Question

The reversible chemical reaction

A+B⇌C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=7.3

What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ?

Answer 1

ICE Table:

                    [A]                 [B]                 [C]                 [D]               

initial             1.0                 2.0                 0                   0                 

change              -1x                 -1x                 +1x                 +1x               

equilibrium         1.0-1x              2.0-1x              +1x                 +1x               

Equilibrium constant expression is
Kc = [C]*[D]/[A]*[B]
7.3 = (1*x)(1*x)/((1-1*x)(2-1*x))
7.3 = (1*x^2)/(2-3*x + 1*x^2)
14.6-21.9*x + 7.3*x^2 = 1*x^2
14.6-21.9*x + 6.3*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 6.3
b = -21.9
c = 14.6

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 1.117*10^2

roots are :
x = 2.577 and x = 0.8993

x can't be 2.577 as this will make the concentration negative.so,
x = 0.8993

At equilibrium:
[D] = x = 0.8993 M

Answer: 0.899 M