Multiple choice

For the following equilibrium reaction, $N_{2}O_{4}(g) \rightleftharpoons 2NO_{2} (g)$ $NO_{2}$ is $50\%$ of the total volume at a given temperature. Hence vapour density of the equilibrium mixture is:

  1. $34.5$
  2. $25.0$
  3. $23.0$
  4. $20.0$
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A Correct answer
Explanation

The reaction is N2O4 -> 2NO2. If NO2 is 50% of the volume, the mole fraction of NO2 is 0.5 and N2O4 is 0.5. The average molar mass is (0.5 * 46) + (0.5 * 92) = 23 + 46 = 69. Vapour density is molar mass / 2, which is 69 / 2 = 34.5.

AI explanation

Because NO2 is 50 percent of the total volume, the molar ratio of NO2 to N2O4 in the mixture is 1 to 1. The average molecular weight of the mixture is (46 plus 92) divided by 2, which equals 69 g/mol. Vapour density is half the average molecular weight, so 69 divided by 2 equals 34.5.

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