Consider the formation of hydrogen fluoride: H2(g) + F2(g) ↔ 2HF(g) If a 3.9 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0052 M H2 is connected to a 3.3 L container filled with 0.034 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.

The molar concentration of HF at equilibrium is

Explanation:

To calculate the number of moles for given molarity, we use the equation:

      ......(1)

Molarity of hydrogen gas solution = 0.0052 M

Volume of solution = 3.9 L

Putting values in equation 1, we get:

Molarity of fluorine solution = 0.034 M

Volume of solution = 3.3 L

Putting values in equation 1, we get:

For the given chemical equation:

Initial:          0.0203   0.1122

At eqllm:    0.0203-x    0.1122-x            2x

Volume of the solution = [3.3 + 3.9] L = 7.2 L

Relation of with is given by the formula:

where,

= equilibrium constant in terms of partial pressure =

= equilibrium constant in terms of concentration = ?

R = Gas constant =

T = temperature

= change in number of moles of gas particles =

Putting values in above equation, we get:

The expression of K_c for above equation follows:

We are given:

Putting values in above expression, we get:

Neglecting the value of x = 0.1122 because equilibrium concentration of hydrogen gas will become negative, which is not possible

So, equilibrium moles of

Now, calculating the molarity of HF by using equation 1,

Moles of HF = 0.0406 moles

Volume of solution = 7.2 L

Putting values in equation 1, we get:

Hence, the molar concentration of HF at equilibrium is

your answer would be “dry”

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