Question

In: Chemistry

At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is...

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?

Solutions

Expert Solution

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

queen_honey_blossom answered 2 years ago
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