Question

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.16), \(K_{p}\) for the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ is about \(1 \times 10^{-13}\). Assuming that the exhaust gas (total pressure \(1 \mathrm{~atm}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2}\), and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the above 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?

Short answer

The exhaust gas system is not at equilibrium, as the reaction quotient (Q) is greater than the equilibrium constant (Kp): \(Q \approx 8.33 \times 10^{-5}\) and \(Kp = 1 \times 10^{-13}\). Since the reaction will proceed to the left (reverse direction) to reach equilibrium, the CO concentration will decrease, while the CO2 concentration will increase. Therefore, a catalyst that speeds up the reaction would decrease the CO concentration in the exhaust gas.

Step-by-step solution

Calculate the Partial Pressures of CO, CO2, and O2

Using the given percentages, we will calculate the partial pressures for each gas: For CO: 0.2% of 1 atm = (0.2/100) * 1 atm = 0.002 atm For CO2: 12% of 1 atm = (12/100) * 1 atm = 0.12 atm For O2: 3% of 1 atm = (3/100) * 1 atm = 0.03 atm

Calculate the Reaction Quotient (Q)

Using the partial pressures of CO, CO2, and O2, we will calculate the reaction quotient Q: \[ Q = \dfrac{[CO]^2 [O_2]}{[CO_2]^2} \] Plugging in the partial pressures: \[ Q = \dfrac{(0.002)^2 (0.03)}{(0.12)^2} \] Now, solve for Q: \[ Q \approx 8.33 \times 10^{-5} \]

Compare Q to Kp

Compare the reaction quotient (Q) to the equilibrium constant (Kp) to determine if the system is at equilibrium: Kp = 1 × 10^{-13} Q ≈ 8.33 × 10^{-5} Since Q > Kp, the system is not at equilibrium, and the reaction will proceed to the left (reverse direction) to reach equilibrium.

Predict the Change in CO Concentration

Since the reaction will proceed to the left (reverse direction) to reach equilibrium, the CO concentration will decrease while the CO2 concentration will increase. Thus, a catalyst that speeds up the reaction would decrease the CO concentration in the exhaust gas.

Key concepts

Reaction Quotient

The reaction quotient, often represented by the symbol \(Q\), is a useful tool in chemistry that helps us understand the status of a reaction at any given time. It is similar to the equilibrium constant, \(K\), but is used for systems that are not yet at equilibrium. To calculate \(Q\), one uses the current concentrations or partial pressures of the reactants and products involved in the chemical equation.

• Formula: \(Q = \frac{[Products]}{[Reactants]}\).
• If \(Q < K\), the reaction will move forward to produce more products.
• If \(Q > K\), the reaction will shift backward to create more reactants.
• When \(Q = K\), the system is at equilibrium.

In our exercise, we found \(Q\) to be much larger than \(K_p\), indicating that the product concentrations are too high, and the system will shift in the reverse direction to establish equilibrium.

Partial Pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases. It is an essential concept in chemical equilibrium calculations, where knowing the part each gas plays is crucial as reactions can involve multiple gases.

• Calculated from the total pressure using mole fractions.
• Formula: \(P_{gas} = \text{Mole Fraction of Gas} \times \text{Total Pressure}\).
• Used to calculate both \(Q\) and \(K_p\).

In the exercise, the total pressure of the gases was given as 1 atm, making the calculation straightforward and based on their volume percentages. This clarity in calculating partial pressures allows for accurate determination of \(Q\). Understanding how to calculate partial pressures ensures accuracy in finding reaction quotients and in analyzing whether a system is at equilibrium.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time.

• Dynamic process: reactions continue to occur, but no net change is observed.
• Characterized by the equilibrium constant \(K\).

In our specific scenario, the equilibrium constant \(K_p\) is extraordinarily small \((1 \times 10^{-13})\), suggesting that at equilibrium, very little \(CO\) and \(O_2\) are present compared to \(CO_2\). Since \(Q\) was much greater than \(K_p\), the system must adjust by forming more \(CO_2\) to reach equilibrium. This shift emphasizes the importance of \(K\) in determining the direction of reaction adjustments required for equilibrium.

Catalysis

Catalysis refers to the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst. Catalysts are unique because they speed up reactions without being consumed themselves.

• Lower the activation energy for a reaction.
• Do not affect the equilibrium position, just the rate at which equilibrium is achieved.
• Can be applied repeatedly without changing in amount or form.

In the context of the automobile exhaust reaction, a catalyst would help the system reach equilibrium faster but would not change the equilibrium concentrations of the components. As determined, speeding up the reaction towards equilibrium would result in decreased \(CO\) concentration, as the system shifts to form more \(CO_2\). This demonstrates how catalysts are used pragmatically to reduce certain emissions in industrial applications.