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Chapter 14: Problem 10

You study the effect of temperature on the rate of two reactions and graph the natural logarithm of the rate constant for each reaction as a function of \(1 / T\). How do the two graphs compare (a) if the activation energy of the second reaction is higher than the activation energy of the first reaction but the two reactions have the same frequency factor, and \((b)\) if the frequency factor of the second reaction is higher than the frequency factor of the first reaction but the two reactions have the same activation energy? [Section 14.5\(]\)

Short Answer

Expert verified
(a) If the activation energy of the second reaction is higher than that of the first reaction but both reactions have the same frequency factor, the slope of the second graph will be steeper in the negative direction compared to the first graph, and both graphs will have the same y-intercepts. (b) If the frequency factor of the second reaction is higher than that of the first reaction but both reactions have the same activation energy, the y-intercept of the second graph will be higher than that of the first graph, and both graphs will have the same slope.

Step by step solution

01

Write the Arrhenius equation

The Arrhenius equation establishes the relationship between the rate constant (k), activation energy (Ea), frequency factor (A), and temperature (T) as follows: \[k = A \cdot e^ \frac{-Ea}{RT}\], where R is the gas constant. Take the natural logarithm of both sides: \[\ln k = \ln A - \frac{Ea}{RT}\]. Since we are asked to compare the graphs based on the natural logarithm of the rate constant as a function of \(1/T\), we rewrite the equation: \[\ln k = \ln A - Ea \left( \frac{1}{RT} \right)\].
02

Compare the graphs in case (a)

In this case, the activation energy of the second reaction is higher than that of the first reaction, but both reactions have the same frequency factor (A). To compare the graphs, we analyze the effect of a higher activation energy on the graph of the second reaction: Reaction 1: \(\ln k_1 = \ln A - Ea_1 \left( \frac{1}{RT} \right)\) Reaction 2: \(\ln k_2 = \ln A - Ea_2 \left( \frac{1}{RT} \right)\), where \(Ea_1 < Ea_2\) As the activation energy of the second reaction is higher, the slope of the second graph will be steeper in the negative direction compared to the first graph. Both graphs will have the same y-intercepts, since they have the same frequency factor.
03

Compare the graphs in case (b)

In this case, the frequency factor of the second reaction is higher than that of the first reaction, but both reactions have the same activation energy. To compare the graphs, we analyze the effect of a higher frequency factor on the graph of the second reaction: Reaction 1: \(\ln k_1 = \ln A_1 - Ea \left( \frac{1}{RT} \right)\) Reaction 2: \(\ln k_2 = \ln A_2 - Ea \left( \frac{1}{RT} \right)\), where \(A_1 < A_2\) As the frequency factor of the second reaction is higher, the y-intercept of the second graph will be higher than that of the first graph (since \(\ln A_2 > \ln A_1\)). Both graphs will have the same slope, since they have the same activation energy.

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Key Concepts

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

Activation Energy
Activation energy is a key concept in chemical kinetics. It refers to the minimum amount of energy that reactant molecules need to overcome to transform into products. Think of it as a hill that molecules must climb over.

In reactions, if the activation energy is higher, it means the molecules need more energy to react. That often results in slower reactions because fewer molecules will have enough energy to reach the peak of the hill at any given moment. This is why reactions with high activation energies are usually slower.

When comparing graphs, a reaction with a higher activation energy will show a steeper slope when graphed as \( ln k \) against \( rac{1}{T} \). It tells us that changes in temperature have a more profound effect on reactions with higher activation energies.
Temperature and Reaction Rate
The temperature of a system plays a crucial role in determining the rate of a chemical reaction. Typically, as the temperature increases, the rate of reaction also increases. This happens because higher temperatures give molecules more energy, enabling more of them to reach the activation energy needed to react.

According to the Arrhenius equation, the rate constant \(k\) increases exponentially with an increase in temperature. This is seen as increases in the rate of reaction graphically, showing a sharp rise in \(ln k\) with increasing \(1/T\). The principle is that a hotter environment makes molecules move faster and collide more vigorously, facilitating the making and breaking of bonds.

In classifying reactions by temperature dependence, those sensitive to temperature changes (with steep slopes in plots) have high activation energies, making them more reactive to temperature fluctuations.
Frequency Factor
The frequency factor, often denoted by the letter \(A\), is a component of the Arrhenius equation representing how often molecules collide with the correct orientation to react. In simple terms, it's about the likelihood of collisions leading to a reaction.
  • Better orientation of molecules during collisions means a higher frequency factor.
  • It can be thought of as a measure of the opportunities for reactions to occur.

In graph terms, when reactions have different frequency factors while maintaining the same activation energy, their graphs will differ in the positioning of their y-intercepts. A reaction with a higher frequency factor will have a higher y-intercept on a \(ln k \) versus \(1/T \) plot.

The frequency factor assumes molecules are properly aligned during collisions, emphasizing that not every collision results in a reaction. Efficiency in every collision can drastically change the reaction rate. This is what makes frequency factor a vital aspect of kinetics.

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Most popular questions from this chapter

(a) Two reactions have identical values for \(E_{a} .\) Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a larger rate constant than the other. Account for these observations.

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\), what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. (b) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathrm{c}) \mathrm{What}\) is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M},\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M}\) ?

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right)\). (a) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

(a) Explain the importance of enzymes in biological systems. (b) What chemical transformations are catalyzed (i) by the enzyme catalase, \((i i)\) by nitrogenase? (c) Many enzymes follow this generic reaction mechanism, where \(\mathrm{E}\) is enzyme, \(\mathrm{S}\) is substrate, ES is the enzyme-substrate complex (where the substrate is bound to the enzyme's active site), and \(\mathrm{P}\) is the product: 1\. \(\mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES}\) 2\. \(\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P}\) What assumptions are made in this model with regard to the rate of the bound substrate being chemically transformed into bound product in the active site?
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