Question

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.

Answer 1

To find the equilibrium concentration of Cl2 in the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), we use the equilibrium constant and initial concentrations to set up and solve a quadratic equation. The resulting equilibrium concentration of Cl2 is 0.054 M.

Step-by-step explanation:

To calculate the equilibrium concentration of Cl2 from the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), let's denote the change in concentration of Cl2 as 'x' when the system reaches equilibrium. Since Cl2 and PCl3 start at 0.0571 M and PCl5 starts at 0 M, the equilibrium concentrations will be 0.0571 - x for PCl3 and Cl2, and x for PCl5.

The equilibrium constant expression is Kc = [PCl5]/[PCl3][Cl2]. Plugging in the equilibrium concentrations and the given Kc value of 0.021, we get:
0.021 = x / (0.0571 - x)2. After solving this quadratic equation, we find that the equilibrium concentration of Cl2 rounded to the nearest thousandth is 0.054 M.

Answer 2

Concentration of Cl₂ = 0.0255 M

Step-by-step 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