/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 Suppose that a certain biologica... [FREE SOLUTION] | 91影视

91影视

Log out

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掳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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

A flask is charged with 0.100 mol of A and allowed to react to form \(B\) according to the hypothetical gas-phase reaction \(A(g) \longrightarrow \mathrm{B}(g) .\) The following data are collected:(a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that \(\mathrm{A}\) is cleanly converted to \(\mathrm{B}\) with no intermediates. (b) Calculate the average rate of disappearance of A for each 40 s interval in units of mol/s. (c) Which of the following would be needed to calculate the rate in units of concentration per time: (i) the pressure of the gas at each time, (ii) the volume of the reaction flask, (iii) the temperature, or (iv) the molecular weight of A?

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at 330 \(\mathrm{K}\) , \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c)\) is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is 0.0477 \(\mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M},\) the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{nm}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at \(30.0 \mathrm{~min}\). Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbondioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of 3.924 A. (a) Estimate how many platinum atoms would fit into a 2.0 -nm sphere; the volume of a sphere is \((4 / 3) \pi r^{3} .\) Recall that \(1 \hat{\mathrm{A}}=1 \times 10^{-10} \mathrm{m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{m} .\) (b) Estimate how many platinum atoms are on the surface of a \(2.0-\mathrm{nm}\) Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of 2.8 A. (c) Using your results from (a) and (b), calculate the percentage of Pt atoms that are on the surface of a 2.0 -nm nanoparticle. (d) Repeat these calculations for a 5.0 -nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

The gas-phase decomposition of ozone is thought to occur by the following two- step mechanism. \(\begin{array}{ll}{\text { Step } 1 :} & {\mathrm{O}_{3}(g) \Longrightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g) \text { (fast) }} \\ {\text { Step } 2 :} & {\mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(g) \quad(\text { slow })}\end{array}\) (a) Write the balanced equation for the overall reaction. (b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.) (c) Is O a catalyst or an intermediate? (d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

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.