Explanation Given The rate constant of uncatalyzed reaction is 0.039 s − 1 . The temperature of the reaction is 25 ° C . The rate constant of enzyme-catalyzed reaction is 1.0 × 10 6 s − 1 . The conversion of temperature from ° C into Kelvin is done as, T ( K ) = T ( ° C ) + 273 Hence, the conversion of temperature from 25 ° C into Kelvin is, T ( K ) = ( 25 + 273 ) K = 298 K The difference in activation energy of the uncatalyzed and catalyzed reaction at the same temperature is calculated by the formula, ln k 1 k 2 = E a ( 2 ) − E a ( 1 ) R T E a ( 2 ) − E a ( 1 ) = R T ln ( k 1 k 2 ) Where, k 2 is the rate constant of uncatalyzed reaction

14th Edition

ISBN: 9780134863016

Author: Brown

Publisher: PEARSON

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Chapter 14, Problem 85E

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To determine: The difference in the activation energies of uncatalyzed and enzyme-catalyzed reaction.

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Chapter 14, Problem 84E

Chapter 14, Problem 86E

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Chapter 14 Solutions

CHEMISTRY THE CENTRAL SCIENCE 14TH EDI

Ch. 14.2 - If the experiment in Figure 14.2 is run for 60 s,...Ch. 14.2 - Prob. 14.1.2PE Ch. 14.2 - Which of the following could be the instantaneous...Ch. 14.2 - Using Figure 14.3, determine the instantaneous...Ch. 14.2 - At a certain time in a reaction, substance A is...Ch. 14.2 - Prob. 14.3.2PE Ch. 14.3 - Suppose the rate law for the reaction in this...Ch. 14.3 - Assuming that rate = k[A][B], rank the mixtures...Ch. 14.3 - Prob. 14.5.1PE Ch. 14.3 - Prob. 14.5.2PE

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