Question

Consider the formation of hydrogen fluoride:

H₂(g) + F₂(g) ↔ 2HF(g)

1) If a 1.5 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0077 M H₂ is connected to a 4.0 L container filled with 0.029 M F₂. The equilibrium constant, K_p, is 7.8 x 10¹⁴ (Hint, this is a very large number, what does that imply?) Calculate the molar concentration of HF at equilibrium.

2) Suppose a 3.00 L nickel reaction container filled with 0.0098 M H₂ is connected to a 4.00 L container filled with 0.229 M F₂. Calculate the molar concentration of H₂ at equilibrium.

Answer 1

H₂(g) + F₂(g) ↔ 2HF(g)

Moles of H₂ present = molarity*volume of solution in litres = 0.0077*1.5 = 0.01155

Moles of F₂ = molarity*volume of solution in litres = 0.029*4 = 0.116

Now, after the containers are joined, total volume of the solution = 5.5 L

molar concentration of H₂ = 0.01155/5.5 = 0.0021

molar concentration of F₂ = 0.116/5.5 = 0.021

Now, K_p = K_c*(R*T)ⁿ ; where n = moles of gaseous products - moles of gaseous reactnts in the balanced reaction = 0

Thus, K_p = K_c

Now, let at eqb. [H₂] = [F₂] = 0.021 - x ; [HF] = 2x

Thus, K_c = [HF]²/{[H₂]*[F₂]}

or, 7.8*10¹⁴ = (x/0.021-x)²

or, x = 0.02099

Thus, [H₂] at eqb. = 0.0001 M