Question

Experiments during a recent summer on a number of fireflies (small beetles, Lampyridae photinus) showed that the average interval between flashes of individual insects was 16.3 s at 21.0⁰C and 13.0 s at 27.8⁰C.

1. What is the apparent activation energy of the reaction that controls the flashing?
2. What would be the average interval between flashes of an individual firefly at 30.0⁰C?
3. 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.

Short answer

1. $2.14\times {10}^4$.
2. 12 sec.
3. 21⁰C, 28⁰C, 30⁰C.

Step-by-step solution

First calculate the rate constant for both the flashes.

a. Average interval between flashes of individual insect at 21⁰C = 16.3 s

Average interval between flashes of individual insect at 27⁰C = 13.0 s

If the interval between flashes is 16.3s, then the rate is:

$$\begin{gathered} \frac{1\mathrm{flash}}{16.3}=\mathrm{k} \\ \mathrm{k}=6.13\times {10}^{-2}\mathrm{s} \end{gathered}$$

Doing the same calculation for 13.0s

$$\begin{gathered} \frac{1\mathrm{flash}}{13.0}=\mathrm{k} \\ \mathrm{k}=7.69\times {10}^{-2}\mathrm{s} \end{gathered}$$

Calculate Ea. with this formula :

$$\begin{gathered} \mathrm{n}\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right) \\ \ln \frac{7.69\times {10}^{-2}}{6.13\times {10}^{-2}}=\frac{{\mathrm{E}}_{\mathrm{a}}}{8.314}\left(\frac{1}{294}-\frac{1}{300}\right) \\ {\mathrm{E}}_{\mathrm{a}}=2.14\times {10}^4 \end{gathered}$$

Again, use the same formula to calculate rate constant: 

b.$\ln \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right)$

Where k₁ is the Rate of reaction of reaction at T₁ temperature

K₂ is the rate of reaction at T₂ temperature

Upon substituting the values in the formula,

$$\begin{gathered} \ln \left(\frac{k}{6.13\times {10}^{-2}}\right)=\frac{2.14}{8.314}\left(\frac{1}{21+273}-\frac{1}{30+273}\right) \\ \ln \left(\frac{k}{6.13\times {10}^{-2}}\right)=\frac{2.14\times {10}^4}{8.314}\left(\frac{1}{294}-\frac{1}{303}\right) \\ \ln \left(\frac{k}{6.13\times {10}^{-2}}\right)=29.0\times {10}^3\left(1.01\times {10}^{-4}\right) \\ \ln \left(\frac{k}{6.13\times {10}^{-2}}\right)=0.293 \\ \left(\frac{k}{6.13\times {10}^{-2}}\right)=e^{-0.293} \\ \left(\frac{k}{6.13\times {10}^{-2}}\right)=1.340 \end{gathered}$$

$$\begin{gathered} int\mathrm{ervel}=\frac{1}{\mathrm{k}} \\ \frac{1}{8.21\times {10}^{-2}}=12\mathrm{seconds} \end{gathered}$$

The comparison between the rule is below; 

c.

T	INTERVAL	54-2 (interval)
21⁰C	16.3	[54-2(16.3)] ⁰C= 21⁰C
27.8⁰C	13.0	[54-2(13.0)] ⁰C=28⁰C
30.0⁰C	12	[54-29(12)] ⁰C=30⁰C

The rule of thumb is in agreement to two significant figures.