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Chapter 16: Problem 89

16.89 A slightly bruised apple will rot extensively in about 4 days at room temperature \(\left(20^{\circ} \mathrm{C}\right)\). If it is kept in the refrigerator at \(0^{\circ} \mathrm{C}\), the same extent of rotting takes about 16 days. What is the activation energy for the rotting reaction?

Short Answer

Expert verified
The activation energy for the rotting reaction is approximately 45.75 kJ/mol.

Step by step solution

01

Determine the temperatures in Kelvin

Convert both temperatures from Celsius to Kelvin. Use the formula: \[ T(K) = T(掳C) + 273.15 \]For room temperature:\[ T_1 = 20 + 273.15 = 293.15 \text{ K} \]For the refrigerator:\[ T_2 = 0 + 273.15 = 273.15 \text{ K} \]
02

Record the rate constants

Use the given times to determine the relative rates. The rate constant is inversely proportional to time since more time means slower rot, which can be represented as follows:\[ k_1 \text{ (room temperature)} = \frac{1}{4\text{ days}} \]\[ k_2 \text{ (refrigerator)} = \frac{1}{16\text{ days}} \]
03

Use the Arrhenius Equation

The Arrhenius equation relates the rate constant to temperature and activation energy:\[ k = A e^{-E_a/(RT)} \]For two different temperatures, the equation can be rearranged to find the activation energy (E_a):\[ \frac{k_1}{k_2} = \frac{e^{-E_a/(R T_1)}}{e^{-E_a/(R T_2)}} \]Simplifying gives:\[ \frac{k_1}{k_2} = e^{E_a/R \times (1/T_2 - 1/T_1)} \]
04

Taking the natural logarithm

Take the natural logarithm of both sides to solve for the activation energy:\[ \text{ln}\bigg(\frac{k_1}{k_2}\bigg) = \frac{E_a}{R} ( \frac{1}{T_2} - \frac{1}{T_1}) \]Substitute the values for \(k_1\), \(k_2\), \(T_1\) and \(T_2\):\[ \text{ln}\bigg(\frac{1/4}{1/16}\bigg) = \frac{E_a}{8.314} \bigg( \frac{1}{273.15} - \frac{1}{293.15} \bigg) \]\[ \text{ln}(4) = \frac{E_a}{8.314} \bigg( \frac{1}{273.15} - \frac{1}{293.15} \bigg) \]
05

Solve for activation energy

Calculate the values step-by-step:\[ \text{ln}(4) \text{ is approximately } 1.386 \]Then calculate the temperatures' reciprocal difference:\[ \frac{1}{273.15} - \frac{1}{293.15} \text{ is approximately } 2.52 \times 10^{-4}\]Finally solve for \(E_a\):\[ 1.386 = \frac{E_a}{8.314} \times 2.52 \times 10^{-4} \]\[E_a = \frac{1.386 \times 8.314}{2.52 \times 10^{-4}}\]\[E_a \text{ is approximately } 45,750 \text{ J/mol or } 45.75 \text{ kJ/mol} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arrhenius Equation
The Arrhenius Equation is a powerful formula that shows how the rate constant (k) of a chemical reaction changes with temperature. It's written as: \[ k = A e^{-E_a/(RT)} \] where:
  • A is the frequency factor (indicates how often molecules collide in the correct orientation)
  • E_a is the activation energy (energy barrier that must be overcome for the reaction to occur)
  • R is the universal gas constant (8.314 J/mol路K)
  • T is the temperature in Kelvin
This equation helps us understand why reactions proceed faster at higher temperatures. By rearranging the formula and comparing rate constants at different temperatures, we can calculate the activation energy as shown in the original solution.
Temperature Conversion to Kelvin
Temperature plays a crucial role in chemical kinetics, and the Arrhenius equation requires temperatures to be in Kelvin. To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature using the formula:\[ T(K) = T(掳C) + 273.15 \]For example, room temperature of 20掳C converts to: \[ 20 + 273.15 = 293.15 \text{ K} \]And a refrigerator temperature of 0掳C converts to: \[ 0 + 273.15 = 273.15 \text{ K} \]This standardizes the measurement, ensuring accurate calculations in the Arrhenius equation.
Natural Logarithm in Rate Equations
The natural logarithm (ln) is essential when working with the Arrhenius equation, especially when comparing rates at different temperatures. By taking the ln of both sides of \[ \frac{k_1}{k_2} = \frac{e^{-E_a/(R T_1)}}{e^{-E_a/(R T_2)}} \] we simplify it to: \[ \text{ln}\bigg(\frac{k_1}{k_2}\bigg) = \frac{E_a}{R} \bigg( \frac{1}{T_2} - \frac{1}{T_1} \bigg) \]This form makes it easier to isolate the activation energy (E_a). In the original solution, this approach helped solve for E_a using the given rate constants and temperatures.
Reaction Rate Constants
The rate constant (k) quantifies the speed of a reaction. Smaller k means a slower reaction, and k is inversely proportional to the time required for the reaction. For instance, in the apple rotting example: \[ k_1 \text{ (room temperature)} = \frac{1}{4 \text{ days}} \]\[ k_2 \text{ (refrigerator)} = \frac{1}{16 \text{ days}} \]A larger k (shorter time) at room temperature indicates a faster rotting process compared to the refrigerator. Such constants are crucial for computing the activation energy using the Arrhenius equation.
Chemical Kinetics
Chemical kinetics is the study of reaction rates and the steps that occur during a reaction. Factors influencing the rates include:
  • Temperature: higher temperatures generally increase reaction rates.
  • Concentration: higher concentration of reactants typically increases the rate.
  • Presence of catalysts: catalysts lower the activation energy, speeding up reactions.
Understanding how these factors interplay helps us predict and control reactions in various fields, from food preservation to drug development. The original exercise is a practical application of chemical kinetics principles, illustrating how temperature influences the apple rotting rate and allowing calculation of the activation energy.

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Most popular questions from this chapter

In a study of ammonia production, an industrial chemist discovers that the compound decomposes to its elements \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in a first-order process. She collects the following data: $$ \begin{array}{llll} \text { Time (s) } & 0 & 1.000 & 2.000 \\ {\left[\mathrm{NH}_{3}\right](\mathrm{mol} / \mathrm{L})} & 4.000 & 3.986 & 3.974 \end{array} $$ (a) Use graphical methods to determine the rate constant. (b) What is the half-life for ammonia decomposition?

The effect of substrate concentration on the first-order growth rate of a microbial population follows the Monod equation: \(\mu=\frac{\mu_{\max } S}{K_{\mathrm{s}}+S}\) where \(\mu\) is the first-order growth rate \(\left(\mathrm{s}^{-1}\right), \mu_{\max }\) is the maximum growth rate \(\left(\mathrm{s}^{-1}\right), S\) is the substrate concentration \(\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) and \(K_{\mathrm{s}}\) is the value of \(S\) that gives one-half of the maximum growth rate (in \(\mathrm{kg} / \mathrm{m}^{3}\) ). For \(\mu_{\max }=1.5 \times 10^{-4} \mathrm{~s}^{-1}\) and \(K_{\mathrm{s}}=0.03 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Plot \(\mu\) vs. \(S\) for \(S\) between 0.0 and \(1.0 \mathrm{~kg} / \mathrm{m}^{3}\). (b) The initial population density is \(5.0 \times 10^{3}\) cells \(/ \mathrm{m}^{3}\). What is the density after \(1.0 \mathrm{~h}\), if the initial \(S\) is \(0.30 \mathrm{~kg} / \mathrm{m}^{3} ?\) (c) What is it if the initial \(S\) is \(0.70 \mathrm{~kg} / \mathrm{m}^{3}\) ?

Like any catalyst, palladium, platinum, or nickel catalyzes both directions of a reaction: addition of hydrogen to (hydrogenation) and its elimination from (dehydrogenation) carbon double bonds. (a) Which variable determines whether an alkene will be hydrogenated or dehydrogenated? (b) Which reaction requires a higher temperature? (c) How can all-trans fats arise during hydrogenation of fats that contain some double bonds with a cis orientation?

You are studying the reaction \(\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow 2 \mathrm{AB}(g)\) to determine its rate law. Assuming that you have a valid experimental procedure for obtaining \(\left[\mathrm{A}_{2}\right]\) and \(\left[\mathrm{B}_{2}\right]\) at various times, explain how you determine (a) the initial rate, (b) the reaction orders, and (c) the rate constant.

Aqua regia, a mixture of \(\mathrm{HCl}\) and \(\mathrm{HNO}_{3},\) has been used since alchemical times to dissolve many metals, including gold. Its orange color is due to the presence of nitrosyl chloride. Consider this one-step reaction for the formation of this compound: \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \operatorname{NOCl}(g)+\mathrm{Cl}(g) \quad \Delta H^{\circ}=83 \mathrm{~kJ}\) (a) Draw a reaction energy diagram, given \(E_{\text {affwd }}=86 \mathrm{~kJ} / \mathrm{mol}\). (b) Calculate \(E_{\text {arrev }}\). (c) Sketch a possible transition state for the reaction. (Note: The atom sequence of nitrosyl chloride is \(\mathrm{Cl}-\mathrm{N}-\mathrm{O} .)\)
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