Consider the reaction below. The initial concentrations of PCl3 and Cl2 are each 0.0571 M, and the initial concentration of PCl5 is 0 M. If the equilibrium constant is Kc=0.021 under certain conditions, what is the equilibrium concentration (in molarity) of Cl2? PCl5 (g) ↽--⇀PCl3 (g) + Cl2 (g) Remember to use correct significant figures in your answer (round your answer to the nearest thousandth). Do not include units in your response.

Concentration of Cl₂ = 0.0255 M

Explanation:

For the chemical reaction of the decomposition of PCl₅, with initial concentration of PCl₅ = 0 M, PCl₃ = 0.0571 M and Cl₂ = 0.0571 M, and the value of equilibrium constant (kc) = 0.021

The ICE table can be written as -

PCl₅ (g) ⇄ PCl₃ (g) + Cl₂ (g)

inital moles 0 0.0571 0.0571

at equilibrium x (0.0571 - x) (0.0571 - x)

The equilibrium constant (Kc) for the reaction can be written as the -

Kc = [PCl₃][Cl₂] / [PCl₅]

Kc = (0.0571 - x) (0.0571 - x) / x

Kc = (0.0571 - x) ² / x

0.021 = 0.00326041 - 0.1142x + x²

x² - 0.1352 x + 0.00326041 = 0

Solving the quadratic equation,

x = 0.031415

At equilibrium, the concentration of Cl₂ = (0.0571 - x) = (0.0571 - 0.031415)

the concentration of Cl₂ = 0.0255 M

Hence, The concentration of Cl₂ at equilibrium = 0.0255M