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Chapter 15: Problem 93

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 \(\%=\mathrm{mol} \%\).

Short Answer

Expert verified
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 by step solution

01

- 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: \(P_{O_2} = 0.03 × 101.3 ~kPa = 3.039 ~kPa\)
02

- 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: \[Q = \frac{ (0.2026)^2 \cdot 3.039}{(12.156)^2} \]
03

- 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.
04

- 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.

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Most popular questions from this chapter

For the equilibrium $$\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \operatorname{BrCl}(g) $$ at \(400 \mathrm{~K}, K_{c}=7.0 .\) If $0.25 \mathrm{~mol}\( of \)\mathrm{Br}_{2}\( and \)0.55 \mathrm{~mol}$ of \(\mathrm{Cl}_{2}\) are introduced into a 3.0-L container at \(400 \mathrm{~K}\), what will be the equilibrium concentrations of $\mathrm{Br}_{2}, \mathrm{Cl}_{2}\(, and \)\mathrm{BrCl}$ ?

Which of the following statements are true and which are false? (a) The equilibrium constant can never be a negative number. (b) In reactions that we draw with a single-headed arrow, the equilibrium constant has a value that is very close to zero. (c) As the value of the equilibrium constant increases, the speed at which a reaction reaches equilibrium increases.

The following equilibria were measured at \(823 \mathrm{~K}\) : $$\begin{array}{l} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{array}$$ (a) Use these equilibria to calculate the equilibrium constant, \(K_{c},\) for the reaction $\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)$ \(+\mathrm{CO}_{2}(g)\) at \(823 \mathrm{~K}\). (b) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K} ?(\mathbf{c})\) If you were to place \(5.00 \mathrm{~g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of \(250 \mathrm{~mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of $101.3 \mathrm{kPa}\( and a temperature of \)298 \mathrm{~K},$ what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). (d) If the reaction vessel from part (c) is heated to $823 \mathrm{~K}$ and allowed to come to equilibrium, how much \(\operatorname{CoO}(s)\) remains?

Consider the reaction $$\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)$$ At \(25^{\circ} \mathrm{C}\), the equilibrium constant is $K_{c}=2.4 \times 10^{-5}\( for this reaction. (a) If excess \)\operatorname{CaSO}_{4}(s)$ is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4},\) what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{~L},\) what is the minimum mass of \(\operatorname{CaSO}_{4}(s)\) needed to achieve equilibrium?

Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$ \begin{array}{l} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) \quad K_{c}=2.0 \\ 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=2.1 \times 10^{30} \end{array} $$
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