1.0 mol of nitrogen oxide NO and 1.0 mol of oxygen were mixed in a container and heated to 450 oC. At equilibrium the number of moles of oxygen was found to be 0.70 mol. The total pressure in the vessel was 4.0 atm. Calculate the value of Kp for the reaction:
2NO (g) + O2 (g) 2 NO2 (g)
2 answers
I NEED ANSWER PLEAS
I would approach the problem this way.
You know equilibrium moles O2 = .7 which means 0.3 mole O2 was used. Then mole NO2 formed must be 2 x 0.3 = 0.6 and moles NO must be 1-0.6 = 0.4.
Total moles = moles NO + moles O2 + moles NO2.
mole fraction O2, for example = moles O2/total moles.
Then partial pressure O2 = mole fraction O2 x total pressure.
The NO and NO2 are done the same way.
You know equilibrium moles O2 = .7 which means 0.3 mole O2 was used. Then mole NO2 formed must be 2 x 0.3 = 0.6 and moles NO must be 1-0.6 = 0.4.
Total moles = moles NO + moles O2 + moles NO2.
mole fraction O2, for example = moles O2/total moles.
Then partial pressure O2 = mole fraction O2 x total pressure.
The NO and NO2 are done the same way.
1 answer