what is the concentration of B at equilibrium?

Question

Consider the reaction.A(aq)↽−−⇀3B(aq) c=8.50×10−6 at 500 K

If a 2.50 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

[B]=??

Answer 1

A(aq) <==> 3B(aq) Kc = 8.50x10^-6

2.5..............0.........Initial

-x..............+3x.......Change

2.5-x..........3x........Equilibrium

Kc = 8.50x10^-6 = [B]³ / [A]

8.50x10^-6 = (3x)³ / 2.5-x (assume x is small relative to 2.5 and ignore it in the denominator)

8.50x10^-6 = 27x³ / 2.5

27x³ = 2.125x10^-5

x³ = 7.87x10^-7

x = 9.23x10^-3 M (note: this is 0.5% of 2.5 so above assumption was valid)

[B] at equilibrium = 3x = (3)(9.23x10^-3 M) = 0.0277 M