Question

The equilibrium constant Kc for the reaction

C <---> D + E

is 6.30*10^-5. The initial composition of the reaction mixture is [C] = [D] = [E] = 1.30*10^-3 M. What is the equilibrium concentration of C,D,E?

Answer 1

given

initally

[C] = [D] =[E] = 1.3 x 10-3

now

consider the given reaction

C ---> D + E

using ICE table

at equilibrium

[C] = 1.3 x 10-3 - y

[D] = 1.3 x 10-3 + y

[E] = 1.3 x 10-3 + y

now

Kc = [D] [E] / [C]

so

6.3 x 10-5 = [ 1.3 x 10-3 + y] [1.3 x 10-3 + y]/ [1.3 x 10-3 - y]

6.3 x 10-5 = [1.3 x 10-3 + y]^2 / [1.3 x 10-3 - y]

solving

we get

y = -9.255 x 10-4

so

at equilibrium

[D]= [E] = 1.3 x 10-3 - 9.255 x 10-4 = 3.745 x 10-4 M

[C] = 1.3 x 10-3 + 9.255 x 10-4 = 2.2255 x 10-3 M