Question

the equilibrium constant Kc for the reaction below is is 5.877×10-5 . c->d+e The initial composition of the reaction mixture is [C] = [D] = [E] = 1.995×10-3  M.

What is the equilibrium concentration of C?

Answer 1

Given Kc = 5.877 * 10^-5

Given reaction is

C -----> D + E

Q =  [E] [D] / [C] = 1.995 * 10^-3 * 1.995 * 10^-3 / 1.995 * 10^-3 = 1.995 * 10^-3

Q > Kc

means more products hence reverse reaction is favoured

C D E

Intial 1.995 * 10^-3 1.995 * 10^-3 1.995 * 10^-3

converted x -x -x

equilibrium 1.995 * 10^-3 + x 1.995 * 10^-3 - x 1.995 * 10^-3 - x

Given Kc = 5.877 * 10^-5

Kc = ( [E] [D] / [C] )eq = (1.995 * 10^-3 - x) (1.995 * 10^-3 - x) / (1.995 * 10^-3 + x) = 5.877 * 10^-5

3.98 * 10^-6 - 3.99 * 10^-3 x + x² = 1.172 * 10^-7 + 5.877 * 10^-5 * x

x² - 4.05 * 10^-3 * x + 3.863 * 10^-6 = 0

solving the above equation

x = 2.51 * 10^-3 or x = 1.54 * 10^-3

first solution is not possible as it cannot go beyond intial concentration

hence equilibrium concentration [C] = 1.54 * 10^-3 M Answer