Problem 140 - General Chemistry: Principles and Modern Applications

Question

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(23^{\circ} \mathrm{C}\) and a barometric pressure of 751 Torr. The vapor pressure of water at \(23^{\circ} \mathrm{C}\) is \(21 \mathrm{mmHg}\). The partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in the sample collected is (a) \(21 \mathrm{mmHg}_{i}\) (b) 751 Torr; \((\mathrm{c}) 0.96 \mathrm{atm} ;\) (d) \(1.02 \mathrm{atm}\).

Step-by-Step Solution

Answer

The partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in the sample collected is 0.96 atm (answer c).

Step 1: Conversion of units

Before moving ahead, ensure all the units are consistent. Here, both the barometric pressure is given in Torr and water vapour pressure is given in mmHg. Thankfully, 1 Torr is equal to 1 mmHg. Thus, no conversion is required for this problem.

Step 2: Application of Dalton's Law of Partial Pressures

As per Dalton's Law, the total pressure is the sum of the individual partial pressures.In this scenario, total pressure (P total) is the barometric pressure, and this consists of two partial pressures - the pressure from O2 gas (P O2) and the vapor pressure of water (P water). We know, P total = P O2 + P water. Thus, re-arranging, we get P O2 = P total - P water. Given, P total = 751 mmHg and P water = 21 mmHg, substituting these values into the equation gives P O2 = 751 mmHg - 21 mmHg = 730 mmHg.

Step 3: Convert mmHg to atm

Finally, we want to convert the pressure of O2 from mmHg to atm, as the options are given in atm. We use the conversion factor that 1 atm equals 760 mmHg. Thus, to get the pressure in atm, we divide the pressure in mmHg by 760. Hence, P O2 in atm = 730 mmHg / 760 mmHg/atm = 0.96 atm.

Key Concepts

Gas Collection Over Water

When collecting a gas over water, it’s important to understand that the gas is not pure due to the presence of water vapor. Water is constantly evaporating, which means that any gas collected will include some water vapor, altering the total pressure.

In practical terms, this means the measured pressure doesn’t just come from the gas you’re interested in. It’s a combination of the desired gas and the water vapor’s pressure. Therefore, when you’re interested in isolating the pressure of just the \( ext{O}_2\) gas, you need to account for and subtract the water vapor pressure from the total pressure.

This situation is common in chemistry labs and requires an understanding of Dalton’s Law, which helps you handle these types of calculations effectively.

Vapor Pressure

Vapor pressure is a crucial concept when dealing with gases collected over water. It's the pressure exerted by a vapor in equilibrium with its liquid form at a given temperature.

• This pressure varies with temperature; as temperature goes up, so does vapor pressure.
• The vapor pressure of water at \({23}^{\circ} \text{C}\) is given as 21 mmHg.

For accuracy in experiments involving gas collection over water, it’s necessary to know the vapor pressure of water at the specific temperature of your experiment. This allows you to subtract it from the total pressure to find the exact pressure of the collected gas.

By reflecting the amount of water vapor in the air, vapor pressure plays a key role in many scientific calculations, ensuring that measured pressures reflect the actual amounts of gases present.

Pressure Unit Conversion

In many scientific contexts, it’s vital to convert pressure units for consistency and comparison. Here, converting from mmHg (or Torr) to atm is a frequent requirement, especially in chemistry problems.

• 1 atm is equivalent to 760 mmHg.
• Torr and mmHg are used interchangeably as they represent the same pressure unit.

In the given exercise, the pressure of \( ext{O}_2\) was originally calculated in mmHg. To convert this to atm, you divide the pressure in mmHg by 760.

For example, if \( ext{O}_2\) has a partial pressure of 730 mmHg, converting this to atmospheres involves \( \frac{730\, \text{mmHg}}{760\, \text{mmHg/atm}} = 0.96\, \text{atm} \).

Making these conversions allows scientists to use standard units, making results more universally understandable and comparable.