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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

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

Consider the reaction of peroxydisulfate ion $\left(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\right)\( with iodide ion \)\left(\mathrm{I}^{-}\right)$ in aqueous solution: $$ \mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}(a q)+3 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{SO}_{4}^{2-}(a q)+\mathrm{I}_{3}^{-}(a q) $$ At a particular temperature, the initial rate of disappearance of \(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\) varies with reactant concentrations in the following manner: $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\right](M)} & {\left[\mathrm{I}^{-}\right](M)} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.018 & 0.036 & 2.6 \times 10^{-6} \\ 2 & 0.027 & 0.036 & 3.9 \times 10^{-6} \\ 3 & 0.036 & 0.054 & 7.8 \times 10^{-6} \\ 4 & 0.050 & 0.072 & 1.4 \times 10^{-5} \\ \hline \end{array} $$ (a) Determine the rate law for the reaction and state the units of the rate constant. (b) What is the average value of the rate constant for the disappearance of \(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\) based on the four sets of data? (c) How is the rate of disappearance of $\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\( related to the rate of disappearance of \)\mathrm{I}^{-} ?(\mathbf{d})\( What is the rate of disappearance of \)\mathrm{I}^{-}$ when \(\left[\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\right]=0.025 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=0.050 \mathrm{M} ?\)

(a) A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{b})\) Another first-order reaction also has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?(\mathrm{c})\) What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

(a) What is meant by the term molecularity? (b) Why are termolecular elementary reactions so rare? (c) What is an intermediate in a mechanism?

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in NO and first order in \(\mathrm{O}_{2}\). When [NO] \(=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M},\) the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8 ?\)

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