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(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

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To check the effectiveness of automobile catalytic converters, one should look for reduced levels of carbon monoxide (CO), hydrocarbons (HC), and nitrogen oxides (NOx) in the car exhaust. High temperatures in catalytic converters can be advantageous due to higher reaction rates and enhanced catalyst activity, but can also be disadvantageous due to potential catalyst degradation and thermal stress. The rate of flow of exhaust gases over a catalytic converter is important for maintaining sufficient reaction time, ensuring optimal catalyst temperature, and preventing overheating.

Step by step solution

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a) Substances to look for in the car exhaust

To determine the effectiveness of automobile catalytic converters, one should look for substances that indicate the presence of pollutants. Generally, catalytic converters are designed to reduce harmful emissions of the following substances: 1. Carbon monoxide (CO): A toxic, colorless, and odorless gas produced from the incomplete combustion of fuel. 2. Hydrocarbons (HC): Unburned or partially burned fuel molecules that contribute to smog and are harmful to health. 3. Nitrogen oxides (NOx): A group of gases contributing to smog, acid rain, and respiratory illnesses. The lower the levels of these substances in the car exhaust, the more effective the catalytic converter is at reducing harmful emissions.
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b) Advantages and disadvantages of high temperatures

A catalytic converter works at high temperatures due to the hot exhaust gases that stream through it. The following are advantages and disadvantages of this situation: Advantages: 1. Higher reaction rates: The high temperature accelerates chemical reactions, allowing the catalytic converter to more efficiently break down harmful substances. 2. Enhanced catalyst activity: At high temperatures, catalysts are generally more active, increasing their ability to transform pollutants into less harmful compounds. Disadvantages: 1. Potential catalyst degradation: Prolonged exposure to high temperatures can cause the catalyst material to degrade over time, reducing its effectiveness in reducing emissions. 2. Thermal stress: High temperatures can cause thermal stress on the catalytic converter's materials and surrounding components, potentially leading to failure or damage.
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c) Importance of the rate of flow of exhaust gases

The rate of flow of exhaust gases over a catalytic converter is important for several reasons: 1. Reaction time: To allow sufficient time for the catalytic reactions to occur, the exhaust gases must flow over the catalyst at an appropriate rate. If the flow rate is too high, the pollutants may not have enough contact time with the catalyst, resulting in incomplete conversions and higher emissions. 2. Catalyst temperature: If the flow rate is too low, the catalyst temperature may not reach the optimal range for efficient reactions, leading to reduced performance. 3. Preventing overheating: Maintaining a proper rate of flow can also help prevent excessive temperatures, which may lead to catalyst degradation or component failure. By controlling the rate of flow of exhaust gases over a catalytic converter, one can optimize its performance and maintain its effectiveness in reducing harmful emissions.

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Most popular questions from this chapter

Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines. The decomposition of nitric oxide \((\mathrm{NO})\) to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is second order with a rate constant of \(0.0796 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(737^{\circ} \mathrm{C}\) and \(0.0815 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(947{ }^{\circ} \mathrm{C}\). Calculate the activation energy for the reaction.

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(28{ }^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.0 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2} \mathrm{~s} \mathrm{} 0.025 \mathrm{M}\), what is the rate of formation of \(\mathrm{Cl}^{-}\)?

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) \(125^{\circ} \mathrm{C}\) ?

Heterogeneous catalysts that perform hydrogenation reactions, as illustrated in Figure 14.24, are subject to "poisoning," which shuts down their catalytic ability. Compounds of sulfur are often poisons. Suggest a mechanism by which such compounds might act as poisons.

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at \(330 \mathrm{~K}_{2} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(\)alc \()\), is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is \(0.0477 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 M\), the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

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