Posted by Robby on .
At 900.0K, the equilibrium comstant (kp) for the following reaction is 0.345
2SO2 (g) + O2 (g) yields 2SO3 (g)
At equilibrium the partial pressure of SO2 is 35.0 atm and that of O2 is 15.9 atm. What is the partial pressure of SO3 in atm?
6.20 X 10-4 ( I think it is this one)
I don't think so.
Why do you think that is the one?
Kp = 0.345 = (p^2)SO3/(p)O2*(p^2)SO2
0.345 = (p^2)SO3 = 15.9*(35.0)^2