A 1.9 L reaction vessel, initially at 298 K, contains nitrogen gas at a partial pressure of 337 mmHg and oxygen gas at a partial pressure of 580 mmHg . What is the pressure of N2O4 in the reaction vessel after the reaction?

Therefore since by Avogadro's law equal volumes of all gases contain equal numbers of molecules, there fore as the gases are within the same vessel, thier partial pressure is equivalent to their concentration

from the reaction, 1 mole of N₂ react with 2 moles of O₂ to produce 1 mole of N₂O₄

Thus

1 mmHg of N₂ react with 2 mmHg of O₂ to produce 1 mmHg of N₂O₄

337 mmHg N₂ ×(1 mmHg of N₂O₄/ 1 mmHg of N₂) = 337 mmHg N₂O₄

580 mmHg O₂ ×(1 mmHg of N₂O₄/ 2 mmHg of O₂) = 290 mmHg N₂O₄

As seen from the above calculation, the limting reactant is oxygen and the partial pressure of N₂O₄ = 290 mmHg

adefioyelaoye adefioyelaoye · Answer 2 · 2022-02-03T14:22:16+01:00

The pressure of N₂O₄ in the reaction vessel after the reaction is

290 mmHg.

What is Avogadro's law?

This states that equal volumes of all gases contain equal

numbers of molecules.We can infer that the gases within the

same vessel have partial pressure which is equivalent to their

concentration.

The reaction is given below:

N₂ (g) + 2O₂ (g) → N₂O₄ (g)

1 mole of N₂ react with 2 moles of O₂ to produce 1 mole of N₂O₄

1 mmHg of N₂ reacts with 2 mmHg of O₂ to produce 1 mmHg of N₂O₄

337 mmHg N₂ ×(1 mmHg of N₂O₄/ 1 mmHg of N₂) = 337 mmHg N₂O₄

580 mmHg O₂ ×(1 mmHg of N₂O₄/ 2 mmHg of O₂) = 290 mmHg N₂O₄

The partial pressure of N₂O₄ in the reaction vessel after the reaction = 290 mmHg.

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