Question

The equilibrium constant for the decomposition AB into its elements A2 and B2 has an equilibrium constant of 0.0900 at a given temperature. What is the equilibrium concentration of AB if 0.050 M AB is allowed to decompose?

2 AB(g) ~ A2(g) + B2(g)

Answer 1

The reaction taking place is”

2 AB <-> A2 + B2

ICE Table:

[AB] initial 0.05 change -2x equilibrium 0.05-2x

Equilibrium constant expression is

Kc = [A2]*[B2]/[AB]^2

0.09 = (1*x)^2/(5*10^-2-2*x)^2

sqrt(0.09) = (1*x)/(5*10^-2-2*x)

0.3 = (1*x)/(5*10^-2-2*x)

1.5*10^-2-0.6*x = 1*x

1.5*10^-2-1.6*x = 0

x = 0.009375

At equilibrium:

[AB] = 0.05-2x = 0.05-2*0.009375 = 0.03125 M

[A2] = +1x = +1*0.009375 = 0.009375 M

[B2] = +1x = +1*0.009375 = 0.009375 M

Answer: [AB] = 0.031 M