2020 H2 Chem P1 Q7 — Chemical equilibrium Solution

The answer is B (0.79 atm).

Let \(x\) mol of \(\text{H}_2\) (and \(x\) mol of \(\text{I}_2\)) be formed at equilibrium.
• Initial: \(\text{H}_2\) = 0, \(\text{I}_2\) = 0, \(\text{HI}\) = 0.040 mol
• Change: \(+x\), \(+x\), \(-2x\)
• Equilibrium: \(x\), \(x\), \(0.040-2x\)

Total amount = 0.040 mol at all times and the total pressure is 1.0 atm, so each partial pressure = (amount / 0.040) \(\times\) 1.0 atm.

\(K_p = \frac{p_{\text{HI}}^2}{p_{\text{H}_2}\,p_{\text{I}_2}} = \frac{\left(\frac{0.040-2x}{0.040}\right)^2}{\left(\frac{x}{0.040}\right)\left(\frac{x}{0.040}\right)} = \frac{(0.040-2x)^2}{x^2} = 54\)

\(\sqrt{54} = 7.348 = \frac{0.040-2x}{x}\)
\(0.040 = 9.348x\), so \(x = 4.28 \times 10^{-3}\)

\(p_{\text{HI}} = \frac{0.040-2(4.28 \times 10^{-3})}{0.040} \times 1.0 = 0.786 \approx 0.79\text{ atm}\)
Check: \(p_{\text{H}_2} = p_{\text{I}_2} = 0.107\text{ atm}\), and \(\frac{0.786^2}{0.107^2} = 54\).

• A: if \(p_{\text{HI}} = 0.06\text{ atm}\), then \(p_{\text{H}_2} = p_{\text{I}_2} = 0.47\text{ atm}\) and \(\frac{0.06^2}{0.47^2} = 0.016\), not 54.
• C: if \(p_{\text{HI}} = 0.88\text{ atm}\), then \(p_{\text{H}_2} = p_{\text{I}_2} = 0.06\text{ atm}\) and \(\frac{0.88^2}{0.06^2} = 215\), not 54.
• D: if \(p_{\text{HI}} = 0.96\text{ atm}\), then \(p_{\text{H}_2} = p_{\text{I}_2} = 0.02\text{ atm}\) and \(\frac{0.96^2}{0.02^2} = 2304\), not 54.