5.6 x 10-6 mol of A and 5 x 10-5 mol of B are mixed in a 200 mL flask. The system is represented by the equation:

Question

2A(G) + B(G) <--> 3C(G)

At equilibrium, there is 4.8 x 10-5 mol of B. Calculate the value of the equilibrium constant.

Answer 1

(A) = 5.6E-6/0.2 = 2.8E-5M
(B) = 5E-5/0.2 = 2.5E-4M

.........2A + B ==> 3C
I.....2.8E-5..2.5E-4..0
C.......-2x...-x....3x
E...2.8E-5-2x..2.5E-4-x...3x

We know B at equilibrium = 4.8E-5 mol/0.2L = 2.4E-4; therefore,
2.5E-4-x = 2.4E-4
Solve for x which lets you work out A and C at equilibrium. Then substitute into Keq expression and solve for K.