Question

2 NO2(g) <---------> 2 NO(g) + O2(g) Starting with pure NO2 , the flask is heated...

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.

PNO ____________________atm

PNO2____________________atm

Ptotal____________________atm

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,

PO2 = (mole fraction of O2)*Ptotal

Given PO2 = 0.133 atm at equilibrium,

0.133 atm = (1/5)*Ptotal

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

PNO2 = (mole fraction of NO2)*Ptotal = (2/5)I*0.665 atm = 0.266 atm

PNO = (mole fraction of NO)*Ptotal = (2/5)*0.665 atm = 0.266 atm

Ptotal = 0.665 atm (ans).

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