Problem 87 - Chemistry: The Central Science

Question

Hydrogen has two naturally occurring isotopes, \({ }^{1} \mathrm{H}\) and \({ }^{2} \mathrm{H}\). Chlorine also has two naturally occurring isotopes, \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\). Thus, hydrogen chloride gas consists of four distinct types of molecules: \({ }^{1} \mathrm{H}^{35} \mathrm{Cl},{ }^{1} \mathrm{H}^{37} \mathrm{Cl},{ }^{2} \mathrm{H}^{35} \mathrm{Cl}\), and \({ }^{2} \mathrm{H}^{37} \mathrm{Cl}\). Place these four molecules in order of increasing rate of effusion.

Step-by-Step Solution

Answer

The order of increasing rate of effusion for the four hydrogen chloride gas molecules is: \({ }^{2} \mathrm{H}^{37} \mathrm{Cl} < { }^{1} \mathrm{H}^{37} \mathrm{Cl} < { }^{2} \mathrm{H}^{35} \mathrm{Cl} < { }^{1} \mathrm{H}^{35} \mathrm{Cl}\)

Step 1: Determine the molar mass of each molecule

To apply Graham's law of effusion, we first need to determine the molar mass of each hydrogen chloride gas molecule. We can do this by adding the molar mass of hydrogen (or deuterium, the heavier isotope of hydrogen) and the molar mass of chlorine. For \({ }^{1} \mathrm{H}^{35} \mathrm{Cl}\): Molar mass = 1 (for hydrogen) + 35 (for chlorine) = 36 g/mol For \({ }^{1} \mathrm{H}^{37} \mathrm{Cl}\): Molar mass = 1 (for hydrogen) + 37 (for chlorine) = 38 g/mol For \({ }^{2} \mathrm{H}^{35} \mathrm{Cl}\): Molar mass = 2 (for deuterium) + 35 (for chlorine) = 37 g/mol For \({ }^{2} \mathrm{H}^{37} \mathrm{Cl}\): Molar mass = 2 (for deuterium) + 37 (for chlorine) = 39 g/mol

Step 2: Apply 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. We can express this relationship as: \(Rate \propto \frac{1}{\sqrt{Molar \, Mass}}\) Now we'll compare the rates of effusion for each molecule by comparing the inverse square root of their molar masses.

Step 3: Calculate the inverse square roots of molar masses

For \({ }^{1} \mathrm{H}^{35} \mathrm{Cl}\): Inverse square root = \( \frac{1}{\sqrt{36}} = \frac{1}{6} \) For \({ }^{1} \mathrm{H}^{37} \mathrm{Cl}\): Inverse square root = \( \frac{1}{\sqrt{38}} \) For \({ }^{2} \mathrm{H}^{35} \mathrm{Cl}\): Inverse square root = \( \frac{1}{\sqrt{37}} \) For \({ }^{2} \mathrm{H}^{37} \mathrm{Cl}\): Inverse square root = \( \frac{1}{\sqrt{39}} \)

Step 4: Arrange the molecules in order of increasing rate of effusion

Comparing the inverse square roots of the molar masses, we can determine the order of increasing rate of effusion: 1. \({ }^{2} \mathrm{H}^{37} \mathrm{Cl}\) (smallest inverse square root) 2. \({ }^{1} \mathrm{H}^{37} \mathrm{Cl}\) 3. \({ }^{2} \mathrm{H}^{35} \mathrm{Cl}\) 4. \({ }^{1} \mathrm{H}^{35} \mathrm{Cl}\) (largest inverse square root) So, the order of increasing rate of effusion for the four hydrogen chloride gas molecules is as follows: \({ }^{2} \mathrm{H}^{37} \mathrm{Cl} < { }^{1} \mathrm{H}^{37} \mathrm{Cl} < { }^{2} \mathrm{H}^{35} \mathrm{Cl} < { }^{1} \mathrm{H}^{35} \mathrm{Cl}\)

Key Concepts

Isotopes

Isotopes are versions of the same chemical element that differ in neutron numbers, and consequently in nucleon numbers. This means they have the same number of protons but a different number of neutrons. Take hydrogen as an example: it has two naturally occurring isotopes, namely,

• **Protium ( ** ¹H**): with no neutrons and a mass number of 1
• **Deuterium ( ²H**): with one neutron and a mass number of 2

The variation in neutron number results in different atomic masses for these isotopes. Similarly, chlorine has two isotopes, ** ³⁵Cl** and ** ³⁷Cl**, making the molecules they form with hydrogen atoms varied in mass. The presence of different isotopes leads to unique properties, especially when calculating properties that depend on atomic mass. For example, in the context of Graham's law of effusion, the type of isotopes affects the overall molar mass of the molecules.

Molar Mass Calculation

Calculating the molar mass is fundamental when determining the behavior of molecules during chemical reactions or processes like effusion. To find the molar mass of a compound, simply add up the atomic masses of each atom within the molecule. Here is how it's done:

• Identify each element within the molecule and the number of its atoms present.
• Look up the atomic mass of each element. For isotopes, this will be specific to the isotope's form.
• Multiply the atomic mass by the number of atoms to get the total mass of that element within the molecule.
• Sum up the results for each element to get the total molar mass of the molecule.

For example, in hydrogen chloride (HCl) molecules, we see combinations like ¹H³⁵Cl **which has a molar mass calculated as follows:
Molar mass = 1 (for ¹H) + 35 (for ³⁵Cl) = 36 g/mol. Different combinations result in different overall molecular weights, guiding how they behave under physical processes.

Rate of Effusion

The rate of effusion, as described by Graham's law, is inversely proportional to the square root of a molecule's molar mass. The formula is expressed as:\[ Rate \propto \frac{1}{\sqrt{\text{Molar Mass}}} \]This implies that lighter molecules, with lower molar masses, effuse faster than heavier ones. Graham's law is a helpful tool in predicting and comparing rates of effusion. As illustrated in the original exercise, different isotope combinations of hydrogen chloride result in varying molar masses:

• For ¹H³⁵Cl: Molar mass = 36 g/mol
• For ¹H³⁷Cl: Molar mass = 38 g/mol
• For ²H³⁵Cl: Molar mass = 37 g/mol
• For ²H³⁷Cl: Molar mass = 39 g/mol

By comparing the inverse square roots of these molar masses, we ascertain the ranking of molecules from fastest to slowest effusion as:

• ¹H³⁵Cl (quickest)
• ²H³⁵Cl
• ¹H³⁷Cl
• ²H³⁷Cl (slowest)

Such insights are vital in fields requiring gas separation or when understanding chemical kinetics.