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In an automobile catalytic converter the catalysis is accomplished on a surface consisting of platinum and other precious metals. The metals are deposited as a thin layer on a ceramic support that is a fine grid (see the photo). (a) Why is the ceramic support arranged in the grid-like geometry? (b) Why are the metals deposited on the ceramic surface instead of being used as strips or rods?

Short Answer

Expert verified
The ceramic grid maximizes area for efficient chemical reactions, while thin metal layers reduce costs and increase surface efficiency.

Step by step solution

01

Understanding the Function of the Ceramic Grid

The ceramic support in a catalytic converter serves to maximize the surface area on which reactions can occur. By being arranged in a grid or honeycomb structure, this ceramic support exposes more surface area of the precious metals to the exhaust gases. A larger surface area allows for more efficient contact between the exhaust gases and the catalyst, which in turn maximizes the conversion of harmful gases into less harmful emissions.
02

Importance of Maximized Surface Area

The significance of having a large surface area is directly linked to the efficiency of the catalytic reaction. The larger the surface area, the more active sites are available for the catalysis to take place. This means more exhaust gas molecules can simultaneously react, leading to a faster and more complete conversion of pollutants.
03

Reason for Using a Thin Layer of Metal

Metals like platinum are very expensive, so it is imperative to use them efficiently. Depositing these metals as a thin layer on the ceramic substrate, rather than using solid strips or rods, saves material costs while still providing a broad active surface for the catalytic reaction. This thin layer covers the extensive surface area provided by the grid-like ceramic, ensuring that the catalytic reaction can occur over a large area without the need for bulkier, costlier solid metal pieces.
04

Economic and Practical Considerations

The decision to use a thin layer of metal also stems from practical and economic considerations. Using metal strips or rods would substantially increase the cost and weight of the catalytic converter, and it would reduce the efficiency due to a decreased surface area. By using minimal quantities of metal in a dispersed format, the same catalytic effect is achieved more economically and effectively.

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Key Concepts

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

Chemical Catalysis
Catalytic converters play a crucial role in reducing vehicle emissions by transforming harmful pollutants into less harmful emissions through chemical catalysis. Catalysis involves the acceleration of a chemical reaction by a substance, usually called a catalyst, which remains unchanged at the reaction's completion. In the case of catalytic converters, precious metals such as platinum and palladium act as the catalyst. These metals facilitate the conversion of toxic gases like carbon monoxide and hydrocarbons into non-toxic emissions like carbon dioxide and water vapor.
Some advantages of using chemical catalysis in catalytic converters include:
  • Improved air quality by significantly reducing harmful emissions.
  • Increased fuel efficiency due to the acceleration of chemical reactions.
  • Longevity of the catalyst, as it doesn't get consumed during the reaction process.
Understanding chemical catalysis helps to appreciate the intricate design and functioning of catalytic converters in emissions reduction.
Surface Area Maximization
Surface area plays a fundamental role in the effectiveness of catalytic converters. A larger surface area provides more active sites for reactions, enhancing the converter's efficiency. This is achieved by designing the ceramic support in a grid-like or honeycomb structure.
The benefits of maximizing surface area include:
  • Increased interaction between exhaust gases and catalytic material, leading to a higher rate of conversion.
  • A more complete transformation of pollutants into less harmful substances.
  • Enhanced overall durability and lifespan of the catalytic converter due to a more efficient catalytic process.
By exposing more of the catalyst material to the exhaust gases, surface area maximization is crucial in maintaining optimal converter performance without increasing the amount of precious metals used.
Precious Metals in Catalysis
Precious metals like platinum, palladium, and rhodium are critical in the function of catalytic converters due to their exceptional catalytic properties. These metals are highly effective in accelerating the breakdown of pollutants but come with a significant cost, making their efficient use essential.
Here’s why precious metals are chosen:
  • They have the ability to withstand high temperatures and harsh environments.
  • Their chemical stability helps in catalyzing reactions without undergoing deterioration.
  • Their effectiveness in breaking down a wide range of pollutants quickly and efficiently.
To economize on these costly materials, they are deposited as thin layers on ceramic supports, maximizing their efficacy while minimizing costs.
Ceramic Support Structure
The ceramic support structure, usually in a grid or honeycomb shape, is an essential component of catalytic converters. This structure provides several advantages that enhance the converter’s efficiency.
Key functions of ceramic support include:
  • Supplying a stable and durable base for precious metals to be deposited upon.
  • Increasing the exposed surface area, allowing more reactions to occur simultaneously.
  • Ensuring even distribution of the precious metals, optimizing their catalytic function.
The ceramic's high thermal resistance also means it can function effectively under the intense heat produced by vehicle exhausts. This arrangement not only increases the converter’s efficiency but also ensures long-term reliability by supporting both the catalytic material and structural integrity under various operating conditions.

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Most popular questions from this chapter

Many biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and \(X\) is the reactant) is Step 1: reversible HA(aq) \(\rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq}) \quad\) fast Step 2: reversible \(\mathrm{X}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{XH}^{+}(\mathrm{aq}) \quad\) fast Step 3: \(\mathrm{XH}^{+}(\mathrm{aq}) \longrightarrow\) products slow (a) Derive the rate law from this mechanism. (b) Determine the order of reaction with respect to HA. (c) Determine how doubling the concentration of HA would affect the rate of the reaction.

When substrates are present at relatively high concentration and are catalyzed by enzymes, the effect on reaction rate of changing substrate concentration can be described by zeroth-order kinetics. Calculate by what factor the rate of an enzyme-catalyzed reaction changes when the substrate concentration is changed from \(1.5 \times 10^{-2} \mathrm{M}\) to \(4.5 \times 10^{-2} \mathrm{M}\)

In a time-resolved picosecond spectroscopy experiment, Sheps, Crowther, Carrier, and Crim (Journal of Physical Chemistry \(A,\) Vol. 110,\(2006 ;\) pp. \(3087-3092\) ) generated chlorine atoms in the presence of pentane. The pentane was dissolved in dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\). The chlorine atoms are free radicals and are very reactive. After a nanosecond the chlorine atoms have reacted with pentane molecules, removing a hydrogen atom to form \(\mathrm{HCl}\) and leaving behind a pentane radical with a single unpaired electron. The equation is \(\mathrm{Cl} \cdot(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{12}(\mathrm{dcm}) \longrightarrow \mathrm{HCl}(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{11} \cdot(\mathrm{dcm})\) where \((\mathrm{dcm})\) indicates that a substance is dissolved in dichloromethane. Measurements of the concentration of chlorine atoms were made as a function of time at three different concentrations of pentane in the dichloromethane. These results are shown in the table. of Chlorine Atoms (M) \begin{tabular}{clcc} & \multicolumn{3}{c} { Concentration of Chlorine Atoms (m) } \\ Time (ps) & \multicolumn{4}{c} { for Different \(\mathrm{C}_{5} \mathrm{H}_{12}\) Concentrations } \\ \cline { 3 - 5 } & \(0.15-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.30-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.60-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) \\ \hline 100.0 & \(4.42 \times 10^{-5}\) & \(3.11 \times 10^{-5}\) & \(1.77 \times 10^{-5}\) \\ 140.0 & \(3.823 \times 10^{-5}\) & \(2.39 \times 10^{-5}\) & \(1.15 \times 10^{-5}\) \\\ 180.0 & \(3.38 \times 10^{-5}\) & \(1.94 \times 10^{-5}\) & \(8.48 \times 10^{-6}\) \\\ 220.0 & \(3.03 \times 10^{-5}\) & \(1.49 \times 10^{-5}\) & \(6.04 \times 10^{-6}\) \\\ 260.0 & \(2.68 \times 10^{-5}\) & \(1.19 \times 10^{-5}\) & \(4.12 \times 10^{-6}\) \\\ 300.0 & \(2.42 \times 10^{-5}\) & \(9.45 \times 10^{-6}\) & \(3.14 \times 10^{-6}\) \\\ 340.0 & \(2.08 \times 10^{-5}\) & \(7.75 \times 10^{-6}\) & \(2.38 \times 10^{-6}\) \\\ 380.0 & \(1.91 \times 10^{-5}\) & \(6.35 \times 10^{-6}\) & \(1.75 \times 10^{-6}\) \\\ 420.0 & \(1.71 \times 10^{-5}\) & \(4.58 \times 10^{-6}\) & \(1.61 \times 10^{-6}\) \\\ 460.0 & \(1.53 \times 10^{-5}\) & \(3.77 \times 10^{-6}\) & \(9.98 \times 10^{-7}\) \\\ \hline \end{tabular} (a) Determine the order of the reaction with respect to chlorine. (b) Determine whether the reaction rate depends on the concentration of pentane in dichloromethane. If so, determine the order of the reaction with respect to pentane. (c) Explain why the concentration of pentane in dichloromethane does not affect the data analysis that you performed in part (a). (d) Write the rate law for the reaction and calculate the rate of reaction for a concentration of chlorine atoms equal to \(1.0 \mu \mathrm{M}\) and a pentane concentration of \(0.23 \mathrm{M}\). (e) Sheps, Crowther, Carrier, and Crim found that the rate of formation of \(\mathrm{HCl}\) matched the rate of disappearance of Cl. From this they concluded that there were no intermediates and side reactions were not important. Explain the ic

Measurements of the initial rate of hydrolysis of benzenesulfonyl chloride in aqueous solution at \(15^{\circ} \mathrm{C}\) in the presence of fluoride ion yielded the results in the table for a fixed concentration of benzenesulfonyl chloride of \(2 \times\) \(10^{-4} \mathrm{M}\). The reaction rate is known to be proportional to the concentration of benzenesulfonyl chloride. \begin{tabular}{cc} \hline & Initial Rate \(\times 10^{7}\) \\ {\(\left[\mathrm{~F}^{-}\right] \times 10^{2}(\mathrm{~mol} / \mathrm{L})\)} & \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 0 & 2.4 \\ 0.5 & 5.4 \\ 1.0 & 7.9 \\ 2.0 & 13.9 \\ 3.0 & 20.2 \\ 4.0 & 25.2 \\ 5.0 & 32.0 \\ \hline \end{tabular} Note that some reaction must be occurring in the absence of any fluoride ion, because at zero concentration of fluoride the rate is not zero. This residual rate should be subtracted from each observed rate to give the rate of the reaction being studied. (a) Derive the complete rate law for the reaction. (b) Calculate the rate constant \(k\) and express it in appropriate units.

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