Question

At a particular temperature the equilibrium constant for the reaction:
H₂(g) + F₂(g) ⇔ 2HF(g)
is K = 64.0. A reaction mixture in a 10.00-L flask contains 0.23 moles each of hydrogen and fluorine gases plus 0.50 moles of HF. What will be the concentration of H₂ when this mixture reaches equilibrium?

Answer 1

Equilibrium Quotient expression is
Qc = [HF]^2/[H2]*[F2]
= (0.05^2)/(0.023^1)*(0.023^1)
=4.73

comparing Q and k
since Q is lesser than K
reaction will move in forward direction
ICE Table:

                    [H2]                [F2]                [HF]              

initial             0.023               0.023               0.05              

change              -1x                 -1x                 +2x               

equilibrium         0.023-1x            0.023-1x            0.05+2x           

Equilibrium constant expression is
Kc = [HF]^2/[H2]*[F2]
64.0 = (0.05 + 2*x)^2/(0.023-1*x)^2
sqrt(64.0) = (0.05 + 2*x)/(0.023-1*x)
8.0 = (0.05 + 2*x)/(0.023-1*x)
0.184-8*x = 0.05 + 2*x
0.134-10*x = 0
x = 0.0134

At equilibrium:
[H2] = 0.023-1x = 0.023-1*0.0134 = 0.0096 M

Answer: 0.0096 M