Equilibria: Question 2

Worked solution

Part (a): Writing the Kc expression

For a homogeneous gas equilibrium, $K_c$ is written as products over reactants, each raised to the power of its stoichiometric coefficient:

$K_c=\frac{[\text{PCl}_3(g)][\text{Cl}_2(g)]}{[\text{PCl}_5(g)]}$

Part (b): Finding the equilibrium concentrations

The amount of $\text{PCl}_5$ that has decomposed is the difference between the initial and equilibrium amounts: $n(\text{PCl}_5 \text{ reacted}) = 0.400 - 0.280 = 0.120\ \text{mol}$

Since the stoichiometry is $1:1:1$, this also equals the amount of $\text{PCl}_3$ and $\text{Cl}_2$ formed: $n(\text{PCl}_3) = n(\text{Cl}_2) = 0.120\ \text{mol}$

Dividing each amount by the fixed container volume, $2.00\ \text{dm}^3$, to get concentrations: $[\text{PCl}_5] = \frac{0.280}{2.00} = 0.140\ \text{mol dm}^{-3}$ $[\text{PCl}_3] = \frac{0.120}{2.00} = 0.0600\ \text{mol dm}^{-3}$ $[\text{Cl}_2] = \frac{0.120}{2.00} = 0.0600\ \text{mol dm}^{-3}$

Part (c): Calculating Kc

Substituting the concentrations from part (b) into the expression from part (a): $K_c=\frac{(0.0600)\times(0.0600)}{0.140}=\frac{0.00360}{0.140}=0.02571...\ \text{mol dm}^{-3}$

To 3 significant figures: $K_c = 0.0257\ \text{mol dm}^{-3}$

(Check: recomputing the division independently, $3.60\div140=0.02571$, consistent with the value above. Units: $\text{mol dm}^{-3}\times\text{mol dm}^{-3}\div\text{mol dm}^{-3}=\text{mol dm}^{-3}$, since the reaction has 2 mol of gas on the product side and 1 mol on the reactant side.)

Final answers

• (a) $K_c=\dfrac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}$
• (b) $[\text{PCl}_5]=0.140\ \text{mol dm}^{-3}$, $[\text{PCl}_3]=[\text{Cl}_2]=0.0600\ \text{mol dm}^{-3}$
• (c) $K_c=0.0257\ \text{mol dm}^{-3}$ (3 s.f.)