Question

What is the ratio of effusion rates for the lightest gas, \(\mathrm{H}_{2}\), and the heaviest known gas, UF \(_{6} ?\)

Short answer

The ratio of effusion rates for H\(_2\) and UF\(_6\) is approximately 13.27.

Step-by-step solution

Understand Graham's Law of Effusion

According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The law is given by: \[ \text{Rate of Effusion} \propto \frac{1}{\sqrt{M}} \]

Write the Relation for Two Gases

The ratio of effusion rates for two gases can be written as follows using Graham's Law: \[ \frac{\text{Rate of Effusion of Gas 1}}{\text{Rate of Effusion of Gas 2}} = \sqrt{\frac{M_2}{M_1}} \] where \( M_1 \) and \( M_2 \) are the molar masses of the two gases.

Find Molar Masses of H\(_2\) and UF\(_6\)

Calculate the molar masses of H\(_2\) and UF\(_6\): - Molar mass of H\(_2\): \( 2 \) g/mol (since each H atom has a mass of approximately 1 g/mol) - Molar mass of UF\(_6\): \( 238 + (6 \times 19) = 238 + 114 = 352 \) g/mol (Uranium has a mass of 238 g/mol and each Fluorine atom has a mass of 19 g/mol).

Apply Graham's Law to Find the Ratio

Substitute the molar masses into Graham’s Law: \[ \frac{\text{Rate of Effusion of H}_2}{\text{Rate of Effusion of UF}_6} = \sqrt{\frac{352}{2}} = \sqrt{176} \] Simplify the expression: \[ \sqrt{176} \approx 13.27 \]

Key concepts

Effusion Rates

Effusion is the process by which gas particles pass through a tiny opening from one container to another. Here, 'effusion rate' refers to how quickly a gas escapes through this small hole. Understanding effusion rates is important because it helps us compare the behavior of different gases. According to Graham's Law of Effusion, lighter gases effuse faster than heavier gases because they move more quickly. This happens because lighter gas molecules have higher velocities at the same temperature.
So if you have a gas like hydrogen (H\(_2\)), it will effuse much faster than a heavier gas like uranium hexafluoride (UF\( \)6). This difference in effusion rates is what Graham's Law calculates.

Gas Molar Mass

The molar mass of a gas plays a crucial role in its effusion rate. Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Knowing the molar mass allows us to use it in mathematical equations to find how quickly gases like hydrogen and uranium hexafluoride will effuse. For instance, the molar mass of hydrogen, H \( \)2, is 2 g/mol. Each hydrogen atom has an approximate mass of 1 g/mol, and since there are two atoms, the total is 2 g/mol.
Conversely, uranium hexafluoride, UF \( \)6, has a much higher molar mass. Uranium's molar mass is 238 g/mol, and each fluorine atom's molar mass is 19 g/mol. UF \( \)6 includes one uranium atom and six fluorine atoms, bringing the total molar mass to 352 g/mol.

Inverse Proportionality

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. In mathematical terms, it is written as: \[ \text{Rate of Effusion} \propto \frac{1}{\sqrt{M}} \]. This means that if we increase the molar mass of a gas, its rate of effusion will decrease and vice versa.
The law can also be applied to compare two different gases. The ratio of their effusion rates can be found using the formula: \[ \frac{\text{Rate of Effusion of Gas 1}}{\text{Rate of Effusion of Gas 2}} = \sqrt{\frac{M_2}{M_1}} \] Here, \( M_1 \) and \( M_2 \) are the molar masses of the two gases. For example, to find the ratio of effusion rates for hydrogen and uranium hexafluoride, we use their molar masses (2 g/mol for H \( \)2 and 352 g/mol for UF \( \)6). Substituting these values in: \[ \frac{\text{Rate of Effusion of H}_2}{\text{Rate of Effusion of UF}_6} = \sqrt{\frac{352}{2}} = \sqrt{176} \approx 13.27 \].
This shows that hydrogen effuses about 13.27 times faster than uranium hexafluoride.