H₂ (g) + F₂ (g) ⇌ 2 HF (g) (25)

Question

An empty steel container is filled with 4.40 atm of H₂ and 4.40 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

Hi Hunter D.,

Do this by "ICE" = initial change equilibrium as a tabular form.

where change of each of H2 and F2 is -x, change of HF is +2x (why?)

and set equilibrium concs into the Kc expression, then solve for x.

It should probably be noted that putting HF + F2 into a pressurized steel container is likely to corrode the container, as noted per the data in (for example) https : / / apps .dtic. mil /dtic /tr / fulltext /u2 /423551 . pdf . (remove all spaces from the link before googleing it -- this format is required to not trigger Wyzant atomatic filters against websites).

-- Cheers, -- Mr. d.