Question

2 NO₂(g) <---------> 2 NO(g) + O₂(g)

Starting with pure NO₂ , the flask is heated and he partial pressure of O2 at equlibrium is 0.133 am. Calculate the partial pressures of NO and NO₂ at equlibrium and the total pressure in the flask at equlibrium.

P_NO ____________________atm

P_NO2____________________atm

P_total____________________atm

Answer 1

The given reaction is

2 NO₂ (g) -------> 2 NO (g) + O₂ (g)

As per the balanced stoichiometric equation, the molar ratio of the reactants and products is 2:2:1.

Therefore, as per the molar equation, the mole fraction of O₂ in the gaseous mixture (both reactants and products combined) is 1 mole/(2 + 2 + 1) mole = 1/5.

Assuming all the gases to behave ideally and applying Dalton’s law, we must have,

P_O2 = (mole fraction of O₂)*P_total

Given P_O2 = 0.133 atm at equilibrium,

0.133 atm = (1/5)*P_total

====> P_total = (0.133 atm)*5 = 0.665 atm

P_NO2 = (mole fraction of NO₂)*P_total = (2/5)I*0.665 atm = 0.266 atm

P_NO = (mole fraction of NO)*P_total = (2/5)*0.665 atm = 0.266 atm

P_total = 0.665 atm (ans).