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 \(\left(\mathrm{Si} \mathrm{O}_{2}\right)\) (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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In conclusion: a) Depositing metallic catalysts as thin films on high surface area substances like alumina or silica is more effective than using powdered metals, as it provides a higher surface area, improved catalyst utilization, and enhanced stability. b) The surface area affects the rate of reaction by allowing more reactant molecules to interact with the catalyst, increasing the chances of productive collisions and a faster reaction rate.

Step by step solution

01

Understanding Catalysis

Catalysis is a process in which a substance called a catalyst speeds up a chemical reaction without being consumed in the reaction. Catalysts work by lowering the activation energy required for a reaction to occur, allowing the reaction to happen faster and more efficiently.
02

Knowing the role of Surface Area

Surface area plays a critical role in the rate of reaction, especially in heterogeneous catalysis, where the catalyst is in a different phase than the reactants. The larger the surface area, the more reactant molecules can come in contact with the catalyst, allowing for more simultaneous reaction sites and, hence, a faster overall reaction rate.
03

Discussing Thin Films vs. Powdered Metals

Depositing metallic catalysts as thin films on high surface area substances such as alumina or silica is more effective than using powdered metals for several reasons: 1. Higher surface area: By depositing the catalyst as a thin film, the catalyst's overall surface area is increased, which allows more reactant molecules to come in contact with the catalyst, leading to a faster reaction rate. 2. Improved catalyst utilization: A thin film of catalyst allows reactants to access the catalyst more easily compared to powdered metals, ensuring the catalyst material is being used more efficiently. 3. Enhanced stability: Thin films are less likely to become contaminated or degraded, ensuring a more consistent performance and better longevity of the catalyst.
04

Explaining the effect of surface area on reaction rate

As discussed earlier, the surface area of the catalyst directly affects the reaction rate in heterogeneous catalysis. A larger surface area means that there are more available sites on the catalyst where reactant molecules can interact. This effectively increases the chances of fruitful collisions between reactant molecules and catalyst, thus enhancing the reaction rate. In essence, the larger the surface area of the catalyst, the more efficient and faster the reaction will be. In conclusion: a) Depositing metallic catalysts as thin films on high surface area substances is more effective than using powdered metals because it provides a higher surface area for reactants to interact and ensures more efficient catalyst utilization with better stability. b) The surface area directly affects the rate of reaction by providing more available sites for reactant molecules to interact with the catalyst, increasing the chances of fruitful collisions and a faster reaction rate.

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Key Concepts

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

Heterogeneous Catalysis
Heterogeneous catalysis is a process wherein the catalyst exists in a different state (solid, liquid, or gas) from the reactants. The catalyst typically is a solid with reactants as gases or liquids, creating a scenario where the magic of the reaction occurs on the surface of the solid catalyst.

Imagine having countless tiny helpers, each occupying a minuscule portion of space but equipped to help transform reactants into products efficiently. That's akin to what a heterogeneous catalyst does—it's not used up or changed during the reaction, which also makes it a sustainable choice.

Why Heterogeneous Catalysts?

They're popular because of their ease of separation from the reactants and products; since they’re in different states, you often can just filter them out! Plus, their reusability makes them economically attractive in industrial processes.
Surface Area and Reaction Rate
Surface area is essentially the 'playground' where chemical reactions take place. The bigger the playground, the more room there is for reactant molecules to run around and interact with the catalyst, much like children in a play area. The larger the surface area, the more molecules that can 'play' at the same time, speeding up the reaction.

From a scientific standpoint, having a large surface area increases the number of active sites available for the reactions to take place. These active sites are special spots on the catalyst surface where reactants adhere temporarily, transforming into new products through a series of steps. It's akin to having more checkout counters open at a grocery store—the more counters available, the faster customers (or reactants) can complete their transactions (or reactions).
Catalyst Utilization
Effective catalyst utilization means getting the most out of your catalyst for as long as possible. In the world of chemistry, this translates to how well a catalyst performs its role over time without deteriorating or becoming less efficient.

Think of it as getting the best value out of a high-quality purchase. When catalysts are used optimally, the expenses involved in these chemical processes can be dramatically reduced. This concept is pivotal in industries reliant on catalysis, as it can be the difference between a profitable operation and one that hemorrhages resources.

Best Practices for Utilization

Methods to enhance catalyst utilization include proper catalyst selection, maintaining the ideal reaction conditions, and ensuring the catalyst is not wasted or overused.
Thin Film Catalysts
Thin film catalysts are the chemical equivalent of a large mural painted on a wall versus a standalone sculpture. By spreading the catalyst in a thin layer over a large area, much like paint, the catalyst provides a vast canvas for reactions to occur.

This 'spread out' approach allows for more reactant molecules to come in direct contact with the catalyst at the same time. It's a method that's both resource-smart and space-efficient, perfectly embodying the saying 'less is more'—less material for more chemical reaction 'real estate'.

The Innovation of Thin Films

These ultrathin coatings can be applied to various substrates using advanced techniques like physical vapor deposition, enhancing not only the reaction kinetics but also potentially introducing new properties to the catalyst, such as resistance to poisoning and thermal stability.

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Most popular questions from this chapter

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: $$ \begin{array}{l}{\text { Step } 1 : \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}} \\ {\text { Step } 2 : \mathrm{A}+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}}\end{array}$$ \(X\) 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. (ii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

You perform a series of experiments for the reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) and find that the rate law has the form rate \(=k[\mathrm{A}]^{x}\) . Determine the value of \(x\) in each of the following cases: (a) There is no rate change when \([\mathrm{A}]_{0}\) is tripled. (b) The rate increases by a factor of 9 when \([\mathrm{A}]_{0}\) is tripled. (c) When \([\mathrm{A}]_{0}\) is doubled, the rate increases by a factor of \(8 .\)

\(\begin{array}{l}{\text { (a) What is meant by the term elementary reaction? }} \\ {\text { (b) What is the difference between a unimolecular }} \\\ {\text { and a bimolecular elementary reaction? (c) What is a }}\end{array}\) \(\begin{array}{l}{\text {reaction mechanism?}(\mathbf{d}) \text { What is meant by the term rate- }} \\ {\text { determining step? }}\end{array}\)

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) .\) In water, without the enzyme, the reaction proceeds with a rate constant of 0.039 \(\mathrm{s}^{-1}\) at \(25^{\circ} \mathrm{C}\) . In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{s}^{-1}\) at \(25^{\circ} \mathrm{C}\) . Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in 0.1 \(\mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)$$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 M,\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\) , (a) What is the rate constant, \(k ?\) units of \(s^{-1}\) . (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to 0.100\(?\)

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