AQA A Level Chemistry

http://www.tomred.org/uploads/7/7/8/3/778329/chem_u4_old_qp_jan_2007.pdf

For Q2 How did they work out part f please?
How would you know to do the partial pressure of Cl2 over PCl5?

Reply 1

basilpl4nt

because the Kp expression is

Kp= p(products) / p(reactants)

so the Kp expression for this equilibrium wil always be

Kp= (p(Cl2) x p(PCl3)) / p(PCl5)

we've been given the partial pressure for Cl2, which is the mole ratio x total pressure. total pressure will be constant, and the mole ratio you can expect to be the same as PCl3 because if we did an ICE table, we'd find that both values started at 0 and went up by the same amounts, as they have a 1:1 molar ratio. so therefore we can assume that p(Cl2)=p(PCl3) for this reaction.

therefore

Kp= p(Cl2)^2/p(PCl5). plug your numbers in and there ya go!