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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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Most popular questions from this chapter

The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{NO}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g) \\ \mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism. (c) Identify any intermediates in the mechanism. (d) The observed rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\). If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{array}{l} \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ \mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{array} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-}\). This rapid reaction gives the following rate data: $$ \begin{array}{lll} \hline\left[\mathrm{OCl}^{-}\right](M) & {\left[I^{-}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1.5 \times 10^{-3} & 1.5 \times 10^{-3} & 1.36 \times 10^{-4} \\ 3.0 \times 10^{-3} & 1.5 \times 10^{-3} & 2.72 \times 10^{-4} \\ 1.5 \times 10^{-3} & 3.0 \times 10^{-3} & 2.72 \times 10^{-4} \\ \hline \end{array} $$ (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{OCl}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}\)

The reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). Using the following kinetic data, determine the magnitude and units of the first order rate constant: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { Pressure } \mathrm{SO}_{2} \mathrm{Cl}_{2} \text { (atm) } \\ \hline 0 & 1.000 \\ 2,500 & 0.947 \\ 5,000 & 0.895 \\ 7,500 & 0.848 \\ 10,000 & 0.803 \\ \hline \end{array} $$

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?
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