Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 88

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\) -fold increase in the reaction rate?

Short Answer

Expert verified
To achieve a \(1 \times 10^{5}\)-fold increase in the reaction rate, the enzyme must lower the activation energy of the reaction by approximately 169.6 kJ/mol.

Step by step solution

01

Write down the Arrhenius equation

The Arrhenius equation is given by: \(k = Ae^{-\frac{E_a}{RT}}\) where: - \(k\) is the rate constant of the reaction - \(A\) is the pre-exponential factor (also known as the collision factor) - \(E_a\) is the activation energy of the reaction - \(R\) is the gas constant, approximately equal to 8.314 J/(mol*K) - \(T\) is the temperature in Kelvin
02

Convert the temperature to Kelvin

The given temperature is 37°C. To convert it to Kelvin, we add 273.15: \(T = 37 + 273.15 = 310.15 \ \mathrm{K}\)
03

Set up the ratio of rate constants

We want to find the reduction in activation energy needed for a 100,000-fold increase in the reaction rate. We will call the initial activation energy \(E_{a1}\) and the modified activation energy (with the enzyme) \(E_{a2}\). Then, the desired ratio of rate constants is: \(\frac{k_2}{k_1} = 1 \times 10^5\) Using the Arrhenius equation, we can write: \(\frac{Ae^{-\frac{E_{a2}}{RT}}}{Ae^{-\frac{E_{a1}}{RT}}} = 1 \times 10^5\)
04

Simplify and solve for delta E

Since the collision factor (pre-exponential factor) A remains the same, we can simplify the equation: \(e^{\frac{E_{a1} - E_{a2}}{RT}} = 1 \times 10^5\) Now we will solve for the difference in activation energies \(\Delta E = E_{a1} - E_{a2}\): \(\Delta E = RT \cdot \ln{(1 \times 10^5)}\) Plugging in the values for R and T: \(\Delta E = (8.314 \ \mathrm{\frac{J}{mol\cdot K}})(310.15 \ \mathrm{K})\ln{(1 \times 10^5)}\) Calculating the result: \(\Delta E \approx 1.696 \times 10^5 \ \mathrm{J/mol}\)
05

Express the result in kJ/mol

To express the result in kJ/mol, we divide by 1000: \(\Delta E \approx \frac{1.696 \times 10^5}{1000} \ \mathrm{kJ/mol} = 169.6 \ \mathrm{kJ/mol}\) So, the enzyme must lower the activation energy of the reaction by approximately 169.6 kJ/mol to achieve a 100,000-fold increase in the reaction rate.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

As described in Exercise 14.41 , the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at 660 \(\mathrm{K}\) is \(4.5 \times 10^{-2} \mathrm{s}^{-1}\) .half-life for this reaction? (b) If you start with 0.050\(M \mathrm{I}_{2}\) at this temperature, how much will remain after 5.12 s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

Consider the hypothetical reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\) . The following two-step mechanism is proposed for the reaction: $$ \begin{array}{l}{\text { Step } 1 : \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}} \\ {\text { Step } 2 : \mathrm{A}+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}}\end{array}$$ \(X\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (ii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

Consider the following reaction: $$\mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q)$$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-} .\) When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M},\) the reaction rate at 298 \(\mathrm{K}\) is 0.0432 \(\mathrm{M} / \mathrm{s}\) . (a) What is the value of the rate constant? (\mathbf{b} )What are the units of the rate constant? (c) What would happen to the rate if the concentration of OH \(^{-}\) were tripled? (d) What would happen to the rate if the concentration of both reactants were tripled?

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are 154 \(\mathrm{kJ} / \mathrm{mol}\) and 136 \(\mathrm{kJ} / \mathrm{mol}\) , respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\) . (b) What is the activation energy for the reverse reaction?

You have studied the gas-phase oxidation of HBr by \(\mathrm{O}_{2}\) : $$4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g)$$ You find the reaction to be first order with respect to HBr and first order with respect to \(\mathrm{O}_{2}\) . You propose the following mechanism: $$ \begin{array}{c}{\operatorname{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{HOOBr}(g)} \\ {\operatorname{HOOBr}(g)+\operatorname{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)} \\\ {\operatorname{HOBr}(g)+\operatorname{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)}\end{array}$$ (a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free