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

The activation energy (Ea) for the first temperature drop (77°F to 59°F) is 77.5 kJ/mol, and 76.2 kJ/mol for the second temperature drop (77°F to 41°F). These values are consistent as they are similar. Using the activation energy of 76.2 kJ/mol obtained from the 36° temperature drop, and the given half-life of 2.7 years at \(25^{\circ} \mathrm{C}\), the half-life at \(-15^{\circ} \mathrm{C}\) for cellulose and starch breakdown is approximately 490 years.

Step-by-step solution

Convert temperatures to Kelvin

We need to convert the given Fahrenheit and Celsius temperatures into Kelvin: 1. First temperature change (\(77^{\circ} \mathrm{F}\) to \(59^{\circ} \mathrm{F}\)): Find the corresponding temperatures in Kelvin using the conversion formula: \(T(K) = (\frac {T(°F) - 32}{1.8}) + 273.15 \) 2. Second temperature change (77°F to 41°F): Convert these temperatures using the same procedure as above. 3. Third temperature for half-life calculation (\(25^{\circ} \mathrm{C}\) and \(-15^{\circ} \mathrm{C}\)): Convert these to Kelvin using the formula \(T(K) = T(°C) + 273.15\)

Calculate activation energy (Ea) for both temperature drops

We can use the Arrhenius equation to find the activation energy (Ea) based on the rate constant (k) ratio and temperature changes. 1. Ratio of reaction rates for the first temperature drop: 6 2. Ratio of reaction rates for the second temperature drop: 40 We can then solve for activation energy (Ea) using the equation: \(ln\frac{k_2}{k_1} = \frac{-Ea}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\) Calculate the activation energy for both temperature drops using this equation.

Compare activation energies

Next, compare the calculated activation energies (Ea) for the two temperature drops. If they are similar, they are consistent. #Step 2: Calculate the Half-Life for Breakdown at \(-15^{\circ} \mathrm{C}\)#

Find the rate constant at \(25^{\circ} \mathrm{C}\)

Use the given half-life of 2.7 years at \(25^{\circ} \mathrm{C}\) to find the rate constant (k) at that temperature for a first-order reaction. Use the formula \(t_{1/2} = \frac{ln2}{k}\) to calculate 'k'.

Calculate the rate constant at \(-15^{\circ} \mathrm{C}\)

Using the calculated activation energy from the 36-degree drop, the Arrhenius equation, and the rate constant value found at \(25^{\circ} \mathrm{C}\), solve for the rate constant (k') at a temperature of \(-15^{\circ} \mathrm{C}\).

Calculate the half-life at \(-15^{\circ} \mathrm{C}\)

Using the rate constant (k') at \(-15^{\circ} \mathrm{C}\) and the first-order reaction half-life formula, calculate the half-life at that temperature.

Key concepts

Activation Energy

Activation energy (\( E_a \) is the minimum amount of energy required for a chemical reaction to occur. It's like pushing a ball up a hill before it can roll down; the initial push is analogous to the activation energy needed to start a reaction. In terms of chemical kinetics, a reaction will only proceed if the reactant molecules have enough energy to overcome this energy barrier. The higher the activation energy, the slower the reaction rate, as fewer molecules will have sufficient energy to react at a given temperature.

For instance, a reaction with high activation energy will proceed significantly slower than one with low activation energy under the same conditions. This concept is crucial for understanding how temperature affects reaction rates because increasing the temperature provides more energy to the molecules, increasing the chances that reactants will have enough energy to overcome the activation energy barrier.

Arrhenius Equation

The Arrhenius equation mathematically relates the rate constant (\( k \)) of a reaction to the temperature and activation energy. It is expressed as:
\( k = A e^{\frac{-E_a}{RT}} \)

where \( A \) is the pre-exponential factor (a constant that includes factors such as the frequency of collisions with proper orientation), \( E_a \) is the activation energy, \( R \) is the universal gas constant (8.314 J/(mol*K)), and \( T \) is the absolute temperature in Kelvin.

The equation shows that the rate constant increases exponentially with an increase in temperature and decreases with an increase in activation energy. Essentially, it quantifies how a change in temperature affects the reaction rate by altering reactant molecules' energy and their ability to surpass the activation energy.

Reaction Rate

Reaction rate refers to how quickly a chemical reaction proceeds, commonly measured as the change in concentration of a reactant or product per unit time. Reaction rates can vary dramatically depending on the reaction and conditions.

A key factor affecting reaction rate is temperature; as the temperature increases, so does the kinetic energy of moleclues. This results in more frequent and more energetic collisions between reactant molecules, enhancing the probability of successful reactions that overcome the activation energy barrier. As such, even a small increase in temperature can lead to a significant increase in reaction rate, demonstrating a sensitive dependency on this variable. The New York Times article mentioned in the exercise exemplifies this sensitivity: an 18-degree Fahrenheit drop resulted in a sixfold decrease, while a 36-degree drop caused a fortyfold decrease in reaction rate.

Temperature Conversion

Temperature conversion is necessary when working with the Arrhenius equation since it requires the absolute temperature in Kelvin (K). The formulas for temperature conversion from Celsius to Kelvin and Fahrenheit to Kelvin are straightforward but essential for accurate computation.

For Celsius to Kelvin, the formula is:
\( T(K) = T(°C) + 273.15 \)
For Fahrenheit to Kelvin, it's a two-step process. Convert Fahrenheit to Celsius first, then to Kelvin:
\( T(K) = (\frac{T(°F) - 32}{1.8}) + 273.15 \)

Using these conversions allows us to apply temperatures from different scales consistently into the Arrhenius equation, which is crucial for calculating reaction rates and activation energies accurately.