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Chapter 14: Problem 75

Briefly comment on the effect of a catalyst on each of the following: (a) activation energy, (b) reaction mechanism, (c) enthalpy of reaction, (d) rate of forward step, (e) rate of reverse step.

Short Answer

Expert verified
A catalyst decreases the activation energy, changes the reaction mechanism, has no effect on enthalpy, and increases both the rate of the forward and reverse steps.

Step by step solution

01

Implication of a Catalyst on Activation Energy

A catalyst provides an alternate path or mechanism for a reaction to follow, which has a lower activation energy than the uncatalyzed reaction. Therefore, a catalyst decreases the activation energy of a reaction.
02

Implication of a Catalyst on Reaction Mechanism

A catalyst doesn't change the reaction itself but provides a new mechanism or reaction path to proceed, that is, it changes the intermediary stages of the reaction while the initial and final stages remain the same.
03

Implication of a Catalyst on Enthalpy of Reaction

Since a catalyst doesn't change the states of the reactants and products, it has no effect on the enthalpy or heat content of the reaction. The enthalpy change (\( \Delta H\)) for a reaction remains the same whether a catalyst is used or not.
04

Implication of a Catalyst on Rate of Forward Step

By lowering the activation energy, a catalyst increases the rate of the forward step of a reaction, so the reaction reaches equilibrium more quickly.
05

Implication of a Catalyst on Rate of Reverse Step

A catalyst accelerates both the forward and reverse reactions equally, because it merely lowers the activation energy barrier. So, the rate of reverse reaction also increases when a catalyst is involved.

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Most popular questions from this chapter

The reaction \(\mathrm{H}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2}+\mathrm{H}\) has been studied for many years. Sketch a potential-energy-versusreaction-progress diagram for this reaction.

Thallium(I) is oxidized by cerium(IV) as follows: $$ \mathrm{Tl}^{+}+2 \mathrm{Ce}^{4+} \longrightarrow \mathrm{Tl}^{3+}+2 \mathrm{Ce}^{3+} $$ The elementary steps, in the presence of \(\mathrm{Mn}(\mathrm{II}),\) are as follows: $$ \begin{aligned} \mathrm{Ce}^{4+}+\mathrm{Mn}^{2+} & \longrightarrow \mathrm{Ce}^{3+}+\mathrm{Mn}^{3+} \\ \mathrm{Ce}^{4+}+\mathrm{Mn}^{3+} \longrightarrow \mathrm{Ce}^{3+}+\mathrm{Mn}^{4+} \\ \mathrm{Tl}^{+}+\mathrm{Mn}^{4+} \longrightarrow \mathrm{Tl}^{3+}+\mathrm{Mn}^{2+} \end{aligned} $$ (a) Identify the catalyst, intermediates, and the ratedetermining step if the rate law is given by rate \(=\) \(k\left[\mathrm{Ce}^{4+}\right]\left[\mathrm{Mn}^{2+}\right]\) (b) Explain why the reaction is slow without the catalyst. (c) Classify the type of catalysis (homogeneous or heterogeneous).

The following expression shows the dependence of the half-life of a reaction \(\left(t_{\frac{1}{2}}\right)\) on the initial reactant concentration \([\mathrm{A}]_{0}:\) $$ t_{\frac{1}{2}} \propto \frac{1}{[\mathrm{~A}]_{0}^{n-1}} $$ where \(n\) is the order of the reaction. Verify this dependence for zero-, first-, and second-order reactions.

What is meant by the order of a reaction?

The bromination of acetone is acid-catalyzed: \(\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{Br}_{2} \frac{\mathrm{H}^{+}}{\text {catalyst }} \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{Br}+\mathrm{H}^{+}+\mathrm{Br}\) The rate of disappearance of bromine was measured for several different concentrations of acetone, bromine, and \(\mathrm{H}^{+}\) ions at a certain temperature: $$ \begin{array}{lcllc} & & & & {\text { Rate of }} \\ & & & & \text { Disappearance } \\ & {\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]} & {\left[\mathrm{Br}_{2}\right]} & {\left[\mathrm{H}^{+}\right]} & \text {of } \mathrm{Br}_{2}(\mathrm{M} / \mathrm{s}) \\ \hline \text { (a) } & 0.30 & 0.050 & 0.050 & 5.7 \times 10^{-5} \\ \text {(b) } & 0.30 & 0.10 & 0.050 & 5.7 \times 10^{-5} \\ \text {(c) } & 0.30 & 0.050 & 0.10 & 1.2 \times 10^{-4} \\ \text {(d) } & 0.40 & 0.050 & 0.20 & 3.1 \times 10^{-4} \\ \text {(e) } & 0.40 & 0.050 & 0.050 & 7.6 \times 10^{-5} \end{array} $$ (a) What is the rate law for the reaction? (b) Determine the rate constant.
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