Solution Found!
A gaseous mixture of 10.0 volumes of CO2,1.00 volume of
Chapter 17, Problem 73P(choose chapter or problem)
An equilibrium mixture of car exhaust gases consisting of 10.0 volumes of \(\mathrm{CO}_{2}\), 1.00 volume of unreacted \(\mathrm{O}_{2}\), and 50.0 volumes of unreacted \(N_{2}\) leaves the engine at 4.0 atm and 800. K.
(a) Given this equilibrium, what is the partial pressure of CO?
\(2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+O_{2}(g) \quad K_{p}=1.4 \times 10^{-28} \text { at } 800 . \mathrm{K}\)
(b) Assuming the mixture has enough time to reach equilibrium, what is the concentration in picograms per liter (pg/L) of CO in the exhaust gas? (The actual concentration of CO in car exhaust is much higher because the gases do not reach equilibrium in the short transit time through the engine and exhaust system.).
Equation Transcription:
at 800. K
Text Transcription:
CO_2
O_2
N_2
2CO_2(g) ⇌ 2CO(g) + O2(g) K_p=1.4x10^-28 at 800. K
Questions & Answers
QUESTION:
An equilibrium mixture of car exhaust gases consisting of 10.0 volumes of \(\mathrm{CO}_{2}\), 1.00 volume of unreacted \(\mathrm{O}_{2}\), and 50.0 volumes of unreacted \(N_{2}\) leaves the engine at 4.0 atm and 800. K.
(a) Given this equilibrium, what is the partial pressure of CO?
\(2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+O_{2}(g) \quad K_{p}=1.4 \times 10^{-28} \text { at } 800 . \mathrm{K}\)
(b) Assuming the mixture has enough time to reach equilibrium, what is the concentration in picograms per liter (pg/L) of CO in the exhaust gas? (The actual concentration of CO in car exhaust is much higher because the gases do not reach equilibrium in the short transit time through the engine and exhaust system.).
Equation Transcription:
at 800. K
Text Transcription:
CO_2
O_2
N_2
2CO_2(g) ⇌ 2CO(g) + O2(g) K_p=1.4x10^-28 at 800. K
ANSWER:
Step 1 of 4
The activity of individual gas components in a combination of varying gases is provided by the partial pressures. The partial pressures of all gases constitute together to give total pressure of the gas.
Dalton's law is employed in the case of partial pressures.
(a)
Given Data:
*The volume of carbon dioxide is 10.0 volume.
*The volume of unreacted oxygen is 1.0 volume.
*The volume of unreacted nitrogen is 50.0 volume.
*The total pressure is 4.0 atm.
*The temperature is 800 K.
*The value of 
The volume is the sum of the volumes of all the gases.
Therefore, it becomes 61 volumes.
The mole fraction can be calculated using individual volume divided by total volume of the gases.
The partial pressure for individual gases is calculated using the mole fraction multiplied by the total pressure.
For carbon dioxide,
The partial pressure becomes,
