Chemical equilibrium — Question 122

Answer
$[Y]\approx 9.49\times {10}^{-4}\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}$; approximately 0.316% of X converts, so the criterion is met.

Working
Let $[Y]=[Z]=x$ and $[X]=0.300-x$. The equilibrium law is $K_c=x^2/(0.300-x)$. Assume $0.300-x\approx 0.300$. Then $x=\sqrt{(3.0\times {10}^{-6})(0.300)}=9.49\times {10}^{-4}$. The percentage converted is $100x/0.300=0.316\%$, well below 5%.

Marking criteria

• Writes Kc = x²/(0.300 − x) with equal product concentrations. [1 mark]
• States the approximation [X]equilibrium ≈ 0.300 mol L⁻¹. [1 mark]
• Calculates [Y] ≈ 9.49 × 10⁻⁴ mol L⁻¹. [1 mark]
• Calculates the percentage conversion ≈ 0.316%. [1 mark]
• Compares it with 5% and accepts the approximation. [1 mark]