Question

Question 1   :
We have a solution that has a concentration of 0.321 M for A(aq) and 0.618 M for B(aq). There are no other solutes initially. The reaction 2 A(aq) + B(aq) <---> 2 C(aq) + D(aq) takes place. At equilibrium, the concentration of C(aq) is 0.158 M. What is the equilibrium constant of this reaction?

Answer 1

The reaction with ICE TABLE

2 A(aq) + B(aq) <---> 2 C(aq) + D(aq)

I 0.321 0.618

C - 2x - x +2x +x

E 0.321-2x 0.618-x 2x x

At equilibrium

concentration of C(aq) = 2x = 0.158 M

x = 0.079 M

At equilibrium

concentration of A(aq) = 0.321-2x = 0.321 - 2*0.079

= 0.163 M

At equilibrium

concentration of B(aq) = 0.618-x = 0.618 - 0.079

= 0.539 M

At equilibrium

concentration of D(aq) = x = 0.079 M

Equilibrium constant expression of the reaction

Kc = [D] [C]^2 / [B] [A]^2

= (0.079)*(0.158*0.158) / (0.539)*(0.163*0.163)

= 0.001972156 / 0.014320691

= 0.1377