Question

A 1-L container has a mixture of 1.02 mole of A, 1.11 mole of B and 0.561 mole of C at a fixed
temperature. The mixture was left to reach equilibrium and it was found that the number of
moles of A is now 1.79. Based on the below reaction, calculate the equilibrium constant, Kc
2�(�) ↔ 2�(�) + �(�)

Answer 1

The equilibrium constant, Kc, is given by the formula:

Kc = [C]^c/[A]^a[B]^b

Where [C], [A], [B] are the molar concentrations of C, A, and B respectively, and a, b, c are the coefficients of the balanced chemical equation.

In this case, the coefficients are a=2, b=2, and c=1. We are given the initial moles of A, B, and C, as well as the moles of A at equilibrium.

Initial moles:
A = 1.02 mole
B = 1.11 mole
C = 0.561 mole

Moles of A at equilibrium:
A = 1.79 mole

Since the reaction is 2A ↔ 2B + C, we see that at equilibrium, moles of B = 2(1.02 - 1.79) = 0.44 mole, and moles of C = 1.11 - 0.561 = 0.549 mole.

Now we can calculate the molar concentrations:
[A] = 1.79/1 = 1.79 M
[B] = 0.44/1 = 0.44 M
[C] = 0.549/1 = 0.549 M

Now we can plug these values into the equilibrium constant formula:
Kc = (0.549)^1/(1.79)^2(0.44)^2
Kc = 0.549/3.204
Kc = 0.171

Therefore, the equilibrium constant Kc for the reaction 2A ↔ 2B + C is 0.171.