Question

Consider the reaction. A(aq) = 2 B(aq) Kc = 6.90 x 10 6 at 500 K If a 3.00 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

Answer 1

Answer:

0.004548 M is the concentration of B at equilibrium at 500 K.

Explanation:

                         A(aq) ⇆ 2 B(aq)

Initially               3.00 M

At equilibrium   3.00 -x    2x

Equilibrium constant of the reaction at 500 K =$K_c=6.90\times 10^{-6}$

Concentration of A at 500 K at equilibrium , [A] = (3.00 -x )M

Concentration of B at 500 K at equilibrium,[B]= 2x

An expression of equilibrium constant is given as:

$K_c=\frac{[B]^2}{[A]}$

$6.90\times 10^{-6}=\frac{4x^2}{(3.00-x)}$

On solving for x:

x = 0.002274 M

[B] = 2 x = 2 × 0.002274 M = 0.004548 M

[A] = (3-x) = 3 M - 0.002274 M =2.997726 M

0.004548 M is the concentration of B at equilibrium.