Question

Consider the reaction and associated equilibrium constant: aA(g)⇌bB(g), Kc = 1.9

Find the equilibrium concentrations of A and B for a=2 and b=2. Assume that the initial concentration of A is 1.0 M and that no B is present at the beginning of the reaction.

Find the equilibrium concentrations of A and B for a = 2 and b = 1. Assume that the initial concentration of A is 1.0 M and that no B is present at the beginning of the reaction.

Answer 1

if a=2 and b=2 :
2 A(g)   < ------>   2 B(g)
1M                                0    (initial)
1-2x                              2x    (at equilibrium)

use:
Kc = [B]^2 / [A]^2
Kc = {[B] / [A]}^2
1.9 = (2x / 1-2x) ^2
2x / (1-2x) = sqrt (1.9)
2x /(1-2x) = 1.378
2x = 1.378 - 2.756*x
x = 0.29 M
So equilibrium concentration of A = 1-2x = 1-2*0.29 = 0.42 M
equilibrium concentration of B = 2x = 2*0.29 = 0.58 M
-----------------------------------------------------------
if a=2 and b=1 :
2 A(g)   < ------>   1 B(g)
1M                                0    (initial)
1-2x                              1x    (at equilibrium)

use:
Kc = [B] / [A]^2
1.9 = x / (1-2x) ^2
1.9*(1 - 2x)^2 = x
1.9*(1 + 4x^2 - 4x) = x
1.9 + 7.6*x^2 - 7.6*x = x
7.6*x^2 - 8.6*x + 1.9 = 0
x = 0.83M and x = 0.3 M
2x has to be less than 0.5 otherwise equilibrium concetration of A will become negative
so, x = 0.3 M

So equilibrium concentration of A =1-2x = 1-2*0.3 = 0.4 M
equilibrium concentration of B = x =0.3 M