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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 \(\left(\mathrm{SiO}_{2}\right)\). (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

Expert verified
Depositing metallic catalysts as thin films on high surface area substances like alumina (Al2O3) or silica (SiO2) is more effective compared to using powdered metals due to better surface area utilization, reduced costs, and enhanced stability. A higher surface area directly impacts the rate of reaction by providing more reaction sites, increasing the probability of successful collisions, and facilitating better adsorption and desorption of reactants and products.

Step by step solution

01

Answer (a): Reasons for depositing catalysts as thin films

Depositing metallic catalysts (especially precious metals) on high surface area substances like alumina (Al2O3) or silica (SiO2) is more effective compared to using powdered metals because of the following reasons: 1. Surface area utilization: A higher surface area of the catalyst means that more reaction sites are available for the reacting molecules, leading to faster reaction rates. Thin films of the catalyst on these high surface area substances increase the number of catalyst sites, thus making the process more efficient. 2. Reduction of costs: Precious-metal catalysts are often expensive, so utilizing them in the form of thin films can significantly reduce the amount of catalyst material needed for the process. This reduces the overall cost of the process and makes it more economically feasible. 3. Enhanced stability: Depositing catalysts on supports like alumina or silica also helps in stabilizing the catalyst particles, preventing them from interacting with each other and forming larger particles, which would reduce the surface area available for the reaction.
02

Answer (b): Surface area effects on reaction rate

The surface area of a catalyst directly impacts the rate of reaction due to the following reasons: 1. Higher surface area leads to more reaction sites: When the surface area of a catalyst is increased, there are more reaction sites available for the molecules of the reactants to interact with the catalyst. This leads to faster reaction rates. 2. Increased probability of collisions: A catalyst with a larger surface area offers more chances for reactant molecules to interact with its surface, thus increasing the probability of a successful collision and thereby enhancing the reaction rate. 3. Facilitates better adsorption and desorption of reactants: A higher surface area enables more reactants to be adsorbed onto the catalyst's surface, which can facilitate the activation of bonds and the formation of activated complexes. Likewise, a higher surface area also allows for better desorption of the products, making room for more reactant molecules to interact with the catalyst. In conclusion, an increased catalyst surface area can significantly enhance the rate of reaction by providing more reaction sites, increasing the probability of successful collisions, and facilitating better adsorption and desorption of reactants and products.

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Key Concepts

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

Surface Area
The surface area of a catalyst plays a vital role in chemical reactions, especially when dealing with thin films of precious metals deposited on high surface area supports like alumina \((\mathrm{Al}_2\mathrm{O}_3)\) or silica \((\mathrm{SiO}_2)\). Increasing the surface area means there are more places or "sites" for the reaction to occur.
This increase in available sites enhances the likelihood of reactant molecules finding a place to react.
- More contact points: A larger surface area offers more opportunities for molecules to come into contact with the catalyst.- Enhanced efficiency: By maximizing the exposed surface, the reaction can proceed more efficiently and quickly.Ultimately, an expansive surface area in catalysts makes for a greater likelihood of successful reactions as there are more places for molecules to interact.
Reaction Rate
The reaction rate is directly influenced by the surface area of a catalyst. With more surface area, the reaction rate can increase significantly because more molecules can interact with the catalyst at the same time.
This leads to a higher frequency of successful collisions, which is crucial for speeding up chemical processes. - Frequency of Collisions: With more surface area, there is a higher probability that reactant molecules will collide with the catalyst, leading to enhanced reaction rates. - Adsorption Benefits: Greater surface area enhances the catalyst's ability to adsorb reactants, promoting the necessary bond formations for the reaction. In summary, increasing the surface area of a catalyst not only increases the number of sites for reactions but also boosts the overall speed of these chemical processes.
Precious Metals
Precious metals such as platinum, palladium, and gold are often used as catalysts due to their ability to facilitate chemical reactions effectively.
These metals are highly valued not only for their allure but also for their efficiency in catalysis.
- High reactivity: Precious metals can aid in reactions by providing active sites that are exceptional for facilitating chemical transformations. - Cost Considerations: Since these metals are costly, they are typically used in very small amounts. Depositing them as thin films on supports maximizes their use while minimizing costs. Overall, precious metals remain a popular choice for catalysis due to their unique properties and effectiveness, despite their high price point.
Catalyst Supports
Catalyst supports, like alumina or silica, are materials used to extend the efficiency and stability of metal catalysts by providing a high-surface-area base.
These supports help to disperse the active metal into very small particles, maximizing the surface area. - Stability Enhancement: Supports help prevent the catalyst particles from agglomerating and losing surface area. - Cost Efficiency: By using catalysts in conjunction with supports, the amount of precious metal needed is reduced, providing a more economical solution. Using catalyst supports allows for maximizing the active sites available for reactions, thereby enhancing the overall effectiveness of the catalytic material.

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Most popular questions from this chapter

You study the effect of temperature on the rate of two reactions and graph the natural logarithm of the rate constant for each reaction as a function of \(1 / T\). How do the two graphs compare (a) if the activation energy of the second reaction is higher than the activation energy of the first reaction but the two reactions have the same frequency factor, and \((b)\) if the frequency factor of the second reaction is higher than the frequency factor of the first reaction but the two reactions have the same activation energy? [Section 14.5\(]\)

When chemists are performing kinetics experiments, the general rule of thumb is to allow the reaction to proceed for 4 half-lives. (a) Explain how you would be able to tell that the reaction has proceeded for 4 half-lives. (b) Let us suppose a reaction \(\mathrm{A} \rightarrow \mathrm{B}\) takes 6 days to proceed for 4 half-lives and is first order in A. However, when your lab partner performs this reaction for the first time, he does not realize how long it takes, and he stops taking kinetic data, monitoring the loss of A, after only 2 hours. Your lab partner concludes the reaction is zero order in A based on the data. Sketch a graph of [A] versus time to convince your lab partner the two of you need to be in the lab for a few days to obtain the proper rate law for the reaction.

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1} .\) Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C},\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g) .\) If the reaction is first order with a half-life of 56.3 min at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).

(a) A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{b})\) Another first-order reaction also has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?(\mathrm{c})\) What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

The isomerization of methyl isonitrile \(\left(\mathrm{CH}_{3} \mathrm{NC}\right)\) to acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) was studied in the gas phase at \(215^{\circ} \mathrm{C},\) and the following data were obtained: $$ \begin{array}{rl} \hline \text { Time (s) } & {\left[\mathrm{CH}_{3} \mathrm{NC}\right](\boldsymbol{M})} \\ \hline 0 & 0.0165 \\ 2,000 & 0.0110 \\ 5,000 & 0.00591 \\ 8,000 & 0.00314 \\ 12,000 & 0.00137 \\ 15,000 & 0.00074 \\ \hline \end{array} $$ (a) Calculate the average rate of reaction, in \(M / s\), for the time interval between each measurement. (b) Calculate the average rate of reaction over the entire time of the data from \(t=0\) to \(t=15,000 \mathrm{~s}\). (c) Graph [CH \(\left._{3} \mathrm{NC}\right]\) versus time and determine the instantaneous rates in \(M /\) s at \(t=5000 \mathrm{~s}\) and \(t=8000 \mathrm{~s}\).

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