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
See your other post on this.
To find the equilibrium constant "Kp" at 200°C for the given reaction, you can use the relationship Kp = Kc(RT)^(delta n). Here's how you can proceed step by step:
1. Determine the values of Kc for the relevant reactions:
- The first reaction: P4(s) + 10 Cl2(g) ⇌ 4 PCl5(s) (Kc = 8.12)
- The second reaction: PCl3(l) + Cl2(g) ⇌ PCl5(s) (Kc = 0.771)
2. Find the value of "delta n" for the overall reaction:
- From the balanced equation: P4(s) + 6 Cl2(g) ⇌ 4 PCl3(l)
- Comparing the stoichiometric coefficients of reactants and products, the change in the number of moles of gas is (0 - 6) = -6. Therefore, delta n = -6.
3. Convert the temperature to Kelvin (T = 200°C + 273.15).
4. Plug in the values into the equation: Kp = Kc(RT)^(delta n).
- Kc = (8.12)(R)(T)^(delta n) and T is in Kelvin.
- Since Kp is in terms of pressure, you need to convert Kc to Kp by multiplying it with the appropriate gas constant value (R) raised to the power of delta n.
5. Calculate Kp.
- Use the ideal gas constant R = 0.08206 L.atm/mol.K.
- Calculate Kp: Kp = (8.12)(0.08206 L.atm/mol.K)(T in Kelvin)^(-6).
After substituting the temperature in Kelvin into the equation, you will find that Kp ≈ 6.7 x 10^(-9).
Note: The Kc values given for the other reactions are not directly used to find Kp for the target reaction. They are included to provide additional information but are not necessary for calculating Kp in this case.