If one starts with pure NO2 (g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2NO2 (g) 2NO (g) + O2 (g) reaches equilibrium is 0.674 atm. Calculate the equilibrium partial pressure of NO2.

The partial pressure of NO2 = 0.152 atm

Explanation:

Step 1: Data given

Pressure NO2 = 0.500 atm

Total pressure at equilibrium = 0.674 atm

Step 2: The balanced equation

2NO2 (g) → 2NO (g) + O2 (g)

Step 3: The initial pressure

pNO2 = 0.500 atm

pNO = 0 atm

p O2 = 0 atm

Step 4: Calculate pressure at the equilibrium

For 2 moles NO2 we'll have 2 moles NO and 1 mol O2

pNO2 = 0.500 - 2x atm

pNO = 2x atm

pO2 = xatm

The total pressure = p (total) = p (NO2) + p (NO) + p (O2)

p (total) = (0.500 - 2x) + 2x + x = 0.674 atm

0.500 + x = 0.674 atm

x = 0.174 atm

This means the partial pressure of NO2 = 0.500 - 2*0.174 = 0.152 atm