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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 \(1 \times 10^{-13}\) . 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 CO \(_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\operatorname{mol} \% .\)

Short Answer

Expert verified
The exhaust gas is not at equilibrium with respect to the CO₂ reaction, as the reaction quotient (Q_p) is greater than the equilibrium constant (K_p). A catalyst that speeds up the CO₂ reaction would help the system reach equilibrium faster, leading to a decrease in CO concentration in the exhaust gas.

Step by step solution

01

Calculate the Mole Fractions of Each Gas

We have the following volume percentages: 0.2% CO, 12% CO₂, and 3% O₂. Since the volume percent is equal to the mole percent at a fixed temperature and pressure, we can convert them into mole fractions by dividing by 100 (total volume). CO: \(0.2\% = \frac{0.2}{100} = 0.002 \) CO₂: \(12\% = \frac{12}{100} = 0.12\) O₂: \(3\% = \frac{3}{100} = 0.03\)
02

Write the Reaction Quotient Expression

The reaction quotient (Q) can be expressed as: \(Q_p = \frac{[CO]^2[O_2]}{[CO_2]^2}\) To find Q, we can insert the mole fractions of each gas into the expression.
03

Calculate Q

Now, we substitute the mole fractions into the Q expression: \(Q_p = \frac{(0.002)^2(0.03)}{(0.12)^2} \) Calculating Q, we get: \(Q_p ≈ 8.33 \times 10^{-5}\)
04

Compare Q to K_p

We are given that: \(K_p = 1 \times 10^{-13}\) Comparing Q to K_p: \(Q_p = 8.33 \times 10^{-5} > K_p = 1 \times 10^{-13}\)
05

Determine if the System is at Equilibrium and the Effect of a Catalyst on CO Concentration

Since Q > K, the reaction will proceed to the left. The system is not at equilibrium with respect to the CO₂ reaction, and the CO concentration in the exhaust gas is higher than it would be if the reaction were at equilibrium. A catalyst that speeds up the CO₂ reaction would help the system reach equilibrium faster. In this case, since the reaction will proceed to the left, decreasing the CO concentration and increasing the CO₂ concentration. Thus, a catalyst would decrease the CO concentration in the exhaust gas.

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Most popular questions from this chapter

Two different proteins \(X\) and \(Y\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\) . The proteins bind in a \(1 : 1\) ratio to form \(X Y . A\) solution that is initially 1.00 \(\mathrm{mM}\) in each protein is allowed to reach equilibrium. At equilibrium, 0.20 \(\mathrm{mM}\) of free \(\mathrm{X}\) and 0.20 \(\mathrm{mM}\) of free Y remain. What is \(K_{c}\) for the reaction?

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{COCl}(g)+\mathrm{Cl}(g)$$ At \(25^{\circ} \mathrm{C},\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} M^{-1} \mathrm{s}^{-1}\) and \(9.3 \times 10^{10} M^{-1} \mathrm{s}^{-1}\) respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium?

The equilibrium constant for the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ is \(K_{c}=1.3 \times 10^{-2}\) at 1000 \(\mathrm{K}\) . (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2},\) or does it favor NOBr? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) (c) Calculate \(K_{c}\) for \(\operatorname{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\)

Consider the reaction \(10_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\) If you start with 25.0 \(\mathrm{mL}\) of a 0.905 \(\mathrm{M}\) solution of \(\mathrm{NaIO}_{4},\) and then dilute it with water to 500.0 \(\mathrm{mL}\) , what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}\) at equilibrium?

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2} :\) $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use thermochemical data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?
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