## 2nd Final 2007 Q4C

Marie_Bae_3M
Posts: 25
Joined: Wed Sep 21, 2016 2:57 pm

### 2nd Final 2007 Q4C

The equil. constant Kc equals 0.045 at 250C for the decomposition reaction PCl5 <--> PCl3 + Cl2
Calculate the percentage of PCl5 that disassociates if 0.05 mole of PCl5 is placed in a closed vessel (constant volume) at 250C abd 2.00atm pressure..

what i tried to do was find molar concentration by finding volume and dividing the initial moles 0.05 by it and use the ICE table to find the value x (amount dissociated??)?? Can someone explain the method used in the book??
Also i didn't understand the 0.045=x^2/(0.05-x) part.. where did the 1.072 go from the previously shown equation? I know that 1.072 was cancelled by the 1.072 on the bottom but in the numerator 1.072 is squared?? So shouldnt there be 1.072 remaining in the equation then??

Thank you :)

Chem_Mod
Posts: 17839
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 406 times

### Re: 2nd Final 2007 Q4C

The way that you solved the problem should work as well for finding the equilibrium concentrations, and then to find the percent dissociated you simply divide x by the initial concentration of $PCl_{5}$ and multiply by 100%. The course reader takes a different approach of not actually dividing by the volume in the calculations and leaving everything in moles instead. Since the volume is constant the "L" that you divide each amount of moles by is the same, this means that if you do the ICE chart with the moles instead of M you will just find the moles of x then if you divide by the L you will get the M. I believe both ways should work the same but if it's easier for you to just do the whole calculations in M (I think this is easier since this was how we taught it) then you can definitely do that and get the same answer. We won't take points off if you do the problem a different way than the answer key so long as you show your work and your reasoning in why you did your way makes sense.