Experiments during a recent summer on a number of fireflies (small beetles, Lampyridaes photinus) showed that the average interval between flashes of individual insects was 16.3 \(\mathrm{s}\) at \(21.0^{\circ} \mathrm{C}\) and 13.0 \(\mathrm{s}\) at \(27.8^{\circ} \mathrm{C}\) a. What is the apparent activation energy of the reaction that controls the flashing? b. What would be the average interval between flashes of an individual firefly at \(30.0^{\circ} \mathrm{C} ?\) c. Compare the observed intervals and the one you calculated in part b to the rule of thumb that the Celsius temperature is 54 minus twice the interval between flashes.

Since the Arrhenius equation requires temperature to be measured in Kelvin, we must convert the given temperatures from Celsius to Kelvin: \(T_K = T_C + 273.15\) At \(T_C = 21.0^{\circ}C\): \(T_1 = 21.0 + 273.15 = 294.15 K\) At \(T_C = 27.8^{\circ}C\): \(T_2 = 27.8 + 273.15 = 300.95 K\)

To determine the activation energy, we will use the Arrhenius equation. Let's rewrite the equation to find the ratio of reaction rates: \(\frac{k_1}{k_2} = \frac{Ae^{-\frac{E_a}{R T_1}}}{Ae^{-\frac{E_a}{R T_2}}}\) Simplify the equation by canceling out the pre-exponential factor A: \(\frac{k_1}{k_2} = e^{\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\) We are given the intervals between firefly flashes at each respective temperature. We can assume that the flash intervals are inversely proportional to the reaction rates, meaning that we can write the ratio of reaction rates as the inverse ratio of the flash intervals: \(\frac{k_1}{k_2} = \frac{t_2}{t_1}\) Now, using the given intervals \(t_1 = 16.3s\) and \(t_2 = 13.0s\), and the temperature values we calculated, we can solve for \(E_a\): \(E_a = \frac{R \ln{\frac{t_2}{t_1}}}{\frac{1}{T_2} - \frac{1}{T_1}}= \frac{8.314 \ln{\frac{13.0}{16.3}}}{\frac{1}{300.95} - \frac{1}{294.15}} \approx 4.59 \times 10^4 J/mol\) So, the apparent activation energy is approximately \(4.59 \times 10^4 J/mol\).

Now, we will use the activation energy we found to calculate the average interval between flashes at \(T_C = 30.0^{\circ}C\). First, convert the temperature to Kelvin: \(T_3 = 30.0 + 273.15 = 303.15 K\) Next, use the Arrhenius equation, replacing one of the reaction rate ratios with the inverse ratio of the flash intervals: \(\frac{t_3}{t_1} = e^{\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_3}\right)}\) Solve for \(t_3\): \(t_3 = t_1e^{\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_3}\right)} = 16.3 \times e^{\frac{4.59 \times 10^4}{8.314}\left(\frac{1}{294.15} - \frac{1}{303.15}\right)} \approx 11.6 s\) So, the average interval between flashes at \(30.0^{\circ}C\) would be approximately 11.6 seconds.

Finally, we will compare our calculated interval to the given rule of thumb: \(T_C = 54 - 2(\Delta t)\). Using the intervals at each temperature, calculate the predicted temperatures using the rule of thumb: For \(t_1 = 16.3s\): \(T_{C1} = 54 - 2(16.3) = 21.4^{\circ}C\) For \(t_2 = 13.0s\): \(T_{C2} = 54 - 2(13.0) = 28^{\circ}C\) For \(t_3 = 11.6s\): \(T_{C3} = 54 - 2(11.6) = 30.8^{\circ}C\) Comparing these values to the original temperatures (\(21.0^{\circ}C\), \(27.8^{\circ}C\), and \(30.0^{\circ}C\)), we observe that the rule of thumb produces reasonably accurate predictions for the temperature, particularly for the second and third intervals. However, it should be noted that the rule of thumb may only be accurate within a certain range of temperatures and flash intervals, and caution should be taken when using this rule for specific situations.