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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? (b) How does the surface area affect the rate of reaction?

Short Answer

Expert verified
Depositing metallic catalysts as thin films on substances with high surface area per unit mass increases the efficiency of the catalyst by exposing more catalyst sites to the reactants, enabling more reactions to occur simultaneously. The surface area of a catalyst directly impacts the rate of reaction; a larger surface area allows for more reactant-catalyst interactions and thus a faster rate of reaction.

Step by step solution

01

(a) Reason for Depositing Metallic Catalysts as Thin Films on Substances with High Surface Area per Unit Mass

Metallic catalysts are often deposited as thin films on substances with high surface area per unit mass to increase the efficiency of the catalyst. In a catalytic reaction, the catalyst provides an alternative reaction pathway with lower activation energy, which speeds up the reaction without being consumed in the process. The effectiveness of a catalyst depends on its surface area, as it is the surface where the reactants come into contact with the catalyst and form temporary bonds. By depositing the catalyst material as a thin film on substances like alumina or silica with high surface area, more catalyst sites are exposed to the reactants. Hence, more reaction can take place simultaneously, and the catalyst material is used more efficiently.
02

(b) Surface Area's Effect on the Rate of Reaction

The surface area of the catalyst directly affects the rate of reaction in a catalytic process. When the surface area increases, more active sites are exposed to the reactants, which leads to an increase in the number of effective collisions between reactant molecules and the catalyst surface. As a result, this will cause an increase in the rate of the reaction. Conversely, when the surface area of the catalyst decreases, fewer active sites are available for the reactants to come into contact with, hence reducing the number of collisions and lead to a slower rate of reaction. In summary, a larger surface area of the catalyst is associated with a faster rate of reaction, as it allows more reactant molecules to interact with the catalyst at any given time.

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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 is a key factor in determining how effective a catalyst can be. Catalysts often work by providing a surface on which reactants can adhere and react more easily. Essentially, the more surface area available, the more space the reactants have to interact with the catalyst. Thin films on materials like alumina (Al₂O₃) or silica (SiO₂) are used because these substances have a high surface area per unit mass. This method exposes more of the catalyst to the reactants, making the reaction process more efficient.
  • A larger surface area allows more molecules to bind at once.
  • Increased exposure of catalyst sites enhances reaction opportunities.
  • Efficient use of the catalyst material minimizes costs and waste.
By maximizing the surface area, we can ensure that more molecules are interacting with the catalyst at any given time, inevitably speeding up the reaction.
Reaction Rate
The reaction rate in catalytic processes is heavily reliant on the surface characteristics of the catalyst. A catalyst with a high surface area enables a higher number of reactant molecules to collide with its active sites at any moment. This boost in collisions significantly increases the rate of reaction. The relationship between surface area and reaction rate can be simplified as follows:
  • More surface area = more active sites available.
  • More active sites = more effective collisions.
  • More effective collisions = faster reaction rate.
When catalysts are used as thin films, they provide many locations for reactants to interact. If this area is limited, fewer reactant molecules can come into contact with the catalyst, restricting the frequency of effective collisions and slowing down the rate. Thus, increasing the surface area by utilizing thin film deposition methods significantly enhances how quickly a reaction can proceed.
Activation Energy
Activation energy is the minimum amount of energy that reactant molecules need to undergo a chemical reaction. Catalysts, including those used in thin film form, play a crucial role in lowering the activation energy of a reaction. How does a catalyst reduce activation energy?
  • Provides an alternative reaction pathway requiring less energy.
  • Helps in forming intermediate compounds that consume less energy.
  • Lowers the energy barrier, making it easier for reactants to transform into products.
By lowering the activation energy, catalysts make it easier for reactions to proceed at a faster rate, even at lower temperatures. When more surface area is available, as discussed, this property is enhanced because more reactant molecules can be brought into close proximity with less energy input, thus speeding up the reaction even further. This reduction in required energy and increased activity translates directly into more efficient industrial processes, where time and energy are conserved.

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Most popular questions from this chapter

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is catalyzed by \(\mathrm{NO}_{2}\). Thereaction proceeds as follows: $$ \begin{aligned} &\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ &2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{aligned} $$ (a) Show that the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3}\). (Hint: The top reaction must be multiplied by a factor so the \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) cancel out.) (b) Why do we consider \(\mathrm{NO}_{2}\) a catalyst and not an intermediate in this reaction? (c) Is this an example of homogeneous catalysis or heterogeneous catalysis?

(a) The gas-phase decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) \(\longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\), is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). At \(600 \mathrm{~K}\) the half-life for this process is \(2.3 \times 10^{5} \mathrm{~s}\). What is the rate constant at this temperature? (b) At \(320^{\circ} \mathrm{C}\) the rate constant is \(2.2 \times 10^{-5} \mathrm{~s}^{-1}\). What is the half-life at this temperature?

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\). When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M}\), the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8\) ?

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) : At \(23^{\circ} \mathrm{C}\) and in \(0.5 \mathrm{M} \mathrm{HCl}\), the following data were obtained for the disappearance of sucrose: $$ \begin{array}{cl} \hline \text { Time }(\mathrm{min}) & \left.\mathrm{IC}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](M) \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \end{array} $$ (a) Is the reaction first order or second order with respect to \(\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] ?\) (b) What is the value of the rate constant?

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 \(14.5\)

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