Problem 93 - Chemistry: The Central Science in SI Units, Global Edition

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 \(1 \times 10^{-11}\). Assuming that the exhaust gas (total pressure \(101.3 \mathrm{kPa}\) ) 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 \(\mathrm{CO}_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).

Step-by-Step Solution

Answer

The system is not at equilibrium, as the calculated reaction quotient (\(Q \approx 8.069 \times 10^{-7}\)) is greater than the given equilibrium constant (\(K_p = 1 \times 10^{-11}\)). The reaction will shift towards the reactants to reach equilibrium, thus favoring the consumption of CO and the production of CO₂. Adding a catalyst would speed up the conversion of CO to CO₂, ultimately decreasing the CO concentration in the exhaust gas.

Step 1 - Convert the given volume percentages to partial pressures.

To find the partial pressures of each gas, we need to multiply the total pressure by the volume percentage, which is equivalent to the mole percentage. Partial pressure of CO₂: \(P_{CO_2} = 0.12 × 101.3 ~kPa = 12.156 ~kPa\) Partial pressure of CO: \(P_{CO} = 0.002 × 101.3 ~kPa = 0.2026 ~kPa\) Partial pressure of O₂: \(P_{O_2} = 0.03 × 101.3 ~kPa = 3.039 ~kPa\)

Step 2 - Calculate the reaction quotient (Q).

The reaction quotient (Q) is calculated using the formula for the given equilibrium reaction: \[Q = \frac{P^{2}_{CO} \cdot P_{O_2}}{P^{2}_{CO_2}}\] Plug in the values of partial pressures: \[Q = \frac{ (0.2026)^2 \cdot 3.039}{(12.156)^2} \]

Step 3 - Compare Q with Kp.

Now, calculate the value of Q: \[Q \approx 8.069 \times 10^{-7}\] The given value of \(K_p\) is 1 × 10⁻¹¹ Now compare Q with Kp: Since \(Q > K_p\), the reaction is not at equilibrium, and it will shift towards the reactants to reach equilibrium. The forward reaction will be favored to consume CO and produce CO₂.

Step 4 - Determine the effect of a catalyst.

A catalyst speeds up both the forward and reverse reactions, but it does not affect the equilibrium constant. In this case, adding a catalyst would speed up the conversion of CO to CO₂, ultimately decreasing the CO concentration in the exhaust gas.

Key Concepts

Reaction Quotient

The reaction quotient, commonly denoted as \( Q \), helps us evaluate the state of a chemical reaction in relation to equilibrium. In essence, \( Q \) is calculated the same way as the equilibrium constant (\( K \)), but with one major difference: it uses the current concentrations or partial pressures, instead of those at equilibrium.
When you compute \( Q \) for a gaseous reaction, you use the equation:

• \( Q = \frac{P_{products}}{P_{reactants}} \)

For the reaction \( 2 \, \text{CO}_2(g) \rightleftharpoons 2 \, \text{CO}(g) + \text{O}_2(g) \), the specific equation becomes:

• \( Q = \frac{P_{CO}^2 \cdot P_{O_2}}{P_{CO_2}^2} \)

Once you have calculated \( Q \), you can compare it to \( K_p \):

• If \( Q = K_p \), the system is at equilibrium.
• If \( Q > K_p \), the reaction shifts toward the reactants.
• If \( Q < K_p \), the reaction shifts toward the products.

This comparison tells us whether reactants or products will be consumed more to reach equilibrium.

Partial Pressure

Partial pressure is an essential concept in understanding gases. Each gas in a mixture exerts its own pressure, as if it were alone in the container. The total pressure of the mixture is the sum of the partial pressures of all the gases present. This principle is known as Dalton’s Law of Partial Pressures.
In this exercise, you determine the partial pressure of each gas using the formula:

• \( P_i = \left( \text{Volume Percentage} \right) \times \text{Total Pressure} \)

For example, in the given problem where the total pressure is 101.3 kPa:

• CO₂: \( P_{CO_2} = 0.12 \times 101.3 \, \text{kPa} = 12.156 \, \text{kPa} \)
• CO: \( P_{CO} = 0.002 \times 101.3 \, \text{kPa} = 0.2026 \, \text{kPa} \)
• O₂: \( P_{O_2} = 0.03 \times 101.3 \, \text{kPa} = 3.039 \, \text{kPa} \)

Calculating partial pressures allows us to substitute these values into the equation for the reaction quotient \( Q \). This helps assess how close the system is to equilibrium and predicts the direction of change necessary to reach that state.

Equilibrium Constant

The equilibrium constant, denoted as \( K \) or specifically \( K_p \) for pressure, is an invaluable quantity in chemical reactions. It is a measure of the ratio of product concentrations to reactant concentrations at equilibrium. Unlike \( Q \), which varies depending on the current state of the reaction, \( K \) is constant at a given temperature. This remains true regardless of how much reactant or product is initially present.
For the reaction at hand, \( K_p \) is used because we're considering gases and pressures:

• \( K_p = \frac{P_{products}^2 \cdot P_{O_2}}{P_{CO_2}^2} \)

In the original problem, \( K_p \) is given as \( 1 \times 10^{-11} \). When we calculated \( Q \) and found that \( Q \) is much larger than \( K_p \), it indicates that there are more products than would be present at equilibrium. Therefore, the reaction naturally shifts toward the reactants. A catalyst, though it doesn’t alter \( K_p \), can expedite the process of reaching this equilibrium point. This means faster conversion and a decrease in the current product concentrations.