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Ch.14 - Chemical Kinetics
All textbooksBrown 14th EditionCh.14 - Chemical KineticsProblem 104
Chapter 14, Problem 104

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.

Verified step by step guidance
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Step 1: Use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and activation energy (Ea): k = A * exp(-Ea/(R*T)), where A is the pre-exponential factor and R is the gas constant (8.314 J/mol·K).
Step 2: Take the natural logarithm of both sides of the Arrhenius equation to linearize it: ln(k) = ln(A) - Ea/(R*T). This equation is in the form of y = mx + c, where y = ln(k), m = -Ea/R, and x = 1/T.
Step 3: Plot ln(k) versus 1/T using the given data points. The slope of the resulting line will be equal to -Ea/R.
Step 4: Calculate the slope of the line from the plot. Use the formula for the slope of a line: slope = (ln(k2) - ln(k1)) / (1/T2 - 1/T1) for any two data points.
Step 5: Solve for the activation energy (Ea) using the slope from Step 4: Ea = -slope * R. Convert the result from J/mol to kJ/mol by dividing by 1000.
Related Practice
Textbook Question

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at 520 nm. The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5.60 * 103 M-1 cm-1 at 520 nm. (d) How long does it take for the absorbance to fall to 0.100?

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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?
Textbook Question

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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Open 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 _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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