Chemical equilibrium — Question 125

Answer
Initially $Q_c=0.444$, so the net reaction is forward. At equilibrium $[H_2]=[I_2]=0.133$ and $[HI]=0.533\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}$.

Working
$Q_c={0.200}^2/{0.300}^2=0.444<16.0$. Let x mol L⁻¹ of each reactant be consumed. Then $16.0=(0.200+2x{)}^2/(0.300-x{)}^2$. Concentrations are positive, so $4=(0.200+2x)/(0.300-x)$. This gives $x=1/6$. The concentrations are 0.1333, 0.1333 and 0.5333 mol L⁻¹.

Marking criteria

• Calculates initial Qc = 0.444. [1 mark]
• Compares Qc with Kc and predicts the forward reaction. [1 mark]
• Uses the correct equilibrium changes −x, −x and +2x. [1 mark]
• Solves the equilibrium relation to obtain x = 0.1667 mol L⁻¹. [1 mark]
• Reports all three equilibrium concentrations correctly. [1 mark]