The equilibrium composition for the reaction PCl₃+Cl₂⇌ PCl₅ at 298 K is given below:
[PCl₃]_(eq)=0.2 mol L⁻¹,[Cl₂]_(eq)=0.1 mol L⁻¹,[PCl₅]_(eq)=0.40 mol L⁻¹
If 0.2 mol of Cl₂ is added at the same temperature, the equilibrium concentrations of PCl₅ is __________ × 10⁻² mol L⁻¹
Given : K_c for the reaction at 298 K is 20

Solution

Chemistry ma’am · 4 steps

Extra Cl₂ pushes the reaction forward. Let x react, then make Q equal K_c again.

1. After adding Cl2. Take the volume as 1 L, so 0.2 mol of Cl₂ raises [Cl₂] from 0.1 to 0.3 mol L⁻¹. Now Q = 0.40/(0.2 × 0.3) = 6.67, which is less than K_c = 20, so the reaction moves forward.
2. Set up the shift. Let x mol L⁻¹ of PCl₃ react with x of Cl₂. At the new equilibrium, [PCl₃] = 0.2 - x, [Cl₂] = 0.3 - x and [PCl₅] = 0.40 + x.

   (0.40 + x)/(0.2 - x)(0.3 - x) = 20 ⇒ 20x² - 11x + 0.8 = 0

3. Solve for x. x = (11 ± √(121 - 64))/40 = (11 ± 7.55)/40. The plus sign gives x = 0.46, which would make [PCl₃] negative, so x = 0.0863.
4. So. [PCl₅] = 0.40 + 0.0863 = 0.486 mol L⁻¹ = 48.6 × 10⁻² mol L⁻¹, which rounds to 49.

The common slip: 40. Reporting the old [PCl₅] of 0.40 gives 40. The added Cl₂ shifts the equilibrium forward, so PCl₅ must increase.

The idea to keep. When you disturb an equilibrium, write the new starting values, let x shift in the direction Q demands, and use the same K_c. Keep only the root that leaves every concentration positive.

Answer from the NTA final answer key. Solution by PrepWiser.