Question

1)Consider the formation of hydrogen fluoride:

H₂(g) + F₂(g) ? 2HF(g)

If a 2.8 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0085 M H₂ is connected to a 3.6 L container filled with 0.030 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 5.00 L nickel reaction container filled with 0.0052 M H₂ is connected to a 3.00 L container filled with 0.287 M F₂. Calculate the molar concentration of H₂ at equilibrium.

Answer 1

1) First we calculate the amount of initial moles:

.0085 mol/L * (2.8L) = .0238 mol of H₂

.030 mol/L * (3.6L) = .108 mol of F₂

Then we generate our diagram:

            H₂        +       F₂        <->   2HF

B   .0238 mol        .108 mol          0 mol

C       x/2                   x/2               0 mol

E   .0238 - x/2       .108 - x/2         x mol

So we now substitute on the equilibrium constant:

K = [HF]²/ [H₂][F₂]

7.8x10¹⁴ =

x = .0476 moles of HF at equilibrium

[HF] = .0476moles/2.8L (assuming 2.8L reaction container) = .017 M = [HF]

If we take whole combined volume -> .0476/6.4 = .0074M

2) Same procedure:

First determine moles:

.0052 mol/L * (5L) = .026 mol of H₂

.287 mol/L * (3L) = .861 mol of F₂

Then we generate our diagram:

            H₂        +       F₂        <->   2HF

B   .026 mol        .861 mol          0 mol

C       x/2                   x/2               0 mol

E   .026 - x/2       .861 - x/2         x mol

So we now substitute on the equilibrium constant:

K = [HF]²/ [H₂][F₂]

7.8x10¹⁴ =

x = .052 mol of HF

To get the concentrationf of H₂ at equilibrium, we subtract initial moles minus x/2, and divide by volume:

.026 - .052/2 = 0

[H₂] = Very close to 0M