Problem 93 - Chemistry The Central Science

Question

At \(1200 \mathrm{K},\) the approximate temperature of automobile exhaust gases (Figure 15.15), \(K_{p}\) 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 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 \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the CO \(_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\operatorname{mol} \% .\)

Step-by-Step Solution

Answer

The exhaust gas is not at equilibrium with respect to the CO₂ reaction, as the reaction quotient (Q_p) is greater than the equilibrium constant (K_p). A catalyst that speeds up the CO₂ reaction would help the system reach equilibrium faster, leading to a decrease in CO concentration in the exhaust gas.

Step 1: Calculate the Mole Fractions of Each Gas

We have the following volume percentages: 0.2% CO, 12% CO₂, and 3% O₂. Since the volume percent is equal to the mole percent at a fixed temperature and pressure, we can convert them into mole fractions by dividing by 100 (total volume). CO: \(0.2\% = \frac{0.2}{100} = 0.002 \) CO₂: \(12\% = \frac{12}{100} = 0.12\) O₂: \(3\% = \frac{3}{100} = 0.03\)

Step 2: Write the Reaction Quotient Expression

The reaction quotient (Q) can be expressed as: \(Q_p = \frac{[CO]^2[O_2]}{[CO_2]^2}\) To find Q, we can insert the mole fractions of each gas into the expression.

Step 3: Calculate Q

Now, we substitute the mole fractions into the Q expression: \(Q_p = \frac{(0.002)^2(0.03)}{(0.12)^2} \) Calculating Q, we get: \(Q_p ≈ 8.33 \times 10^{-5}\)

Step 4: Compare Q to K_p

We are given that: \(K_p = 1 \times 10^{-13}\) Comparing Q to K_p: \(Q_p = 8.33 \times 10^{-5} > K_p = 1 \times 10^{-13}\)

Step 5: Determine if the System is at Equilibrium and the Effect of a Catalyst on CO Concentration

Since Q > K, the reaction will proceed to the left. The system is not at equilibrium with respect to the CO₂ reaction, and the CO concentration in the exhaust gas is higher than it would be if the reaction were at equilibrium. A catalyst that speeds up the CO₂ reaction would help the system reach equilibrium faster. In this case, since the reaction will proceed to the left, decreasing the CO concentration and increasing the CO₂ concentration. Thus, a catalyst would decrease the CO concentration in the exhaust gas.

Key Concepts

Reaction Quotient

The reaction quotient, often denoted as \(Q\), is a mathematical expression used to determine whether a chemical reaction is at equilibrium. It's similar to the equilibrium constant \(K\), but it can be calculated at any point during the reaction, not just at equilibrium.

When calculating \(Q\), we use the current concentrations (or partial pressures, in the case of gas reactions) of the reactants and products. The expression for \(Q\) looks much like that for \(K\):

• For gases, \(Q_p = \frac{[products]^{coefficients}}{[reactants]^{coefficients}}\)

The provided exercise focuses on determining if the automobile exhaust reaction
2 CO_2(g) \rightleftharpoons 2 CO(g) + O_2(g)
is at equilibrium by comparing \(Q_p\) to \(K_p\). We calculate \(Q_p\) using mole fractions because the given problem is under constant pressure and high temperature conditions, where mole fractions translate directly into partial pressures. In this exercise, \(Q_p = 8.33 \times 10^{-5}\).

Once we have \(Q\), we can determine the direction in which the reaction will proceed to reach equilibrium. If \(Q\) is greater than \(K\), the system will shift left, favoring the formation of reactants to decrease product concentration.

Equilibrium Constant

The equilibrium constant, \(K\), is a number that expresses the ratio of the concentrations of products to reactants at equilibrium. For gases, this is denoted as \(K_p\), where 'p' stands for partial pressures. At high temperatures, the value of \(K_p\) can tell us the extent to which a reaction will proceed.

In the exercise, the given \(K_p\) is \(1 \times 10^{-13}\) at \(1200 \mathrm{K}\). This small value suggests that, at equilibrium, the concentration of products (\(CO\) and \(O_2\)) is much lower than the concentration of reactants (\(CO_2\)).

• When \(Q > K\), the reaction shifts to the left to form more reactants.
• When \(Q < K\), the reaction shifts to the right to form more products.
• When \(Q = K\), the system is already at equilibrium.

By comparing \(Q\) from the earlier step to \(K_p\), we can tell whether the exhaust gases are in equilibrium. In this case, because \(Q_p\) is greater than \(K_p\), the system is not at equilibrium, leading us to expect the reaction will produce more \(CO_2\) and less \(CO\) and \(O_2\) if left to reach equilibrium.

Catalysts

Catalysts are substances that increase the rate at which a reaction reaches equilibrium without being consumed in the process. They work by lowering the activation energy needed for the reaction, allowing for faster progress in both the forward and reverse directions, thus helping the reaction to achieve equilibrium more quickly.

Importantly, catalysts do not change the position of equilibrium. They simply make it faster to reach it. In the context of the exercise, adding a catalyst to the exhaust gas system would result in the reaction shifting faster towards equilibrium. Given our condition where \(Q_p > K_p\), the catalyst would speed up the shift to produce more \(CO_2\) and use up \(CO\), effectively decreasing the \(CO\) concentration in the exhaust gas.

In summary, a catalyst in this particular reaction would:

• Increase the rate at which equilibrium is achieved.
• Accelerate the shift towards producing more reactants from products, as it speeds up both the forward and backward reactions.

Gas Mixtures

Gas mixtures are an integral part of many chemical equilibria and reactions, especially in automotive exhaust systems. In such systems, the behavior of gases under varying conditions of temperature and pressure is of specific interest.

At constant temperature and pressure, as in the given exercise, the volume percentage of a gas is equivalent to its mole fraction. This is particularly useful because it allows us to calculate reaction quotients \(Q\) based on volume percentages, simplifying the process.

The composition of a gas mixture can significantly affect the reaction behavior. For instance:

• A higher concentration of a reactant gas tends to drive the reaction forward to produce more products, provided it's not at equilibrium.
• Conversely, a higher concentration of product gases supports the reverse reaction.

In the discussed exercise, the mixture contains 0.2% \(CO\), 12% \(CO_2\), and 3% \(O_2\). Understanding the composition helps in calculating \(Q\) and determining whether the gas mixture is at equilibrium. When dealing with gas mixtures, it's important to remember that these calculations are sensitive to the conditions specified, such as high temperatures typically found in car exhausts.