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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. Explain 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, enhancing catalytic efficiency.

Step by step solution

01

Understanding Catalysis Efficiency

Catalysts increase the rate of a chemical reaction without being consumed in the process. The effectiveness of a catalyst depends on its ability to provide a surface for the reaction to occur.
02

Surface Area of Catalysts

Catalysts with greater surface area facilitate more reaction sites for the reactants. A finely divided metal, like rhodium, has a much larger surface area compared to the same amount of catalyst in a solid block form.
03

Surface Area to Volume Ratio

The increased surface area to volume ratio of finely divided metals allows more molecules of the reactants to interact with the catalyst at the same time, thus speeding up the reaction.
04

Catalytic Efficiency

Finely divided rhodium provides more active sites per unit of mass, significantly improving the efficiency of the catalysis in hydrogenation reactions. This is why finely divided rhodium is more efficient than a small block.

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Key Concepts

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

Hydrogenation Reactions
Hydrogenation is a key chemical reaction that involves the addition of hydrogen (\(\mathrm{H}_2\)) to a molecule. This process is crucial in various industries, including food processing and pharmaceuticals, as it can alter the properties of substances. For example, converting unsaturated fats to saturated fats. The beauty of hydrogenation lies in its ability to change a molecule’s structure and function with the help of catalysts.

Catalysts are essential to hydrogenation because they accelerate the reaction without being consumed. This is vital because pure \(\mathrm{H}_2\) on its own tends to react very slowly. In the context of hydrogenation reactions, finely divided metals like rhodium are commonly used. The finely divided nature of these metals plays a significant role in maximizing their reaction capacity.
Surface Area to Volume Ratio
The concept of surface area to volume ratio is vital for understanding the efficiency of catalysts in hydrogenation reactions. When a metal like rhodium is finely divided, it increases its total surface area without a significant increase in volume. This property means that a larger number of reactant molecules can come into contact with the catalyst's surface at once.

Think of it this way, more surface area equals more spots for reactions to happen. More spots mean more simultaneous reactions and, therefore, a quicker overall process.
  • This increased surface area helps maximize the use of the catalyst's material, allowing it to facilitate numerous reactions at once.
  • Finely divided catalysts ensure that even a small amount of metal can have a large impact.
The high surface area to volume ratio is what makes a finely divided catalyst much more effective than a large block of the same metal.
Catalytic Efficiency
Catalytic efficiency refers to how effectively a catalyst speeds up a reaction. In hydrogenation reactions, finely divided metals like rhodium showcase significant catalytic efficiency through increased active sites. Active sites are regions on the catalyst where the reactants bind and the reaction takes place.

By having more of these active sites, finely divided catalysts are able to:
  • Provide more opportunities for reactant molecules to interact with the catalyst.
  • Enable quicker bond transformations and increase reaction rates.
  • Reduce the overall energy needed for the reaction, making the process more feasible.
Therefore, the ability of finely divided rhodium to provide numerous active sites leads to a substantial improvement in reaction speed and efficiency compared to using a larger block. In practical terms, this means reactions can occur faster and with less catalyst material, proving cost-effective in industrial applications.

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Most popular questions from this chapter

The decomposition of nitrogen dioxide at a high temperature $$ \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ is second-order in this reactant. (a) Determine the rate constant for this reaction if it takes 1.76 min for the concentration of \(\mathrm{NO}_{2}\) to fall from 0.250 mol/L to 0.100 mol/L. (b) If the chemical equation is written as $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ what is the value of the rate constant?

Hypofluorous acid, HOF, is very unstable, decomposing in a first-order reaction to give HF and \(\mathrm{O}_{2},\) with a half-life of \(30 .\) minutes at room temperature: $$ \mathrm{HOF}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ If the partial pressure of HOF in a 1.00-L flask is initially \(1.00 \times 10^{2} \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C},\) what are the total pressure in the flask and the partial pressure of HOF after exactly 30 minutes? After 45 minutes?

Calculate the activation energy, \(E_{a}\) for the reaction $$ 2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ from the observed rate constants: \(k\) at \(25^{\circ} \mathrm{C}=\) \(3.46 \times 10^{-5} \mathrm{s}^{-1}\) and \(k\) at \(55^{\circ} \mathrm{C}=1.5 \times 10^{-3} \mathrm{s}^{-1}\).

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 B is halved, what happens to the reaction rate?

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 pH buffered solution? (In a pH buffered solution, the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) is a constant.) (b) Which of the following mechanisms is the most appropriate for the interpretation of this rate law? Explain. (Note that when writing the expression for \(K,\) the equilibrium constant, \(\left.\left[\mathrm{H}_{2} \mathrm{O}\right] \text { is not involved. See Chapter } 15 .\right)\) Mechanism 1 \(\mathrm{NO}_{2} \mathrm{NH}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) Mechanism 2 $$ \mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+} \frac{k_{2}}{\overleftarrow{k_{2}^{\prime}}} \mathrm{NO}_{2} \mathrm{NH}_{3}^{+}+\mathrm{H}_{2} \mathrm{O} $$ \(\mathrm{NO}_{2} \mathrm{NH}_{3}^{+} \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{3} \mathrm{O}^{+} \quad\) (rate-limiting step) Mechanism 3 \(\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \frac{k_{4}}{k_{4}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}^{-}+\underset{(\text { rapid equilibrium })}{\mathrm{H}_{3} \mathrm{O}^{+}}\) \(\mathrm{NO}_{2} \mathrm{NH}^{-} \stackrel{k_{5}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{OH}^{-} \quad\) (rate-limiting step) \(\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \stackrel{k_{6}}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{O} \quad\) (very fast reaction) (c) Show the relationship between the experimentally observed rate constant, \(k\), and the rate constants in the selected mechanism. (d) Based on the experimental rate law, will the reaction rate increase or decrease if the pH of the solution is increased?

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