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Chapter 14: Problem 76

(a) If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would you want to look for in the car exhaust? (b) Automobile catalytic converters have to work at high temperatures, as hot exhaust gases stream through them. In what ways could this be an advantage? In what ways a disadvantage? (c) Why is the rate of flow of exhaust gases over a catalytic converter important?

Short Answer

Expert verified
To assess the effectiveness of automobile catalytic converters, we need to check for substances like carbon monoxide (CO), hydrocarbons (HC), and nitrogen oxides (NO and NO2) in the car exhaust. High temperatures can be both advantageous, improving conversion efficiency and faster startup performance, and disadvantageous, causing material degradation, thermal stress, and increased fuel consumption. The exhaust gas flow rate is crucial for ensuring optimum reaction rates, maintaining the catalyst's optimal operating temperature, and avoiding catalyst fouling.

Step by step solution

01

(a) Substances to check in car exhaust

To check the effectiveness of automobile catalytic converters, it's important to look for the following substances in the car exhaust: 1. Carbon Monoxide (CO): A harmful gas that should be converted into carbon dioxide (CO2) by the catalytic converter. 2. Hydrocarbons (HC): Unburnt fuel particles that should be oxidized into water (H2O) and carbon dioxide (CO2) by the catalytic converter. 3. Nitrogen Oxides (NO and NO2): These harmful pollutants should be reduced to nitrogen (N2) and oxygen (O2) by the catalytic converter.
02

(b) High temperatures in catalytic converters: Advantages

High temperatures in automobile catalytic converters could be advantageous because: 1. Improved conversion efficiency: Chemical reactions occurring in the catalytic converter require activation energy. Higher temperatures provide this energy, allowing reactions to occur faster and more efficiently. 2. Faster startup performance: A catalytic converter heats up faster at high temperatures, allowing it to reach its optimal operating temperature quickly and start removing pollutants from the exhaust gases sooner.
03

(b) High temperatures in catalytic converters: Disadvantages

High temperatures in automobile catalytic converters could be disadvantageous because: 1. Material degradation: Prolonged exposure to high temperatures can cause the materials in the catalytic converter to degrade, resulting in reduced performance and a shorter service life. 2. Thermal stress: The high temperatures can cause thermal stress on the catalytic converter and nearby components, potentially leading to premature failure. 3. Increased fuel consumption: In order to maintain high temperatures, the engine may need to use more fuel, leading to increased fuel consumption and emissions.
04

(c) Importance of exhaust gas flow rate over a catalytic converter

The rate of flow of exhaust gases over a catalytic converter is important because: 1. Optimum reaction rates: A suitable flow rate ensures that pollutants in the exhaust gases spend enough time in contact with the converter's catalysts, allowing the chemical reactions to occur efficiently. 2. Temperature management: A proper exhaust gas flow rate helps maintain the catalyst's optimal operating temperature, ensuring efficient performance. 3. Avoiding catalyst fouling: If the flow rate is too low, it could lead to incomplete combustion and a buildup of pollutants on the catalyst surface, which can reduce the catalytic converter's efficiency and service life.

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Most popular questions from this chapter

NO catalyzes the decomposition of \(\mathrm{N}_{2} \mathrm{O}\), possibly by the following mechanism: $$ \begin{array}{r} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{array} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Why is NO considered a catalyst and not an intermediate? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism? If you think not, suggest what might be going on.

The rate of the reaction \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow\) $$ \mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) $$ was measured at several temperatures, and the following data were collected: \begin{tabular}{ll} \hline Temperature \(\left({ }^{\circ} \mathrm{C}\right)\) & \(k\left(\boldsymbol{M}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 15 & \(0.0521\) \\ 25 & \(0.101\) \\ 35 & \(0.184\) \\ 45 & \(0.332\) \\ \hline \end{tabular} Using these data, graph \(\ln k\) versus \(1 / T\). Using your graph, determine the value of \(E_{a}\)

The activation energy of an uncatalyzed reaction is \(95 \mathrm{~kJ} / \mathrm{mol}\). The addition of a catalyst lowers the activation energy to \(55 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C}\), (b) \(125^{\circ} \mathrm{C}\) ?

(a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) What information is necessary to relate the rate of disappearance of reactants to the rate of appearance of products?

Consider the following reaction: $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M}\), the reaction rate at \(298 \mathrm{~K}\) is \(0.0432 \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of \(\mathrm{OH}^{-}\) were tripled?
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