Explanation Given: Ea = 249 kJ/mol; A = 1.6 × 10 14 s -1 ; T = 710 K Formula used: k = A ⋅ e − E a R T [ A ] = [ A ] 0 ⋅ e − k t ln k 1 k 2 = ( 1 T 2 − 1 T 1 ) ⋅ E a R Calculation: We can use the Arrhenius equation and the first order kinetics equation to solve this problem. First, we need to solve for the rate constant k . k = A ⋅ e − E a R T k = ( 1.6 × 10 14 ) ⋅ e − 249 ( 8.314 × 10 − 3 ) ( 710 ) = 7.67 × 10 − 5 s -1 Simply inserting the variables into the equation will provide us with the rate constant which is found to be 7.67 × 10 −5 s −1 . Now we can use the first order equation to find the percentage of ethyl chloride remaining after 15 minutes. We need to convert 15 minutes into seconds which is 900 seconds...

Chemistry: Structure and Properties (2nd Edition)

2nd Edition

ISBN: 9780134293936

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 14, Problem 112E

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To determine: The rate constant at T = 710 K, the percentage of ethyl chloride remaining after 15 minutes and the temperature where the rate is doubled.

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Chapter 14, Problem 111E

Chapter 14, Problem 113E

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Chapter 14 Solutions

Chemistry: Structure and Properties (2nd Edition)

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The rate constant at T = 710 K, the percentage of ethyl chloride remaining after 15 minutes and the temperature where the rate is doubled.

Solution Summary: The author explains how to determine the rate constant at T = 710 K, the percentage of ethyl chloride remaining after 15 minutes, and the temperature where it doubles.

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To determine: The rate constant at T = 710 K, the percentage of ethyl chloride remaining after 15 minutes and the temperature where the rate is doubled.

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Chapter 14 Solutions