Question

A 500.-mL sample of \(\mathrm{O}_{2}\) gas at 24 ' \(\mathrm{C}\) was prepared by decomposing a \(3 \%\) aqueous solution of hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) in the presence of a small amount of manganese catalyst by the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) $$ The oxygen thus prepared was collected by displacement of water. The total pressure of gas collected was \(755 \mathrm{~mm} \mathrm{Hg}\). What is the partial pressure of \(\mathrm{O}_{2}\) in the mixture? How many moles of \(\mathrm{O}_{2}\) are in the mixture? (The vapor pressure of water at 24 " is \(23 \mathrm{~mm} \mathrm{Hg}\).)

Short answer

The partial pressure of \(\mathrm{O}_2\) in the mixture is 732 mm Hg. There are approximately 0.0202 moles of \(\mathrm{O}_2\) in the 500 mL gas sample.

Step-by-step solution

Use Dalton's Law of Partial Pressures to find the partial pressure of \(\mathrm{O}_2\)

Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is equal to the sum of the partial pressures of the individual gases. In this case, the total pressure is composed of the partial pressures of \(\mathrm{O}_2\) and water vapor. We can represent this relationship as: \(P_\text{total} = P_{\mathrm{O}_2} + P_{\mathrm{H}_2\mathrm{O}}\) We are given the total pressure (\(P_\text{total}\) = 755 mm Hg) and the vapor pressure of water (\(P_{\mathrm{H}_2\mathrm{O}}\) = 23 mm Hg) at 24°C. We can solve for the partial pressure of \(\mathrm{O}_2\): \(P_{\mathrm{O}_2} = P_\text{total} - P_{\mathrm{H}_2\mathrm{O}}\) \(P_{\mathrm{O}_2} = 755 \, \mathrm{mm \, Hg} - 23 \, \mathrm{mm \, Hg}\) \(P_{\mathrm{O}_2} = 732 \, \mathrm{mm \, Hg}\) Thus, the partial pressure of \(\mathrm{O}_2\) in the mixture is 732 mm Hg.

Use the Ideal Gas Law to find the number of moles of \(\mathrm{O}_2\)

We can use the Ideal Gas Law, represented by the equation \(PV = nRT\), to calculate the number of moles of \(\mathrm{O}_2\) present in the gas mixture. In this equation: - \(P\) is the partial pressure of \(\mathrm{O}_2\) (which we found in Step 1) - \(V\) is the volume of the gas sample (given as 500 mL, or 0.5 L) - \(n\) is the number of moles of the gas (which we want to find) - \(R\) is the ideal gas constant (we'll use the value 0.0821 L atm / mol K) - \(T\) is the temperature (given as 24°C, or 297 K) First, we need to convert the partial pressure of \(\mathrm{O}_2\) to atm: \(P_{\mathrm{O}_2} = \frac{732 \, \mathrm{mm \, Hg}}{760 \, \mathrm{mm \, Hg/atm}} = 0.963 \, \mathrm{atm}\) Now, let's plug the values into the Ideal Gas Law equation and solve for the number of moles of \(\mathrm{O}_2\): \(0.963 \, \mathrm{atm} \times 0.5 \, \mathrm{L} = n \times 0.0821 \, \frac{\mathrm{L \, atm}}{\mathrm{mol \, K}} \times 297 \, \mathrm{K}\) \(n = \frac{0.963 \, \mathrm{atm} \times 0.5 \, \mathrm{L}}{0.0821 \, \frac{\mathrm{L \, atm}}{\mathrm{mol \, K}} \times 297 \, \mathrm{K}}\) \(n = 0.0202 \, \mathrm{mol}\) There are approximately 0.0202 moles of \(\mathrm{O}_2\) in the 500 mL gas sample.

Key concepts

Dalton's Law of Partial Pressures

When dealing with gas mixtures, Dalton's Law of Partial Pressures is a key principle to understand. This law states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas component. In simple terms, each gas in a mixture exerts its own pressure as if it were the only gas present. This means that the total pressure can be calculated by adding the individual pressures of the gases present.

In the provided exercise, the overall pressure of the gas mixture, which consists of oxygen and water vapor, was given. By knowing the vapor pressure of water at a certain temperature, you could separate the contribution of water vapor from the total pressure to find the pressure exerted solely by the oxygen. In this way, Dalton's Law allowed us to find the partial pressure of \(\mathrm{O}_2\) in the mixture by using \(P_{\mathrm{O}_2} = P_{\text{total}} - P_{\mathrm{H}_2\mathrm{O}}\).

Partial Pressure

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. It's important because it helps in understanding how gases behave when mixed. Each gas contributes its part to the total pressure of the mixture, and this is what we call its partial pressure.

In the scenario given in the exercise, the total pressure of the gas mixture, which included both oxygen and water vapor, was known to be 755 mm Hg. To determine the partial pressure of \(\mathrm{O}_2\), we needed to subtract the vapor pressure of water (23 mm Hg) from the total pressure. This gave us a partial pressure for \(\mathrm{O}_2\) of 732 mm Hg. Understanding partial pressures is crucial for correctly working with gas laws in chemistry.

Mole Calculation

Mole calculation uses the Ideal Gas Law to determine the number of moles of a gas, which is a fundamental measure in chemistry for quantifying substances. The Ideal Gas Law equation is \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature.

In the context of this exercise, we used the partial pressure of \(\mathrm{O}_2\), as previously found, which was 732 mm Hg converted to 0.963 atm. The volume was provided as 0.5 L, and the temperature was converted to Kelvin as 297 K. By plugging these values into the Ideal Gas Law equation, along with the constant 0.0821 L atm/mol K, we calculated the moles of \(\mathrm{O}_2\) to be approximately 0.0202 mol. This calculation helps illustrate how we can quantify the amount of a gas in a sample.

Vapor Pressure

Vapor pressure is a measure of the pressure exerted by a vapor in equilibrium with its liquid or solid form. It is an essential concept when studying how liquids evaporate and how gases persist in mixtures. The vapor pressure depends on the temperature of the liquid; as the temperature increases, so does the vapor pressure.

In the context of the exercise, the vapor pressure of water at 24°C was given as 23 mm Hg. This information is critical because it allowed us to determine the partial pressure of \(\mathrm{O}_2\) by subtracting it from the total pressure measured. Vapor pressure significantly affects calculations involving gas collections over water because the water vapor contributes to the total pressure reading. Understanding how vapor pressure works helps in better handling and predicting the behavior of gases in various situations.