For the chemical reaction shown below, Kc = 6.7x10^-4. N2 + O2 2NO. If 1.0 mol of N2 and O2 were placed in a 1.0dm3 container at 250 degrees Celsius, what would the equilibrium partial pressure of NO?

Question

For the chemical reaction shown below, Kc = 6.7x10^-4. N2 + O2 <--> 2NO. If 1.0 mol of N2 and O2 were placed in a 1.0dm3 container at 250 degrees Celsius, what would the equilibrium partial pressure of NO?

anonymous · Answer

I'm not sure what equilibrium partial pressure means im sorry, but what I would do is you know there is a 1 : 2 molar ratio between N2 and NO or O2 and NO. So that gives us 2 mols of NO. Also, we know the temperature 250 degrees celcius = 523 Kelvin. And we also know the volume which is 1.0 L since dm3 is litres. Therefore, we can use:
$$PV = nRT$$ and rearrange for P. 
We get.. $$P = nRT/V$$ and so we get a value for P. Oh and by the way, R is the gas constant. Im not sure where the equilibrium fits in im afraid.. sorry