2025 H2 Chem P1 Q10 — Chemical equilibrium Solution

The answer is A (0.16 atm).
Ni is a solid, so only CO and \(\text{Ni(CO)}_4\) contribute to the pressure. Let the partial pressure of CO that reacts be \(x\) atm.
• Initial: \(p_{\text{CO}} = 1\), \(p_{\text{Ni(CO)}_4} = 0\)
• Change: \(-x\) and \(+\tfrac{x}{4}\) (4 CO form 1 \(\text{Ni(CO)}_4\))
• Equilibrium: \(1 - x\) and \(\tfrac{x}{4}\)
Total pressure: \(1 - x + \tfrac{x}{4} = 0.37\), so \(\tfrac{3x}{4} = 0.63\) and \(x = 0.84\).
\(p_{\text{CO}} = 1 - 0.84 = 0.16\) atm
• B is wrong: 0.21 atm is the partial pressure of \(\text{Ni(CO)}_4\) (\(\tfrac{x}{4}\)), not of CO.
• C is wrong: 0.37 atm is the total pressure of the equilibrium mixture.
• D is wrong: 0.63 atm is the fall in total pressure (\(\tfrac{3x}{4}\)), not the partial pressure of CO.