2NO(g)+O2(g)<->2NO2(g) has KP = 2.77 × 1011 at 25 °C. Suppose a mixture of the reactants is prepared at 25 °C by transferring 0.943 g of NO and 589 mL of O2 measured at 30.7 °C and 829 torr into a 1 L vessel.

Question

When the mixture comes to equilibrium, what will be the concentrations of each of the three gases?

Answer 1

[NO]=
[O2]=
[NO2]=

Answer 2

mols NO = grams/molar mass and convert that to pressure with PV = nRT. I get approximately 0.8 atm but you doit more accurately.
USe PV = nRT to solve for mols O2 and convert that to pressure in the 1L container using PV = nRT. I get approximately 0.6 atm.

...........2NO + O2 ==> 2NO2
I..........0.8...0.6.....0
C..........-2x....-x.....+2x
E.......0.8-2x.0..6-x....2x

Kp = 2.77E11 = p^2NO2/p^2NO*pO2
Solve for x and evaluate the three gas pressures. Post your work if you get stuck.

Answer 3

What about the 829 torr? Where does that go?

Answer 4

The problem isn't worded that clearly; however, I think the 829 torr is the pressure of the 589 mL O2 and that is used to find mols O2. Then you must use PV = nRT again, with the n for O2, to find the pressure in the 1L flask. Alternatively you can use (P1V1/T1 = (P2V2/T1) and convert the 589 mL O2 at the conditions listed to the conditions in the 1 L flask.