Problem 85 - 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 \(\mathrm{Cl}\) and37 Cl. Thus, hydrogen chloride gas consists of four distinct types of molecules: \(^{1} \mathrm{H}^{35} \mathrm{Cl},^{2} \mathrm{H}^{37} \mathrm{Cl},^{2} \mathrm{H}^{35} \mathrm{Cl},\) 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 hydrogen chloride gas molecules is as follows: \(^{2}\mathrm{H}^{37}\mathrm{Cl}\), \(^{1}\mathrm{H}^{37}\mathrm{Cl}\), \(^{2}\mathrm{H}^{35}\mathrm{Cl}\), and \(^{1}\mathrm{H}^{35}\mathrm{Cl}\) (from lowest to highest rate).

Step 1: Determine the Molar Masses of Each Molecule

To determine the molar mass of each molecule, we need to add the molar mass of hydrogen and the molar mass of chlorine isotope. The molar masses for each isotope should be provided either in the question or in a periodic table, but for this exercise we will assume they are as follows: - \(^{1}\mathrm{H}\): 1 g/mol - \(^{2}\mathrm{H}\): 2 g/mol - \(^{35}\mathrm{Cl}\): 35 g/mol - \(^{37}\mathrm{Cl}\): 37 g/mol Using these values, we can now determine the molar masses for each of the four molecules: 1. \(^{1}\mathrm{H}^{35}\mathrm{Cl}\): 1 + 35 = 36 g/mol 2. \(^{1}\mathrm{H}^{37}\mathrm{Cl}\): 1 + 37 = 38 g/mol 3. \(^{2}\mathrm{H}^{35}\mathrm{Cl}\): 2 + 35 = 37 g/mol 4. \(^{2}\mathrm{H}^{37}\mathrm{Cl}\): 2 + 37 = 39 g/mol

Step 2: Apply Graham's Law of Effusion

Graham's Law states that the rate of effusion is inversely proportional to the square root of the molar mass. Mathematically, this means that: Rate of effusion \(∝ \dfrac{1}{\sqrt{Molar\ Mass}}\) Now, we can compare the rates of effusion for the four molecules based on their molar masses: 1. \(^{1}\mathrm{H}^{35}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{36}}\) = \(\dfrac{1}{6}\) 2. \(^{1}\mathrm{H}^{37}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{38}}\) 3. \(^{2}\mathrm{H}^{35}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{37}}\) 4. \(^{2}\mathrm{H}^{37}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{39}}\) It should be noted that the values don't need to be simplified, as we're only interested in comparing the rates. Since larger molar mass results in a smaller rate of effusion and smaller molar mass results in a larger rate, we can now put the molecules in order of increasing rate of effusion: 1. \(^{2}\mathrm{H}^{37}\mathrm{Cl}\) - the lowest rate (highest molar mass) 2. \(^{1}\mathrm{H}^{37}\mathrm{Cl}\) 3. \(^{2}\mathrm{H}^{35}\mathrm{Cl}\) 4. \(^{1}\mathrm{H}^{35}\mathrm{Cl}\) - the highest rate (lowest molar mass)

Key Concepts

Isotopes

Isotopes are different forms of the same element that have the same number of protons but a different number of neutrons in the nucleus. This leads to different atomic masses. Thus, isotopes of an element share chemical properties, but their physical properties might vary due to the difference in mass.

For example, in the original exercise, hydrogen has two isotopes: \(^{1}\mathrm{H}\) and \(^{2}\mathrm{H}\). \(^{1}\mathrm{H}\) has one proton and no neutrons, while \(^{2}\mathrm{H}\) has one proton and one neutron, which makes it heavier. Similarly, chlorine has two isotopes, \(^{35}\mathrm{Cl}\) and \(^{37}\mathrm{Cl}\), with different neutron numbers contributing to their distinct masses.

Understanding isotopes is essential when studying molecular behavior, as their differing masses affect physical phenomena such as the rate of effusion.

Molar Mass

Molar mass is a critical property in chemistry, representing the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. It is expressed in grams per mole (g/mol). Molar mass is important when dealing with chemical reactions and equations, as it allows us to convert between the mass of a compound and the amount in moles.

To find the molar mass of a compound, you add up the molar masses of its constituent elements. In the exercise, hydrogen chloride molecules are composed of hydrogen and chlorine isotopes. We calculated the molar masses for each variant:

• \(^{1}\mathrm{H}^{35}\mathrm{Cl}\): 36 g/mol
• \(^{1}\mathrm{H}^{37}\mathrm{Cl}\): 38 g/mol
• \(^{2}\mathrm{H}^{35}\mathrm{Cl}\): 37 g/mol
• \(^{2}\mathrm{H}^{37}\mathrm{Cl}\): 39 g/mol

The differing molar masses are instrumental in determining the rate of effusion as per Graham's Law, since heavier molecules tend to effuse more slowly.

Effusion Rate

The effusion rate of a gas is a measure of how quickly gas molecules pass through a small opening. Graham's Law of Effusion states that the rate of effusion is inversely proportional to the square root of the molar mass of the gas molecules. Mathematically, it can be expressed as:
\[\text{Rate of effusion} \propto \frac{1}{\sqrt{\text{Molar Mass}}}\]

This implies that lighter molecules effuse faster than heavier ones. In the exercise, the molar masses of the different hydrogen chloride molecules help predict their effusion rates. Smaller molar mass leads to a higher rate as calculated:

• \(^{1}\mathrm{H}^{35}\mathrm{Cl}\): the highest rate
• \(^{2}\mathrm{H}^{35}\mathrm{Cl}\)
• \(^{1}\mathrm{H}^{37}\mathrm{Cl}\)
• \(^{2}\mathrm{H}^{37}\mathrm{Cl}\): the lowest rate

By evaluating the effusion rates, students can enhance their understanding of gas behavior and the practical implications of molar mass variations due to isotopes.