At 1200 K, the approximate temperature of vehicle exhaust gases, Kp for the reaction
<-
2CO2(g) -> 2CO(g) + O2(g)
is about 1*10E-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?
To determine whether the system is at equilibrium with respect to the given reaction, we need to compare the calculated Kp value with the given Kp value at the same temperature.
First, let's calculate the partial pressures of CO2, CO, and O2 using the given volume percentages and total pressure:
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 plug these values into the Kp expression:
Kp = (P(CO)^2 * P(O2)) / P(CO2)^2
Kp = (0.002^2 * 0.03) / (0.12^2)
Kp = 1.25 * 10^-7
Comparing this calculated Kp to the given Kp (1 * 10^-13), we can see that the calculated Kp is significantly larger. Therefore, the system is not at equilibrium with respect to the given reaction.
Now, let's analyze the effect of a catalyst that speeds up the reaction. A catalyst increases the rate of the forward and reverse reactions without being consumed in the process. It does not affect the position of equilibrium.
Since the catalyst speeds up the reaction, both the forward and reverse reactions will increase in rate. However, the forward reaction converts CO2 to CO and O2, leading to a decrease in the concentration of CO2 and an increase in the concentration of CO. Therefore, the CO concentration in the exhaust will be increased by a catalyst that speeds up the reaction.
In summary, the system is not at equilibrium with respect to the reaction, and a catalyst that speeds up the reaction would increase the CO concentration in the exhaust.