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Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right) .(\mathbf{a})\) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

Short Answer

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(a) Depositing metallic catalysts as thin films on high surface area materials like alumina or silica is effective because it provides better dispersion of the catalyst, maximizing the number of active sites for reactants to interact with. This not only enhances the reactivity of the catalyst but also improves its stability, prolonging its lifespan. (b) A larger surface area results in more available active sites on the catalyst surface, leading to an increased probability of reactant-catalyst interactions. As the number of active sites and the frequency of reactant-catalyst interaction increase, the rate of reaction is boosted, speeding up the overall chemical process.

Step by step solution

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(a) Advantages of thin films compared to powdered metals

The effectiveness of a catalyst is largely determined by its capacity to provide an increased number of active sites where reactant molecules can interact, thus promoting the desired reaction to proceed at a faster rate. Depositing the catalyst material as a thin film on a substrate with a high surface area per unit mass, like alumina or silica, helps maximize its efficiency in several ways: 1. Improved Dispersion: Creating a thin film allows for a better dispersion of the catalyst material over the surface of the support. This ensures the optimal use of the catalytic material since it becomes more evenly distributed and provides a larger number of active sites. In contrast, using powdered metals could limit the dispersion of the catalyst, deactivating some of its molecules by preventing them from being accessible to reactants. 2. High Surface Area: A high surface area support, such as alumina or silica, increases the number of available active sites for reactants to interact with. This occurs because a larger surface area means that there is more surface for the catalyst to cover, creating more potential reaction sites and enhancing the reactivity of the catalyst. 3. Stability: The precious-metal catalysts are often expensive and rare, making it important to use them sparingly and prolong their lifespan. Depositing a thin film of the catalyst on a stable material can help preserve the material's properties and degradation due to environmental factors, thus enhancing its long-term effectiveness.
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(b) Effect of surface area on the rate of reaction

The surface area of a catalyst has a direct impact on the rate of reaction. As previously discussed, the effectiveness of a catalyst lies in its ability to provide active sites for reactant molecules to interact. A larger surface area means more available active sites on the catalyst surface, which leads to an increased probability of reactant molecules encountering and interacting with the catalyst. As the number of active sites and the frequency of reactant-catalyst interaction increase, the rate of reaction is boosted. The reaction takes place more rapidly, producing the desired products faster and speeding up the overall chemical process. Therefore, using catalysts with a high surface area or depositing catalysts on high surface area supports can greatly enhance the efficiency and rate of reactions.

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Key Concepts

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

Metallic Catalysts
Metallic catalysts play a crucial role in speeding up chemical reactions by lowering the activation energy required for the process to occur. These catalysts often involve precious metals like platinum or palladium. Their unique properties and the ability to facilitate reactions make them indispensable in industrial applications. However, these metals are generally costly and scarce, which necessitates efficient usage.

Using metal catalysts dispersed as thin films allows for greater exposure of their active sites. By minimizing the amount of metal required and maximizing its exposure for reaction, these catalysts effectively increase the rate of reactions without the need for large amounts of the metal. Thin films ensure optimal usage of expensive metallic catalysts, making them economically viable and sustainable. This method prevents wastage and ensures that each available metal atom can play its part in facilitating the reaction.
  • Efficient use of precious metals
  • Increases reaction rate by exposing more active sites
  • Cost-effective and sustainable
Thin Films
Thin films offer a highly efficient way to use solid materials, especially in the case of catalysts. When metallic catalysts are applied as thin films, the surface area is maximized, while using minimal material. This phenomenon increases the chances of successful interactions between the reactant molecules and the catalyst.

The primary advantage of thin films in catalytic processes is their improved dispersion over a supporting material, like alumina or silica. This setup allows for a uniform distribution of the catalytic sites, ensuring that each part of the film is actively involved in the reaction. By contrast, powdered catalysts might clump together, which reduces the number of exposed active sites, potentially deactivating some parts of the catalyst.
  • Maximizes surface exposure
  • Promotes even distribution of active sites
  • Prevents catalyst deactivation
Surface Area
Surface area plays a pivotal role in determining the efficiency of a catalyst. A higher surface area means more active sites are available for the interaction between the catalyst and the reactants, significantly increasing the probability of successful collisions.

The rate of a chemical reaction is directly proportional to the surface area of the catalyst. More surface area allows for a greater number of interactions, leading to a faster reaction rate. High surface area supports like alumina or silica offer an expansive terrain for the catalyst to populate, effectively increasing the reaction's speed without requiring additional catalyst material.

Overall, increasing the surface area of a catalyst or using supports with high surface areas is a strategic approach to accelerating chemical reactions, enhancing their efficiency while minimizing costs.
  • Increases available active sites
  • Boosts reaction rate
  • Optimizes catalyst usage

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Most popular questions from this chapter

Consider the following reaction between mercury(II) chloride and oxalate ion: $$ 2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) $$ The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the following rate data were obtained for the rate of disappearance of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) : $$ \begin{array}{llll} \hline \text { Experiment } & {\left[\mathrm{HgCl}_{2}\right](M)} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right](M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1 & 0.164 & 0.15 & 3.2 \times 10^{-5} \\ 2 & 0.164 & 0.45 & 2.9 \times 10^{-4} \\ 3 & 0.082 & 0.45 & 1.4 \times 10^{-4} \\ 4 & 0.246 & 0.15 & 4.8 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is \(0.25 \mathrm{M}\) if the temperature is the same as that used to obtain the data shown?

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{nm} .\) (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at \(30.0 \mathrm{~min} .\) Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(392.4 \mathrm{pm} .(\mathbf{a})\) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3} .\) Recall that \(1 \mathrm{pm}=1 \times 10^{-12} \mathrm{~m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m} .(\mathbf{b})\) Estimate how many platinum atoms are on the surface of a 2.0-nm Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of \(280 \mathrm{pm}\) (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a 5.0-nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\). The following two-step mechanism is proposed for the reaction: $$ \begin{array}{l} \text { Step } 1: \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X} \\\ \text { Step } 2: \mathrm{A}+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D} \end{array} $$ \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21^{\circ} \mathrm{C}\). (a) Write out the balanced equation for the reaction catalyzed by urease. \((\mathbf{b})\) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C}\), what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

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