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Equilibria Help

In this question:

During an experiment, 2 moles I2(g), 1.5 moles H2(g) and 3.2 moles of HI(g) were mixed together at 445oC and were allowed to reach equilibrium. On reaching equilibrium, it was noted that the amount of HI decreased by 11.0%. Give the expression for the equilibrium constant, Kp, and calculate the value of Kp at this temperature.

Would I2 and H2 have decreased by 5.5% (each) at equilibrium or would you have to find what 11% of 3.2 is (i.e. 0.352 moles) then divide that by two according to the stoichiometry of the equation (i.e. 0.176 moles) and subtract it from the original amounts of I2 and H2 respectively?

I'm 99% sure it's the second one; I started working it out with the first version but I felt it didn't make much sense :s-smilie:
Reply 1
Avatar for Pigster
Pigster
20
11% of 3.2. You should have been 100% sure. Although H2 and I2 will increase.
Original post by Pigster
11% of 3.2. You should have been 100% sure. Although H2 and I2 will increase.


Oh yep sorry I was meant to write increase :redface: I get mixed up with these problems sometimes haha but thank you :smile:
I have a second part to the question:

At 500oC, the equilibrium constant Kp, was found to be 50. Find the partial pressure of HI (g) formed at equilibrium if H2 and I2 are mixed together in equimolar amounts at a total pressure of 6 atm.

Would you start with x moles of H2 and I2 and have x-y moles of each and 2y moles of HI at equilibrium or can you start with 1 mole of each and have 1-x moles of each at equilibrium? I don't think you can directly assume that the initial number of moles was 1 though :curious:

If you had to use x-y you'd end up with: 50=(2y)2(x−y)250=\frac{(2y)^2}{(x-y)^2} (the total pressure and the total number of moles would cancel out) and it wouldn't be solvable since you'd have two unknowns.

(I do believe I'm overcomplicating things and that there probably is a much simpler answer to this :rolleyes:)
(edited 9 years ago)
Reply 4
Avatar for Pigster
Pigster
20
Initial partial pressures must be 2 atm each.

New pp of HI is 2+x
New pp of H2 and I2 is 2-x

Kp = (2+x)2/(2-x)2 = 50

x = 1.504 (or 2.659)

i.e. pp HI = 3.50 atm.
Original post by Pigster
Initial partial pressures must be 2 atm each.

New pp of HI is 2+x
New pp of H2 and I2 is 2-x

Kp = (2+x)2/(2-x)2 = 50

x = 1.504 (or 2.659)

i.e. pp HI = 3.50 atm.


Would it be possible to explain where the 2 atm came from please? :smile:

Thank you for answering my questions by the way: greatly appreciate it :redface:
Reply 6
Avatar for Pigster
Pigster
20
Ooops, I need new glasses. I read it as HI, H2 and I2 were all added together.

Since H2 and I2 are equimolar and total P =6 atm, pp of each =3 atm.

Kp = x2/(3-x)2 = 50

x = 2.628 (or 3.494) = pp HI
Original post by Pigster
Ooops, I need new glasses. I read it as HI, H2 and I2 were all added together.

Since H2 and I2 are equimolar and total P =6 atm, pp of each =3 atm.

Kp = x2/(3-x)2 = 50

x = 2.628 (or 3.494) = pp HI


Oh thank you for your help :smile: It's much clearer now to be honest :redface:

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