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?

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 =

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:

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.

the answer is d. 74.02