Problem 106 - Chemistry: The Central Science

Question

The following is a quote from an article in the August 18, 1998 , issue of The New York Times about the breakdown of cellulose and starch: "A drop of 18 degrees Fahrenheit [from \(77^{\circ} \mathrm{F}\) to \(59^{\circ} \mathrm{F}\) ] lowers the reaction rate six times; a 36-degree drop [from \(77^{\circ} \mathrm{F}\) to \(41^{\circ} \mathrm{F}\) ] produces a fortyfold decrease in the rate." (a) Calculate activation energies for the breakdown process based on the two estimates of the effect of temperature on rate. Are the values consistent? (b) Assuming the value of \(E_{a}\) calculated from the \(36^{\circ}\) drop and that the rate of breakdown is first order with a half-life at \(25^{\circ} \mathrm{C}\) of \(2.7 \mathrm{yr}\), calculate the half-life for breakdown at a temperature of \(-15^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The activation energies for the breakdown process based on the two estimates are approximately 43.5 kJ/mol and 48.3 kJ/mol. Considering the approximation in rate values, the difference is acceptable, and the values are relatively consistent. Assuming the activation energy value calculated from the 36°F drop, the half-life for breakdown at a temperature of -15°C is approximately 11.23 years.

Step 1: Temperature Conversion to Kelvin

Temperature conversion formulas: From Fahrenheit to Celsius: \(C = \frac{5}{9}(F-32)\) From Celsius to Kelvin: \(K = C + 273.15\) Converting given temperatures to Kelvin: \(T_1 = 77^{\circ} \mathrm{F} = \frac{5}{9}(77-32) = 25^{\circ} \mathrm{C} = 298.15 \mathrm{K}\) \(T_2 = 59^{\circ} \mathrm{F} = \frac{5}{9}(59-32) = 15^{\circ} \mathrm{C} = 288.15 \mathrm{K}\) \(T_3 = 41^{\circ} \mathrm{F} = \frac{5}{9}(41-32) = 5^{\circ} \mathrm{C} = 278.15 \mathrm{K}\)

Step 2: Ratio of rate constants

Given: 18°F drop -> 6 times decrease in reaction rate (k1) 36°F drop -> 40 times decrease in reaction rate (k2) Let initial rate constant at \(T_1\) be \(k\). So at \(T_2\), rate constant = \(k1 = \frac{k}{6}\), and at \(T_3\), rate constant = \(k2 = \frac{k}{40}\).

Step 3: Calculate activation energy using the Arrhenius equation

Using the Arrhenius equation to form ratios of rate constants, we can eliminate the pre-exponential factor A. For \(k2\): \(\frac{k}{k2} = \frac{Ae^{\frac{-E_a}{RT_1}}}{Ae^{\frac{-E_a}{RT_3}}}=40\) Taking natural logarithm on both sides: \(\ln40 = \frac{E_a}{R}(\frac{1}{T_3} - \frac{1}{T_1})\) Similarly, for \(k1\): \(\frac{k}{k1} = \frac{Ae^{\frac{-E_a}{RT_1}}}{Ae^{\frac{-E_a}{RT_2}}}=6\) Taking natural logarithm on both sides: \(\ln6 = \frac{E_a}{R}(\frac{1}{T_2} - \frac{1}{T_1})\) We now have two simultaneous equations. Solving them to find \(E_a\), we can check if the values are consistent.

Step 4: Calculate Ea and check consistency

Solving the above equations, we get: \(E_a^{(1)} = R \frac{\ln6}{\frac{1}{T_2} - \frac{1}{T_1}} = 8.314 \frac{\ln6}{\frac{1}{288.15} - \frac{1}{298.15}} \approx 43.5\,\text{kJ/mol}\) \(E_a^{(2)} = R \frac{\ln40}{\frac{1}{T_3} - \frac{1}{T_1}} = 8.314 \frac{\ln40}{\frac{1}{278.15} - \frac{1}{298.15}} \approx 48.3\,\text{kJ/mol}\) The values are not exactly the same; however, they are relatively close. Since we have only approximated rate values, this difference may be considered acceptable. (b) Calculate the half-life for breakdown at a temperature of -15°C. Given: Half-life at 25°C (\(T_1 = 298.15 K\)) is 2.7 years Rate of breakdown is first-order \(E_a\) from 36°F drop: \(E_a^{(2)} = 48.3\,\text{kJ/mol}\) Required: Half-life at -15°C

Step 1: Convert temperature to Kelvin

-15°C = -15 + 273.15 = 258.15 K

Step 2: Express k1 and k2 in terms of Ea and T

Rate constant for breakdown at 25°C (\(T_1\)): \(k_1 = Ae^{\frac{-E_a^{(2)}}{RT_1}}\) Rate constant for breakdown at -15°C: \(k_2 = Ae^{\frac{-E_a^{(2)}}{RT_2}}\)

Step 3: Use first-order half-life relationship

For first-order reactions, half-life (\(t_{1/2}\)) is related to the rate constant as: \(t_{1/2} = \frac{0.693}{k}\) We are given half-life at 25°C: \(t_{1/2}^{(1)} = 2.7 \,years\) Therefore, \(k_1 = \frac{0.693}{2.7} \, year^{-1}\) We want to find the half-life at -15°C (\(t_{1/2}^{(2)}\)), so we need to find the value of \(k_2\).

Step 4: Ratio of rate constants and solve for k2

Using the Arrhenius equation to find the ratio of rate constants: \(\frac{k_2}{k_1} = \frac{Ae^{\frac{-E_a^{(2)}}{RT_2}}}{Ae^{\frac{-E_a^{(2)}}{RT_1}}}\) \(\frac{k_2}{k_1} = e^{\frac{E_a^{(2)}}{R}(\frac{1}{T_1} - \frac{1}{T_2})}\) \(k_2 = k_1 e^{\frac{E_a^{(2)}}{R}(\frac{1}{T_1} - \frac{1}{T_2})}\) Plug in known values and solve for \(k_2\): \(k_2 = \frac{0.693}{2.7} e^{\frac{48300}{8.314}(\frac{1}{298.15} - \frac{1}{258.15})} = 0.0617\, year^{-1}\)

Step 5: Calculate the half-life at -15°C

Now use the first-order half-life relationship to find \(t_{1/2}^{(2)}\): \(t_{1/2}^{(2)} = \frac{0.693}{k_2} = \frac{0.693}{0.0617} \approx 11.23 \, years\) The half-life for breakdown at a temperature of -15°C is about 11.23 years.

Key Concepts

Arrhenius Equation

Understanding the Arrhenius equation is crucial for students grappling with the fundamental concepts of chemical kinetics. This equation provides a mathematical relationship between the rate constant of a reaction and temperature, revealing how reaction rates increase as the temperature rises. In simpler terms, the higher the temperature, the faster the molecules move, making them more likely to collide and react. The equation is given by:
\[\begin{equation}k = Ae^{\left(-\frac{E_a}{RT}\right)}\end{equation}\]where:

• k is the reaction rate constant,
• A is the pre-exponential factor (also known as the frequency factor),
• E_a is the activation energy of the reaction,
• R is the gas constant, and
• T is the temperature in Kelvin.

When temperature changes, the reaction rate also changes, and this relationship is quantified by the equation. In the exercise, to calculate the activation energy (E_a) for a reaction, we use experimental data about how the reaction rate varies with temperature. Understanding this equation allows students to predict how changing the temperature will affect the speed of a chemical reaction.

Chemical Kinetics

At the heart of reaction rate analysis lies chemical kinetics, a subject that reveals the speed at which chemical reactions occur and the factors that influence this speed. Chemical kinetics focuses on reaction rates, the changes in concentration of reactants or products over time, and mechanisms, the step-by-step pathway from reactants to products. Students should appreciate that kinetics is essential for everything from designing industrial processes to understanding biological systems.
Factors affecting reaction rates include:

• Concentration of reactants,
• Presence of a catalyst,
• Physical state of the reactants, and
• Temperature.

The temperature factor ties back to the Arrhenius equation, which illustrates how even a small temperature change can greatly impact reaction rates. This concept is exemplified in the given textbook problem, where the rate of cellulose and starch breakdown significantly changes with temperature.

Temperature Effects on Reaction Rate

It's widely noted that temperature has a pronounced effect on the rate at which chemical reactions occur, which is a cornerstone of the study of chemical kinetics. An increase in temperature usually leads to a higher reaction rate. This is because, at higher temperatures, molecules have more kinetic energy and move faster, increasing the frequency and energy of collisions between them.
The Arrhenius equation encapsulates this idea by indicating that for many reactions, a 10-degree Celsius increase in temperature can double or triple the reaction rate. In the real-world scenario provided in the exercise, a decrease in temperature leads to a considerable drop in the reaction rate—the rate slows down as the temperature falls.
Students will find that understanding temperature dependence is not just an academic exercise; it's essential for practical applications such as preserving food, optimizing industrial processes, and even managing medications that may have temperature-sensitive chemical properties. The principles delineated in chemical kinetics provide the basis for harnessing the power of temperature to control reaction rates efficiently.