Question

At a certain temperature, the equilibrium pressures of NO2 and N2O4 are 1.4 bar and 0.46 bar, respectively. If the volume of the container is doubled at constant temperature, what would be the partial pressures of the gases when equilibrium is re-established?

Answer 1

The partial pressure of NO₂ at equilibrium is 0.70 bar and the partial pressure of N₂O₄, at equilibrium is 0.23 bar.

Let x be the mole fraction of NO₂ and x' be the mole fraction of N₂O₄.

The total pressure according to Dalton's law of partial pressure is the sum of the partial pressures of each gas.

Let P be the total pressure of the gases, P' be the partial pressure of NO₂ = 1.4 bar and P" be the partial pressure of N₂O₄ = 0.46 bar.

So, P = P' + P"

= 1.4 bar + 0.46 bar

= 1.86 bar

Since P' = xP

x = P'/P

= 1.4 bar/1.86 bar

= 0.753

Also, P" = xP

x' = P"/P

= 0.46 bar/1.86 bar

= 0.247

Now, since the volume of the container is doubled at constant temperature, we use Boyle's law to determine the new pressure. P₁.

Boyle's law states that the pressure of a given mass of gas is inversely proportional to its volume provided the temperature remains constant. It is written mathematicaly as PV = constant

So, PV = P₁V₁

P₁ = (V/V₁)P

Since V/V₁ = 1/2

P₁ = (V/V₁)P

P₁ = P/2

P₁ = 1.86/2 bar

P₁ = 0.93 bar

So, the new partial pressure of NO₂, P₂ = xP₁

= 0.753 × 0.93 bar

= 0.70 bar

The new partial pressure of N₂O₄, P₃ = x'P₁

= 0.247 × 0.93 bar

= 0.23 bar

So, the partial pressure of NO₂ at equilibrium is 0.70 bar and the partial pressure of N₂O₄, at equilibrium is 0.23 bar.

Learn more about partial pressure here:

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