Solid ammonium chloride dissociates according to NH4Cl(s) – NH3(g) + HCl(g). The dissociation vapour pressure of NH4Cl at 427 °C is 608 kPa but at 459 °C it has risen to 1115 kPa. Calculate (i) the equilibrium constant, (ii) the standard reaction Gibbs energy, (iii) the standard enthalpy, (iv) the standard entropy of dissociation, all at 427 °C. Assume that the vapour behaves as a perfect gas and that ?rH° and ?rS° are independent of temperature in the range given.
The Correct Answer and Explanation is:
We are given the dissociation reaction:
$$
\text{NH}_4\text{Cl}(s) \rightleftharpoons \text{NH}_3(g) + \text{HCl}(g)
$$
And the vapor pressures (which equal the total pressure of NH₃ and HCl) at:
- 427 °C = 700.15 K → P₁ = 608 kPa
- 459 °C = 732.15 K → P₂ = 1115 kPa
(i) Equilibrium constant $K_p$ at 427 °C:
From the reaction, each mole of NH₄Cl produces 1 mole of NH₃ and 1 mole of HCl:
So:
$$
K_p = P_{\text{NH}3} \times P{\text{HCl}} = (P/2)^2 = \frac{P^2}{4}
$$
At 427 °C, $P = 608 \, \text{kPa}$:
$$
K_p = \frac{608^2}{4} = \frac{369664}{4} = \boxed{92416 \, \text{kPa}^2}
$$
(ii) Standard Gibbs energy change $\Delta_r G^\circ$:
Use the formula:
$$
\Delta_r G^\circ = -RT \ln K_p
$$
Where:
- $R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}$
- $T = 700.15 \, \text{K}$
- $K_p = 92416$
$$
\Delta_r G^\circ = -8.314 \times 700.15 \times \ln(92416)
$$
$$
\ln(92416) \approx 11.434
$$
$$
\Delta_r G^\circ = -8.314 \times 700.15 \times 11.434 \approx \boxed{-66586 \, \text{J mol}^{-1} \text{ or } -66.6 \, \text{kJ/mol}}
$$
(iii) Standard Enthalpy change $\Delta_r H^\circ$:
Using van’t Hoff equation:
$$
\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta_r H^\circ}{R} \left(\frac{1}{T_2} – \frac{1}{T_1}\right)
$$
First calculate $K_2 = \frac{1115^2}{4} = 310056.25 \, \text{kPa}^2$
Then:
$$
\ln\left(\frac{K_2}{K_1}\right) = \ln\left(\frac{310056.25}{92416}\right) = \ln(3.354) \approx 1.209
$$
$$
\frac{1}{T_2} – \frac{1}{T_1} = \frac{1}{732.15} – \frac{1}{700.15} \approx -0.0000628 \, \text{K}^{-1}
$$
$$
\Delta_r H^\circ = -\frac{1.209 \times 8.314}{-0.0000628} \approx \boxed{160096 \, \text{J/mol} = 160.1 \, \text{kJ/mol}}
$$
(iv) Standard Entropy change $\Delta_r S^\circ$:
From:
$$
\Delta_r G^\circ = \Delta_r H^\circ – T \Delta_r S^\circ
$$
$$
\Delta_r S^\circ = \frac{\Delta_r H^\circ – \Delta_r G^\circ}{T} = \frac{160096 – (-66586)}{700.15} = \frac{226682}{700.15} \approx \boxed{323.8 \, \text{J mol}^{-1} \text{K}^{-1}}
$$
Final Answers:
- $K_p = 92416 \, \text{kPa}^2$
- $\Delta_r G^\circ = -66.6 \, \text{kJ/mol}$
- $\Delta_r H^\circ = 160.1 \, \text{kJ/mol}$
- $\Delta_r S^\circ = 323.8 \, \text{J/mol·K}$
