Question

For the equiLiBrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 4.0 at 228°C. If pure PCl₅ is placed in a 1.00-L container and allowed to come to equiLiBrium, and the equiLiBrium concentration of PCl₅(g) is 0.19 M, what is the equiLiBrium concentration of PCl₃?

1) 0.095 M
2) 0.4 M
3) 0.19 M
4) 0.87 M
5) 0.009 M

Answer 1

The equilibrium concentration of PCl₃ can be found using the equilibrium constant expression. With the given equilibrium constant Kc = 4.0 and PCl₅ concentration of 0.19 M, the calculated concentration of PCl₃ at equilibrium is approximately 0.87 M.

Step-by-step explanation:

The equilibrium concentration of PCl₃(g) can be calculated using the provided equilibrium constant (Kc) and the equilibrium concentration of PCl₅(g) for the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Given that Kc = 4.0 at 228°C and the equilibrium concentration of PCl₅(g) is 0.19 M, we can set up an expression using the equilibrium constant formula:

Kc = [PCl₃][Cl₂] / [PCl₅]

Since the reaction starts with pure PCl₅, the initial concentrations of PCl₃ and Cl₂ are 0. At equilibrium, for every mole of PCl₅ that dissociates, 1 mole each of PCl₃ and Cl₂ are formed. Thus, if x is the change in concentration of PCl₅, then x is also the concentration of PCl₃ and Cl₂ at equilibrium. The equation becomes:

4.0 = x² / (0.19 - x)

Approximating that 0.19 - x is close to 0.19, as x will be small with respect to 0.19 M, we get:

4.0 = x² / 0.19, which simplifies to x = √(4.0 × 0.19) = √0.76 ≈ 0.87. Therefore, the equilibrium concentration of PCl₃ is approximately 0.87 M.