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Ch.14 - Chemical Kinetics
All textbooksBrown 14th EditionCh.14 - Chemical KineticsProblem 106
Chapter 14, Problem 106

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.

Verified step by step guidance
1
Step 1: Convert the given temperatures from Fahrenheit to Celsius. Use the formula: \( T(\degree C) = \frac{5}{9} (T(\degree F) - 32) \).
Step 2: Use the Arrhenius equation to relate the rate constants at different temperatures: \( k = A e^{-\frac{E_a}{RT}} \). Since the rate decreases by a factor of 40, set up the equation \( \frac{k_2}{k_1} = 40 \) and solve for \( E_a \) using the temperatures in Kelvin.
Step 3: Convert the temperatures from Celsius to Kelvin using \( T(K) = T(\degree C) + 273.15 \).
Step 4: Use the first-order kinetics formula for half-life: \( t_{1/2} = \frac{0.693}{k} \). Calculate the new rate constant \( k \) at -15 \degree C using the Arrhenius equation and the previously calculated \( E_a \).
Step 5: Calculate the half-life at -15 \degree C using the new rate constant \( k \) and the first-order half-life formula.
Related Practice
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.
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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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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Textbook Question

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O3(g) ⇌ O2(g) + O(g) (fast)

Step 2: O(g) + O3(g) → 2 O2 (slow)

(b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.)

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