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Consider the formation of hydrogen fluoride: H 2 (g) + F 2 (g) ? 2HF(g) If a 4.3

ID: 788951 • Letter: C

Question

Consider the formation of hydrogen fluoride:
H2(g) + F2(g) ? 2HF(g)

If a 4.3 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0085 M H2 is connected to a 4.2 L container filled with 0.030 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.


b) Suppose a 4.00 L nickel reaction container filled with 0.0071 M H2 is connected to a 4.00 L container filled with 0.295 M F2. Calculate the molar concentration of H2 at equilibrium.

Explanation / Answer

Initial H2: (.0085 mol / L) * 4.3L = 0.03655 mol H2

Initial F2: (.030 mol / L ) * 4.2L = .126 mol F2


When the two containers are connected, the volumes combine. Total volume = 4.3 + 4.2 = 8.5 L

When containers combine, but before the two gases begin reacting, the concentrations of the reactants in the equation are as follows:


H2: .03655 mol / 8.5 L = .0043 M

F2: .126 mol / 8.5 L = .0148 M


Equilibrium expression: K = products / reactants = [HF]^2 / {[F2][H2]} = 7.8 * 10^14

(Hint: the large K value indicates that the concentration of products will be very high in comparison to the concentration of reactants)


Solving for [HF]: 7.8 * 10^14 = x^2 / { .0043 * .0148}. x = [HF] = 222799M


b.) Initial H2: (.0071 mol / L) * 4.00 L = .0284 mol H2

Initial F2: (.295 mol / L) * 4.00 L = 1.18 mol F2


When the two containers are combined, the volumes combine. Total volume= 4+4 = 8L

When containers combine, but before the two gases begin reacting, the concentrations of the reactants in the equation are as follows:


H2: .0284 mol / 8 L = .00355 M

F2: 1.18 mol / 8 L = .1475 M


Equilibrium expression: K = products / reactants = [HF]^2 / {[F2][H2]} = 7.8 * 10^14


Solving for [HF]: 7.8 * 10^14 = x^2 / { .00355 * .1475}. x = [HF] = 639083M


Now, we can relate [HF] to [H2] using stoicheometry:


(639083 mol HF / L) * (1 mol H2 / 2 mol HF) = 319542 mol H2 / L