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 =[tex]K_c=6.90\times 10^{-6}[/tex]
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:
[tex]K_c=\frac{[B]^2}{[A]}[/tex]
[tex]6.90\times 10^{-6}=\frac{4x^2}{(3.00-x)}[/tex]
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.
