Question

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.

Short answer

The apparent activation energy of the reaction that controls the firefly flashing is $1.450\times {10}^4\mathrm{J}/\mathrm{mol}$. The average interval between flashes of an individual firefly at ${30.0}^{\circ }\mathrm{C}$ is $11.6\mathrm{s}$. The rule of thumb provides a rough estimate for the average intervals between flashes but is not exact. Using the Arrhenius equation yields a more accurate prediction.

Step-by-step solution

Write down the Arrhenius equation

We will apply the Arrhenius equation to this problem. The equation is given by: $$k=A{e}^{-\frac{E_a}{RT}}$$ Here, $k$ is the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature in Kelvin. Since we are given ∆t (the average interval between flashes), we will use the reciprocal of ∆t (1/∆t), which is proportional to the rate constant, $k$.

Find the relationship between rate constants at different temperatures

We are given two different scenarios with known temperatures and average intervals between flashes. We will use the Arrhenius equation to derive the relationship between these scenarios. Let the first scenario have constant $k_1$ at temperature $T_1={21.0}^{\circ }\mathrm{C}=294.15\mathrm{K}$ and interval $\mathrm{\Delta }{t}_1=16.3\mathrm{s}$. The second scenario has constant $k_2$ at temperature $T_2={27.8}^{\circ }\mathrm{C}=300.95\mathrm{K}$ and interval $\mathrm{\Delta }{t}_2=13.0\mathrm{s}$. Then, we can write the two equations as: $$k_1=A{e}^{-\frac{E_a}{R{T}_1}}$$ $$k_2=A{e}^{-\frac{E_a}{R{T}_2}}$$ We can now find the relationship between these rate constants.

Solve for the activation energy, Ea

Divide the two equations to eliminate the pre-exponential factor, A. $$\frac{k_1}{k_2}=\frac{A{e}^{-\frac{E_a}{R{T}_1}}}{A{e}^{-\frac{E_a}{R{T}_2}}}$$ $$\frac{k_1}{k_2}=e^{\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})}$$ Now, take the natural logarithm of both sides: $$\ln (\frac{k_1}{k_2})=\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ We can now plug in the values of $\frac{1}{\mathrm{\Delta }{t}_1}$ and $\frac{1}{\mathrm{\Delta }{t}_2}$ as the rate constants, along with the temperatures and the gas constant R (8.314 J/mol·K) to solve for the activation energy $E_a$.

Calculate the activation energy, Ea

Let's plug in the values and solve for the activation energy, $E_a$. $$\ln (\frac{1/\mathrm{\Delta }{t}_1}{1/\mathrm{\Delta }{t}_2})=\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ $$\ln (\frac{1/16.3}{1/13.0})=\frac{E_a}{8.314}(\frac{1}{300.95}-\frac{1}{294.15})$$ We can now calculate the apparent activation energy $E_a$, yielding: $$E_a=1.450\times {10}^4\mathrm{J}/\mathrm{mol}$$

Predict the average interval at ${30.0}^{\circ }\mathrm{C}$

We can now use the derived activation energy to predict the interval $\mathrm{\Delta }{t}_3$ at temperature $T_3={30.0}^{\circ }\mathrm{C}=303.15\mathrm{K}$. Let's plug in again the values of the Arrhenius equation: $$\frac{1/\mathrm{\Delta }{t}_3}{1/\mathrm{\Delta }{t}_1}=e^{\frac{E_a}{R}(\frac{1}{T_3}-\frac{1}{T_1})}$$ We can rearrange and solve for $\mathrm{\Delta }{t}_3$: $$\mathrm{\Delta }{t}_3=\mathrm{\Delta }{t}_1{e}^{\frac{E_a}{R}(\frac{1}{T_3}-\frac{1}{T_1})}$$ Using the calculated activation energy $E_a$, $\mathrm{\Delta }{t}_1$, and the temperatures, we can find $\mathrm{\Delta }{t}_3$: $$\mathrm{\Delta }{t}_3=16.3e^{\frac{1.450\times {10}^4}{8.314}(\frac{1}{303.15}-\frac{1}{294.15})}$$ This yields: $$\mathrm{\Delta }{t}_3=11.6\mathrm{s}$$

Compare with the rule of thumb

The rule of thumb states that the Celsius temperature is 54 minus twice the interval between flashes. Let's compare the observed intervals and the interval calculated in part b with this rule: - At ${21.0}^{\circ }\mathrm{C}$, the rule predicts $54-2\cdot 16.3={21.4}^{\circ }\mathrm{C}$, which is close but not exactly equal to the observed value. - At ${27.8}^{\circ }\mathrm{C}$, the rule predicts $54-2\cdot 13={28}^{\circ }\mathrm{C}$, which is also close but not exactly equal to the observed value. - At ${30}^{\circ }\mathrm{C}$, the rule predicts $54-2\cdot 11.6={30.8}^{\circ }\mathrm{C}$, which is somewhat close to our calculated interval. While the rule of thumb may provide a rough estimate of the average intervals between flashes for fireflies at different temperatures, it is not exact, and using the Arrhenius equation provides a more accurate prediction.

Key concepts

Understanding Activation Energy

The concept of activation energy (E_a) is critical in understanding why chemical reactions occur at different rates. In the realm of chemical kinetics, activation energy is the minimum energy required for reactants to transform into products during a chemical reaction. Picture activation energy as a barrier that particles must overcome for a reaction to occur. If the energy of the reacting molecules is lower than the activation energy, the reaction will not proceed. On the other hand, if the molecules possess energy equal to or higher than the activation energy, the reaction is likely to take place.

Visualizing the Energy Landscape

In imagining this concept, you can think of activation energy as a mountain pass that hikers (reactant molecules) must cross to reach their destination (products). The height of the pass represents the energy barrier they must overcome. Situations that lower the barrier, akin to finding a tunnel through the mountain, make it easier for the reaction to occur. Catalysts work by offering such 'tunnels', reducing the activation energy needed.

The Arrhenius equation presented in the exercise and its application to biological systems, like the flashing activity of fireflies, helps in quantifying this concept by relating the frequency of molecular collisions with sufficient energy to the temperature and the activation energy.

Rate Constant and its Importance

The rate constant (k) is another essential element in the Arrhenius equation and pertains to the speed at which a reaction proceeds. In simple terms, the rate constant is the 'speedometer' of a chemical reaction—higher values of k indicate a faster reaction rate, while lower values signify a slower reaction. It's influenced by several factors, including temperature and the presence of a catalyst.

Factors Affecting the Rate Constant

Several factors can alter the rate constant. For instance, higher temperatures generally increase the kinetic energy of molecules, leading to more frequent and energetic collisions, which in turn raises the rate constant. The presence of a catalyst provides an alternative pathway with a lower activation energy, which also increases the rate constant. Importantly, the rate constant isn't affected by reactant concentrations, but it does affect how quickly a reaction will reach completion when those reactants are present. In studying fireflies, the rate constant aligns with the average interval between their flashes, giving us a measurable expression of how environmental conditions influence biological processes. Understanding the rate constant equips us with the ability to predict reaction rates and design experiments effectively.

Temperature Dependency of Reaction Rates

Temperature plays a pivotal role in the field of chemistry, especially when it comes to the temperature dependency of reaction rates. Just as a car moves faster on a hot day compared to a cold one, chemical reactions tend to speed up when the temperature rises. This phenomenon can be understood through the Arrhenius equation, which mathematically links the rate constant (k) of a chemical reaction with the temperature (T).

Impact of Temperature on Molecular Behavior

When the temperature increases, molecules within a substance move more vigorously, leading to a higher frequency of collisions and greater chances that those collisions will have sufficient energy to overcome the activation energy barrier. Consequently, as illustrated in the firefly example, a higher temperature can significantly decrease the time between their light flashes because the underlying chemical reactions happen faster.

It is intriguing to see this concept of temperature dependence exhibited in biological systems. By applying the Arrhenius equation to calculate the intervals between flashes at different temperatures, we gain insights into how organisms adapt and respond chemically to their environment. The calculations provided in the exercise not only offer a scientific perspective but also validate the natural observations encapsulated in the rule of thumb regarding the relationship of firefly flash intervals with temperature.