Question

When 0.040 mol of PCl5 is heated to 250 oC in a 1.0 L vessel, an equilibrium is established in which the concentration of Cl2 is 0.025 M. Find the equilibrium constant for the reaction. PCl5(g) ⇌ PCl3(g) + Cl2(g) (3 marks)

Answer 1

Find the amount of substance for Cl_2:

$$n(\text{Cl}_2)=CV=0.025\cdot1=0.025\text{ mol}.$$

We see that n(PCl₃) is 0.025 mol as well (the ratio of products is 1:1).

Hence find n(PCl5) in the vessel:

$$n(\text{PCl}_5)=0.4-n(\text{Cl}_2)=0.375\text{ mol}.$$

Find the equilibrium constant:

$$K=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}=\frac{0.025\cdot0.025}{0.375}=1.67\cdot10^{-3}.$$