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Chapter 10: Problem 87

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 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 by step solution

01

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
02

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.
03

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 = \( \frac{1}{\sqrt{39}} \)
04

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: \({ }^{2} \mathrm{H}^{37} \mathrm{Cl} < { }^{1} \mathrm{H}^{37} \mathrm{Cl} < { }^{2} \mathrm{H}^{35} \mathrm{Cl} < { }^{1} \mathrm{H}^{35} \mathrm{Cl}\)

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Key Concepts

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

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 鹿H) + 35 (for 鲁鈦礐l) = 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.

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Most popular questions from this chapter

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 mixture containing \(0.765 \mathrm{~mol} \mathrm{He}(\mathrm{g}), 0.330 \mathrm{~mol} \mathrm{Ne}(\mathrm{g})\), and \(0.110 \mathrm{~mol} A r(g)\) is confined in a \(10.00-\mathrm{L}\) vessel at \(25^{\circ} \mathrm{C}\). (a) Calculate the partial pressure of each of the gases in the mixture. (b) Calculate the total pressure of the mixture.

A glass vessel fitted with a stopcock valve has a mass of \(337.428 \mathrm{~g}\) when evacuated. When filled with Ar, it has a mass of \(339.854 \mathrm{~g}\). When evacuated and refilled with a mixture of Ne and Ar, under the same conditions of temperature and pressure, it has a mass of \(339.076 \mathrm{~g}\). What is the mole percent of Ne in the gas mixture?

A sample of \(5.00 \mathrm{~mL}\) of diethylether \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OC}_{2} \mathrm{H}_{5}\right.\) ? density \(=0.7134 \mathrm{~g} / \mathrm{mL}\) ) is introduced into a \(6.00-\mathrm{L}\) vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), whose partial pressures are \(P_{\mathrm{N}_{2}}=0.751 \mathrm{~atm}\) and \(P_{\mathrm{O}_{1}}=0.208 \mathrm{~atm}\). The temperature is held at \(35.0^{\circ} \mathrm{C}\), and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.

Determine whether each of the following changes will increase, decrease, or not affect the rate with which gas molecules collide with the walls of their container: (a) increasing the volume of the container, (b) increasing the temperature, (c) increasing the molar mass of the gas.
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