The equilibrium composition for the reaction PCl 3 + Cl 2 ⇌ PCl 5 at 298 K is given below: PCl 3 eq = 0 . 2 mol L - 1 , Cl 2 eq = 0 . 1 mol L - 1 , PCl 5 eq = 0 . 40 mol L - 1 If 0 . 2 mol of Cl 2 is added at the same temperature, the equilibrium concentrations of PCl 5 is _____ × 10 – 2 mol L – 1 Given: K c for the reaction at 298 K is 20

Step-by-step solution

PCl 3 g 0 . 2 + Cl 2 g 0 . 1 ⇌ PCl 5 g 0 . 4 K c = 0 . 4 0 . 2 × 0 . 1 = 20 If 0 . 2 moles of Cl 2 is added. When reactant is added the reaction takes place in forward direction. 20 = K c = 0 . 4 + X 0 . 3 - X 0 . 2 - X ⇒ 0 . 4 + X = 20 0 . 3 - X 0 . 2 - X ∴   0 . 4 + X = 20 0 . 06 + X 2 - 0 . 5 X 0 . 4 + X = 1 . 2 + 20 X 2 - 10 X 20 X 2 - 11 X + 0 . 8 = 0 X = 11 ± 121 - 64 40 = 11 - 7 . 55 40 ≃ 0 . 08625 ∴   PCl 5 = 0 . 48625 ≃ 48 . 625 × 10 - 2 or 49 &#