Consider the reaction. A(aq)↽−−⇀2B(aq) 𝐾c=5.90×10−6 at 500 K. If a 2.00 M sample of A is heated to 500 K, 500 K, concentration of B at equilibrium?

Answer 1

A(aq) <==> 2B(aq) ... Kc = 5.90x10^-6

2.00.............0...........Initial

-x................+2x........Change

2.00-x..........2x.........Equilibrium

Kc = 5.90x10^-6 = [B]² / [A]

5.90x10^-6 = (2x)² / (2.00-x) if we assume x is small relative to 2.00, we can ignore it in the denominator

5.90x10^-6 = 4x² / 2

4x² = 1.18x10^-5

x² = 2.95x10^-6

x = 1.72x10^-3 M (this is less than 1% of 2.00 so above assumption was valid and we avoid using quadratic)

[B] @ equilibrium = 2x = (2)(1.72x10^-3 M) = 3.44x10^-3 M