Consider the reaction. A(aq)↽−−⇀3B(aq)𝐾c=1.80×10−6at 500 K If a 4.80 M sample of A is heated to 500 K , what is the concentration of B at equilibrium? [B]= M

Answer 1

A(aq) <==> 3B(aq) Kc = 1.80x10^-6

4.80.............0.......Initial

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

4.8-x............3x....Equilibrium

Kc = [B]³ / [A]

1.80x10^-6 = (3x)³ / 4.8-x

Solve for x. Multiply that value by 3 to get the [B] at equilibrium