Question

Consider the following reaction. A(aq)<---> 2B(aq) Kc= 8.19*10^-6 at 500K If a 2.90 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

Answer 1

consider the given reaction

A --> 2B

the equilibrium constant is

Kc = [B]^2 / [A]

now consider the reaction

A ---> 2B

using ICE table

inital conc of A , B are 2.9 , 0

change in conc of A , B are -x , 2x

equilibrium conc of A , B are 2.9-x , 2x

now

Kc = [B]^2 / [A]

so

8.19 * 10-6 = [2x]^2 / [2.9-x]

x <<< 2.9

so

2.9 - x = 2.9

so

8.19 * 10-6 = [2x]^2 / 2.9

x = 2.43675 * 10-3

so

[B]eq =2x = 2 * 2.43675 * 10-3

[B]eq = 4.8735 * 10-3

so

the conc of B at equilibrium is 4.8735 * 10-3 M