Problem 82 - Chemistry A Molecular Approach

Question

Calculate the ratio of effusion rates for each pair of gases. a. \({ }^{238} \mathrm{UF}_{6}\) and \(\mathrm{ClF}\) b. \({ }^{238} \mathrm{UF}_{6}\) and \({ }^{235} \mathrm{UF}_{6}\)

Step-by-Step Solution

Answer

The ratio of effusion rates for \(^{238}UF_6\) and \(ClF\) is approximately \(\sqrt{54.5 / 352}\), and for \(^{238}UF_6\) and \(^{235}UF_6\) it is approximately \(\sqrt{349 / 352}\).

Step 1: Understand 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 is given by the formula \(r_1 / r_2 = sqrt(M_2 / M_1)\), where \(r_1\) and \(r_2\) are the effusion rates of gases 1 and 2, and \(M_1\) and \(M_2\) are the molar masses of the gases respectively.

Step 2: Calculate the ratio for \(^{238}UF_6\) and \(ClF\)

First, we find the molar masses of \(^{238}UF_6\) and \(ClF\). The molar mass of \(^{238}UF_6\) is \(238 + 6(19) = 352\) u, and the molar mass of \(ClF\) is \(35.5 + 19 = 54.5\) u. Using Graham's Law: \(ratio = sqrt(M_{ClF} / M_{^{238}UF_6}) = sqrt(54.5 / 352)\).

Step 3: Calculate the ratio for \(^{238}UF_6\) and \(^{235}UF_6\)

Since both gases are isotopes of the same compound, we only need to consider the difference in their molar masses due to the isotopes. The molar mass of \(^{235}UF_6\) is \(235 + 6(19) = 349\) u. Using Graham's Law: \(ratio = sqrt(M_{^{235}UF_6} / M_{^{238}UF_6}) = sqrt(349 / 352)\).

Key Concepts

Effusion Rate Ratios

When discussing the behavior of gases, effusion rate ratios play a crucial role in understanding how gases move through tiny openings. Effusion is a process by which gas particles passthrough a small hole into a vacuum without collisions between particles.

According to Graham's Law of Effusion, the rate of effusion for a gas is inversely proportional to the square root of its molar mass. This means that lighter gases effuse more quickly than heavier gases. For example, when comparing hydrogen gas to oxygen gas, hydrogen with a much smaller molar mass will effuse much faster. If we were to calculate effusion rate ratios between two gases with known molar masses, we would use the formula \(\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\), where \(r_1\) and \(r_2\) represent the effusion rates of the first and second gas, respectively, and \(M_1\) and \(M_2\) their molar masses.

Molar Mass

Understanding molar mass is pivotal when it comes to exploring the effusion of gases. Molar mass is defined as the mass of one mole of a substance and is typically expressed in units of grams per mole (g/mol) or atomic mass units (u).

The molar mass of a compound can be determined by summing the molar masses of its individual elements. For example, for water (H₂O), we would add twice the molar mass of hydrogen to the molar mass of oxygen. In the context of Graham's Law, the molar mass of each gas being compared is an essential variable, as it directly affects the rate at which the gas effuses. It is important to use accurate molar masses for precise calculations of effusion rates.

Gas Effusion

Gas effusion is a fundamental concept in physical chemistry that describes the movement of gas molecules through a small opening, usually into a vacuum. This movement occurs without collisions between gas molecules, which differentiates it from diffusion – the spread of gas molecules through random motion and collisions.

The rate of effusion can reveal much about the nature of the gas involved. For instance, in a vacuum system, a light gas like helium will effuse quickly, while a heavy gas like xenon will do so more slowly. Understanding gas effusion has practical implications as well, such as in separating isotopes in uranium enrichment processes or in detecting leaks in vacuum systems.

Isotopes

Isotopes are variants of a particular chemical element that differ in neutron number, and consequently in nuclear mass, but not in atomic number. Where isotopes are concerned, their different masses can affect physical properties like the rate of effusion.

In the exercise involving uranium hexafluoride (UF₆), we compared the effusion rates of two isotopes, \(^{238}UF_6\) and \(^{235}UF_6\). Their slight mass difference is due to the different isotopes of uranium they contain. Such differences are significant in industries like nuclear power where isotopic separation is used to obtain fuel-grade uranium, typically enriched in the lighter \(^{235}U\) isotope. Graham's Law can be applied here to predict the extent to which one isotope will effuse faster than the other.