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

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

Step-by-step 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* 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=([B]^2)/([A])`

`6.90* 10^(-6)=(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.