Question

For an equilibrium reaction at a particular temperature, 2 A+B=20 a) One equilibrium mixture has the following concentrations: [A] = 0.30 M, [B] - 0.20 M, and [C] = 0.50 M. Calculate the value of K. b) Another reaction mixture has the following concentrations: [A] = 0.40 M, [B] = 0.40 M, [C] = 0.40 M. Is this mixture at equilibrium? If not which direction will the reaction move to reach equilibrium?

Answer 1

Solution 1

Concentration of [A] = 0.30 M

Concentration of [B] = 0.20 M

Concentration of [C] = 0.50 M

These are concentration of mixture at equilibrium

Given reaction is

2A + B = 2C

Equilibrium constant for above reaction is

K = [C]² / [A]² [B]

K = [0.30]² [0.20] / [0.50]²

K = 13.89

Equilibrium constant for above reaction is (K) = 13.89

Solution 2

Concentration of [A] = 0.40 M

Concentration of [B] = 0.40 M

Concentration of [C] = 0.40 M

Given reaction is

2A + B = 2C

reaction quotient Q for for above reaction is

Q = [C]² / [A]² [B]

Q = [0.40]² [0.40] / [0.40]²

Q = 0.40

Here reaction quotient Q = 0.40 which is less than equilibrium constant K = 13.89

Hence K > Q then reaction favors the product

The ratio of products to reactants is less than that for the system at equilibrium.

The concentration of the reactants is greater than the concentration the products.

The reaction tends toward reach equilibrium, the system shifts to the right to make more products.