Exercise 10 | Chemical Equilibrium

K_C = $\frac{{\left[{\mathrm{PCl}}_3\right]}_{\mathrm{eq}} {\left[{\mathrm{Cl}}_2\right]}_{\mathrm{eq}}}{{\left[{\mathrm{PCl}}_5\right]}_{\mathrm{eq}}}$

[Cl₂]₀ = [PCl₃]₀ = $\frac{0.40}{1.5}$ = 2.7 x 10^-1 mol.L^-1

From the stoichiometry of the chemical equation: 1 mole of Cl₂ reacts with one mole of PCl₃ to form one mole of PCl₅:

[Cl₂]_eq = [Cl₂]₀ – x     with x = change in the number of moles of Cl₂ per liter

[PCl₃]_eq = [PCl₃]₀ – x = [Cl₂]₀ – x

[PCl₅]_eq = x
 

K_C = $\frac{{\left[{\mathrm{PCl}}_3\right]}_{\mathrm{eq}} {\left[{\mathrm{Cl}}_2\right]}_{\mathrm{eq}}}{{\left[{\mathrm{PCl}}_5\right]}_{\mathrm{eq}}}$ = $\frac{{\left({\left[{\mathrm{Cl}}_2\right]}_0 - \mathrm{x}\right)}^2}{\mathrm{x}}$

K_C x = [Cl₂]₀² – 2 [Cl₂]₀ x + x²

⇒ x² – (2[Cl₂]₀ + K_C) x + [Cl₂]₀² = 0

⇒ x = 0.40 or x = 0.18

So [Cl₂]_eq = [Cl₂]₀ – x = 0.27 - 0.40  < 0 ⇒ this is impossible

or [Cl₂]_eq = [Cl₂]₀ – x = 0.27 - 0.18 = 0.08 mol.L^-1