chemistry
posted by l .
What is the equilibrium constant "Kp" at 200 C for the reaction below:
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 nongasses count for "delta n" or not)?
Thank you in advance

Tricky, 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 = ?
:).