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Ch.14 - Chemical Kinetics
Chapter 14, Problem 105

At 28 C, raw milk sours in 4.0 h but takes 48 h to sour in a refrigerator at 5 C. Estimate the activation energy in kJ>mol for the reaction that leads to the souring of milk.

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Identify the given temperatures and times: T1 = 28°C, t1 = 4.0 h; T2 = 5°C, t2 = 48 h.
Convert the temperatures from Celsius to Kelvin: T1 = 28 + 273.15 K, T2 = 5 + 273.15 K.
Use the Arrhenius equation in the form of \( \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \), where \( k \) is the rate constant, \( E_a \) is the activation energy, and \( R \) is the gas constant (8.314 J/mol·K).
Assume the rate constants are inversely proportional to the times: \( k_1 = \frac{1}{t_1} \) and \( k_2 = \frac{1}{t_2} \).
Substitute the values into the Arrhenius equation and solve for \( E_a \) in kJ/mol.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Activation Energy

Activation energy is the minimum energy required for a chemical reaction to occur. It represents the energy barrier that reactants must overcome to transform into products. In the context of the souring of milk, a higher activation energy indicates that the reaction proceeds more slowly at lower temperatures, which is why souring takes longer in the refrigerator.
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Arrhenius Equation

The Arrhenius equation relates the rate of a chemical reaction to temperature and activation energy. It is expressed as k = A * e^(-Ea/RT), 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. This equation helps in estimating activation energy by comparing reaction rates at different temperatures.
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Temperature Dependence of Reaction Rates

The rate of chemical reactions typically increases with temperature due to the increased kinetic energy of molecules, which leads to more frequent and effective collisions. This principle is crucial for understanding why raw milk sours faster at higher temperatures, as the increased thermal energy facilitates the reaction that causes souring.
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Related Practice
Open Question
A colored dye compound decomposes to give a colorless product. The original dye absorbs at 608 nm and has an extinction coefficient of 4.7 * 10^4 M^-1 cm^-1 at that wavelength. You perform the decomposition reaction in a 1-cm cuvette in a spectrometer and obtain the following data: Time (min) Absorbance at 608 nm 0 1.254 30 0.941 60 0.752 90 0.672 120 0.545. From these data, determine the rate law for the reaction 'dye → product' and determine the rate constant.
Open Question
Cyclopentadiene (C5H6) reacts with itself to form dicyclopentadiene (C10H12). A 0.0400 M solution of C5H6 was monitored as a function of time as the reaction 2 C5H6 → C10H12 proceeded. The following data were collected: Time (s) | [C5H6] (M) 0.0 | 0.0400 50.0 | 0.0300 100.0 | 0.0240 150.0 | 0.0200 200.0 | 0.0174 Plot [C5H6] versus time, ln[C5H6] versus time, and 1/[C5H6] versus time. (b) What is the value of the rate constant?
Open Question
The first-order rate constant for the reaction of a particular organic compound with water varies with temperature as follows: Temperature (K) Rate Constant (s⁻¹) 300 3.2 × 10⁻¹¹, 320 1.0 × 10⁻⁹, 340 3.0 × 10⁻⁸, 355 2.4 × 10⁻⁷. From these data, calculate the activation energy in units of kJ/mol.
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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 _x001E_F to 59 _x001E_F] lowers the reaction rate six times; a 36-degree drop [from 77 _x001E_F to 41 _x001E_F] produces a fortyfold decrease in the rate.” (b) Assuming the value of Ea calculated from the 36 _x001E_ drop and that the rate of breakdown is first order with a half-life at 25 _x001E_C of 2.7 yr, calculate the half-life for breakdown at a temperature of -15 _x001E_C.
Textbook Question

The following mechanism has been proposed for the reaction of NO with H2 to form N2O and H2O: NO1g2 + NO1g2¡N2O21g2 N2O21g2 + H21g2¡N2O1g2 + H2O1g2 (d) The observed rate law is rate = k3NO423H24. If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

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Textbook Question

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: Cl1g2 + O31g2¡ClO1g2 + O21g2 ClO1g2 + O1g2¡Cl1g2 + O21g2 (b) What is the catalyst in the reaction?

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