Post a New Question


posted by .

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 non-gasses count for "delta n" or not)?

Thank you in advance

  • chemistry -

    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 = ?

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question