Question

At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is K = 49.0. A reaction mixture in a 10.00-L flask contains 0.25 moles each of hydrogen and fluorine gases plus 0.39 moles of HF. What will be the concentration of H2 when this mixture reaches equilibrium?

Answer 1

K = [HF]^2/[H2][F2]

K = 49

V = 10 L

n = 0.25 mol of H2

n = 0.25 mol o fF2

n = 0.39 mol of HF

[H2] = 0.25/10 = 0.025

[F2] = 0.025

[HF] = 0.039

in equilibrium

[H2] = 0.25/10 = 0.025 - x

[F2] = 0.025 - x

[HF] = 0.039 + 2x

Then

K = ( 0.039 + 2x)^2/((0.025 - x)(0.025 - x))

sqrt(49) = sqrt(( 0.039 + 2x)^2/((0.025 - x)(0.025 - x))

7 = (0.039 + 2x)/(0.025-x)

solve for x

(0.025-x)*7 =  (0.039 + 2x)

0.175 - 7x = 0.039 + 2x

-9x = 0.039-0.175

x = (0.039-0.175)/-9 = -0.01511

Therefore, equilibrium goes backward...

i.e.

2HF <--> H2 + F2

K = 1/K1 = 1/49 = 0.2048

Try once for K

K = [H2][F2]/[HF]^2

NOTE that now:

[H2] = 0.025 + x

[F2] = 0.025 + x

[HF] = 0.039 - 2x

Substitute in K

K = [H2][F2]/[HF]^2

0.0204 = ( 0.025 + x)( 0.025 + x) /( 0.039 - 2x)^2

Then

sqrt(0.0204) = (0.025 + x) / ( 0.039 - 2x)

0.1428*(0.039 -2x) = 0.025 + x

0.00557 - 0.2857x = 0.025 + x

x = -0.0151

Then

[H2] = 0.025 + x = 0.025 - 0.0151 = 0.0099

[F2] = 0.025 + x = 0.025 - 0.0151 = 0.0099

[HF] = 0.039 - 2x = 0.039 -2*-0.0151 = 0.0692

Then

[H2] = 0.0099