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

i just learnt this recently but i need to look through my notes again