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Chapter 14: Problem 63

Explain why most metals used in catalysis are transition metals.

Short Answer

Expert verified
Transition metals are commonly used in catalysis mainly because of their ability to adopt multiple and stable oxidation states, which enable them to transfer electrons with the reactants, thereby lowering the activation energy of the reaction. In addition, they can form complexes, bringing the reactants closer to each other, which further speeds up the reaction.

Step by step solution

01

Understanding Catalysis

Catalysis is a phenomenon where a substance, called a catalyst, speeds up a chemical reaction without being consumed in the process. They work by providing an alternative reaction pathway with a lower activation energy.
02

Characteristics of Transition Metals

Transition metals are elements whose atom has a partially filled d sub-shell. They exhibit unique features such as the ability to form stable, multiple oxidation states and the capability to form complexes.
03

Linking Transition Metals properties with Catalysis

Owing to their ability to adopt multiple and stable oxidation states, transition metals can transfer electrons with species involved in the reaction, thereby lowering the reaction's activation energy. The formation of complexes is also crucial in bringing the reaction participants close together, further enhancing the rate of reaction.

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Most popular questions from this chapter

When methyl phosphate is heated in acid solution, it reacts with water: $$ \mathrm{CH}_{3} \mathrm{OPO}_{3} \mathrm{H}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}_{3} \mathrm{PO}_{4} $$ If the reaction is carried out in water enriched with \({ }^{18} \mathrm{O}\), the oxygen- 18 isotope is found in the phosphoric acid product but not in the methanol. What does this tell us about the bond-breaking scheme in the reaction?

Consider this mechanism for the enzyme-catalyzed reaction $$ \mathrm{E}+\mathrm{S} \stackrel{k_{1}}{\rightleftharpoons_{-1}} \mathrm{ES} \quad \text { (fast equilbrium) } $$ $$ \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \quad(\text { slow }) $$ Derive an expression for the rate law of the reaction in terms of the concentrations of \(\mathrm{E}\) and \(\mathrm{S}\). (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.)

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