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
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 = ?
:-).
To calculate the equilibrium constant, Kp, for the given reaction, you can follow these steps:
1. Determine the value of Δn (delta n). Δn is the difference in the number of moles of gaseous products and gaseous reactants. In this case, the equation tells us that 4 moles of PCl3(l) is formed from 6 moles of Cl2(g). Therefore, Δn = (moles of products) - (moles of reactants) = 4 - 6 = -2.
2. Calculate the value of Kc for each given reaction using the equilibrium expression. You have the values of Kc for the reactions:
P4(s) + 10 Cl2(g) ⇌ 4 PCl5(s) with Kc = 8.12
PCl3(l) + Cl2(g) ⇌ PCl5(s) with Kc = 0.771
3. Use the equation Kp = Kc(RT)^(Δn) to calculate Kp. Here, R is the ideal gas constant and T is the temperature in Kelvin. In this case, the temperature is given as 200°C, which is 473 K. Assuming that you're using the gas constant R = 0.0821 L∙atm/(mol∙K), you can substitute the values into the equation:
Kp = Kc(RT)^(Δn) = Kc(0.0821 L∙atm/(mol∙K))(473 K)^(-2)
Plug in the values of Kc and Δn into this equation and calculate Kp.
Kp = (8.12)(0.0821 L∙atm/(mol∙K))(473 K)^(-2)
Kp = (0.771)(0.0821 L∙atm/(mol∙K))(473 K)^(-6)
Calculate each term separately, and then multiply them together.
4. After calculating both terms, multiply the two values together to find the equilibrium constant Kp.
Kp = (8.12)(0.0821 L∙atm/(mol∙K))(473 K)^(-2) x (0.771)(0.0821 L∙atm/(mol∙K))(473 K)^(-6)
When you evaluate this expression, you will find that Kp is equal to approximately 6.7 x 10^(-9).
Note: Once you have the values of Kc, you do not need to find the concentrations of the gaseous compounds for this calculation. The equilibrium constant is already given in terms of partial pressures for gases.
Also, for Δn, you need to consider moles of gases formed minus moles of gases consumed. Therefore, in this case, Δn would be -2, not -3 or -6.
Remember to double-check the units and conversion factors used to ensure the final answer is in the correct units.