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How is it possible for a catalyst to give products from a reaction mixture that are different from those obtained when no catalyst or a different catalyst is used? Give an example.

Short Answer

Expert verified
Catalysts alter reaction pathways, resulting in different products; for example, platinum catalyzing hydrogenation versus acid catalyzing polymerization of alkenes.

Step by step solution

01

Understand the Role of Catalysts

A catalyst is a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change. Catalysts work by providing an alternative reaction pathway with a lower activation energy than the uncatalyzed mechanism. This means that in a catalyzed reaction, the reaction can proceed faster or via a different pathway.
02

Recognize the Impact on Reaction Pathways

Different catalysts can affect the reaction mechanism, often leading to the formation of different intermediates. This can result in the production of a different set of products compared to the uncatalyzed reaction or one catalyzed by a different substance. It's the specific nature of the catalyst that might prefer certain reaction pathways over others.
03

Example of Catalytic Effects

Consider the reaction of alkenes with hydrogen gas. When a platinum catalyst is used, the reaction commonly leads to the addition of hydrogen across the double bond, resulting in alkane products (hydrogenation). However, if a different catalyst like an acid is used, it might lead to polymerization of the alkene instead, forming different products entirely.

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Key Concepts

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

Catalyst Mechanism
A catalyst is a special ingredient in chemistry that makes reactions happen faster or in a different way without being consumed itself. Imagine it as a helper that speeds up a task without changing itself. Catalysts achieve this by offering a new route for the reaction, known as a reaction mechanism, which requires less energy than normal. This is because they lower the activation energy needed for the reaction.
The catalyst can hold onto molecules, change their shape or orientation, and create temporary states that make it easier for reactions to occur. By facilitating these changes, catalysts help convert reactants into products more efficiently.
Importantly, though catalysts speed up reactions, they don't alter the final balance of the reaction. They simply help it reach that balance more quickly.
Reaction Pathway
A reaction pathway is the series of steps that transform reactants into products. Each step along the path involves breaking and forming chemical bonds. The pathway chosen by a reaction can determine how quickly products are formed and even what products are formed.
In catalyzed reactions, the presence of a catalyst creates an alternate reaction pathway. This alternate path has a lower energy barrier, which allows the reaction to proceed faster than it would otherwise.
  • The pathway can be visualized as a mountain trail: without a catalyst, you have to climb over the highest peak. With a catalyst, there's a tunnel through the mountain, making the journey much easier and faster.
  • Different catalysts may open up different pathways, which is why switching the catalyst can lead to different products.
Chemical Intermediates
Within a reaction pathway, chemical intermediates are the temporary species formed in between the start and end points of a reaction. They're like intermediate checkpoints in a long journey. These intermediates can tell a lot about how a reaction is proceeding.
When a catalyst is involved, the formation and stability of intermediates can be dramatically different from the uncatalyzed reaction or one involving another catalyst. These differences can lead to pathways with unique intermediates that ultimately result in different products.
For example, in the hydrogenation of alkenes with a platinum catalyst, intermediates bind to the catalyst surface, easing the addition of hydrogen atoms. Conversely, an acid catalyst might create a carbocation intermediate, which can lead to entirely different final products such as in polymerization.

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Most popular questions from this chapter

The equilibrium-constant expression for a reaction is $$ K_{c}=\frac{\left[\mathrm{NO}_{2}\right]^{4}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]^{2}} $$ What is the equilibrium-constant expression when the equation for this reaction is halved and then reversed?

At \(25^{\circ} \mathrm{C}\) in a closed system, ammonium hydrogen sulfide exists as the following equilibrium: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ a. When a sample of pure \(\mathrm{NH}_{4} \mathrm{HS}(s)\) is placed in an evacuated reaction vessel and allowed to come to equilibrium at \(25^{\circ} \mathrm{C}\), total pressure is \(0.660 \mathrm{~atm} .\) What is the value of \(K_{p} ?\) b. To this system, sufficient \(\mathrm{H}_{2} \mathrm{~S}(g)\) is injected until the pressure of \(\mathrm{H}_{2} \mathrm{~S}\) is three times that of the ammonia at equilibrium. What are the partial pressures of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) ? C. In a different experiment, \(0.750\) atm of \(\mathrm{NH}_{3}\) and \(0.500\) atm of \(\mathrm{H}_{2} \mathrm{~S}\) are introduced into a \(1.00-\mathrm{L}\) vessel at \(25^{\circ} \mathrm{C}\). How many moles of \(\mathrm{NH}_{4} \mathrm{HS}\) are present when equilibrium is established?

At high temperatures, a dynamic equilibrium exists between carbon monoxide, carbon dioxide, and solid carbon. $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) ; \Delta H^{\circ}=172.5 \mathrm{~kJ} $$ At \(850^{\circ} \mathrm{C}, K_{c}\) is \(0.153\). a. What is the value of \(K_{p} ?\) b. If the original reaction system consisted of just carbon and \(1.50 \mathrm{~atm}\) of \(\mathrm{CO}_{2}\), what are the pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) when equilibrium has been established? c. How will the equilibrium pressure of \(\mathrm{CO}\) change if the temperature is decreased?

At moderately high temperatures, \(\mathrm{SbCl}_{5}\) decomposes into \(\mathrm{SbCl}_{3}\) and \(\mathrm{Cl}_{2}\) as follows: $$ \mathrm{SbCl}_{5}(g) \rightleftharpoons \mathrm{SbCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ A \(65.4-g\) sample of \(\mathrm{SbCl}_{5}\) is placed in an evacuated 5.00-L vessel and it is raised to \(195^{\circ} \mathrm{C}\) and the system comes to equilibrium. If at this temperature \(35.8 \%\) of the \(\mathrm{SbCl}_{5}\) is decomposed, what is the value of \(K_{p}\) ?

Two moles of \(\mathrm{H}_{2}\) are mixed with \(1 \mathrm{~mol} \mathrm{O}_{2}\) at \(25^{\circ} \mathrm{C}\). No observable reaction takes place, although \(K_{c}\) for the reaction to form water is very large at this temperature. When a piece of platinum is added, however, the gases react rapidly. Explain the role of platinum in this reaction. How does it affect the equilibrium composition of the reaction mixture?

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