Consider the reaction. A(aq)↽−−⇀3B(aq)𝐾c=8.52×10−6at 500 K If a 1.50 M sample of A is heated to 500 K , what is the concentration of B at equilibrium? [B]= M

Answer 1

A <==> 3B

Kc = [B]³ / [A]

A <==> 3B

1.5…….0…..Initial

-x……..+3x…Change

1.5-x…..3x….Equilibrium

8.52x10^-6 = (3x)³ / 1.5-x and assuming x is small relative to 1.5 we can ignore and simplify to

8.52x10^-6 = (3x)³ / 1.5

(3x)³ = 1.28x10^-5

9x³ = 1.28x10^-5

x³ = 1.42x10^-6

x = 1.12x10^-2 M which is small relative to 1.5 (<1%) so assumption was valid

[B] = 3x = 3 x 1.12x10^-2

[B] = 3.36x10^-2 M

(be sure to check the math)