Question

If the ratio of masses of \(\mathrm{SO}_{3}\) and \(\mathrm{O}_{2}\) gases confined in a vessel is \(1: 1\), then the ratio of their partial pressures would be (a) \(5: 2\) (b) \(2: 5\) (c) \(2: 1\) (d) \(1: 2\)

Short answer

The ratio of the partial pressures of \(\mathrm{SO}_{3}\) to \(\mathrm{O}_{2}\) would be 1:2, so the correct answer is (d) 1:2.

Step-by-step solution

- Understand Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases. The partial pressure of each gas is proportional to its mole fraction.

- Determine Moles of Gases

Given the ratio of masses of \(\mathrm{SO}_{3}\) and \(\mathrm{O}_{2}\) gases is 1:1, use the molar mass of each to convert this mass ratio into a mole ratio. The molar mass of \(\mathrm{SO}_{3}\) is 80.066 g/mol (32.065 g/mol for S and 16.00 g/mol for each of the three O atoms) and for \(\mathrm{O}_{2}\) is 32.00 g/mol (16.00 g/mol for each O atom). If the mass is the same, then the number of moles of \(\mathrm{SO}_{3}\) is half that of \(\mathrm{O}_{2}\) because \(\mathrm{SO}_{3}\) has twice the molar mass.

- Calculate the Ratio of Partial Pressures

Using the mole ratio obtained from Step 2 as the mole fractions and Dalton's Law from Step 1, the partial pressure of \(\mathrm{SO}_{3}\) will be less because it has fewer moles. Specifically, since the number of moles of \(\mathrm{SO}_{3}\) is half that of \(\mathrm{O}_{2}\), the ratio of partial pressures will also be 1:2 for \(\mathrm{SO}_{3}\) to \(\mathrm{O}_{2}\).

Key concepts

Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures is a fundamental concept in chemistry that describes the behavior of gas mixtures. According to this law, the total pressure of a mixture of non-reacting gases is the sum of the partial pressures of each individual gas.
When gases are combined in a container, each gas exerts pressure as if it were the only gas present, irrespective of the presence of other gases. Therefore, the partial pressure of any single gas in the mixture can be calculated by multiplying the total pressure by the mole fraction of that gas. The mole fraction is a measure of concentration, defined as the number of moles of a particular gas divided by the total number of moles of all gases in the mixture.
Understanding Dalton's Law is crucial when dealing with gaseous reactions and processes, such as those occurring in chemical reactors, the atmosphere, or even in our lungs during respiration.

Mole Fraction

Mole fraction is a dimensionless number that expresses the ratio of the number of moles of a particular component to the total number of moles in the mixture. It is a way to represent the concentration of that component within the mixture.
To calculate the mole fraction, denoted by the symbol \( X \), use the formula: \[ X_i = \frac{n_i}{n_{\text{total}}} \] where \( n_i \) is the number of moles of component \( i \) and \( n_{\text{total}} \) is the total number of moles of all components. Mole fraction is particularly useful in applications of Dalton's Law, as the partial pressure of each gas in a mixture is directly proportional to its mole fraction.

Molar Mass

Molar mass is the mass of one mole of a substance, usually denoted in grams per mole (g/mol). It effectively conveys the mass of individual molecules when considering a discernible amount of material, allowing chemists to transition from the microscopic world to the laboratory scale.
In a mixture of gases, the molar mass directly influences the calculation of the mole fraction and partial pressures. With the knowledge of the molar masses of the components of a gas mixture and their individual masses, one can determine the mole fraction and subsequently use Dalton's Law to find the partial pressures.
In the given exercise, the molar mass of \( \mathrm{SO}_3 \) is twice that of \( \mathrm{O}_2 \). Therefore, for equal masses of these two gases, the gas with the higher molar mass, \( \mathrm{SO}_3 \), will have fewer moles. This results in a smaller mole fraction and therefore a lower partial pressure when compared to \( \mathrm{O}_2 \).