Consider the following reaction. A(aq) <--> 2B(aq) Kc = 6.04 x 10^-6 at 500K

Question

Hello everybody I have been struggling with this and was wondering if somebody could help and explain it to me?
Consider the following reaction. A(aq) <--> 2B(aq) Kc = 6.04 x 10^-6 at 500K If a 5.20 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

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.