what is the equilibrium partial pressure of HF?

Question

An empty steel container is filled with 6.60 atm of H₂ and 6.60 atm of F₂. The system is allowed to reach equilibrium. If Kp = 0.450 for the reaction below, what is the equilibrium partial pressure of HF?

Answer 1

To find the answer to this question, we must first write out what the chemical equation is:

H₂ + F₂ —→ 2 HF

Writing out the equilibrium expression for this equation, we find that it is this:

K_p = [HF]²/([H₂]*[F₂])

We can then set up our ICE (initial, change, end) table like so to find an expression for the equilibrium concentrations of each molecule:

Initial Change End

H₂ 6.60 -x 6.60 - x

F₂ 6.60 -x 6.60 - x

HF 0 +2x 2x

Plugging these values along with the K_p we were given earlier, we get the expression:

.450 = (2x)²/(6.60 - x)(6.60 - x)

Rewritten, we find that:

.450 = (2x)²/(6.60 - x)²

Taking the square root of both sides gives us:

.671 = 2x/(6.60 - x)

(.671)(6.60 - x) = 2x

4.427 - .671x = 2x

4.427 = 2.671x

x = 1.66 atm

Since we know that the amount of HF formed is 2x, we can then find that the final partial pressure of HF is:

P_HF = 3.22 atm