Question

Consider the following reaction.
2A(aq) (Two way arrows) B(aq)
Kp = 5.48*10^-5 at 500K
If a sample of A at 3.10atm is heated to 500K, what is the pressure of B at equilibrium?

I know I need to do the ICE table, but am having trouble solving for x once I plug everything in.

"A" should equal 3.10-2x at equilibrium, and "B" should equal x. How do I solve for x at this point?

I have the equation:
x / (3.10-2x)^2 = 5.48x10^-5

Any help would be greatly appreciated! Thank you in advance!

Answer 1

........2A(aq) <--> B(aq)
I.......3.10........0
C.......-2x.........x
E.....3.10-2x.......x

5.48E-5 = (x)/(3.10-2x)
Clear the denominator.
5.48E-5(3.10-2x) = x
5.48E-5*3.10 - 5.48E-5*2x = x
Complete the multiplication, combine terms, solve for x.