2024 H2 Chem P1 Q13 — Chemical equilibrium Solution

The answer is C (0.0074 mol \(\text{dm}^{-3}\)).
The volume is 1.00 \(\text{dm}^3\), so each concentration equals the amount in mol.
\(\text{H}_2 + \text{CO}_2 \rightleftharpoons \text{H}_2\text{O} + \text{CO}\)
Initial / mol \(\text{dm}^{-3}\): 0.010, 0.010, 0, 0
Change / mol \(\text{dm}^{-3}\): \(-x\), \(-x\), \(+x\), \(+x\)
Equilibrium / mol \(\text{dm}^{-3}\): \(0.010 - x\), \(0.010 - x\), \(x\), \(x\)
\(K_c = \frac{[\text{H}_2\text{O}][\text{CO}]}{[\text{H}_2][\text{CO}_2]} = \frac{x^2}{(0.010 - x)^2} = 0.123\)
Taking the square root: \(\frac{x}{0.010 - x} = 0.3507\), so \(x = \frac{0.003507}{1.3507} = 0.00260\)
\([\text{H}_2] = 0.010 - 0.00260 = 0.0074\) mol \(\text{dm}^{-3}\)
• B is wrong: 0.0026 mol \(\text{dm}^{-3}\) is \(x\), the equilibrium concentration of \(\text{H}_2\text{O}\) (and of CO), not of \(\text{H}_2\).
• A is wrong: 0.0011 mol \(\text{dm}^{-3}\) comes from leaving out the squares, \(\frac{x}{0.010 - x} = 0.123\), and then giving \(x\).
• D is wrong: 0.0089 mol \(\text{dm}^{-3}\) comes from the same unsquared expression, then \(0.010 - 0.0011\).