For #1 you need Kp or Kc for the reaction OR you need the equilibrium pressure of either of the reactants at equilibriium. See below.
2. dG = -RT*ln K
#1. Here is how you do the problem. I assume you have Kp. If not let me know what you have. You can convert Kc to Kp and if you have either product's pressure at equilibiium you can calculate Kp and go from there.
.....................N20(g) + NO2(g) <-> 3NO(g)
I.......................1.1........1.9..................0
C.......................-p.........-p................3p
E....................1.1- p.....1.9 - p...........3p
Kp = (3p)^3/(1.1-p)(1.9-p)
Solve for p = pressure of NO @ equilibrium,
Post your work if you get stuck.
Refer to the equilibrium system below at 500 degrees Celsius.
N20(g) + NO2(g) <-> 3NO(g)
a) only reactants are placed in a rigid container. The initial partial pressure of N2O(g) is 1.1 atm and that of NO2 is 1.9 atm. What is the partial pressure of NO(g) when the system reaches equilibrium?
b) find the standard free energy change, delta G, for the system at 500 degrees Celsius. Include units in your answer
3 answers
I forgot to include that kp= 6.6 x 10^-6 at 500 degrees C!
I assume you can take it from here. If the cubic equation gives you trouble you can find several cubic equation problem solvers by looking on Google.