Determine Equilibrium Concentrations from Kc?

Question

The following equation reacts at 298 K:
2NO (g)<---->N2 (g) + O2 (g)
Kp=2.3e30
Partial Pressure of O2= 0.209 atm.
What is the equalibrium partial pressure of NO?

Kp is actually 2.3x10^30

Answer 1

If you start with NO, the pp of N2= that of O2.
Since pp V= n RT and pp/RT = n/V, we can use pp as a surrogate for moles/liter (n/V)
By equilibrium,
0.209^2/pp(NO)^2 = 2.3x10^30
0.044/2.3x10^30= pp(NO)^2
1.9x10-28 = pp(NO)^2
1.4x10-14 atm= pp(NO) appx.
Remember, this assumes that NO is the initial species.