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Chapter 13: Problem 31

Define activation energy. What role does activation energy play in chemical kinetics?

Short Answer

Expert verified
Activation energy is the minimum energy that reacting particles must have to undergo a specific reaction. It plays an essential role in chemical kinetics by determining the rate of a chemical reaction. High activation energy results in a slower reaction whereas low activation energy results in a faster reaction.

Step by step solution

01

Define Activation Energy

Activation energy, denoted as \(E_a\), is defined as the minimum amount of energy that reacting particles must have to undergo a specific reaction. It's measured in units of energy (joules per mole in the International System of Units).
02

Understand the Concept

Activation energy can be thought of as the 'energy barrier' that must be overcome by the reactants in order for the reaction to proceed. This concept was first introduced by Svante Arrhenius in the late 19th century.
03

Role in Chemical Kinetics

Activation energy plays a vital role in determining the rate at which a chemical reaction proceeds, i.e., its kinetics. If the activation energy is high, fewer molecules have enough kinetic energy to overcome the energy barrier, and the reaction will be slower. Conversely, if the activation energy is low, more molecules can overcome the barrier, and the reaction is faster. Additionally, shifting conditions such as temperature can provide molecules with enough extra energy to overcome activation energy, thereby speeding up the reaction rate.

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Most popular questions from this chapter

Consider the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of \(0.074 M / \mathrm{s}\). (a) At what rate is ammonia being formed? (b) At what rate is molecular nitrogen reacting?

The bromination of acetone is acid-catalyzed: \(\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{Br}_{2} \frac{\mathrm{H}^{+}}{\text {cually }} \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{Br}+\mathrm{H}^{+}+\mathrm{Br}^{-}\) The rate of disappearance of bromine was measured for several different concentrations of acetone, bromine, and \(\mathrm{H}^{+}\) ions at a certain temperature: $$ \begin{array}{lclll} \hline & &{\text { Rate of }} \\ & & & & \text { Disappearance } \\ & {\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]} & {\left[\mathrm{Br}_{2}\right]} & {\left[\mathrm{H}^{+}\right]} & \text {of } \mathrm{Br}_{2}(M / \mathrm{s}) \\ \hline(1) & 0.30 & 0.050 & 0.050 & 5.7 \times 10^{-5} \\ (2) & 0.30 & 0.10 & 0.050 & 5.7 \times 10^{-5} \\ (3) & 0.30 & 0.050 & 0.20 & 1.2 \times 10^{-4} \\ (4) & 0.40 & 0.050 & 0.20 & 3.1 \times 10^{-4} \\ (5) & 0.40 & 0.050 & 0.050 & 7.6 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for the reaction? (b) Determine the rate constant. (c) The following mechanism has been proposed for the reaction: Show that the rate law deduced from the mechanism is consistent with that shown in (a).

How does a catalyst increase the rate of a reaction?

The rate law for the reaction \(2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) is rate \(=k\left[\mathrm{H}_{2}\right][\mathrm{NO}]^{2} .\) Which of the following mechanisms can be ruled out on the basis of the observed rate expression? Mechanism I $$ \begin{array}{c} \mathrm{H}_{2}+\mathrm{NO} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N} \\ \mathrm{N}+\mathrm{NO} \longrightarrow \mathrm{N}_{2}+\mathrm{O} \\ \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \end{array} $$ $$ \begin{aligned} &\text { Mechanism II }\\\ &\begin{array}{l} \mathrm{H}_{2}+2 \mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{array} \end{aligned} $$ Mechanism III $$ \begin{aligned} 2 \mathrm{NO} & \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} \\ \mathrm{~N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} & \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow & \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{aligned} $$

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 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).
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