Calculating Kp for a Gas-Phase Equilibrium Reaction

[rcrx]

A reaction mixture initially contains 2.24 atm Xe and 4.27 atm F₂. If the equilibrium pressure of Xe is 0.34 atm, determine Kp for the reaction.

This question came out of the blue, and all I can think of is that Kp=(PXe)(PF₂), but the answer is supposed to be 25.

I don't get it? Any suggestions? Thanks!

 

[Borek]

Initial pressure of Xe was 2.24 atm, at equilibrium it was 0.34 atm. What have happened to the rest of Xe?

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methods

 

[rcrx]

It must have left the container? I don't know, really :\

 

[Borek]

Question asks about reaction equilibrium constant... What reaction?

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[rcrx]

Oh, sorry. Xe(g) + 2F2(g) ---> XeF4(g)

 

[Gokul43201]

So again, what happened to the rest of the Xe?

 

[rcrx]

The rest of Xe has gone to the formation of XeF4

 

[Gokul43201]

Can you now figure out the partial pressures of all species at equilibrium?

And take another look at the expression you have for Kp (compare with the equation in post#5).