Problem 93 - Chemistry The Central Science In Si Units

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, and a catalyst would decrease the CO concentration.

Step 1: Determine Partial Pressures from Volume Percentages

Start by converting the given volume percentages into partial pressures using the total pressure of the system. Using the formula \( P_{i} = (\text{Volume \% of } i \) \(\times \text{ total pressure})/100 \). Calculate:\( P_{\text{CO}} = (0.2 \% \times 101.3 \text{ kPa})/100 = 0.2026 \text{ kPa} \). Calculate: \( P_{\text{CO}_2} = (12 \% \times 101.3 \text{ kPa})/100 = 12.156 \text{ kPa} \). Calculate: \( P_{\text{O}_2} = (3 \% \times 101.3 \text{ kPa})/100 = 3.039 \text{ kPa} \).

Step 2: Calculate the Reaction Quotient \(Q_{p}\)

The reaction quotient \(Q_{p}\) is calculated using the formula for this equilibrium: \[ Q_{p} = \frac{(P_{\text{CO})^2 \times P_{\text{O}_2}}{(P_{\text{CO}_2})^2} \] Substitute the calculated partial pressures: \[ Q_{p} = \frac{(0.2026)^2 \times 3.039}{(12.156)^2} \] Calculate to find \(Q_{p} = \frac{0.0410}{147.7566} = 2.77 \times 10^{-4} \).

Step 3: Compare \(Q_p\) with \(K_p\)

Compare \(Q_{p}\) and \(K_{p}\) to determine the direction of the reaction: \[ Q_{p} = 2.77 \times 10^{-4} \text{ and } K_{p} = 1 \times 10^{-11} \]. Since \(Q_{p} > K_{p}\), the reaction proceeds towards the reactants (\(\mathrm{CO}_2\)), suggesting the system is not at equilibrium.

Step 4: Predict the Effect of a Catalyst

Since the system is not at equilibrium and \(Q_{p} > K_{p}\), introducing a catalyst would speed up the reaction towards the reactants. This leads to a decrease in \(\mathrm{CO}\) concentration as the reaction shifts to produce more \(\mathrm{CO}_2\).

Key Concepts

Reaction Quotient

The reaction quotient, denoted as \(Q_p\), is an important concept in chemical equilibrium that helps determine the current state of a reaction relative to its equilibrium state.

For a given reaction, like the one we are examining:
\[2 \text{CO}_2(g) \rightleftharpoons 2 \text{CO}(g) + \text{O}_2(g),\]
we calculate \(Q_p\) using the formula:
\[Q_p = \frac{(P_{\text{CO}})^2 \times P_{\text{O}_2}}{(P_{\text{CO}_2})^2},\]
where \(P_{\text{CO}}\), \(P_{\text{CO}_2}\), and \(P_{\text{O}_2}\) are the partial pressures of CO, COyour_dioxide, and O_2, respectively.

• If \(Q_p = K_p\), the system is at equilibrium.
• If \(Q_p > K_p\), the system will shift towards the reactants.
• If \(Q_p < K_p\), the system will shift towards the products.

Understanding this comparison allows us to predict the direction a reaction will take to reach equilibrium when conditions change. Here, with \(Q_p = 2.77 \times 10^{-4}\) being much larger than \(K_p = 1 \times 10^{-11}\), it's clear the reaction is inclined to form more reactants, meaning equilibrium has not been reached.

Partial Pressures

In the context of chemical equilibrium, partial pressures offer a way to express the concentration of gases involved in a reaction.

Instead of dealing with concentrations directly, we use partial pressures, which can be easily derived from volume percentages and total system pressure.

In our exercise:

• CO was at 0.2% volume, resulting in a partial pressure of 0.2026 kPa.
• CO\(_2\) at 12% volume led to 12.156 kPa.
• O\(_2\) at 3% volume brought a partial pressure of 3.039 kPa.

These computations are crucial because they provide the necessary values to determine the reaction quotient (\(Q_p\)).

The partial pressures tell us how much each gas contributes to the total pressure and help us understand the balance of substances in a given state of a chemical system. With these pressures, we can assess whether the chemical reaction in question is at equilibrium or needing a shift to restore balance.

Catalyst Effect

Catalysts play a distinctive role in chemical reactions by lowering the activation energy required to proceed, thus speeding up both forward and reverse reactions equally.

However, it's important to note that catalysts do not affect the position of equilibrium itself; they only assist the system in reaching equilibrium faster.

In the given exercise:

• The reaction quotient \(Q_p\) was greater than \(K_p\), indicating the system was not in equilibrium.
• Introducing a catalyst will accelerate the shift of the reaction towards the reactants, thus potentially reducing CO levels as CO\(_2\) is formed.

This characteristic of catalysts is vital in practical applications like reducing harmful emissions in automobile exhausts, as they can effectively allow a reaction to reach equilibrium conditions more swiftly without changing the actual equilibrium state. By understanding the catalyst effect, we gain insight into how to manipulate reaction speeds for desirable chemical processes.