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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) initiation of the chemical reaction, (3) desorption of products, and (4) regeneration of the catalyst.

Step by step solution

01

Identify the Adsorption of Reactants

The first step in heterogeneous catalysis is the adsorption of reactants on the catalyst surface. The reactants bind to active sites on the surface of the catalyst through physical or chemical adsorption.
02

Initiate the Chemical Reaction

Once adsorbed, the reactants may undergo a chemical reaction. The catalyst provides a surface where reactions can occur more readily by lowering the activation energy required for the reaction.
03

Desorb Products from the Surface

After the reaction, the products formed are desorbed from the catalyst surface. This step is necessary for the catalyst to continue catalyzing subsequent reactions as it frees up the active sites on the catalyst's surface.
04

Regenerate Catalyst

The final step is the regeneration of the catalyst surface. This involves the removal of any residues or poisons that might have been left on the catalyst's active sites, restoring its activity for further reaction cycles.

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Key Concepts

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

Adsorption of Reactants
Understanding the adsorption of reactants is essential in the study of heterogeneous catalysis. This process involves the molecules of reactants adhering to the catalyst's surface. Imagine this as a parking lot where the reactant molecules park on the vacant spots – these are known as active sites. There are two types of adsorption: physical and chemical. Physical adsorption is like a car parked without the handbrake; it's weaker and the car can easily leave (the reactant doesn't bond strongly and can be readily removed). Chemical adsorption is more like parking with a clamp on the wheel; the reactant forms a strong bond with the catalyst, resembling a chemical change.
Catalyst Surface Activity
The real magic of catalysts lies in their surface activity. The active sites on a catalyst's surface are akin to specialized workshops where reactant molecules come to get transformed. These sites have the unique ability to lower the hurdles that reactant molecules typically face when reacting with each other. Just as a workshop provides tools to make a job easier, the catalyst's surface gives reactants what they need to combine or break apart with less effort. What's crucial here is the quality of these active sites. They should not be blocked by impurities, and they must be in the right shape and have the proper electronic properties to make the catalytic reaction as efficient as possible.
Activation Energy
Activation energy is the push needed to start a reaction, much like the energy required to push a ball over a hill before it can roll down the other side. Catalysts are the heroes in this scenario because they effectively make the hill smaller, which means reactants need less of a push (less energy) to react with one another. They achieve this by providing an alternative reaction pathway with a lower peak, enabling reactants to transform into products more comfortably and at a faster rate. This is why, in the presence of a catalyst, reactions can occur at lower temperatures and pressures than would otherwise be possible.
Catalyst Regeneration
Over time, catalysts can lose their charm. Much like a sponge getting clogged with soap scum, a catalyst's surface can get covered with residues or poisons that block the active sites. Catalyst regeneration is like giving that sponge a good rinse, restoring its original texture and absorption ability. This step ensures the longevity and effectiveness of the catalyst so that it can continue to facilitate reactions again and again. Regenerating a catalyst may involve burning off the accumulated residues or using chemical treatments to free up those valuable active sites and keep the catalytic reactions running smoothly.

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Most popular questions from this chapter

The desorption (leaving of the surface) of a single molecular layer of \(n\) -butane from a single crystal of aluminum oxide is found to be first order with a rate constant of \(0.128 / \mathrm{s}\) at \(150 \mathrm{~K}\). a. What is the half-life of the desorption reaction? b. If the surface is initially completely covered with \(n\) -butane at \(150 \mathrm{~K},\) how long will it take for \(25 \%\) of the molecules to desorb (leave the surface)? For \(50 \%\) to desorb? c. If the surface is initially completely covered, what fraction will remain covered after 10 s? After 20 s?

The tabulated data show the rate constant of a reaction mea- sured at several different temperatures. Use an Arrhenius plot to determine the activation barrier and frequency factor for the reaction. $$ \begin{array}{cl} \text { Temperature (K) } & \text { Rate Constant (1/s) } \\ \hline 310 & 0.00434 \\ \hline 320 & 0.0140 \\ \hline 330 & 0.0421 \\ \hline 340 & 0.118 \\ \hline 350 & 0.316 \\ \hline \end{array} $$

Consider the reaction: $$ 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{S}_{8}(g) $$ Complete the table. $$ \begin{array}{lllll} \Delta\left[\mathrm{H}_{2} S\right] / \Delta t & \Delta\left[\mathrm{O}_{2}\right] / \Delta t & \Delta\left[\mathrm{H}_{2} \mathrm{O}\right] / \Delta t & \Delta\left[\mathrm{S}_{8}\right] / \Delta t & \text { Rate } \\ -0.080 \mathrm{M} / \mathrm{s} & & & & \\ \hline \end{array} $$

The rate constant \((k)\) for a reaction was measured as a function of temperature. A plot of \(\ln k\) versus \(1 / T(\) in \(\mathrm{K})\) is linear and has a slope of \(-1.01 \times 10^{4} \mathrm{~K}\). Calculate the activation energy for the reaction.

This reaction is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\) $$ \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{3}(g)+\mathrm{NO}_{2}(g) $$ The rate constant for the reaction at a certain temperature is \(0.053 / \mathrm{s}\) a. Calculate the rate of the reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.055 \mathrm{M}\). b. What would the rate of the reaction be at the concentration indicated in part a if the reaction were second order? Zero order? (Assume the same numerical value for the rate con- stant with the appropriate units.)

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