Initially, 0.920 mol of A is present in a 1.50 L solution. 2A(aq)↽−−⇀2B(aq)+C(aq) At equilibrium, 0.170 mol of C is present. Calculate K. 𝐾=

Answer 1

Again, set up an ICE table.

..........A ======> 2B

I........5.2...............0....

C.......-x...............+2x...

E......5.2-x............2x...

Kc = 6.04x10-6 = [B]2/[A]

6.04x10-6 = (2x)2/5.2-x

4x2 = 3.14x10-5 - 6.04x10-6x

x2 + 1.51x10-6x + 7.85x10-6 = 0

Use quadratic equation to solve for x. Multiply that value by 2 to get the equilibrium concentration of B and subtract the value of x from 5.20 to obtain the equilibrium concentration of A.