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 at equilibrium has been 0.152 atm.

The partial pressure has been the pressure that has been exerted by the single gas in the mixture. The given:

pressure = 0.5 atm.

The total pressure = 0.674 atm.

At the equilibrium, the reaction pressure has been given as:

0.5             0            0      I

-2x             2x         x       C

0.5 - 2x     +2x       x        E

According to the law of conservation:

Total pressure = sum of the partial pressure of each gas:

0.674 atm = 0.5 - 2x + 2x + x

0.674 atm = 0.5 + x

x = 0.174 atm.

The partial pressure of at equilibrium has been 0.5 - 2x

= 0.5 - 2 × 0.174 atm.

= 0.152 atm

The partial pressure of at equilibrium has been 0.152 atm.

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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

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