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March 6, 2015

March 6, 2015

Posted by **l** on Wednesday, July 20, 2011 at 3:25pm.

P4(s) + 6 Cl2(g) --><-- 4 PCL3(l)

Given the following at 200 C

P4(s)+ 10 Cl2(g) --><-- 4 PCl5(s) Kc=8.12

PCl3(l)+ Cl2(g) --><-- PCl5(s) Kc=0.771

Answer: 6.7*[10^(-9)]

not sure what to do.

I know Kp=Kc(RT)^(delta n)

but the fact that they gave me two Kc values is confusing. Am I supposed to use those to somehow find the concentrations of the gaseous compounds?

Also, is "delta n" just the "final n" - "initial n" for the first reaction given (i.e. -3 or -6 depending on whether or not non-gasses count for "delta n" or not)?

Thank you in advance

- chemistry -
**DrBob222**, Wednesday, July 20, 2011 at 3:59pmTricky, tricky.

You will notice that the two Kc values given are for different reactions. What you must do is to multiply equation 2 by 4, reverse it, and add it to equation 1. The result is the equation they want for Kp; i.e., P4(s) + 6Cl2 ==> PCl3(l)

So if you do that, which I did and it works out right, Kc for the final reaction is Kc1 x (1/(Kc2)^4 and for all of that I obtained 22.979 (which is too many s.f., I know, but you can round at the end). That part is usual and there is no trick to it. Standard procedure. The next part is the tricky part.

Kp = Kc(RT)^dn.

Kp = 22.979(R*T)dn.

R = the usual 0.08206

T = 473

Now let's look at the equation.

P4(s) + 6Cl2(g) ==> 4PCl3(l)

This is a heterogeneous reaction. The P4 is a solid and doesn't enter into Kp. PCl3 is a liquid and doesn't enter into Kp. The only gas in the equation is 6Cl2. So n is 0(on the right) - 6(on the left) for dn of -6

Kp = 22.979(0.08206x473)^-6 = ?

:-).

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