The activation energy for a reaction is changed from 184 \(\mathrm{kJ} /\) \(\mathrm{mol}\) to 59.0 \(\mathrm{kJ} / \mathrm{mol}\) at \(600 . \mathrm{K}\) by the introduction of a catalyst. If the uncatalyzed reaction takes about 2400 years to occur, about how long will the catalyzed reaction take? Assume the frequency factor \(A\) is constant, and assume the initial concentrations are the same.

\(k = Ae^{-\frac{E_a}{RT}}\), where \(k\) is the rate constant of the reaction, \(A\) is the frequency factor (which is assumed constant in this problem), \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

Using the Arrhenius equation, we can calculate the rate constant for both the uncatalyzed and catalyzed reactions. Let \(k_1\) be the rate constant for the uncatalyzed reaction with activation energy \(E_{a1} = 184 \; \mathrm{kJ/mol}\) and \(k_2\) be the rate constant for the catalyzed reaction with activation energy \(E_{a2} = 59.0 \; \mathrm{kJ/mol}\). First, let's convert the activation energies to \(\mathrm{J/mol}\): \(E_{a1} = 184 \times 10^3 \; \mathrm{J/mol}\) \(E_{a2} = 59.0 \times 10^3 \; \mathrm{J/mol}\) Since the temperature and frequency factor are the same for both reactions, we can calculate the ratio of rate constants: \(\frac{k_2}{k_1} = \frac{Ae^{-\frac{E_{a2}}{RT}}}{Ae^{-\frac{E_{a1}}{RT}}} = e^{\frac{E_{a1} - E_{a2}}{RT}}\)

Now, we know that \(R = 8.314 \; \mathrm{J / (mol \cdot K)}\), \(T = 600 \; \mathrm{K}\), \(E_{a1} = 184 \times 10^3 \; \mathrm{J/mol}\), and \(E_{a2} = 59.0 \times 10^3 \; \mathrm{J/mol}\). Plug these values into the equation: \(\frac{k_2}{k_1} = e^{\frac{(184 - 59.0)\times 10^3}{(8.314)(600)}} \approx 3.467\times 10^8\)

We know that the time for the uncatalyzed reaction is \(t_1 = 2400 \; \mathrm{years}\). We assume that the initial concentrations of the reactants are the same for the uncatalyzed and catalyzed reactions. So, the time taken for the catalyzed reaction, \(t_2\), can be related to the time for the uncatalyzed reaction by the ratio of the rate constants: \(t_2 = \frac{k_1}{k_2}\times t_1 = \frac{1}{\frac{k_2}{k_1}}\times t_1 = \frac{1}{3.467\times 10^8}\times t_1 \) Now find the time for the catalyzed reaction: \(t_2 = \frac{1}{3.467\times 10^8}\times 2400 \; \mathrm{years} \approx 6.92 \times 10^{-6} \; \mathrm{years}\) Converting to seconds: \(t_2 \approx 6.92 \times 10^{-6} \; \mathrm{years} \times \frac{365 \cdot 24 \cdot 60 \cdot 60}{1 \; \mathrm{year}} \approx 218 \; \mathrm{s}\)

The catalyzed reaction takes approximately 218 seconds to occur assuming the same initial concentrations.