Question

Part A A mixture initially contains A, B, and C in the following concentrations: [A] = 0.700 M , [B] = 1.05 M , and [C] = 0.500 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.600 M and [C] = 0.600 M . Calculate the value of the equilibrium constant, Kc.

Answer 1

A    +   2B   <—>    C

0.700       1.05       0.500       (initial)

0.700-x   1.05-2x   0.500+x   (at equilibrium)

given:

[A] at equilibrium = 0.600

0.700-x = 0.600

x = 0.100 M

So, we have at equilibrium:

[A] = 0.600 M

[B] = 1.05 - 2x = 1.05 - 2*0.100 = 0.850 M

[C] = 0.600 M

Kc = [C] /([A] [B]^2)

Kc = 0.600 / (0.600 * 0.850^2)

Kc = 1.38

Answer: 1.38