Question

Because of their different molar masses, \(\mathrm{H}_{2}\) and \(\mathrm{D}_{2}\) effuse at different rates (Section 5.5 ). (a) If it takes 16.5 min for \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to effuse, how long does it take for \(0.10 \mathrm{~mol}\) of \(\mathrm{D}_{2}\) to do so in the same apparatus at the same \(T\) and \(P ?\) (b) How many effusion steps does it take to separate an equimolar mixture of \(\mathrm{D}_{2}\) and \(\mathrm{H}_{2}\) to \(99 \mathrm{~mol} \%\) purity?

Short answer

(a) 23.3 min (b) Approximately 13-14 effusion steps

Step-by-step solution

Understand the Effusion Rate Formula

The effusion rate of two gases can be compared using Graham's law of effusion: \[ \frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{M_2}{M_1}} \] where \(M_1\) and \(M_2\) are the molar masses of the gases. In this case, gas 1 is \(\mathrm{H}_2\) and gas 2 is \(\mathrm{D}_2\).

Calculate the Molar Masses

\(\mathrm{H}_2\) has a molar mass of approximately \(2.016 \, g/mol\) and \(\mathrm{D}_2\) has a molar mass of approximately \(4.032 \, g/mol\).

Apply Graham's Law

Using Graham's law: \[ \frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{4.032\, g/mol}{2.016\, g/mol}} = \sqrt{2} \] Hence, \[ t_2 = t_1 \cdot \sqrt{2} \]

Calculate Time for \(\mathrm{D}_2\)

Given that \(t_1 = 16.5\) minutes for \(\mathrm{H}_2\), we can find \(t_2\) for \(\mathrm{D}_2\): \[ t_2 = 16.5 \, \text{min} \times \sqrt{2} \approx 16.5 \, \text{min} \times 1.414 \approx 23.3 \, \text{min} \]

Determine Number of Effusion Steps for Separation

We need to calculate the number of effusion steps \(n\) to achieve 99% purity in an equimolar mixture: \[ \left( \frac{99}{1} \right) = \left( \frac{0.99}{0.01} \right) = \left( 99 \right) \] This means that each step increases the ratio of the slower effusing gas (\(\mathrm{D}_2\)) to the faster one (\(\mathrm{H}_2\)).

Apply the Effusion Separation Formula

To find the number of steps required for the desired purity, use: \[ (\text{ratio})^n = \left( \frac{t_2}{t_1} \right)^n = 99 \] \[ (\sqrt{2})^n = 99 \] Taking the logarithm of both sides: \[ n \cdot \log(\sqrt{2}) = \log(99) \] \[ n \cdot \frac{\log(2)}{2} = \log(99) \] \[ n = \frac{2 \cdot \log(99)}{\log(2)} \] Since \(\log(99) \approx 1.996\) and \(\log(2) \approx 0.301\): \[ n \approx \frac{2 \cdot 1.996}{0.301} \approx 13.25 \]

Key concepts

Molar Mass

Molar mass refers to the mass of one mole of a given substance. For gases, the molar mass is crucial in determining various properties, including effusion rates. It is measured in grams per mole (g/mol).

For example, hydrogen gas, \(\text{H}_2\), has a molar mass of approximately 2.016 g/mol, while deuterium gas, \(\text{D}_2\), has a molar mass of about 4.032 g/mol.

Knowing the molar mass allows us to make comparisons, such as how quickly different gases effuse under the same conditions. Understanding molar masses is the first step in solving problems related to gas behavior and characteristics.

Effusion Rate

Effusion is the process by which gas molecules escape through a small hole into a vacuum. The rate at which a gas effuses depends on its molar mass, as described by Graham's law of effusion. This law states that the effusion rate of a gas is inversely proportional to the square root of its molar mass.

Mathematically, Graham's law can be written as: \[\frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{M_2}{M_1}}\] where \(M_1\) and \(M_2\) are the molar masses of the gases.

Let's say we have hydrogen gas (\(\text{H}_2\)) and deuterium gas (\(\text{D}_2\)). Using their molar masses (2.016 g/mol for \(\text{H}_2\) and 4.032 g/mol for \(\text{D}_2\)), we can determine that hydrogen gas effuses faster than deuterium gas by a factor related to the square root of 2—approximately 1.414.

Gas Separation

Gas separation by effusion leverages the differences in effusion rates of gases, based on their molar masses. For instance, if we have an equimolar mixture of \(\text{H}_2\) and \(\text{D}_2\), we can separate these gases by taking advantage of their different effusion rates.

The process involves multiple steps of effusion. Each step incrementally enriches the mixture, increasing the proportion of the gas that effuses slower—in this case, deuterium (\text{D}_2\).

To achieve a purity of 99% deuterium in the mixture, we use the equation: \[(\text{ratio})^n = \( \frac{t_2}{t_1}} )^n = 99\]
After finding the logarithm of both sides and solving for the number of steps \((n)\), it turns out to be approximately 13.25 steps. This means it takes just over 13 rounds of effusion to achieve the desired purity, utilizing the properties described by Graham's law.