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Chapter 10: Problem 85

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.

Short Answer

Expert verified
The order of increasing rate of effusion 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 by step solution

01

Understanding the Problem

We need to order four different molecules based on their rate of effusion. These molecules are formed with isotopes of hydrogen (\(^{1}\mathrm{H}\) and \(^{2}\mathrm{H}\)) and isotopes of chlorine (\(^{35}\mathrm{Cl}\) and \(^{37}\mathrm{Cl}\)). The molecules are: \(^{1}\mathrm{H}^{35}\mathrm{Cl}\), \(^{1}\mathrm{H}^{37}\mathrm{Cl}\), \(^{2}\mathrm{H}^{35}\mathrm{Cl}\), and \(^{2}\mathrm{H}^{37}\mathrm{Cl}\).
02

Applying 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 of the gas. Therefore, molecules with a smaller molar mass will effuse faster than those with a larger molar mass.
03

Calculating Molar Mass of Molecules

Determine the molar mass of each molecule: - \(^{1}\mathrm{H}^{35}\mathrm{Cl}\) has a molar mass of 1 + 35 = 36 g/mol.- \(^{1}\mathrm{H}^{37}\mathrm{Cl}\) has a molar mass of 1 + 37 = 38 g/mol.- \(^{2}\mathrm{H}^{35}\mathrm{Cl}\) has a molar mass of 2 + 35 = 37 g/mol.- \(^{2}\mathrm{H}^{37}\mathrm{Cl}\) has a molar mass of 2 + 37 = 39 g/mol.
04

Ordering by Increasing Molar Mass

Based on the calculated molar masses, arrange the molecules in order of increasing molar mass: \(^{1}\mathrm{H}^{35}\mathrm{Cl}\), \(^{2}\mathrm{H}^{35}\mathrm{Cl}\), \(^{1}\mathrm{H}^{37}\mathrm{Cl}\), \(^{2}\mathrm{H}^{37}\mathrm{Cl}\).
05

Determining the Rate of Effusion

According to Graham's law, the molecule with the smallest molar mass will have the fastest rate of effusion. Therefore, the order of increasing rate of effusion (or decreasing molar mass) will be: \(^{2}\mathrm{H}^{37}\mathrm{Cl}\), \(^{1}\mathrm{H}^{37}\mathrm{Cl}\), \(^{2}\mathrm{H}^{35}\mathrm{Cl}\), \(^{1}\mathrm{H}^{35}\mathrm{Cl}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isotopes
Isotopes are variations of the same chemical element that have identical numbers of protons but different numbers of neutrons. This means isotopes have the same atomic number but different mass numbers. For instance, hydrogen has two common isotopes:
  • 1H (protium) which has one proton and no neutrons.
  • 2H (deuterium) which has one proton and one neutron.
Chlorine also features isotopes like 35Cl and 37Cl. These isotopes impact the mass of molecules, such as hydrogen chloride. Because isotopes affect the total number of neutrons in an atom, they also alter the molecular mass, influencing processes like effusion and diffusion.
Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). To calculate the molar mass of a compound, you sum up the atomic masses of its constituent atoms. For hydrogen chloride molecules with isotopes, it becomes crucial because their molar mass will differ based on the isotopes they contain. Here鈥檚 how you'd calculate the molar mass for chlorinated compounds:
  • 1H35Cl: 1 + 35 = 36 g/mol
  • 1H37Cl: 1 + 37 = 38 g/mol
  • 2H35Cl: 2 + 35 = 37 g/mol
  • 2H37Cl: 2 + 37 = 39 g/mol
This calculation is an essential step in applying Graham's Law for effusion, as it enables you to compare the rates of effusion based on molecular weights.
Effusion Rate
Effusion rate refers to how quickly gas molecules escape through a tiny hole into a vacuum. 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 simpler terms, lighter gases effuse faster than heavier ones. When comparing the hydrogen chloride isotopes, you should recognize that:
  • Lower molar mass leads to faster effusion.
  • Because 1H35Cl has the lowest molar mass (36 g/mol), it effuses the fastest.
  • On the other hand, 2H37Cl, with the highest molar mass (39 g/mol), effuses the slowest.
Understanding the effusion rate is crucial in various chemical processes and separation tasks concerned with isotopic molecules.
Molecular Weight Calculation
Calculating molecular weight mainly involves summing the atomic weights of all atoms present in a molecule. To determine the molecular weight of isotopic molecules like different types of hydrogen chloride, you need to
  • Identify the atomic weight of each atom, considering their isotopes.
  • Sum these weights for all atoms in the molecule.
For example, you add the atomic weights of hydrogen and chlorine isotopes to find the molecular weights of compounds like 1H35Cl and 2H37Cl. This computation is pivotal for understanding how each variant behaves in effusion and other chemical reactions, as molecular weight directly influences these behaviors.

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Most popular questions from this chapter

You have a sample of gas at \(0^{\circ} \mathrm{C}\). You wish to increase the rms speed by a factor of 3 . To what temperature should the gas be heated?

A fixed quantity of gas at \(25^{\circ} \mathrm{C}\) exhibits a pressure of \(99 \mathrm{kPa}\) and occupies a volume of \(4.00 \mathrm{~L}\). (a) Calculate the volume the gas will occupy if the pressure is increased to \(202.6 \mathrm{kPa}\) whilethe temperature is held constant. (b) Calculate the volume the gas will occupy if the temperature is increased to \(100^{\circ} \mathrm{C}\) while the pressure is held constant.

When a large evacuated flask is filled with argon gas, its mass increases by \(3.224 \mathrm{~g}\). When the same flask is again evacuated and then filled with a gas of unknown molar mass, the mass increase is \(8.102 \mathrm{~g}\). (a) Based on the molar mass of argon, estimate the molar mass of the unknown gas. (b) What assumptions did you make in arriving at your answer?

A \(6.53-g\) sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces \(1.72 \mathrm{~L}\) of carbon dioxide gas at \(28^{\circ} \mathrm{C}\) and \(99.06 \mathrm{kPa}\) pressure. (a) Write balanced chemical equations for the reactions that occur between hydrochloric acid and each component of the mixture. (b) Calculate the total number of moles of carbon dioxide that forms from these reactions. (c) Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

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