Experiments have shown the average frequency of chirping of individual snowy tree crickets (Oecanthus fultoni) to be \(178 \mathrm{~min}^{-1}\) at \(25.0^{\circ} \mathrm{C}, 126 \mathrm{~min}^{-1}\) at \(20.3^{\circ} \mathrm{C}\), and \(100 . \mathrm{min}^{-1}\) at \(17.3^{\circ} \mathrm{C}\). a. What is the apparent activation energy of the reaction that controls the chirping? b. What chirping rate would be expected at \(15.0^{\circ} \mathrm{C}\) ? c. Compare the observed rates and your calculated rate from part b to the rule of thumb that the Fahrenheit temperature is 42 plus \(0.80\) times the number of chirps in \(15 \mathrm{~s}\).

The Arrhenius equation is a formula for the temperature dependence of reaction rates. It can be written as: \[k = Ae^{\frac{-E_{a}}{RT}}\] where: - \(k\) is the reaction rate constant - \(A\) is the pre-exponential factor - \(E_{a}\) is the activation energy - \(R\) is the universal gas constant (\(8.314 \mathrm{J/(mol\cdot K)}\)) - \(T\) is the absolute temperature in Kelvin (K)

We are given the reaction rate constants and temperatures, so we can set up the following equations using the Arrhenius equation: \[\frac{k_2}{k_1} = e^{\frac{-E_{a}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}\] \[\frac{k_3}{k_2} = e^{\frac{-E_{a}}{R}\left(\frac{1}{T_3}-\frac{1}{T_2}\right)}\] Here, the given temperature data must first be converted to Kelvin: \(T_1 = 25.0 + 273.15 = 298.15 \mathrm{K}\), \(T_2 = 20.3 + 273.15 = 293.45 \mathrm{K}\), \(T_3 = 17.3 + 273.15 = 290.45 \mathrm{K}\). Now plug in the known values: \[\frac{126}{178} = e^{\frac{-E_{a}}{8.314}\left(\frac{1}{293.45}-\frac{1}{298.15}\right)}\] \[\frac{100}{126} = e^{\frac{-E_{a}}{8.314}\left(\frac{1}{290.45}-\frac{1}{293.45}\right)}\] Solve these equations for the activation energy \(E_{a}\), and we get \(E_{a} \approx 54.26 \mathrm{kJ/mol}\).

Now we want to find the chirping rate (\(k_{15}\)) at \(15^{\circ}\mathrm{C}\) (or \(288.15\mathrm{K}\)). We can use the Arrhenius equation for this with the calculated activation energy: \[k_{15} = 178e^{\frac{-54.26 \times 10^3}{8.314}\left(\frac{1}{288.15}-\frac{1}{298.15}\right)}\] Solving for \(k_{15}\), we get a chirping rate of approximately \(84.63 \min^{-1}\).

According to the rule of thumb, the Fahrenheit temperature is 42 plus 0.80 times the number of chirps in 15 seconds. Let's use this rule to calculate the temperature and compare it to the given data: At \(25^{\circ}\mathrm{C}\) (or \(77^{\circ}\mathrm{F}\)), the number of chirps in 15 seconds is \(178\min^{-1} \times \frac{15}{60} = 44.5\), so the rule of thumb gives: \[77 \approx 42 + 0.80\times 44.5\] At \(15^{\circ}\mathrm{C}\) (or \(59^{\circ}\mathrm{F}\)), the calculated chirping rate is \(84.63\min^{-1}\). Thus, the number of chirps in 15 seconds is \(84.63\min^{-1} \times \frac{15}{60} = 21.16\), so the rule of thumb gives: \[59 \approx 42 + 0.80 \times 21.16\] In both cases, the calculated rates are in good agreement with the rule of thumb, suggesting that our activation energy and chirping rate calculations are reasonable.