11.27

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Tatum Keichline 2B
Posts: 64
Joined: Fri Sep 28, 2018 12:26 am

11.27

Postby Tatum Keichline 2B » Mon Jan 14, 2019 4:24 pm

I am confused on how to do 6th edition 11.27 and solving for the partial pressure of PCL3 using the given partial pressures. Can someone help please?

ChathuriGunasekera1D
Posts: 78
Joined: Fri Sep 28, 2018 12:17 am

Re: 11.27

Postby ChathuriGunasekera1D » Mon Jan 14, 2019 4:33 pm

Hi! Set up the reaction to solve for Q. Given the equation PCl5(g) <----> PCl3(g) + Cl2(g), Q = ((PPCl3)(PCl2))/(PPCl5)). You're given PPCl5 = 1.18 bar, PCl2 5.43 bar, and K=25. You set Q=K and plug in what you have.

[(PPCl3)(5.43)]/(1.18) = 25

PPCl3 is 5.43 bar

Tatum Keichline 2B
Posts: 64
Joined: Fri Sep 28, 2018 12:26 am

Re: 11.27

Postby Tatum Keichline 2B » Mon Jan 14, 2019 8:18 pm

Ok that makes sense, thank you!


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