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# Kp values and partial pressure

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1. Hello everyone,

I am stuck on the last question of the worksheet attached. I keep on getting a value of 22.5 KPa, but the answer is 80 KPa.

I got my answer algebraically - at the start the moles of hydrogen and iodine are X, and at eqm there's 1.5 moles of iodine. Therefore, the change in iodine is -(X-1.5), which is the same for the hydrogen) This also implies HI has increased by 2(x-1.5), giving that many moles at eqm, along with X-(X-1.5) moles of hydrogen.

Are my terms correct - because then I put it in the Kp formula.

Can anyone help?
Attached Files
2. Topic 11 Exercise 4 - Kp (1).doc (30.5 KB, 11 views)
3. Nice Q, made me think for a moment.

Your mathematics are all fine. The thing you're missing is the information given to you by the Kp value.

Kp = pp(HI)2 / (pp(H2) x pp(I2))

= (Ptot x n(HI)/ntot)2 / ((Ptot x n(H2)/ntot) x (Ptot x n(H2)/ntot))

The 2x Ptot and ntot cancel out top and bottom, leaving you with

= n(HI)2 / n(H2) x n(I2)

At equilibrium, we know what n(H2) and n(I2) are, so can work out n(HI) and the rest falls into place.
4. Thanks for the help.

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