Question

The equilibrium constant Kc for the reaction: C <--> D + E is 7.50x10^-5. The initial composition of the reaction mixture is [C] = [D] = [E] = 1.3x10^-3M. What is the equilibrium concentration of C, D, and E?

I've tried to solve it using the ICE method and can't seem to get the right answer.

I have Kc= (1.3*10^-3 + x)^2/(1.3*10^-3 - x) = 7.5*10-5 which gives me the expression:

1x^2 + 0.002675x + 1.593*10-6 = 0 which gives x values (quad equation) of x= 3.315 and -3.317.

I plug the x values and cant seem to get the right answer ( I get C and D = 3.314 and E =3.316M).

Help please.

Answer 1

C <--------------------------> D + E

1.3x10^-3   1.3x10^-3 1.3x10^-3 ------------------> initial

1.3x10^-3-x   1.3x10^-3+x 1.3x10^-3+x --------------> equilibrium

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

7.50x10^-5 = (1.3x10^-3+x )^2 / 1.3x10^-3-x

x^2 + 2.675 x 10^-3 x + 1.5925 x 10^-6 = 0

by solving this

x = -8.95 x 10^-4 , x = -1.78 x 10^-3

x = -1.78 x 10^-3 is not suitable it is more than 1.3 x 10^-3

so x = -8.95 x 10^-4

equilibrium concentrations :

[C] = 1.3 x 10^-3 - x = 1.3 x 10^-3 + 8.95 x 10^-4 = 2.2 x 10^-3 M

[D]= [E] = 1.3 x 10^-3 + x = 1.3 x 10^-3 - 8.95 x 10^-4 = 4.05 x 10^-4 M