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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 \(1 \mathrm{~atm}\) ) 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 \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the \(\mathrm{CO}\) 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 \(\%=\mathrm{mol} \%\).

Short Answer

Expert verified
The initial partial pressures are: \(CO = 0.002 \, \text{atm}, CO_2 = 0.12 \, \text{atm}\), and \(O_2 = 0.03 \, \text{atm}\). The reaction quotient (Q) is calculated as \(Q = \frac{(0.002)^2(0.03)}{(0.12)^2} = 8.33 \times 10^{-6}\). Comparing Q with the given Kp value (\(1 \times 10^{-13}\)), we see that Q > Kp, so the reaction will shift towards reactants to reach equilibrium. Thus, a catalyst that speeds up the CO2 reaction would increase the CO concentration in the exhaust.

Step by step solution

01

Calculate initial partial pressures

To calculate the partial pressures of CO, CO2, and O2, we will use the given volume percentages and the total pressure. Partial pressure of CO: \( (0.2\% \times 1 \, \text{atm}) = 0.002 \, \text{atm} \) Partial pressure of CO2: \( (12\% \times 1 \, \text{atm}) = 0.12 \, \text{atm} \) Partial pressure of O2: \( (3\% \times 1 \, \text{atm}) = 0.03 \, \text{atm} \)
02

Calculate the reaction quotient (Q)

Now that we have the initial partial pressures, let's calculate the reaction quotient (Q). The reaction quotient is given by: \[ Q = \frac{[\mathrm{CO}]^2[\mathrm{O}_2]}{[\mathrm{CO}_2]^2} \] Plugging in the initial partial pressures, we have: \[ Q = \frac{(0.002)^2(0.03)}{(0.12)^2} \] Calculate Q.
03

Compare Q with Kp

Now, we will compare the value of Q with the given Kp value (\(1 \times 10^{-13}\)) to determine whether the system is at equilibrium or not. If Q = Kp, the system is at equilibrium. If Q < Kp, the reaction will shift towards products to reach equilibrium. If Q > Kp, the reaction will shift towards reactants to reach equilibrium. Compare Q with Kp and decide the direction of the reaction.
04

Determine the effect of a catalyst on CO concentration

Based on the comparison of Q with Kp, we can now determine whether the CO concentration in the exhaust would be increased or decreased by a catalyst that speeds up the CO2 reaction: If the reaction shifts towards products, an increase in CO2 conversion will decrease the CO concentration. If the reaction shifts towards reactants, an increase in CO2 conversion will increase the CO concentration. If the system is at equilibrium, the CO concentration will remain the same. Determine the effect of the catalyst on CO concentration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Quotient (Q)
The reaction quotient, often denoted as "Q," is a valuable tool in chemistry for predicting the direction of a reaction at any point, not just at equilibrium. It uses the ratio of the concentrations (or partial pressures in the case of gases) of products to reactants, mirroring how an equilibrium constant (K_p or K_c) is calculated but with values at a particular moment rather than at equilibrium.

For the chemical reaction given:\[ 2 \mathrm{CO}_2(g) \rightleftharpoons 2 \mathrm{CO}(g) + \mathrm{O}_2(g) \]The reaction quotient Q is expressed as:\[ Q = \frac{[\mathrm{CO}]^2[\mathrm{O}_2]}{[\mathrm{CO}_2]^2} \]

The purpose of comparing Q with K_p is to determine how the reaction "behaves":
  • If Q = K_p, the system is at equilibrium, indicating no net shift.
  • If Q < K_p, the system proceeds toward the products to achieve equilibrium, meaning more CO and O_2 will form.
  • If Q > K_p, the system shifts toward the reactants to regain equilibrium, suggesting that more CO_2 will be formed.

In our exercise, initial partial pressures are used to compute Q, guiding predictions about the system's shift necessary to reach equilibrium.
Catalyst Effect
A catalyst is a substance that can increase the rate of a chemical reaction without itself being consumed. Its role is primarily to lower the activation energy required for the reaction, allowing the reaction to proceed more quickly. However, it is crucial to understand that a catalyst does not affect the position of equilibrium.

This means:
  • The same ratio of products to reactants will be achieved, just faster.
  • A catalyst helps a reaction reach equilibrium more quickly without shifting the position of equilibrium.

In the context of our problem, if the chemical system was not at equilibrium, adding a catalyst would accelerate the reaction's adjustment to equilibrium. However, the catalyst does not change the equilibrium concentrations of CO , CO_2 , and O_2 in the exhaust. It only speeds up the process of reaching that equilibrium, providing faster stabilization based on whichever direction (towards reactants or products) is needed according to Q and K_p comparison.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that predicts how a system at equilibrium responds to disturbances or 'stresses'. It states that if a change is applied to a system at equilibrium, the system adjusts in a way that counteracts the change and re-establishes equilibrium.

Disturbances can include:
  • Changes in concentration of reactants or products
  • Changes in pressure or volume (for gaseous reactions)
  • Changes in temperature

Although the exercise primarily focuses on comparing Q and K_p , Le Chatelier's Principle helps explain why the reaction adjusts in a particular way when Q does not equal K_p :
  • If Q < K_p , the system will produce more products to reduce stress by increasing product concentration until Q equals K_p .
  • If Q > K_p , the system will form more reactants to reduce stress by decreasing product concentration.

This principle is key in predicting the behavior of the system when external conditions change, ensuring that students understand how and why shifts in equilibrium composition occur in response to perturbations.

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Most popular questions from this chapter

(a) If \(Q_{c}

In Section 11.5, we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the corresponding expression for \(K_{P}\). (b) By using data in Appendix B, give the value of \(K_{p}\) for this reaction at \(30^{\circ} \mathrm{C}\). (c) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

As shown in Table 15.2, the equilibrium constant for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is \(K_{p}=4.34 \times 10^{-3}\) at \(300^{\circ} \mathrm{C}\). Pure \(\mathrm{NH}_{3}\) is placed in a \(1.00-\mathrm{L}\) flask and allowed to reach equilibrium at this temperature. There are \(1.05 \mathrm{~g} \mathrm{NH}_{3}\) in the equilibrium mixture. (a) What are the masses of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in the equilibrium mixture? (b) What was the initial mass of ammonia placed in the vessel? (c) What is the total pressure in the vessel?

Consider the following equilibrium for which \(\Delta H<0\) $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ How will each of the following changes affect an equilibrium mixture of the three gases: (a) \(\mathrm{O}_{2}(g)\) is added to the system; (b) the reaction mixture is heated; (c) the volume of the reaction vessel is doubled; (d) a catalyst is added to the mixture; (e) the total pressure of the system is increased by adding a noble gas; (f) \(\mathrm{SO}_{3}(g)\) is removed from the system?

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00- \(\mathrm{L}\) flask at \(303 \mathrm{~K}\), \(56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{p}\) for this reaction at \(303 \mathrm{~K}\). (c) According to Le Châtelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture were transferred to a \(15.00\)-L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(15.00\) - \(\mathrm{L}\) vessel at \(303 \mathrm{~K}\).

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