/*! 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 129 Nitrogen dioxide \(\left(\mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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Nitrogen dioxide \(\left(\mathrm{NO}_{2}\right)\) cannot be obtained in a pure form in the gas phase because it exists as a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\). At \(25^{\circ} \mathrm{C}\) and \(0.98 \mathrm{~atm}\), the density of this gas mixture is \(2.7 \mathrm{~g} / \mathrm{L}\). What is the partial pressure of each gas?

Short Answer

Expert verified
The partial pressures are 0.519 atm for \(NO_2\) and 0.461 atm for \(N_2O_4\).

Step by step solution

01

Calculate Molar Mass of the Mixture

With the density \(d = 2.7 \, \mathrm{g/L}\) and pressure \(P = 0.98 \, \mathrm{atm}\), we use the ideal gas law \(PV = nRT\) rearranged to find molar mass \(M = \frac{dRT}{P}\), where \(R = 0.0821 \, \mathrm{L} \,\mathrm{atm} / \mathrm{K} \, \mathrm{mol}\) and \(T = 298 \, \mathrm{K}\). Substituting these into the formula gives us: \[M = \frac{(2.7 \, \mathrm{g/L}) \times (0.0821 \, \mathrm{L} \, \mathrm{atm}/\mathrm{K} \, \mathrm{mol}) \times 298 \mathrm{K}}{0.98 \, \mathrm{atm}} = 67.6 \, \mathrm{g/mol}\]
02

Calculate Molar Masses of Components

Calculate the molar masses of \,\(\mathrm{NO}_{2}\,\) and \,\(\mathrm{N}_{2} \mathrm{O}_{4}\,\): \[\mathrm{M(NO_{2}) = 14 + 2 \times 16 = 46 \,\, \mathrm{g/mol}}\] \[ \mathrm{M(N_{2}O_{4}) = 2 \times 14 + 4 \times 16 = 92 \,\, \mathrm{g/mol}}\]
03

Set up Equation for Mixture's Molar Mass

Let \,\(x\,\) be the mole fraction of \,\(\mathrm{NO}_{2}\,\), then \,\((1-x)\,\) is the mole fraction of \,\(\mathrm{N}_{2} \mathrm{O}_{4}\,\). The equation for molar mass of the mixture becomes: \[67.6 = 46x + 92(1-x)\]
04

Solve for Mole Fraction of Each Gas

Solve the equation from the previous step: \[67.6 = 46x + 92 - 92x \ 67.6 - 92 = 46x - 92x \ -24.4 = -46x \ x = \frac{24.4}{46} = 0.530.\] Thus, \,\(x \, \approx 0.53\,\) for \,\(\mathrm{NO}_{2}\,\) and \,\(1-x = 0.47\,\) for \,\(\mathrm{N}_{2} \mathrm{O}_{4}\,\).
05

Calculate Partial Pressures

Since the partial pressure of a gas is its mole fraction times total pressure: \[P_{\mathrm{NO}_{2}} = 0.53 \times 0.98 = 0.5194 \, \mathrm{atm}\] \[P_{\mathrm{N}_{2} O_{4}} = 0.47 \times 0.98 = 0.4606 \, \mathrm{atm}\]

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental principle used for understanding the properties of gases. It relates the pressure, volume, temperature, and amount of a gas. The equation is expressed as:\[ PV = nRT \]- **P** is the pressure of the gas.- **V** is the volume.- **n** is the number of moles of gas.- **R** is the ideal gas constant, which is 0.0821 L·atm/(K·mol).- **T** is the temperature in Kelvin.
We can rearrange this equation to calculate the molar mass of a gas mixture when its density, temperature, and pressure are known. The rearranged formula looks like:\[ M = \frac{dRT}{P} \]where **M** is the molar mass and **d** is the density of the gas mixture.
Understanding and applying this formula is crucial when dealing with gas mixtures, as it allows us to find important properties like density, pressure, and molar mass.
Mole Fraction
The mole fraction is a measure of the concentration of a particular component in a mixture. It represents how many moles of one component are present compared to the total number of moles in the mixture. The formula for the mole fraction of a component, say \( x \) for nitrogen dioxide \( \text{NO}_2 \), is expressed as:\[ x = \frac{n_{ ext{component}}}{n_{ ext{total}}} \]- **\( n_{\text{component}} \)** is the number of moles of the component.- **\( n_{\text{total}} \)** represents the total moles of all components in the mixture.
This concept is vital when solving problems that require the use of partial pressures in gas mixtures. Knowing the mole fraction of each gas allows you to easily calculate the partial pressures within a mixture, which is a key step in many chemical and physical processes involving gases.
Partial Pressure
Each gas in a mixture exerts its own pressure, known as its partial pressure. The total pressure of the gas mixture is the sum of all individual partial pressures. The partial pressure of a particular gas can be computed using its mole fraction and the total pressure of the mixture:\[ P_{ ext{component}} = x_{ ext{component}} \times P_{ ext{total}} \]- **\( P_{\text{component}} \)** is the partial pressure of the gas.- **\( x_{\text{component}} \)** is the mole fraction of the gas.- **\( P_{\text{total}} \)** is the total pressure of the gas mixture.
For example, in a mixture of nitrogen dioxide \( \text{NO}_2 \) and dinitrogen tetroxide \( \text{N}_2\text{O}_4 \), determining their mole fractions using the relationship based on molar masses allows you to find their respective partial pressures. This method is particularly useful in scenarios where each gas contributes differently to a mixture's properties.
Molar Mass
The molar mass is a critical concept that is defined as the mass of one mole of a substance expressed in grams per mole (g/mol). To calculate it, you need to add together the atomic masses of all elements in the molecule. For example:- **Nitrogen Dioxide (NO\(_2\))**: Consists of 1 nitrogen (14 g/mol) and 2 oxygens (2 × 16 g/mol), resulting in a molar mass of 46 g/mol.- **Dinitrogen Tetroxide (N\(_2\)O\(_4\))**: Comprised of 2 nitrogens (2 × 14 g/mol) and 4 oxygens (4 × 16 g/mol), giving a molar mass of 92 g/mol.
In gas mixtures, the effective molar mass can be derived using the weighted average based on mole fractions of the components. This mediating property between molecular weight and composition is critical for calculating the properties of complex gas mixtures and helps to find mole fractions based on known compositions.

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Most popular questions from this chapter

Apply your knowledge of the kinetic theory of gases to the following situations. (a) Two flasks of volumes \(V_{1}\) and \(V_{2}\left(V_{2}>V_{1}\right)\) contain the same number of helium atoms at the same temperature. (i) Compare the rootmean-square (rms) speeds and average kinetic energies of the helium (He) atoms in the flasks. (ii) Compare the frequency and the force with which the He atoms collide with the walls of their containers. (b) Equal numbers of He atoms are placed in two flasks of the same volume at temperatures \(T_{1}\) and \(T_{2}\left(T_{2}>T_{1}\right) .\) (i) Compare the rms speeds of the atoms in the two flasks. (ii) Compare the frequency and the force with which the He atoms collide with the walls of their containers. (c) Equal numbers of He and neon (Ne) atoms are placed in two flasks of the same volume, and the temperature of both gases is \(74^{\circ} \mathrm{C}\). Comment on the validity of the following statements: (i) The rms speed of He is equal to that of Ne. (ii) The average kinetic energies of the two gases are equal. (iii) The rms speed of each He atom is \(1.47 \times 10^{3} \mathrm{~m} / \mathrm{s}\)

What is the difference between gas diffusion and effusion?

A compound of \(\mathrm{P}\) and \(\mathrm{F}\) was analyzed as follows: Heating \(0.2324 \mathrm{~g}\) of the compound in a \(378-\mathrm{cm}^{3}\) container turned all of it to gas, which had a pressure of \(97.3 \mathrm{mmHg}\) at \(77^{\circ} \mathrm{C}\). Then the gas was mixed with calcium chloride solution, which converted all the \(\mathrm{F}\) to \(0.2631 \mathrm{~g}\) of \(\mathrm{CaF}_{2}\). Determine the molecular formula of the compound.

Lithium hydride reacts with water as follows: $$ \mathrm{LiH}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{LiOH}(a q)+\mathrm{H}_{2}(g) $$ During World War II, U.S. pilots carried LiH tablets. In the event of a crash landing at sea, the \(\mathrm{LiH}\) would react with the seawater and fill their life jackets and lifeboats with hydrogen gas. How many grams of \(\mathrm{LiH}\) are needed to fill a 4.1-L life jacket at 0.97 atm and \(12^{\circ} \mathrm{C}\) ?

Calculate the density of helium in a helium balloon at \(25.0^{\circ} \mathrm{C}\). (Assume that the pressure inside the balloon is \(1.10 \mathrm{~atm} .)\)

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