/*! 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 97 A gaseous mixture of 10.0 volume... [FREE SOLUTION] | 91Ó°ÊÓ

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A gaseous mixture of 10.0 volumes of \(\mathrm{CO}_{2}, 1.00\) volume of unreacted \(\mathrm{O}_{2}\), and 50.0 volumes of unreacted \(\mathrm{N}_{2}\) leaves an engine at 4.0 atm and \(800 .\) K. Assuming that the mixture reaches equilibrium, what are (a) the partial pressure and (b) the concentration (in picograms per liter, \(\mathrm{pg} / \mathrm{L}\) ) of \(\mathrm{CO}\) in this exhaust gas? $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \quad K_{\mathrm{p}}=1.4 \times 10^{-28} \mathrm{at} 800 . \mathrm{K} $$ (The actual concentration of \(\mathrm{CO}\) in exhaust gas is much higher because the gases do not reach equilibrium in the short transit time through the engine and exhaust system.)

Short Answer

Expert verified
The partial pressure of CO is extremely small due to the very small equilibrium constant, and thus the concentration in pg/L is also correspondingly low.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction is: \[ 2\mathrm{CO_{2}(g)} \rightleftharpoons 2\mathrm{CO(g)} + \mathrm{O_{2}(g)} \]
02

Identify initial conditions

Initial volumes of gases: - \(C\mathrm{O_{2}}: 10.0~\text{volumes} \)- \(O_{2}: 1.00~\text{volume} \)- \(N_{2}: 50.0~\text{volumes} \)
03

Calculate total initial pressure

The total initial pressure of the gaseous mixture is given as 4.0 atm. Therefore,\[ P_{\text{total}} = 4.0~\text{atm} \]
04

Define the equilibrium expression

The equilibrium expression for the reaction is based on the partial pressures of the gases:\[ K_p = \frac{(P_{\mathrm{CO}})^2 P_{\mathrm{O_{2}}}}{(P_{\mathrm{CO_{2}}})^2} \]
05

Set up initial partial pressures

Determine the mole fraction for each gas and then convert to initial partial pressures:1 volume = 1 partTotal parts = 10.0 (CO2) + 1.00 (O2) + 50.0 (N2) = 61.0Partial pressures of each gas initially:\[ P_{\mathrm{CO_{2}}}^{\mathrm{initial}} = \frac{10.0}{61.0} \times 4.0~\text{atm} \]\[ P_{\mathrm{O_{2}}}^{\mathrm{initial}} = \frac{1.00}{61.0} \times 4.0~\text{atm} \]\[ P_{\mathrm{N_{2}}}^{\mathrm{initial}} = \frac{50.0}{61.0} \times 4.0~\text{atm} \]
06

Calculate the change in partial pressures at equilibrium

Let 'x' be the change in partial pressure for \(\mathrm{CO_{2}}\) decreasing and \(\mathrm{CO}\) increasing:\[ 2\mathrm{CO_{2}} \rightleftharpoons 2\mathrm{CO} + \mathrm{O_{2}} \]Change at equilibrium:\[P_{\mathrm{CO_{2}}} = P_{\mathrm{CO_{2}}}^{\mathrm{initial}} - 2x \]\[P_{\mathrm{CO}} = 2x\]\[P_{\mathrm{O_{2}}} = P_{\mathrm{O_{2}}}^{\mathrm{initial}} + x\]
07

Solve the equilibrium expression

Substitute the equilibrium values into the \(K_p\) expression:\[ 1.4 \times 10^{-28} = \frac{(2x)^2 (P_{\mathrm{O_{2}}}^{\mathrm{initial}} + x)}{(P_{\mathrm{CO_{2}}}^{\mathrm{initial}} - 2x)^2} \]
08

Approximate and simplify (due to small Kp)

Since \( K_p \) is extremely small, assume the change 'x' is very small:\[ (P_{\mathrm{CO_{2}}}^{\mathrm{initial}} - 2x) \approx P_{\mathrm{CO_{2}}}^{\mathrm{initial}} \]\[ (P_{\mathrm{O_{2}}}^{\mathrm{initial}} + x) \approx P_{\mathrm{O_{2}}}^{\mathrm{initial}} \]So,\[ 1.4 \times 10^{-28} = \frac{(2x)^2 P_{\mathrm{O_{2}}}^{\mathrm{initial}}}{(P_{\mathrm{CO_{2}}}^{\mathrm{initial}})^2} \]
09

Solve for x

Solve for 'x':\[ 1.4 \times 10^{-28} = \frac{4x^2 \times P_{\mathrm{O_{2}}}^{\mathrm{initial}}}{(P_{\mathrm{CO_{2}}}^{\mathrm{initial}})^2} \]\[ x^2 = \frac{(1.4 \times 10^{-28}) \times (P_{\mathrm{CO_{2}}}^{\mathrm{initial}})^2}{4 \times P_{\mathrm{O_{2}}}^{\mathrm{initial}}} \]\[ x = \sqrt{\frac{(1.4 \times 10^{-28}) \times (P_{\mathrm{CO_{2}}}^{\mathrm{initial}})^2}{4 \times P_{\mathrm{O_{2}}}^{\mathrm{initial}}}} \]With values substituted from Step 5, calculate 'x'.
10

Calculate partial pressure of CO

The partial pressure of CO at equilibrium is:\[ P_{\mathrm{CO}} = 2x \]
11

Calculate concentration of CO in pg/L

Use the ideal gas law to convert partial pressure to concentration:\[ \text{Concentration of CO} = \frac{P_{\mathrm{CO}} \times M_{\mathrm{CO}}}{R \times T} \]where \( M_{\mathrm{CO}} \) is the molar mass of CO in \( \text{g/mol} \), \( R = 0.0821 \times \text{ atm} \cdot \text{L}/\text{K} \cdot \text{mol} \), and \( T = 800 \text{K} \). Convert the concentration to \( \text{pg/L} \) (1 g = \(10^{12}\) pg).

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Key Concepts

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

Partial Pressure Calculation
Understanding partial pressure is crucial for solving problems about gases. Partial pressure is the pressure exerted by an individual gas in a mixture. It can be calculated if you know the mole fraction of the gas and the total pressure of the mixture. For example, in a mixture of CO2, O2, and N2 at a total pressure of 4.0 atm, you first determine the mole fraction for each gas. If we have 10 volumes of CO2, 1 volume of O2, and 50 volumes of N2, the total volume parts add up to 61. The partial pressure for each gas can be found by multiplying its mole fraction by the total pressure: \( P_{\text{gas}} = \frac{\text{volume part of gas}}{\text{total volume parts}} \times \text{total pressure} \) So for CO2, it would be \( P_{\text{CO2}} = \frac{10}{61} \times 4.0 \text{ atm} \).
Ideal Gas Law
The Ideal Gas Law is a powerful tool to relate the pressure, volume, temperature, and moles of a gas. It’s given by the equation: \[ PV = nRT \], where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. For example, if you need to find the concentration of CO in a mixture at equilibrium, using the Ideal Gas Law can help. By rearranging the equation to \( n = \frac{P \times V}{R \times T} \), you can calculate the number of moles, which can then be converted to concentration. When pressure \( P_{\text{CO}} \) is known, you can find the concentration of CO: \( \text{Concentration of CO} = \frac{P_{\text{CO}} \times M_{\text{CO}}}{R \times T} \), where \( M_{\text{CO}} \) is the molar mass.
Reaction Equilibrium Constant
The equilibrium constant \( K_p \) helps us understand the ratio of product to reactant partial pressures at equilibrium. It is unique for each reaction and temperature-dependent. For the reaction \( 2\mathrm{CO_{2}(g)} \rightleftharpoons 2\mathrm{CO(g)} + \mathrm{O_{2}(g)} \) with \( K_p = 1.4 \times 10^{-28} \) at 800 K, you can set the equilibrium expression: \[ K_p = \frac{(P_{\text{CO}})^2 P_{\text{O2}}}{(P_{\text{CO2}})^2} \] Using the initial partial pressures and changes in pressure (denoted by 'x'), substitute the values into the equation to solve for \( x \). Given the extremely small \( K_p \), simplify by assuming \( x \) is very small. This leads to: \[ 1.4 \times 10^{-28} = \frac{(2x)^2 (P_{\text{O2}}^{\text{initial}} + x)}{(P_{\text{CO2}}^{\text{initial}} - 2x)^2} \rightarrow \frac{4x^2 P_{\text{O2}}^{\text{initial}}}{P_{\text{CO2}}^{\text{initial}}^2} \]. Finally, solve for \( x \) and consequently calculate \( P_{\text{CO}} = 2x \).

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Most popular questions from this chapter

For the following equilibrium system, which of the changes will form more \(\mathrm{CaCO}_{3} ?\) $$ \begin{array}{r} \mathrm{CO}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-113 \mathrm{~kJ} \end{array} $$ (a) Decrease temperature at constant pressure (no phase change). (b) Increase volume at constant temperature. (c) Increase partial pressure of \(\mathrm{CO}_{2}\). (d) Remove one-half of the initial \(\mathrm{CaCO}_{3}\).

An important industrial source of ethanol is the reaction, catalyzed by \(\mathrm{H}_{3} \mathrm{PO}_{4},\) of steam with ethylene derived from oil: $$ \begin{array}{c} \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-47.8 \mathrm{~kJ} \quad K_{\mathrm{c}}=9 \times 10^{3} \mathrm{at} 600 . \mathrm{K} \end{array} $$ (a) At equilibrium, \(P_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}=200 . \mathrm{atm}\) and \(P_{\mathrm{H}_{0} \mathrm{O}}=400 .\) atm. Cal- culate \(P_{\mathrm{C}_{2} \mathrm{H}_{4}}\). (b) Is the highest yield of ethanol obtained at high or low \(P\) ? High or low \(T ?\) (c) Calculate \(K_{c}\) at \(450 .\) K. (d) In \(\mathrm{NH}_{3}\) manufacture, the yield is increased by condensing the \(\mathrm{NH}_{3}\) to a liquid and removing it. Would condensing the \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) have the same effect in ethanol production? Explain.

The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ At a given temperature, \(K_{\mathrm{p}}=2.7 .\) If \(0.13 \mathrm{~mol}\) of \(\mathrm{CO}, 0.56 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}, 0.62 \mathrm{~mol}\) of \(\mathrm{CO}_{2},\) and \(0.43 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) are put in a \(2.0-\mathrm{L}\) flask, in which direction does the reaction proceed?

At a particular temperature, \(K_{c}=1.6 \times 10^{-2}\) for $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ Calculate \(K_{c}\) for each of the following reactions: (a) \(\frac{1}{2} \mathrm{~S}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)\) (b) \(5 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 5 \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{~S}_{2}(g)\)

For the following reaction, \(K_{c}=115\) at a particular temperature: $$ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g) $$ A container initially holds the following concentrations: \(0.050 \mathrm{M}\) \(\mathrm{H}_{2}, 0.050 \mathrm{M} \mathrm{F}_{2},\) and \(0.10 \mathrm{M} \mathrm{HF} .\) When equilibrium is reached, what is the concentration of HF?

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