Consider the following reaction and associated equilibrium c

Consider the following reaction and associated equilibrium constant: aA(g)bB(g), Kc = 3.8 Find the equilibrium concentrations of A and B for a=1 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. 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 = 1 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.

Solution

The reaction is aA<-> bB,

K= equilibrium constant = [B]^b/ [A]^a =3.8

For a=1 and b=1, K = [B]/[A]= 3.8

Given initial concentration of A= 1M, let x= drop in concentration of A to reach equilibrium

Hence x/(1-x)= 3.8, x= 3.8-3.8x, 4.8x= 3.8, x= 3.8/4.8=0.79

So at Equilibrium [B]=0.79M and [A]=0.21M

When a=2 and b= 2, the reaction is 2A<>2B

K= [B]²/[A]²

Let x= drop in concentration, So at equilibrium, [A]= 1-2x and [B]=2x

K= (2x)²/(1-2x)²= 3.8, x²/(1-2x)²= 3.8/4=0.95, when solved for x, x= 0.3306

So at equilibrium, [A]= 1-2*0.3306=0.3388 and [B]= 2*0.3306=0.6612

When a=1 and b=2, K = [B]²/[A] =3.8

Let x= drop in concentration of A to reach equilibrium, at equilibrium

[A]= 1-x and [B]=2x and K= (2x)²/(1-x)= 3.8, when solved using excel, x=0.61

So at equilibrium; [A]=1-0.61=0.39 and [B]= 2*0.61=1.22