Question

If \(0.10 \mathrm{~mol}\) of \(\mathrm{I}_{2}\) vapor can effuse from an opening in a heated vessel in \(39 \mathrm{~s}\), how long will it take \(0.10 \mathrm{~mol} \mathrm{H}_{2}\) to effuse under the same conditions?

Short answer

It takes approximately 439 seconds for 0.10 mol of H2 to effuse.

Step-by-step solution

Identify the Given Quantities

We are given that the effusion time for 0.10 mol of Iodine (\(\mathrm{I}_2\)) vapor is 39 seconds. We need to find the time taken for the same amount of Hydrogen (\(\mathrm{H}_2\)) to effuse under similar conditions.

Use Graham's Law of Effusion

According to Graham's Law, the rate of effusion is inversely proportional to the square root of molar mass. The formula is given by \[\frac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}} = \sqrt{\frac{M_B}{M_A}}\]where \(M_A\) and \(M_B\) are the molar masses of gases A and B, respectively.

Calculate Molar Masses

The molar mass of \(\mathrm{I}_2\) is roughly 253.8 g/mol and the molar mass of \(\mathrm{H}_2\) is roughly 2.016 g/mol. These values are used to apply Graham's Law.

Set Up the Proportionality Equation

Using Graham's Law and the given data, we have:\[\frac{\text{Time for } \mathrm{H}_2}{\text{Time for } \mathrm{I}_2} = \sqrt{\frac{M_{\mathrm{I}_2}}{M_{\mathrm{H}_2}}}\]Substitute the values of molar masses:\[\frac{t_{\mathrm{H}_2}}{39 \: \mathrm{s}} = \sqrt{\frac{253.8}{2.016}}\]

Solve for the time taken by \(\mathrm{H}_2\)

Calculate the square root:\[\sqrt{\frac{253.8}{2.016}} \approx 11.25\]Hence,\[\frac{t_{\mathrm{H}_2}}{39} = 11.25\]Solving for \(t_{\mathrm{H}_2}\) gives:\[t_{\mathrm{H}_2} = 39 \times 11.25 = 438.75 \: \mathrm{s}\]

Key concepts

Effusion Time

Effusion time refers to the duration required for a gas to move through a small opening from one side of a barrier to the other. This process is distinctly different from diffusion since it involves the passage of gas molecules through minute openings rather than mixing in open space.

• Effusion is influenced by the volume of the container, the temperature, and the characteristics of the gas, such as molar mass.
• Time taken for effusion is an essential metric because it helps us compare how fast different gases undergo this process.

In the context of the exercise, we are tasked to determine the time it would take for hydrogen gas (\(\mathrm{H}_2\)) to effuse, given the data for iodine vapor (\(\mathrm{I}_2\)) effusion. It's important to maintain the same conditions to apply Graham's Law accurately, ensuring factors like temperature don't skew the results.

Molar Mass

Molar mass is a critical factor in determining how swiftly gases effuse. It represents the mass of one mole of a substance and is measured in grams per mole (g/mol). The molar mass is calculated by summing up the atomic masses of all atoms in a molecule.

• Lighter gases, like \(\mathrm{H}_2\) with a molar mass of around 2.016 g/mol, effuse more rapidly due to their lower inertia.
• Heavier gases, such as \(\mathrm{I}_2\) with a molar mass of about 253.8 g/mol, take longer to effuse.

Graham’s Law of Effusion leverages the molar mass to relate the effusion rates of different gases through the equation:\[\frac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}} = \sqrt{\frac{M_B}{M_A}}\]This equation highlights that the rate of effusion is inversely proportional to the square root of the mass of the molecules involved, allowing us to estimate the time required for effusion by comparing the molar masses.

Rate of Effusion

The rate of effusion measures how quickly a gas can pass through a small opening. It is influenced by both the molar mass of the gas and the temperature of the environment.

• The rate is calculated as the quantity of gas escaping per unit time.
• According to Graham’s Law, the rate of effusion is inversely proportional to the square root of the molar mass of the gas: lighter gases have a higher rate of effusion.

In our case, by knowing the effusion time it takes for a known amount of \(\mathrm{I}_2\) vapor to pass through the barrier, we can rearrange the formula to find the time it takes \(\mathrm{H}_2\) under the same conditions:\[\frac{t_{\mathrm{H}_2}}{t_{\mathrm{I}_2}} = \sqrt{\frac{M_{\mathrm{I}_2}}{M_{\mathrm{H}_2}}}\]Here, the calculated square root of the ratio of their molar masses gives us the factor by which the time for hydrogen's effusion will increase compared to iodine's. This helps in applying Graham's Law to make accurate predictions about effusion times.