At 1200 K, the approximate temperature of vehicle exhaust gases, Kp for the reaction 2CO2(g)⇄2CO(g) + O2(g)
is about 1*E-13. Assuming that the exhaust gas (total pressure 1 bar) contains 0.2% CO, 12% CO2 and 3% O2 by volume, is the system at equilibrium with respect to the reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the reaction above?
see mary's question above
To determine whether the system is at equilibrium with respect to the reaction, we need to compare the actual reaction quotient (Q) with the equilibrium constant (Kp).
To calculate Q, we need to know the partial pressures of CO2, CO, and O2 in the exhaust gas. However, the given information provides the percentages by volume.
To convert volume percentages to partial pressures, we need to assume an ideal gas behavior. Since the total pressure of the exhaust gas is given as 1 bar, we can assume that the partial pressures of the gases also add up to 1 bar.
Using these assumptions, we can calculate the partial pressures as follows:
- Partial pressure of CO2 = 0.12 * 1 bar = 0.12 bar
- Partial pressure of CO = 0.002 * 1 bar = 0.002 bar
- Partial pressure of O2 = 0.03 * 1 bar = 0.03 bar
Now, let's calculate Q:
Q = (Partial pressure of CO)^2 * Partial pressure of O2 / (Partial pressure of CO2)^2
Q = (0.002 bar)^2 * 0.03 bar / (0.12 bar)^2
Q = 1.2 * 10^-6 / 0.0144
Q ≈ 8.33 * 10^-5
Now, let's compare Q with Kp. Given that Kp is approximately 1 * 10^-13, we can conclude that Q > Kp. This means that the system is not at equilibrium and the reaction will proceed in the forward direction to reach equilibrium.
A catalyst that speeds up the reaction would shift the equilibrium position to consume more reactants and produce more products. In this case, the catalyst would increase the conversion of CO2 and O2 to form more CO. Therefore, the concentration of CO in the exhaust gas would be increased by the catalyst.