/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 133 Gases such as \(\mathrm{CO}\) ar... [FREE SOLUTION] | 91影视

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Gases such as \(\mathrm{CO}\) are gradually oxidized in the atmosphere, not by \(\mathrm{O}_{2}\) but by the hydroxyl radical, \(\cdot \mathrm{OH},\) a species with one fewer electron than a hydroxide ion. At night, the \cdotOH concentration is nearly zero, but it increases to \(2.5 \times 10^{12}\) molecules \(/ \mathrm{m}^{3}\) in polluted air during the day. At daytime conditions of 1.00 atm and \(22^{\circ} \mathrm{C},\) what is the partial pressure and mole percent of \(\cdot \mathrm{OH}\) in air?

Short Answer

Expert verified
The partial pressure of \(\cdot \text{OH}\) in air is \( 1.00 \times 10^{-13} \text{ atm} , \) and the mole percent is \( 1.00 \times 10^{-11} % . \)

Step by step solution

01

- Calculate Total Concentration of Air Molecules

Use the Ideal Gas Law to find the total concentration of air molecules at 1.00 atm and \(22^\text{掳}\)C.\[PV = nRT\]
02

- Convert Temperature

Convert \(22^\text{掳}\)C to Kelvin.\[ T(K) = 22 + 273.15 = 295.15 \text{ K} \]
03

- Solve for n/V (Moles per Cubic Meter)

Rearrange the Ideal Gas Law to solve for \(n/V\) (the concentration of air molecules in moles per cubic meter).\[\frac{n}{V} = \frac{P}{RT} = \frac{1.00 \text{ atm}}{(0.0821 \text{ L} \cdot \text{atm} / \text{K} \cdot \text{mol}) \cdot 295.15 \text{ K}} = 0.0413 \dfrac{\text{mol}}{\text{L}} = 41.3 \dfrac{\text{mol}}{\text{m}^3}\]
04

- Convert Hydroxyl Radical Concentration to Moles

Convert the concentration of \(\cdot \text{OH}\) from molecules per cubic meter to moles per cubic meter using Avogadro's number \(6.022 \times 10^{23} \text{ molecules} / \text{mol}\).\[\dfrac{2.5 \times 10^{12} \text{ molecules} / \text{m}^3}{6.022 \times 10^{23} \text{ molecules} / \text{mol}} = 4.15 \times 10^{-12} \dfrac{\text{mol}}{\text{m}^3}\]
05

- Calculate Partial Pressure

The partial pressure can be calculated using the mole fraction and the total pressure.\[d_{\cdot \text{OH}} = \dfrac{n_{\cdot \text{OH}}}{n_{\text{total}}} = \dfrac{4.15 \times 10^{-12}\text{ mol}}{41.3 \text{ mol}} = 1.00 \times 10^{-13} \;P_{\cdot \text{OH}} = (1.00 \times 10^{-13}) \times (1.00\, \text{atm}) = 1.00 \times 10^{-13} \text{ atm}\]
06

- Calculate Mole Percent of Hydroxyl Radical

To find the mole percent of \(\cdot \text{OH},\) multiply the mole fraction by 100%.\[\text{Mole percent} = (1.00 \times 10^{-13}) \times 100\text{%} = 1.00 \times 10^{-11}%\]

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Key Concepts

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

Partial Pressure
In simple terms, partial pressure is the pressure that a single gas in a mixture would exert if it were alone in the entire volume. Think of it like this: if you have a mixture of gases in a balloon, each gas has its own contribution to the overall pressure inside the balloon. Partial pressure is crucial when dealing with gas mixtures because each gas behaves independently, even though they share the same space.

You can calculate the partial pressure using the mole fraction of the gas and the total pressure. For example, in the exercise, the partial pressure of the hydroxyl radical \( \cdot \text{OH} \) was found by multiplying its mole fraction (tiny part of the total moles) by the total atmospheric pressure (1 atm). This approach simplifies our understanding of how each gas contributes to the total pressure.
Mole Fraction
Mole fraction represents the ratio of moles of one component to the total number of moles in the mixture. It鈥檚 a handy concept when working with mixtures, as it allows us to express concentrations easily.

Imagine you have a jar with red and blue marbles. The mole fraction of red marbles would be the number of red marbles divided by the total number of marbles. In the given problem, we calculated the mole fraction of \( \cdot \text{OH} \) by taking its number of moles and dividing it by the total number of moles in the air. The formula is easy: \( \frac{{n_{\cdot \text{OH}}}}{{n_{\text{total}}}} \). This fraction then allows us to find other properties, such as partial pressure and mole percent.
Avogadro's Number
Avogadro鈥檚 number \(6.022 \times 10^{23} \) is a fundamental constant in chemistry representing the number of atoms or molecules in one mole of a substance. It鈥檚 like saying, \鈥渙ne dozen\鈥 means 12, except in this case, one mole means \(6.022 \times 10^{23} \) entities.

Avogadro's number is essential for converting between the number of molecules and moles. In the exercise, when we knew the concentration of \( \cdot \text{OH} \) in molecules per cubic meter, we used Avogadro鈥檚 number to convert it to moles per cubic meter. This step made it easier to use the Ideal Gas Law and other calculations. Without Avogadro's number, making sense of molecular quantities in practical terms would be much trickier.
Molecular Concentration
Molecular concentration tells us how many molecules of a substance are present in a certain volume. It鈥檚 typically measured in molecules per cubic meter or moles per cubic meter. This measure is key in fields like chemistry and environmental science, where knowing the exact amount of a gas in a volume is crucial.

For instance, in the exercise, we worked with the molecular concentration of the hydroxyl radical \( \cdot \text{OH} \) which was given as \(2.5 \times 10^{12} \) molecules per cubic meter. This helped us understand how much of this radical was present in the air. Such details can be vital in studying atmospheric chemistry or even in industrial processes where precise concentrations of gases need to be maintained.

Converting between different units, like molecules to moles, is often required for deeper calculations, and that's where Avogadro鈥檚 number comes into play again.

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Most popular questions from this chapter

In a collision of sufficient force, automobile air bags respond by electrically triggering the explosive decomposition of sodium azide \(\left(\mathrm{NaN}_{3}\right)\) to its elements. A \(50.0-\mathrm{g}\) sample of sodium azide was decomposed, and the nitrogen gas generated was collected over water at \(26^{\circ} \mathrm{C}\). The total pressure was \(745.5 \mathrm{mmHg}\). How many liters of dry \(\mathrm{N}_{2}\) were generated?

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The thermal decomposition of ethylene occurs during the compound's transit in pipelines and during the formation of polyethylene. The decomposition reaction is $$\mathrm{CH}_{2}=\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{C}(\mathrm{graphite})$$ If the decomposition begins at \(10^{\circ} \mathrm{C}\) and 50.0 atm with a gas density of \(0.215 \mathrm{~g} / \mathrm{mL}\) and the temperature increases by \(950 \mathrm{~K}\), (a) What is the final pressure of the confined gas (ignore the volume of graphite and use the van der Waals equation)? (b) How does the \(P V / R T\) value of \(\mathrm{CH}_{4}\) compare to that in Figure \(5.23 ?\) Explain.

At a height of \(300 \mathrm{~km}\) above Earth's surface, an astronaut finds that the atmospheric pressure is about \(10^{-8} \mathrm{mmHg}\) and the temperature is \(500 \mathrm{~K}\). How many molecules of gas are there per milliliter at this altitude?

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