dissociation constant of $\text{NH}_3$ at $27^\circ \text{C}$ from the following data:<br><br>$\text{NH}_3 + \text{H}^+ \rightleftharpoons \text{NH}_4^+$; $\Delta H^\circ = -52.21 \text{ kJ/mol}$; $\Delta S^\circ = +1.6 \text{ J K}^{-1} \text{ mol}^{-1}$<br>$\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$; $\Delta H^\circ = 54.70 \text{ kJ/mol}$; $\Delta S^\circ = -76.3 \text{ J K}^{-1} \text{ mol}^{-1}$<br><br>Given: R = 8.3 J/K-mol
💡 Solution & Explanation
The dissociation reaction is $\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$, which is the sum of the two given reactions. Adding the thermodynamic parameters: $\Delta H^\circ_{rxn} = -52.21 + 54.70 = 2.49 \text{ kJ/mol} = 2490 \text{ J/mol}$. $\Delta S^\circ_{rxn} = 1.6 - 76.3 = -74.7 \text{ J K}^{-1} \text{ mol}^{-1}$. At $T = 300 \text{ K}$, $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 2490 - 300(-74.7) = 2490 + 22410 = 24900 \text{ J/mol}$. Using $\Delta G^\circ = -RT \ln K$, we get $24900 = -(8.3)(300) \ln K \implies 24900 = -2490 \ln K \implies \ln K = -10 \implies K = e^{-10}$. Therefore, correct answer is B.