Problem 84 - Chemistry and Chemical Reactivity

Question

In an experiment, you have determined that 0.66 moles of \(\mathrm{CF}_{4}\) effuse through a porous barrier over a 4.8 -minute period. How long will it take for 0.66 moles of \(\mathrm{CH}_{4}\) to effuse through the same barrier?

Step-by-Step Solution

Answer

It will take approximately 11.3 minutes for 0.66 moles of \( \mathrm{CH}_{4} \) to effuse.

Step 1: Understanding Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it's expressed 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 gas 1 and gas 2.

Step 2: Identifying the Gases and Their Molar Masses

We have two gases: \( \mathrm{CF}_{4} \) and \( \mathrm{CH}_{4} \). First, calculate the molar masses: \( \mathrm{CF}_{4} \) has a molar mass of approximately 12 (C) + 4*19 (F) = 88 g/mol, whereas \( \mathrm{CH}_{4} \) has a molar mass of approximately 12 (C) + 4*1 (H) = 16 g/mol.

Step 3: Setting Up the Ratio of Rates

Using Graham's Law, set up the equation: \( \frac{\text{Rate of effusion of } \mathrm{CF}_{4}}{\text{Rate of effusion of } \mathrm{CH}_{4}} = \sqrt{\frac{16}{88}} \). Simplifying this gives us \( \sqrt{\frac{2}{11}} \).

Step 4: Calculating the Time for \( \, \mathrm{CH}_{4} \, \)

Since the number of moles is the same, time taken is inversely proportional to the rate. Therefore, we rearrange the equation to find time for \( \mathrm{CH}_{4} \): \( t_{\mathrm{CH}_{4}} = t_{\mathrm{CF}_{4}} \cdot \sqrt{\frac{11}{2}} \). Substituting \( t_{\mathrm{CF}_{4}} = 4.8 \) minutes, we get \( t_{\mathrm{CH}_{4}} = 4.8 \cdot \sqrt{5.5} \approx 11.3 \) minutes.

Key Concepts

Effusion

Effusion is a fascinating process that involves the movement of gas molecules through a tiny opening. It's like watching sand trickle through an hourglass, but at the molecular level. A gas will effuse through a small hole from one compartment to another much like how water flows through a narrow tube.

The rate at which a gas effuses is influenced by its molar mass. Simply put, lighter gas molecules will effuse faster than heavier ones. This is why lighter gases like hydrogen seem to disappear sooner from a deflated balloon than heavier gases like carbon dioxide.

In experiments and real-life applications, understanding effusion helps predict how different gases behave and explains why certain gases are used for particular purposes, like helium in balloons to make them float.

Molar Mass Calculation

Calculating the molar mass of a gas is crucial when applying Graham's Law of Effusion. Molar mass is a measure of the mass of one mole of a substance and is expressed in grams per mole (g/mol). This can be deduced by summing the atomic masses of the elements in a compound.

For example, if you consider carbon tetrafluoride (\(\mathrm{CF}_{4}\)) and methane (\(\mathrm{CH}_{4}\)), their molar masses are calculated as follows:

• For \(\mathrm{CF}_{4}\): Carbon has a molar mass of 12 g/mol, and fluorine is 19 g/mol. Since \(\mathrm{CF}_{4}\) consists of one carbon atom and four fluorine atoms, you calculate: 12 + 4*19 = 88 g/mol.
• For \(\mathrm{CH}_{4}\): Carbon again has a molar mass of 12 g/mol, and hydrogen is 1 g/mol. With one carbon and four hydrogen atoms, it becomes: 12 + 4*1 = 16 g/mol.

These calculations form the bedrock for determining how fast each gas effuses, as their molar masses directly affect the rate of effusion through Graham's Law.

Rate of Effusion

The rate of effusion is the speed at which gas molecules escape through a porous membrane. Graham's Law provides a helpful formula to compare effusion rates of different gases based on their molar masses. It states that the rate of effusion is inversely proportional to the square root of the molar mass, meaning lighter gases effuse more quickly than heavier ones.

Mathematically, Graham's Law of Effusion can be expressed as:

• \(\frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{M_2}{M_1}}\)

Here, \(M_1\) and \(M_2\) are the molar masses of the gases. Applying this equation helps in predicting how gases will behave in different conditions. For instance, if you know that a heavier gas takes longer to effuse through the same barrier as a lighter gas, you can conclude that the heavier gas has a slower rate of effusion.

In practical terms, this concept helps chemists design experiments and industries choose the right gas for processes where effusion plays a central role.