Kc=(x)(x) /(2-x)
lets check your answer:
Kc=5.6e-5*5.6e-5/(2-5.6e-5)
= 1.25e-9 that answer does not check.
kc=x^2/2-x
(2-x)4e-8 =x^2
x^2+x*4e-8e-8=0
my quadratic solve gets 0.00028282271318173
Your text answer is wrong, I think.
For the reaction, XBr2(g) <--------> X(g) + Br2(g), if a container is charged to an initial concentration, [XBr2] = 2.00 mol/L and the system allowed to react and come to equilibrium, if Kc = 4.0 x 10-8, what is the concentration of X(g) at equilibrium?
This is what I did
I 2.00mol 0 0
C -x x x
E 2.00-x x x
4.0 x 10^-8=2.00-x/x^2
4.0 x 10^-8x^2+x-2.00=0
Then when I use the quadratic equation I don't get the same answer as the answer key.
According to the answer key the answer is 0.000056 M
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