   Chapter 12, Problem 92AE

Chapter
Section
Textbook Problem

# The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the rate constant for the reaction increases by a factor of 2.50 × 103 as compared with the uncatalyzed reaction. Assuming the frequency factor A is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

Interpretation Introduction

Interpretation: The activation energy of a certain uncatalyzed reaction and the increase in the rate constant for the reaction in the presence of catalyst is given. By using these values, the activation energy of the catalyzed reaction is to be calculated.

Concept introduction: Catalyst provides a different pathway for a chemical reaction in which the activation energy is less which results in an increase in the rate of the chemical reaction.

Arrhenius equation shows the relation between the rate constant of the chemical reaction and its activation energy. The expression of Arrhenius equation is as follows:

k=AeEaRT

To determine: The activation energy of catalyzed reaction.

Explanation

Given

The activation energy for uncatalyzed reaction is 50 kJ/mol .

The temperature at which reaction is taking place is 37°C or 310 K .

The rate constant of catalyzed reaction increases by the factor of 2.50×103 .

The conversion of kilojoules (kJ) into joul (J) is done as,

1kJ=103J

Hence, the conversion of 50kJ/mol into J/mol is,

50kJ/mol=50×103J/mol

Formula

Rate constant for the reaction is calculated by using the Arrhenius equation,

k=AeEaRTln(k)=EaRT+lnA

Where,

• k is the rate constant.
• A isthe frequency factor.
• Ea is the activation energy of reaction.
• R is the universal gas constant (8.314J/Kmol) .
• T is temperature.

For uncatalyzed reaction:

Substitute the values of Ea in the above equation.

ln(k)=50kJ/molRT+lnA (1)

For catalyzed reaction:

Substitute the values of rate constant k .

ln(k×2.50×103)=EaRT+lnA (2)

Now, subtract equation (1) from equation (2).

ln(k×2.50×103)ln(k)=EaRT+lnA+50×103J/molRTlnAln(k×2

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