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Chapter 14: Problem 94

(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

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