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Explain the difference between a homogeneous and a heterogeneous catalyst. Give an example of each.

Short Answer

Expert verified
Homogeneous catalysts are in the same phase as reactants, while heterogeneous are in a different phase. Examples are sulfuric acid (homogeneous) and platinum in catalytic converters (heterogeneous).

Step by step solution

01

Define a Catalyst

A catalyst is a substance that speeds up a chemical reaction without being consumed in the process. It lowers the activation energy required for the reaction to occur, making it proceed faster.
02

Homogeneous Catalyst Explained

A homogeneous catalyst is in the same phase as the reactants. This means they are either all in a gaseous state, all in liquid form, or all dissolved in a solution. For example, if both the reactants and the catalyst are in aqueous solution, that is a homogeneous catalyst.
03

Example of Homogeneous Catalyst

A common example of a homogeneous catalyst is sulfuric acid (\(\text{H}_2\text{SO}_4\)) used in the nitration of benzene. Both the catalyst and benzene are in liquid form, demonstrating a homogeneous system.
04

Heterogeneous Catalyst Explained

A heterogeneous catalyst is in a different phase than the reactants. Most commonly, the catalyst is a solid, while the reactants are in liquid or gaseous form. The interface between the different phases allows the reaction to occur.
05

Example of Heterogeneous Catalyst

One example of a heterogeneous catalyst is the use of platinum in catalytic converters in cars. The solid platinum acts as a catalyst for gases in the car's exhaust system, facilitating reactions that convert pollutants to less harmful substances.

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Key Concepts

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

Homogeneous Catalyst
In chemistry, when we refer to a homogeneous catalyst, we're talking about a catalyst that shares the same phase as the reactants it's helping along. This means that everything involved in the reaction is all mixed together in the same state, like all being in a liquid form or all dissolved in a solution. Homogeneous catalysts are beneficial because they can seamlessly interact with the reactants since they are in the same phase. This closeness often allows for a very efficient reaction process.

For example, in the nitration of benzene, a process important in producing many chemical products, sulfuric acid (\(\text{H}_2\text{SO}_4\)) acts as a homogeneous catalyst. Both benzene and sulfuric acid are in a liquid phase. This facilitates a smooth chemical interaction because the catalyst can uniformly distribute throughout the solution, reaching all molecules equally.

One advantage of using homogeneous catalysts is their ability to provide uniform conditions, which can be beneficial in consistently producing predictable results. However, separating these catalysts after the reaction can sometimes require additional steps, such as distillation or other purification methods.
Heterogeneous Catalyst
A heterogeneous catalyst is a catalyst that exists in a different phase than the reactants. Unlike homogeneous catalysts, heterogeneous catalysts often have the form of a solid, while the reactants are gases or liquids. This means that the reaction occurs at the surface where the catalyst and reactants meet. Such a setup often makes heterogeneous catalysts relatively easy to separate from the products since they are not in the same phase.

One of the most well-known examples of a heterogeneous catalyst is platinum used in car catalytic converters. Here, the solid platinum catalyst interacts with the gaseous exhaust emissions. It facilitates important reactions that convert harmful pollutants to less dangerous gases such as nitrogen and water vapor. This surface-area interaction enables the exhaust conversion to proceed efficiently while allowing easy recovery of the catalyst.

Heterogeneous catalysts can be advantageous due to their ease of separation and reuse. Additionally, the surface interactions can be finely tuned by altering the physical characteristics of the catalyst, such as its surface area or pore size.
Activation Energy
Activation energy is a critical concept in understanding how catalysts work. In simple terms, activation energy is the minimum energy required for a reaction to proceed. Every chemical reaction involves an energy barrier that must be overcome for reactants to transform into products. This energy is what the molecules need to successfully collide and react.

Catalysts, whether homogeneous or heterogeneous, work by lowering this activation energy. By providing an alternative pathway for the reaction with a lower energy requirement, catalysts speed up reactions without being consumed themselves in the process.

Imagine activation energy as a hill that reactants must climb over to transform into products. A catalyst effectively makes this hill smaller, allowing reactants to "climb" it more easily and thus speeding up the reaction.
  • Lowering the activation energy
  • Providing an alternative reaction pathway
  • Not consumed in the reaction process
This is why even a small amount of catalyst can affect a large amount of reactant, as it continually provides a way to overcome the energy barrier more easily.

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Most popular questions from this chapter

For the reaction $$ 2 \mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) $$ make qualitatively correct plots of the concentrations of \(\mathrm{NO}_{2}(\mathrm{~g}), \mathrm{NO}(\mathrm{g}),\) and \(\mathrm{O}_{2}(\mathrm{~g})\) versus time. Draw all three graphs on the same axes; assume that you start with \(\mathrm{NO}_{2}(\mathrm{~g})\) at a concentration of \(1.0 \mathrm{~mol} / \mathrm{L}\). Explain how you would determine, from these plots, (a) the initial rate of the reaction. (b) the final rate (that is, the rate as time approaches infinity).

Hydrogenation reactions-processes in which \(\mathrm{H}_{2}\) is added to a molecule-are usually catalyzed. An excellent catalyst is a very finely divided metal suspended in the reaction solvent. Tell why finely divided rhodium, for example, is a much more efficient catalyst than a small block of the metal that has the same mass.

For the reaction of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) at \(660 \mathrm{~K}\), \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\) \begin{tabular}{lcc} \hline \multicolumn{2}{c} { Concentration (mol/L) } & \\ \hline [NO] & {\(\left[\mathrm{O}_{2}\right]\)} & Rate of Disappearance of \(\mathrm{NO}\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 0.010 & 0.010 & \(2.5 \times 10^{-5}\) \\ 0.020 & 0.010 & \(1.0 \times 10^{-4}\) \\ 0.010 & 0.020 & \(5.0 \times 10^{-5}\) \\ \hline \end{tabular} (a) Determine the order with respect to each reactant. (b) Write the rate equation for the reaction. (c) Calculate the rate constant. (d) Calculate the rate when [NO] \(=0.025 \mathrm{~mol} / \mathrm{L}\) and \(\left[\mathrm{O}_{2}\right]=0.050 \mathrm{~mol} / \mathrm{L}\) (e) If \(\mathrm{O}_{2}\) disappears at a rate of \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}, \mathrm{cal}-\) culate the rate at which NO is consumed. Calculate the rate at which \(\mathrm{NO}_{2}\) is formed.

In an automobile catalytic converter the catalysis is accomplished on a surface consisting of platinum and other precious metals. The metals are deposited as a thin layer on a ceramic support that is a fine grid (see the photo). (a) Why is the ceramic support arranged in the grid-like geometry? (b) Why are the metals deposited on the ceramic surface instead of being used as strips or rods?

Assume that each gas-phase reaction occurs via a single bimolecular step. For which reaction would you expect the steric factor to be more important? Why? $$ \begin{aligned} \mathrm{H}_{2} \mathrm{C} &=\mathrm{CH}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3} \text { or } \\ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{HBr} \longrightarrow &\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBr}-\mathrm{CH}_{3} \end{aligned} $$

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