2023 H2 Chem P1 Q13 — Chemical equilibrium Solution

The answer is C (19.6).
At equilibrium, \(p_{\text{NH}_3} = 92\) atm and the total pressure is 100 atm, so \(p_{\text{N}_2} + p_{\text{H}_2} = 100 - 92 = 8\) atm.
\(\text{N}_2\) and \(\text{H}_2\) start in a 1 : 3 ratio and react in a 1 : 3 ratio, so they remain in a 1 : 3 ratio at equilibrium:
\(p_{\text{N}_2} = \frac{1}{4} \times 8 = 2\) atm and \(p_{\text{H}_2} = \frac{3}{4} \times 8 = 6\) atm
\(K_p = \frac{(p_{\text{NH}_3})^2}{(p_{\text{N}_2})(p_{\text{H}_2})^3} = \frac{92^2}{(2)(6^3)} = \frac{8464}{432} = 19.6\text{ atm}^{-2}\)
• A is wrong: 0.0510 is \(\frac{1}{19.6}\), the expression written upside down (reactants over products).
• B is wrong: 7.67 is \(\frac{92}{(2)(6)}\), which leaves out the powers from the stoichiometric coefficients (2 for \(\text{NH}_3\), 3 for \(\text{H}_2\)).
• D is wrong: 313 is \(\frac{92^2}{(1)(3^3)}\), which uses the 1 : 3 mole ratio itself as the partial pressures instead of sharing the remaining 8 atm in that ratio.