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Hydrogenation reactions, processes wherein \(\mathrm{H}_{2}\) 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.

Short Answer

Expert verified
Finely divided rhodium offers more surface area, increasing catalytic efficiency.

Step by step solution

01

Introduction to Catalysts

Catalysts speed up reactions by lowering the activation energy required for the reaction to occur. This makes the reaction proceed faster without the catalyst being consumed.
02

Understanding Surface Area

Catalytic efficiency is influenced by the surface area available for the reaction. The larger the surface area, the more opportunities there are for reactant molecules to interact with the catalyst.
03

Comparing Finely Divided Metals to Bulk Metals

Finely divided rhodium has a much higher surface area compared to a block of rhodium. The particles are smaller and more dispersed, providing more surface area for the hydrogenation reaction to take place.
04

Effect of Surface Area on Reaction Rate

The increased surface area from finely divided rhodium exposes more active sites for the reactants to interact with, thus increasing the reaction rate efficiency compared to a bulk metal block.

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Key Concepts

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

Hydrogenation reactions
In the realm of chemistry, hydrogenation is a critical reaction where a molecule absorbs hydrogen (\( \mathrm{H}_{2}\)). This is commonly observed in processes such as the conversion of unsaturated fats to saturated fats. The hydrogenation process frequently relies on catalysts that assist in overcoming the energy barrier required for the reaction, known as the activation energy. Without a catalyst, hydrogenation would occur too slowly to be practical in many applications. Catalysts do not get consumed in the reaction and can be used repeatedly, speeding up the process substantially. One of the primary applications of hydrogenation is in the food industry, contributing to the texture and shelf-life of products like margarine.
  • Hydrogenation involves the addition of hydrogen atoms to molecules.
  • Catalysts lower the energy needed for hydrogenation to occur.
  • Widely used in various industries, particularly in food processing.
Surface area in catalysis
The surface area is a crucial factor in the effectiveness of a catalyst. Catalysts function by providing a surface where the reaction can take place. The more extensive the surface area, the greater the number of active sites available for the reactant molecules to meet and interact. This increased availability can significantly enhance the reaction rate.
A finely divided metal, such as rhodium, offers a considerable surface area compared to its bulk form. The increase in surface area means that there are more cavities and edges, which are often places where the chemical reactions occur. This increases the number of opportunities for reactant particles to hit the catalyst and undergo the desired transformation.
  • Higher surface area means more sites for reactions to occur.
  • Finely divided metals increase catalytic efficiency.
  • Surface area enhancement directly correlates with improved reaction rates.
Rhodium as a catalyst
Rhodium is a metal renowned for its catalytic properties, making it an ideal choice in many hydrogenation reactions. When used as a catalyst, it can speed up chemical reactions without altering the final outcome. The reason finely divided rhodium is especially effective in catalysis is due to its high surface area, which allows it to provide numerous active sites for the reaction.
The finely divided form means rhodium particles are dispersed throughout the reactant mixture, maximizing contact with the reactants. This dispersion ensures that more reactants are converted to products, thereby increasing efficiency. Rhodium's selectivity also aids in achieving the desired reaction outcomes without unwanted by-products, a significant advantage in applications requiring high purity.
  • Rhodium speeds up reactions through increased active site availability.
  • Finely divided rhodium maximizes contact with reactants.
  • Ensures high efficiency and selectivity in reactions.

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Most popular questions from this chapter

A reaction has the experimental rate equation Rate \(=k[\mathrm{A}]^{2} .\) How will the rate change if the concentration of \(\mathrm{A}\) is tripled? If the concentration of \(A\) is halved?

Nitramide, \(\mathrm{NO}_{2} \mathrm{NH}_{2},\) decomposes slowly in aqueous solution according to the following reaction: $$\mathrm{NO}_{2} \mathrm{NH}_{2}(\mathrm{aq}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ The reaction follows the experimental rate law $$\text { Rate }=\frac{k\left[\mathrm{NO}_{2} \mathrm{NH}_{2}\right]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}$$ (a) What is the apparent order of the reaction in a buffered solution? (b) Which of the following mechanisms is the most appropriate for the interpretation of this rate law? Explain. Mechanism 1 \(\mathrm{NO}_{2} \mathrm{NH}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) Mechanism 2 \(\begin{array}{l}\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+} \frac{k_{2}}{k_{2}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}_{3}^{+}+\mathrm{H}_{2} \mathrm{O} \\\\\mathrm{NO}_{2} \mathrm{NH}_{3}+\stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2}\mathrm{O}+\mathrm{H}_{3} \mathrm{O}^{+}(\text {rapid equilibrium })\end{array}\) Mechanism 3 \(\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \frac{k_{4}}{k_{4}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) \(\mathrm{NO}_{2} \mathrm{NH}^{-} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{OH}^{-} \quad\) (rapid equilibrium) \(\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \stackrel{k_{6}}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{O} \quad\) (very fast reaction) (c) Show the relationship between the experimentally observed rate constant, \(k\), and the rate constants in the selected mechanism. (Note that when writing the expression for \(K\), the equilibrium constant, \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) is not involved. \(\rightarrow\) Chapter \(16 .\) ) (d) Show that hydroxide ions catalyze the decomposition of nitramide.

Using the rate equation "Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}],\) " define the order of the reaction with respect to \(\mathrm{A}\) and \(\mathrm{B}\). What is the total order of the reaction?

The reaction of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) is thought to occur in two steps: Step 1: Slow \(\mathrm{NO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g})\) Step 2: Fast \(\mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Show that the elementary steps add up to give the overall, stoichiometric equation. (b) What is the molecularity of each step? (c) For this mechanism to be consistent with kinetic data, what must be the experimental rate equation? (d) Identify any intermediates in this reaction.

A reaction has the following experimental rate equation: Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}] .\) If the concentration of \(\mathrm{A}\) is doubled and the concentration of \(\mathrm{B}\) is halved, what happens to the reaction rate?

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