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What are the characteristics of a catalyst?

Short Answer

Expert verified
A catalyst is a substance that speeds up a chemical reaction without being consumed by the reaction. It can work in either homogeneous or heterogeneous catalysis, providing a lower-energy pathway for the reaction. Crucially, while catalysts speed up the rate of both the forward and reverse reactions, they do not shift the equilibrium of the reaction.

Step by step solution

01

Identifying Characteristics of a Catalyst

Firstly, we can begin by stating that a catalyst is a substance that speeds up a chemical reaction without being consumed by the reaction. This means that a catalyst can be used repeatedly for the same reaction because it does not get used up.
02

Additional Properties

Secondly, it's important to note that catalysts can be in the same phase as the reactants, known as homogeneous catalysis, or in a different phase, known as heterogeneous catalysis. The catalyst changes the mechanism of the reaction, often by providing a lower-energy pathway for the reaction to occur.
03

Role of a Catalyst

Lastly, catalysts do not change the equilibrium of a reaction. They only affect the rate at which the equilibrium is reached. That is, the catalysts speed up the rate of both the forward and reverse reactions equally, allowing the system to reach equilibrium faster but does not shift the position of the equilibrium.

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Most popular questions from this chapter

Polyethylene is used in many items such as water pipes, bottles, electrical insulation, toys, and mailer envelopes. It is a polymer, a molecule with a very high molar mass made by joining many ethylene molecules (the basic unit is called a monomer) together (see p. 369 ). The initiation step is $$ \mathrm{R}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{R} \cdot \quad \text { initiation } $$ The \(\mathrm{R}\) - species (called a radical) reacts with an ethylene molecule \((\mathrm{M})\) to generate another radical $$ \mathrm{R} \cdot+\mathrm{M} \longrightarrow \mathrm{M}_{1} \cdot $$ Reaction of \(\mathrm{M}_{1}\). with another monomer leads to the growth or propagation of the polymer chain: $$ \mathrm{M}_{1} \cdot+\mathrm{M} \stackrel{k_{\mathrm{p}}}{\longrightarrow} \mathrm{M}_{2} \cdot \quad \text { propagation } $$ This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine $$ \mathrm{M}^{\prime} \cdot+\mathrm{M}^{\prime \prime} \cdot \stackrel{k_{t}}{\longrightarrow} \mathrm{M}^{\prime}-\mathrm{M}^{\prime \prime} \quad \text { termination } $$ (a) The initiator used in the polymerization of ethylene is benzoyl peroxide \(\left[\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2}\right]:\) $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO} \cdot $$ This is a first-order reaction. The half-life of benzoyl peroxide at \(100^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~min} .\) (a) Calculate the rate constant (in \(\min ^{-1}\) ) of the reaction. (b) If the half-life of benzoyl peroxide is \(7.30 \mathrm{~h}\) or \(438 \mathrm{~min},\) at \(70^{\circ} \mathrm{C},\) what is the activation energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) for the decomposition of benzoyl peroxide? (c) Write the rate laws for the elementary steps in the above polymerization process and identify the reactant, product, and intermediates. (d) What condition would favor the growth of long high-molar-mass polyethylenes?

List four factors that influence the rate of a reaction.

A factory that specializes in the refinement of transition metals such as titanium was on fire. The firefighters were advised not to douse the fire with water. Why?

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 the reaction $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \\\ & \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+\mathrm{H}^{+}(a q)+\mathrm{I}^{-}(a q) \end{aligned} $$ How could you follow the progress of the reaction by measuring the electrical conductance of the solution?

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