Question

All of the reagents in the reaction below are in aqueous form. The initial concentration of A is 4.5x10^-4 M. The initial concentration of B is 5.3x10^-3 M. At equilibrium, the concetration of A is 1.2 x 10^-4 M. What is the value of K_eq?

_{A + 2B <----------> C + 3D}

Answer 1

                    [A]                 [B]                 [C]                 [D]               

initial             4.5*10^-4            0.0053              0                   0                 

change              -1x                 -2x                 +1x                 +3x               

equilibrium         4.5*10^-4-1x        0.0053-2x           +1x                 +3x               

Given at equilibrium,
[A] = 1.2*10^-4
4.5*10^-4-1x = 1.2*10^-4
x = 3.3*10^-4

Equilibrium constant expression is
Kc = [C]*[D]^3/[A]*[B]^2
Kc = (x)*(3x)^3/(4.5*10^-4-x)*(0.0053-2x)^2
Kc = (3.3*10^-4)*(3.3*10^-4)/((4.5*10^-4)-(3.3*10^-4))*(0.0053-(2*3.3*10^-4))^2
Kc = 1.24*10^-4

Answer: 1.24*10^-4