Question

A(g) + 3B(s) ⇄ C(g) + D(g)

Kc=1.22*10^-2

What is the equilibrium concentration of D when the initial concentration of A is 0.61M? (Answer in M)

Answer 1

ICE Table:

[A] [C] [D]

initial 0.61 0 0

change -1x +1x +1x

equilibrium 0.61-1x +1x +1x

Equilibrium constant expression is

Kc = [C]*[D]/[A]

0.0122 = (1*x)(1*x)/((0.61-1*x))

0.0122 = (1*x^2)/(0.61-1*x)

7.442*10^-3-1.22*10^-2*x = 1*x^2

7.442*10^-3-1.22*10^-2*x-1*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = -1

b = -1.22*10^-2

c = 7.442*10^-3

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 = 2.992*10^-2

roots are :

x = -9.258*10^-2 and x = 8.038*10^-2

since x can't be negative, the possible value of x is

x = 8.038*10^-2

At equilibrium:

[D] = x = 8.038*10^-2

Answer: 8.0*10^-2