Question

If a 3.40 M sample of A is heated to 500 K, what is the concentration of B at equilibrium? Kc = 4.90x10E-6 at 500K

A(aq) <-----> 2B (aq)

Answer 1

A(aq)       <—>   2 B(aq)

3.40           0       (initial)

3.40-x       2x       (at equilibrium)

Kc = [B]^2 / [A]

4.90*10^-6 = (2x)^2 / (3.40-x)

since Kc is small, x will be small and can be ignored as compared to 3,40

Above expression thus becomes:

4.90*10^-6 = (2x)^2 / (3.40)

(2x)^2 = 1.67*10^-5

2x = 4.08*10^-3 M

[B] = 2x = 4.08*10^-3 M

Answer: 4.08*10^-3 M