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 \(\left.59^{\circ} \mathrm{F}\right]\) lowers the reaction rate six times; a 36 -degree drop [from \(77^{\circ} \mathrm{F}\) to \(\left.41^{\circ} \mathrm{F}\right]\) 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 -degree drop and that the rate of breakdown is first order with a half-life at \(25^{\circ} \mathrm{C}\) of 2.7 years, calculate the half-life for breakdown at a temperature of \(-15^{\circ} \mathrm{C}\).

Short answer

\[E_a = 6.545 \times 10^3 J \: \text{mol}^{-1}\] For the second temperature drop of 36 degrees Fahrenheit, we have: \[T_1 = 77^\circ F \to 298.15 K\] \[T_2 = 41^\circ F \to 278.15 K\] \[\frac{k_1}{k_2} = 40\] Now, we can rewrite the Arrhenius equation as: \[\frac{40}{1} = \frac{\exp(-E_a / (8.314 \times 298.15))}{\exp(-E_a / (8.314 \times 278.15))}\] Solve for the activation energy: \[E_a = 4.386 \times 10^{4} J \: \text{mol}^{-1}\] The activation energies for both cases are not consistent. #tag_title# Step 3: Calculate the Half-Life at -15°C#tag_content# We will assume the value of \(E_a\) from the 36-degree drop: \[E_a = 4.386 \times 10^4 J \: \text{mol}^{-1}\] We are given the half-life at \(25^\circ C\) (298.15 K) as 2.7 years, and we need to find the half-life at \(-15^\circ C\) (258.15 K). We can use the equation for half-life, \(\text{half-life} = \frac{0.693}{k}\), combined with the Arrhenius equation to find the half-life at the lower temperature. First, find the rate constant at \(25^\circ C\): \[k_{25} = \frac{0.693}{2.7 \: \text{years}} = 0.257 \: \text{year}^{-1}\] Now, we can find the rate constant at \(-15^\circ C\): \[\frac{k_{-15}}{k_{25}} = \frac{\exp(-E_a / (8.314 \times 258.15))}{\exp(-E_a / (8.314 \times 298.15))}\] \[k_{-15} = 3.91 \times 10^{-4} \: \text{year}^{-1}\] Finally, calculate the half-life at \(-15^\circ C\): \[\text{Half-Life}_{-15} = \frac{0.693}{3.91 \times 10^{-4} \: \text{year}^{-1}} = 1,774 \: \text{years}\] In conclusion, the activation energies for the two different temperature drops are not consistent, and the half-life for the breakdown at \(-15^\circ C\) is approximately 1,774 years.

Step-by-step solution

Understand the Arrhenius Equation

The Arrhenius equation is given by: \[k = A \exp(-E_a / R T)\] where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. To find the ratio of rate constants at different temperatures, we will use the following form of the equation: \[\frac{k_1}{k_2} = \frac{A \exp(-E_a / RT_1)}{A \exp(-E_a / RT_2)}\]

Calculate Activation Energies

We will first find Ea using the given data. For the first temperature drop of 18 degrees Fahrenheit, we have the following information: \[T_1 = 77^\circ F \to 298.15 K\] \[T_2 = 59^\circ F \to 288.71 K\] \[\frac{k_1}{k_2} = 6\] Now, we can rewrite the Arrhenius equation as: \[\frac{6}{1} = \frac{\exp(-E_a / (8.314 \times 298.15))}{\exp(-E_a / (8.314 \times 288.71))}\] Solve for the activation energy:

Key concepts

Activation Energy

Let's start by unpacking the term 'activation energy'. Activation energy, denoted as \( E_a \), is the minimum amount of energy that reacting molecules must possess for a chemical reaction to occur. Imagine it as a hurdle that reactants need to jump over for them to transform into products. Higher \( E_a \) means molecules need more energy to start reacting, making the reaction slower, whereas a lower \( E_a \) means it's easier for the reaction to take place quickly.

Applying this to the exercise at hand, when we calculate the activation energy for the breakdown of cellulose and starch, we're effectively finding out how big that 'energy hurdle' is. The equation from the Arrhenius equation provides a scientifically sound method for estimating \( E_a \) by examining how the rate of the reaction changes with temperature. It's a crucial value that can predict how a reaction rate will change under different conditions, making it a fundamental concept for chemists and engineers.

Reaction Rates

The rate of a chemical reaction indicates how quickly reactants are transformed into products over time. The reaction rate is dependent on several factors, such as concentration of reactants, surface area, catalysts, and temperature. In the context of our textbook problem on the breakdown of cellulose and starch, we're looking specifically at how temperature affects the rate at which these substances degrade.

The Arrhenius equation contains a rate constant \( k \) which acts as an indicator of the reaction rate. When comparing two different temperatures, as in our exercise, the ratio of the rate constants \( \frac{k_1}{k_2} \) tells us how much faster or slower the reaction occurs. Here, a six-fold decrease in the reaction rate corresponds to a lower temperature, which aligns with the general understanding that reactions tend to slow down as the temperature decreases.

Temperature Effect on Reaction Rate

Temperature plays a vital role in driving chemical reactions. As temperature increases, the molecules involved in the reaction move faster and collide more frequently with greater energy. This generally leads to more successful collisions, meaning those that surpass the activation energy, resulting in an increased reaction rate.

In the exercise, we see evidence of the temperature effect, where a drop in temperature causes a significant reduction in the reaction rate. Specifically, an 18-degree Fahrenheit drop reduces the rate sixfold, and a 36-degree drop decreases it fortyfold. The Arrhenius equation quantitatively links temperature and reaction rate by incorporating the ratio of rate constants and the exponential term that includes temperature. This allows us to calculate the exact impact of changing temperatures on the reaction rate, a vital concept for anyone studying chemical kinetics.