Problem 91 - 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 \(\mathrm{CO}_{2}\) reaction?

Step-by-Step Solution

Answer

The reaction is not at equilibrium, as the calculated reaction quotient (Q) is greater than the given equilibrium constant (Kp), \(Q = 8.33 \times 10^{-5}\) and \(K_p = 1 \times 10^{-13}\). Since the reaction proceeds in the reverse direction to reach equilibrium, a catalyst would speed up the conversion of CO and O2 into CO2, resulting in a decrease in the concentration of CO in the automobile exhaust gas.

Step 1: Write the expression for the Reaction Quotient (Q)

The reaction quotient (Q) for the given reaction can be written as follows: \[Q = \frac{[\mathrm{CO}]^2[\mathrm{O}_{2}]}{[\mathrm{CO}_{2}]^2}\]

Step 2: Calculate molar fractions of gases

We are given the volume percentages of CO, CO2, and O2 in the exhaust gas. We can assume that the molar fraction is equal to the volume fraction because the total pressure is 1 atm and there is no restriction on the volume of each gas. The molar fractions are: \[\mathrm{Molar\,fraction\,of\,CO} = 0.002\] \[\mathrm{Molar\,fraction\,of\,CO_2} = 0.12\] \[\mathrm{Molar\,fraction\,of\,O_2} = 0.03\]

Step 3: Calculate the current reaction quotient (Q) using the given molar fractions

Now we need to substitute the molar fractions of the gases into the reaction quotient expression derived in step 1: \[Q = \frac{(0.002)^2(0.03)}{(0.12)^2} = 8.33 \times 10^{-5}\]

Step 4: Compare Q and Kp

Now, we can compare the calculated value of Q with the given value of Kp: \[Q = 8.33 \times 10^{-5}\] \[K_p = 1 \times 10^{-13}\] Since \(Q > K_p\), the reaction is not at equilibrium. In this case, the reaction will proceed in the reverse direction, converting CO and O2 back into CO2.

Step 5: Analyze the effect of a catalyst on the CO concentration

A catalyst speeds up the reaction without changing the equilibrium constant or the position of the equilibrium. Since we found that the reaction would proceed in the reverse direction to reach equilibrium, a catalyst would speed up the conversion of CO and O2 into CO2, therefore decreasing the concentration of CO in the automobile exhaust gas.

Key Concepts

Reaction Quotient

In chemical equilibrium, the reaction quotient (Q) is a valuable tool used to determine the direction in which a chemical reaction is proceeding. Unlike an equilibrium constant (K_p), which is only defined at equilibrium, the reaction quotient helps us assess the current state of a reaction mixture at any given moment.
For our reaction \(2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\), the reaction quotient is expressed as:

• \(Q = \frac{[\mathrm{CO}]^2[\mathrm{O}_{2}]}{[\mathrm{CO}_{2}]^2}\)

Using the respective molar fractions of each gas allows us to calculate Q and compare it with K_p.
If Q is greater than K_p, the reaction tends to move in the reverse direction—towards the formation of reactants. Conversely, if Q is less than K_p, it shifts forward, producing more products.

Molar Fractions

Molar fractions are important in calculating the reaction quotient. They reflect the ratio of moles of a specific gas component to the total moles of the gas in a mixture. This is especially useful under conditions of 1 atm pressure, as here molar fractions equate directly to volume percentages.
In this exercise, the molar fractions were:

• CO: 0.002
• CO₂: 0.12
• O₂: 0.03

These fractions are critical for stepping into the calculation of the reaction quotient because they help determine Q by replacing concentrations in the reaction expression. For gases, it simplifies the process and guides us to understand the reaction's behavior in relation to equilibrium.

Catalysts

Catalysts are substances that notably accelerate chemical reactions without being consumed in the process. They are crucial in industries where speeding up reactions is necessary without altering the position of equilibrium or K_p.
For the reaction \(2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) in the automobile exhaust, using a catalyst facilitates reaching equilibrium faster, thereby enhancing the conversion efficiency.
Since the reaction was found to be moving in the reverse direction to reach equilibrium (from the comparison of Q and K_p), adding a catalyst would decrease the concentration of CO by accelerating its conversion back into CO₂, improving the exhaust's emission quality.

Le Chatelier's Principle

Le Chatelier's Principle provides valuable insights into how a chemical system at equilibrium responds to environmental changes like pressure, concentration, or temperature shifts. Although catalysts do not change equilibrium positions, this principle is still useful in understanding system adjustments.
If a system is disturbed, the principle suggests it will shift in the direction that counteracts the change. In our context, since Q > K_p , increasing reactant concentration (CO₂) by reversing the reaction is nature's way of balancing the system.
Thus, applying this principle helps anticipate the changes ensuing from equilibrium disturbances, reinforcing that while catalysts speed the reactions, they do not alter these inherent properties dictated by Le Chatelier's Principle.