Explanation Given: Temperature(K) Rate Constant(s −1 ) 310 0.00434 320 0.014 330 0.0421 340 0.118 350 0.316 A plot of “Ln k versus 1/T” will produce a straight line, the slope of this line can be used to find the values for Activation energy and the rate constant Formula used: k = ( A ) e − E a RT Ln k=LnA − E a RT k is the rate constant E a is the activation energy R is the gas constant ( 8.314 JK − 1 mol − 1 ) T is temperature ( K e l v i n ) y = m x + b m = ( y 2 − y 2 ) ( x 2 − x 1 ) y is the dependent variable m is slope b is the intercept Calculation: The first thing to do is to get values for 1/T and Ln K from the original values: 1/T Ln K 0.00225806 -5.439880931 0.003251 -4.268697949 0.003030303 -3.167707538 0.002941176 -2.137070655 0.002857143 -1.152013065 Make a plot using these values

Chemistry: Structure and Properties (2nd Edition)

2nd Edition

ISBN: 9780134293936

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 14, Problem 68E

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To determine:

The activation barrier and frequency factor for the reaction.

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Chapter 14, Problem 67E

Chapter 14, Problem 69E

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Chapter 14 Solutions

Chemistry: Structure and Properties (2nd Edition)

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The activation barrier and frequency factor for the reaction.

Solution Summary: The author explains how to determine the activation barrier and frequency factor for the reaction. A plot of "Ln k versus 1/T" produces a straight line.

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To determine: The activation barrier and frequency factor for the reaction.

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Chapter 14 Solutions

To determine:

The activation barrier and frequency factor for the reaction.