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

The equilibrium partial pressure of NO2 is 0.152 atm

Explanation:

Step 1: Data given

Initial pressure of NO2 = 0.500 Atm

Total pressure inside the vessel at equilibrium = 0.674 atm

Step 2: The balanced equation

2 NO2(g) ⇌ 2 NO(g) + O2(g)

Step 3: The initial pressures

pNO2 = 0.500 atm

pNO = 0 atm

pO2 = 0 atm

Step 4: The pressure at the equilibrium

pNO2 = 0.500 - 2x

pNO = 2x

pO2 = x

Total pressure = 0.674 = (0.500 - 2x) + 2x + x

0.674 = 0.500 + x

x = 0.174

pNO2 = 0.500 - 2*0.174 = 0.152 atm

pNO = 2x = 0.348 atm

pO2 = x = 0.176 atm

The equilibrium partial pressure of NO2 is 0.152 atm

answer; i believe that the correct answer is (a) potassium(k)

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