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.)

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 \( P_{\text{CO2}} = \frac{10}{61} \times 4.0 \text{ atm} \).

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.

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 \( P_{\text{CO}} = 2x \).