Question

Hydrogen has two stable isotopes, \({ }^{1} \mathrm{H}\) and \({ }^{2} \mathrm{H}\), with atomic weights of 1.0078 amu and 2.0141 amu, respectively. Ordinary hydrogen gas, \(\mathrm{H}_{2}\), is a mixture consisting mostly of \({ }^{1} \mathrm{H}_{2}\) and \({ }^{1} \mathrm{H}^{2} \mathrm{H}\). Calculate the ratio of rates of effusion of \({ }^{1} \mathrm{H}_{2}\) and \({ }^{1} \mathrm{H}^{2} \mathrm{H}\) under the same conditions.

Short answer

The ratio of effusion rates of \( { }^{1} \mathrm{H}_{2} \) to \( { }^{1} \mathrm{H}^{2} \mathrm{H} \) is approximately 1.224.

Step-by-step solution

Understanding Graham's Law of Effusion

Graham's Law states that the rate of effusion of gases is inversely proportional to the square root of their molar masses. Mathematically, it is given by \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r_1 \) and \( r_2 \) are the rates of effusion of gas 1 and gas 2 respectively, and \( M_1 \) and \( M_2 \) are their respective molar masses.

Calculate the Molar Masses of Isotopes

For \( { }^{1} \mathrm{H}_{2} \), the molar mass is twice the atomic weight of \( { }^{1} \mathrm{H} \), which is \( 2 \times 1.0078 = 2.0156 \) amu. For \( { }^{1} \mathrm{H}^{2} \mathrm{H} \), the molar mass is the sum of the atomic weights of \( { }^{1} \mathrm{H} \) and \( { }^{2} \mathrm{H} \): \( 1.0078 + 2.0141 = 3.0219 \) amu.

Apply Graham's Law to Calculate the Effusion Rate Ratio

Substitute the molar masses into Graham's Law to find the ratio of effusion rates:\[ \frac{r({ }^{1} \mathrm{H}_{2})}{r({ }^{1} \mathrm{H}^{2} \mathrm{H})} = \sqrt{\frac{3.0219}{2.0156}} \]Calculate the square root to find the ratio.

Perform the Square Root Calculation

Compute the square root:\[ \frac{r({ }^{1} \mathrm{H}_{2})}{r({ }^{1} \mathrm{H}^{2} \mathrm{H})} = \sqrt{1.498} \approx 1.224 \]Therefore, the ratio of the rates of effusion is approximately 1.224.

Key concepts

Graham's Law

Graham's Law of Effusion is a principle that helps us understand how gases behave. Effusion refers to the process of gas molecules escaping through a tiny opening. According to Graham's Law, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. This means that lighter gases effuse faster than heavier ones. The mathematical expression for Graham's Law is given by:

• \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \)

where \( r_1 \) and \( r_2 \) are the effusion rates of two different gases, and \( M_1 \) and \( M_2 \) are their respective molar masses. This equation allows us to compare how quickly two gases will effuse under identical conditions. By understanding and applying Graham's Law, we can make predictions about gas behavior that is crucial in fields such as chemistry and physics.

This concept can seem tricky at first, but visualizing the process can help. Imagine inflating two balloons with different gases. Over time, you'll notice that the one with a lighter gas deflates more quickly due to the faster rate of effusion.

Isotopes

Isotopes are variants of the same chemical element that have the same number of protons but a different number of neutrons. This difference in neutron count results in isotopes of an element having slightly different atomic weights, although chemically they behave similarly.

For example, hydrogen has three primary isotopes:

• Protium (\({ }^1\mathrm{H}\))
• Deuterium (\({ }^2\mathrm{H}\))
• Tritium (\({ }^3\mathrm{H}\))

Protium, with just one proton and no neutrons, is the most common hydrogen isotope. Deuterium, which contains one proton and one neutron, has a greater atomic weight. The rarity of tritium sets it apart as it is radioactive.

Understanding isotopes is important in many scientific fields, from studying chemical reactions to dating ancient artifacts. Chemists and physicists use isotopes to explore molecular dynamics and thermodynamics due to their unique properties and interactions.

Hydrogen Isotopes

Hydrogen isotopes, specifically protium and deuterium, are particularly interesting due to their simplicity and widespread presence in the universe. Protium, \({ }^1\mathrm{H}\), is the most abundant hydrogen isotope, composing about 99.98% of the natural hydrogen found on Earth. In contrast, deuterium, \({ }^2\mathrm{H}\), accounts for only about 0.02% of hydrogen atoms.

Despite this small prevalence, deuterium plays significant roles in various applications, such as nuclear fusion and tracing chemical reactions. The heavier atomic mass of deuterium makes it useful in heavy water reactors and in studies involving isotope exchange reactions. In our context, understanding the effusion rates of hydrogen isotopes helps scientists utilize these substances in practical applications like hydrogen storage and isotope separation.

In the given exercise, calculating the effusion rates of these isotopes allows us to see how variations in atomic mass can affect gaseous behavior, demonstrating how even small differences can be profound.

Molar Mass

Molar mass is crucial in understanding gas behavior and is defined as the mass of one mole of a substance. It's expressed in units such as grams per mole (g/mol). Molar mass combines atomic weights from the periodic table to yield a substance's mass value over one mole.

For instance, to calculate the molar mass of hydrogen gas (\(\mathrm{H}_2 \)), we add the atomic masses of two hydrogen atoms. For protium, the molar mass calculation would look like:

• Two times 1.0078 amu for \({ }^1\mathrm{H}_2\), resulting in 2.0156 amu.

If different isotopes make up a molecule, such as \({ }^1\mathrm{H}^2\mathrm{H} \), the atomic masses are summed based on isotope contribution, leading to a distinct molar mass.

Molar mass influences various properties, including the effusion and diffusion rates of gases. Knowing the molar mass aids in accurate predictions and calculations within chemical reactions, benefiting both academic studies and industrial applications. Understanding molar mass is fundamental to mastering topics related to gas laws and chemical equations.