Chapter 11, Problem 91AE

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 s at 21.0°C and 13.0 sat 27.8°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°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.

Interpretation Introduction

Interpretation: The average interval between flashes of individual insects at two different temperatures is given in an experiment. By using these values, the apparent activation energy of the reaction, average interval between flashes of a firefly at a given temperature is to be calculated and compare to the rule of thumb.

Concept introduction: A chemical reaction occurs inside the bodies of fireflies, which results in the flashing.  This reaction is called bioluminescence and follows first order kinetics.

The threshold energy needed to overcome to produce a chemical reaction is called activation energy.

The activation energy for bioluminescence reaction can be calculated by the following formula:

$\ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

To determine: The activation energy of a chemical reaction that results in the flashing.

Answer to Problem 91AE

The activation energy is $\underset{\_}{24.1kJ/mol}$.

Explanation of Solution

Given

The average interval between the flashes of individual insects at $21°\text{C}$ is $16.3\text{ s}$.

The average interval between the flashing of individual insects at $27.8°\text{C}$ is $13.0\text{ s}$.

Rate constant at temperature $21°\text{C}$ is,

$\begin{array}{c}{k}_1=\frac{1\text{ flash}}{16.3\text{ s}}\\ =6.13\times {10}^{-2}{\text{ s}}^{-1}\end{array}$

Rate constant at temperature $27.8°\text{C}$ is,

$\begin{array}{c}{k}_2=\frac{1\text{ flash}}{13.0\text{ s}}\\ =7.69\times {10}^{-2}{\text{ s}}^{-1}\end{array}$

The activation energy is calculated using the formula,

$\ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Where,

• $k_1$ is rate constant at temperature $T_1$.
• $k_2$ is rate constant at temperature $T_2$.
• $R$ is universal gas constant $\left(8.314\text{J/K}\cdot \text{mol}\right)$.
• $E_a$ is the activation energy.

Substitute the values of $k_1,k_2,T_1,T_2$, and $R$ in the above equation.

$\begin{array}{c}\ln \left(\frac{7.69\times {10}^{-2}{\text{ s}}^{-1}}{6.13\times {10}^{=2}{\text{ s}}^{-1}}\right)=\frac{E_a}{8.314\text{J/K}\cdot \text{mol}}\left(\frac{1}{\left(21+273\right)\text{K}}-\frac{1}{\left(27.8+273\right)\text{K}}\right)\\ \ln \left(\frac{7.69\times {10}^{-2}{\text{ s}}^{-1}}{6.13\times {10}^{=2}{\text{ s}}^{-1}}\right)=\frac{E_a}{8.314\text{J/K}\cdot \text{mol}}\left(\frac{1}{\left(294\right)\text{K}}-\frac{1}{\left(300.8\right)\text{K}}\right)\\ \ln \left(1.25\right)=\frac{E_a}{8.314\text{J/K}\cdot \text{mol}}\left(7.69\times {10}^{-5}\text{/K}\right)\\ 0.223=\frac{E_a}{8.314\text{J/K}\cdot \text{mol}}\left(7.69\times {10}^{-5}\text{/K}\right)\end{array}$

Simplify the above equation.

$\begin{array}{c}{E}_a=\frac{0.223\times 8.314\text{ J/K}\cdot \text{mol}}{7.69\times {10}^{-5}\text{/K}}\\ =24109.5\text{J/mol}\end{array}$

The conversion of joule $\left(\text{J}\right)$ into kilojoules $\left(\text{kJ}\right)$ is done as,

$1\text{J}=\frac{1}{1000}\text{kJ}$

Hence, the conversion of $24109.5\text{J/mol}$ into is,

$\begin{array}{c}24109.5\text{J/mol}=\frac{24109.5}{1000}\text{kJ/mol}\\ =\underset{\_}{24.1kJ/mol}\end{array}$

Conclusion

The activation energy of the chemical reaction that occurs in the bodies of fireflies is $24.1\text{kJ/mol}$.

Interpretation Introduction

Interpretation: The average interval between flashes of individual insects at two different temperatures is given in an experiment. By using these values, the apparent activation energy of the reaction, average interval between flashes of a firefly at a given temperature is to be calculated and compare to the rule of thumb.

Concept introduction: A chemical reaction occurs inside the bodies of fireflies, which results in the flashing.  This reaction is called bioluminescence and follows first order kinetics.

The threshold energy needed to overcome to produce a chemical reaction is called activation energy.

The activation energy for bioluminescence reaction can be calculated by the following formula:

$\ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

To determine: The average time interval between flashes of an individual firefly at $30°\text{C}$.

Answer to Problem 91AE

The average time interval between flashes of an individual firefly at $30°\text{C}$

is $\underset{\_}{12s}$.

Answer: The rate constant of the chemical reaction at temperature $30°\text{C}$ is $\underset{\_}{8.21\times 10^{-2}{s}^{-1}}$.

Explanation of Solution

Given

The average interval between the flashes of individual insects at $21°\text{C}$ is $16.3\text{ s}$.

The activation energy is $24109.5\text{J/mol}$.

Formula

The rate constant of chemical reaction is calculated using the formula,

$\ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Where,

• $k_1$ is rate constant at temperature $T_1$.
• $k_2$ is rate constant at temperature $T_2$.
• $R$ is universal gas constant $\left(8.314\text{J/K}\cdot \text{mol}\right)$.
• $E_a$ is the activation energy.

Substitute the values of $E_a,k_2,T_1,T_2$, and $R$ in the above equation.

$\begin{array}{c}\ln \left(\frac{k_2}{6.13\times {10}^{-2}{\text{ s}}^{-1}}\right)=\frac{24109.5\text{J/mol}}{8.314\text{ J/K}\cdot \text{mol}}\left(\frac{1}{\left(21+273\right)\text{K}}-\frac{1}{\left(30+273\right)\text{K}}\right)\\ \ln \left(\frac{k_2}{6.13\times {10}^{-2}{\text{ s}}^{-1}}\right)=\frac{24109.5\text{J/mol }}{8.314\text{ J/K}\cdot \text{mol}}\left(\frac{1}{\left(294\right)\text{K}}-\frac{1}{\left(303\right)\text{K}}\right)\\ \ln \left(\frac{k_2}{6.13\times {10}^{-2}{\text{ s}}^{-1}}\right)=29.0\times {10}^4\times \left(1.01\times {10}^{-4}\right)\\ \ln \left(\frac{k_2}{6.13\times {10}^{-2}{\text{ s}}^{-1}}\right)=0.293\end{array}$

Simplify the above equation.

$\begin{array}{c}\left(\frac{k_2}{6.13\times {10}^{-2}{\text{ s}}^{-1}}\right)=e^{0.293}\\ \left(\frac{k_2}{6.13\times {10}^{-2}{\text{ s}}^{-1}}\right)=1.340\\ k=6.13\times {10}^{-2}{\text{ s}}^{-1}\times 1.340\\ =\underset{\_}{8.21\times 10^{-2}{s}^{-1}}\end{array}$

Conclusion

The rate constant for the chemical reaction at ${30}^{\circ }\text{C}$ is $8.21\times {10}^{-2}{\text{ s}}^{=1}$

Interpretation Introduction

Interpretation: The average interval between flashes of individual insects at two different temperatures is given in an experiment. By using these values, the apparent activation energy of the reaction, average interval between flashes of a firefly at a given temperature is to be calculated and compare to the rule of thumb.

Concept introduction: A chemical reaction occurs inside the bodies of fireflies, which results in the flashing.  This reaction is called bioluminescence and follows first order kinetics.

The threshold energy needed to overcome to produce a chemical reaction is called activation energy.

The activation energy for bioluminescence reaction can be calculated by the following formula:

$\ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

To determine: Comparison of observed and calculated interval and verify the value of temperature for each interval by using the rule of thumb.

Answer to Problem 91AE

All the average intervals follow the rule of thumb.

Explanation of Solution

The celsius temperature for each intervals are verified by the following formula:

$T=\left(54-2\left(t\right)\right)°\text{C}$

Where,

• $T$ is temperature.
• $t$ is average time interval.

For $T=21°\text{C}$

Substitute the value of $t$ in the above equation.

$\begin{array}{c}T=\left(54-2\left(16.3\right)\right)°\text{C}\\ =\left(54-32.6\right)°\text{C}\\ \text{=21}\text{.4}°\text{C}\end{array}$

For $T=27.8°\text{C}$

Substitute the value of $t$ in the above equation.

$\begin{array}{c}T=\left(54-2\left(13.0\right)\right)°\text{C}\\ =\left(54-26\right)°\text{C}\\ =\text{28}°\text{C}\end{array}$

For $T=30°\text{C}$

Substitute the value of $t$ in the above equation.

$\begin{array}{c}T=\left(54-2\left(12.0\right)\right)°\text{C}\\ =\left(54-24\right)°\text{C}\\ \text{=30}°\text{C}\end{array}$

Conclusion

All the average time intervals (observed and calculated) follow the rule of thumb as the value of celsius temperature for each interval comes to the same as given.