Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What are the four basic steps involved in heterogeneous catalysis?

Short Answer

Expert verified
The four basic steps in heterogeneous catalysis are: 1) Adsorption of reactants, 2) Diffusion of reactants on the surface, 3) Reaction on the surface, and 4) Desorption of products.

Step by step solution

01

Adsorption of reactants

In this initial step, reactant molecules are adsorbed onto the surface of the catalyst. This means that the reactants are held to the surface of the catalyst by physical or chemical bonds, which facilitates the reaction.
02

Diffusion of reactants on the surface

Once the reactants are adsorbed, they will diffuse, or move, across the surface of the catalyst to find an optimal site where the reaction can occur most efficiently.
03

Reaction on the surface

The adsorbed reactants undergo the chemical reaction while on the catalyst's surface. The arrangement of the catalyst surface allows for the reactants to come together in a favorable orientation, which lowers the activation energy and makes the reaction proceed more quickly.
04

Desorption of products

After the reaction occurs, the product molecules are not as strongly bound to the catalyst surface as the reactants were. Thus, they desorb, or leave the surface of the catalyst, freeing up space for new reactant molecules to adsorb and go through the same process.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Adsorption of Reactants
Picture a bustling city square where numerous people are walking around, looking for the perfect spot to settle down, and that's similar to the scenario in the initial step of heterogeneous catalysis, called adsorption. To kick-start a reaction, molecules of reactants must first find a place on the catalyst's surface, where they are held in place by forces ranging from weak (physical adsorption) to strong (chemical adsorption).

This step is crucial because if the reactants don't attach to the catalyst, the subsequent reaction steps cannot occur. Consider the catalyst like a specialized dock, where reactant ‘boats’ come in and ‘tie up’ before unloading their cargo—the process of the reaction. The better the 'dock’ can accommodate various 'boats', the more efficient the process will be. The strength and nature of the adsorption can greatly influence not only the rate but also the pathway of the reaction, leading to different products.
Diffusion of Reactants
Once the reactants find a prime spot on the catalyst's surface, they're not necessarily ready to react. They may need to travel across the surface to find the 'goldilocks' zone—a site that's just right for the reaction to occur. This movement is called diffusion, and it's comparable to commuters navigating through crowded streets to reach their offices.

Just as a smooth traffic flow is essential in a city, in catalysis, the ease with which reactants can move over the catalyst surface can greatly affect the reaction rate. Some catalysts even have a unique structure that resembles a maze, with specific pathways that lead reactants directly to the perfect reaction sites, enhancing the overall efficiency of the reaction.
Surface Reaction
When reactants finally come together at the optimal site on the surface, the magic of the surface reaction happens. Think of it like a dance floor where every couple needs to find the right rhythm to dance together smoothly. On the catalyst's surface, reactants must be aligned just right to react and form the desired products.

During this stage, the catalyst lowers the energy hurdle that reactants have to jump over to transform into products, commonly known as the activation energy. In a sense, the catalyst 'choreographs' the reactants into the right orientation, which ensures that they react more efficiently and rapidly. This principle is so effective that even reactions that normally take a lot of energy to proceed can occur at much lower temperatures or at a faster rate with the help of a catalyst.
Desorption of Products
After the reactants have come together and formed the new products, it's time for the final act – desorption. This step can be thought of as guests leaving a party. Now that the reactants have transformed into products, their affinity for the catalyst surface diminishes, and they detach or 'desorb' from the surface.

Because the bond to the catalyst was stronger when they were reactants than now as products, the newly formed molecules can leave the surface relatively easily. This departure is essential because it frees up space on the catalyst for new reactant molecules to dock and go through the reaction cycle. Moreover, if the products stayed attached to the catalyst, they could block other reactants from accessing the active sites, reducing the catalyst's effectiveness.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write integrated rate laws for zero-order, first-order, and second-order reactions of the form \(A \longrightarrow\) products.

A particular reaction, \(A \longrightarrow\) products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found to depend on the initial concentration of A. Determine whether each statement is like- ly to be true or false for this reaction. \begin{equation} \begin{array}{l}{\text { a. A doubling of the concentration of A doubles the rate of the reaction. }} \\ {\text { b. A plot of } 1 /[\mathrm{A}] \text { versus time is linear. }} \\ {\text { c. The half-life of the reaction gets longer as the initial concentration of }} \\ {\text { A increases. }} \\\ {\text { d. A plot of the concentration of A versus time has a constant slope. }}\end{array} \end{equation}

The rate constant \((k)\) for a reaction is measured as a function of tem- perature. A plot of ln \(k\) versus 1\(/ T(\) in \(\mathrm{K})\) is linear and has a slope of \(-7445 \mathrm{K}\) . Calculate the activation energy for the reaction.

How is the order of a reaction generally determined?

Indicate the order of reaction consistent with each observation. \begin{equation} \begin{array}{l}{\text { a. A plot of the concentration of the reactant versus time yields a }} \\ {\text { straight line. }} \\ {\text { b. The reaction has a half-life that is independent of initial }} \\ {\text { c. A plot of the inverse of the concentration versus time yiclds a }} \\ {\text { straight line. }}\end{array} \end{equation}

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free