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(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 higher rate constant than the other. Account for these observations.

Short Answer

Expert verified
\(E_a\) values being identical does not guarantee the same rate constant for two reactions at the same temperature, as the pre-exponential factor (A) may differ. Two reactions with the same rate constant at \(25^{\circ} \mathrm{C}\) but different rate constants at \(35^{\circ} \mathrm{C}\) could arise due to differences in activation energies or pre-exponential factors, which are more significant at higher temperatures due to the exponential term in the Arrhenius equation.

Step by step solution

01

Part (a)

Step 1: Recall the Arrhenius equation The Arrhenius equation is given by: \[k = Ae^{-\frac{E_a}{RT}}\] where k is the rate constant, A is the pre-exponential factor, \(E_a\) is the activation energy, R is the gas constant, and T is the temperature in Kelvin. Step 2: Determine if the rate constants will be the same We are given that both reactions have identical values for \(E_a\). Since both reactions are run at the same temperature, T is the same for both reactions. Thus, the only factor that can be different in the two reactions in the Arrhenius equation is the pre-exponential factor (A). If the value of A is the same for both reactions, then they will have the same rate constant, but if A is different, then their rate constants will be different as well. Therefore, having identical values for \(E_a\) does not ensure that the reactions will have the same rate constant if run at the same temperature.
02

Part (b)

Step 1: Use the Arrhenius equation We will use the Arrhenius equation to relate the rate constants and temperatures of the two reactions. We know that the rate constant is the same for both reactions at \(25^{\circ} \mathrm{C}\) and different at \(35^{\circ} \mathrm{C}\). \[k_1 = Ae^{-\frac{E_{a1}}{R(25+273)}}\] \[k_2 = Ae^{-\frac{E_{a2}}{R(25+273)}}\] At \(35^{\circ} \mathrm{C}\): \[k_1' = Ae^{-\frac{E_{a1}}{R(35+273)}}\] \[k_2' = Ae^{-\frac{E_{a2}}{R(35+273)}}\] Step 2: Account for the observations At \(25^{\circ} \mathrm{C}\), since the rate constants are the same, the activation energies (\(E_{a1}\) and \(E_{a2}\)) and pre-exponential factors (A) for both reactions may be similar or their combinations balance each other out. However, at the higher temperature of \(35^{\circ} \mathrm{C}\), the effect of the activation energy differences becomes more significant due to the exponential term, hence one reaction has a higher rate constant than the other. This indicates that the activation energies of the two reactions or their pre-exponential factors, which account for factors like the frequency of molecular collisions, are not identical at different temperatures.

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Most popular questions from this chapter

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{gathered} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{gathered} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? How do you know? (c) What is the intermediate in the reaction? How do you distinguish it from the catalyst?

Consider two reactions. Reaction (1) has a constant halflife, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

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?

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 \(14.5\)

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