For the equilibrium PCl₅(g) PCl₃(g)+ Cl₂(g),K_c = 2.0 × 10¹ at 240°C.If pure PCl₅ is placed in a 1.00-L container and allowed to come to equilibrium,and the equilibrium concentration of PCl₃(g)is 0.50 M,what is the equilibrium concentration of PCl₅(g)?

Question

Quizplus · Accepted Answer

First, write the expression for the equilibrium constant: K_c = [PCl₃][Cl₂]/[PCl₅] Substitute the equilibrium concentrations, and the given value of the equilibrium constant: 2.0 × 10¹ = (0.50 M)(Cl₂])/[PCl₅] Solve for [Cl₂]: [Cl₂] = (2.0 × 10¹)([PCl₅])/(0.50 M) Substitute back into the equilibrium expression and simplify: 2.0 × 10¹ = (0.50 M)(2.0 × 10¹)([PCl₅])/(0.50 M) [PCl₅] = 0.012 M Therefore, the equilibrium concentration of PCl₅ is 0.012 M, and the correct answer is D.

For the Equilibrium PCl5(g) PCl3(g)+ Cl2(g),Kc = 2

For the Equilibrium PCl₅(g) PCl₃(g)+ Cl₂(g),K_c = 2