A reaction rate doubles when an investigator increases the temperature by 10 C, from 298 to 308

Question:

A reaction rate doubles when an investigator increases the temperature by 10 °C, from 298 to 308 K . What is the activation energy?

Strategy

Start with Equation 13.11:

-E. In ( 41 ) = 2 + (-1/17 RT 1 T

To solve this problem, we need to recognize that the ratio k1/k2 is 1:2 because we are told that the rate doubles. Because we are given T1 and T2, and R is a constant that we know, we can solve for the activation energy, Ea.

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Related Book For  book-img-for-question

Chemistry Principles And Practice

ISBN: 9780534420123

3rd Edition

Authors: Daniel L. Reger, Scott R. Goode, David W. Ball

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