Calculating Kp for a Gas-Phase Equilibrium Reaction

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To calculate Kp for the reaction Xe(g) + 2F2(g) → XeF4(g), the initial pressures of Xe and F2 are given as 2.24 atm and 4.27 atm, respectively, with Xe's equilibrium pressure at 0.34 atm. The decrease in Xe indicates that some has reacted to form XeF4, which is essential for determining the equilibrium pressures of all species. The correct expression for Kp should consider the stoichiometry of the reaction, specifically Kp = (P_XeF4) / (P_Xe * (P_F2)^2). By determining the partial pressures of XeF4 and F2 at equilibrium, Kp can be calculated, which is expected to be 25. Understanding the changes in pressure and the reaction's stoichiometry is crucial for solving the problem accurately.
rcrx
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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!
 
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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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It must have left the container? I don't know, really :\
 
Question asks about reaction equilibrium constant... What reaction?

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Oh, sorry. Xe(g) + 2F2(g) ---> XeF4(g)
 
So again, what happened to the rest of the Xe?
 
The rest of Xe has gone to the formation of XeF4
 
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).
 
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