1.) Consider the formation of hydrogen fluoride: H 2 (g) + F 2 (g) 2HF(g) If a 4

Question

1.) Consider the formation of hydrogen fluoride:

H2(g) + F2(g) 2HF(g)

If a 4.4 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0067 M H2 is connected to a 2.9 L container filled with 0.035 M F2. The equilibrium constant, Kp, is 7.8 x 1014 (Hint, this is a very large number, what does that imply?) Calculate the molar concentration of HF at equilibrium.

A further hint is provided after the first attempt in the feedback.

2.) BONUS Suppose a 2.00 L nickel reaction container filled with 0.0080 M H2 is connected to a 3.00 L container filled with 0.240 M F2. Calculate the molar concentration of H2 at equilibrium.

Explanation / Answer

Total V = 4.4 + 2.9 = 7.3 L

total mol:

mol of H2 = MV = 4.4*0.0067 = 0.02948

mol of F2 = MV = 0.035*2.9 = 0.1015

recalculate concentrations

[H2] = 0.02948/7.3 = 0.004038

[F2] = 0.1015/7.3 = 0.0139041

Recall that

Kp = Kc*(RT)^(dn)

dn = change in moles = 2 mol of HF - 1 mol of H2 - 1 mo of F2 = 0

so

Kp = Kc*(RT)^0

Kp = Kc

so

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

initially

[H2] = 0.004038

[F2] = 0.0139041

[HF] = 0

in equilibrium

[H2] = 0.004038 - x

[F2] = 0.0139041 - x

[HF] = 0 + 2x

so

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

7.9*10^14 = (2x)^2 / ((0.004038 - x)(0.0139041 - x))

(7.9*10^14) * (0.004038 *0.0139041) - (0.004038+0.0139041)x + x^2 = 4x^2

solve for x

(7.9*10^14)*(5.614*10^-5) - (0.0179421)(7.9*10^14) x -3x^2 = 0

x = 0.00312

[H2] = 0.004038 - 0.00312 = 0.000918

[F2] = 0.0139041 - 0.00312 = 0.0107841

[HF] = 0 + 2*0.00312 = 0.00624

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