/*! 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
The enzyme must lower the activation energy by approximately \(16.1 \mathrm{kJ \ mol^{-1}}\) for a \(100,000\)-fold increase in the reaction rate at physiological temperature.

Step by step solution

01

Write down the Arrhenius equation

The Arrhenius equation describes the relationship between the reaction rate constant \(k\) and activation energy \(E_a\), \(A\) is the pre-exponential factor, which is related to the collision factor, \(T\) is the temperature, and \(R\) is the gas constant: \(k = A e^{\frac{-E_a}{RT}}\)
02

Write down the equation for the uncatalyzed and catalyzed reactions

\(\) For the uncatalyzed reaction, we have: \(k_{1} = A e^{\frac{-E_{a1}}{RT}}\) For the catalyzed reaction, we have: \(k_{2} = A e^{\frac{-E_{a2}}{RT}}\)
03

Calculate the reaction rate constant ratio

\(\) The given increase in reaction rate between the uncatalyzed and catalyzed reactions is \(1 \times 10^{5}\). So, the ratio of the reaction rate constants is: \(\frac{k_{2}}{k_{1}} = 1 \times 10^{5}\)
04

Substitute the expressions for \(k_1\) and \(k_2\) into the equation for the rate constant ratio

\(\) We can now substitute the expressions for the reaction rate constants from Step 2 into the equation obtained in Step 3: \(\frac{A e^{\frac{-E_{a2}}{RT}}}{A e^{\frac{-E_{a1}}{RT}}} = 1 \times 10^{5}\)
05

Simplify the equation and solve for the difference in activation energy

\(\) We can simplify the equation by cancelling out the pre-exponential factor \(A\) and taking the natural logarithm of both sides: \(\frac{e^{\frac{-E_{a2}}{RT}}}{e^{\frac{-E_{a1}}{RT}}} = 1 \times 10^{5}\) Taking the natural logarithm, we get: \(\frac{-E_{a2}}{RT} - \frac{-E_{a1}}{RT} = \ln{(1 \times 10^{5})}\) Now, we can solve for the difference in activation energy: \((E_{a1} -E_{a2}) = RT\ln{(1 \times 10^{5})}\) We are given a temperature of \(37^{\circ} \mathrm{C}\), which we must convert to Kelvin by adding 273.15: \(T = 37^{\circ} \mathrm{C} + 273.15 = 310.15 K \) Using \(R = 8.314 \mathrm{J \ mol^{-1} \ K^{-1}}\): \((E_{a1} -E_{a2}) = (310.15 K)(8.314 \frac{\mathrm{J}}{\mathrm{mol \cdot K}}) \ln{(1 \times 10^{5})}\) Calculating the difference: \((E_{a1} -E_{a2}) = 16,120 \mathrm{J \ mol^{-1}} \approx 16.1 \mathrm{kJ \ mol^{-1}}\)
06

Final answer

\(\) Thus, the enzyme must lower the activation energy by approximately 16.1 kJ/mol for a 100,000-fold increase in the reaction rate at physiological temperature.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arrhenius equation
The Arrhenius equation is a fundamental formula that describes how the temperature and activation energy affect the rate of a chemical reaction. It helps predict how quickly a reaction will proceed based on these factors. By understanding the components of this equation, we can unravel the complexities of reaction kinetics.

The equation is expressed as follows:
  • \( k = A e^{\frac{-E_a}{RT}} \)
Here:
  • \(k\) is the reaction rate constant, a measure of how fast the reaction occurs.
  • \(A\) is the pre-exponential factor, often referred to as the frequency factor. It relates to the frequency of collisions between molecules.
  • \(E_a\) is the activation energy, the minimum energy required for the reaction to happen.
  • \(R\) is the universal gas constant, with a value of 8.314 J/mol路K.
  • \(T\) is the absolute temperature in Kelvin.
According to the Arrhenius equation, increasing the temperature or decreasing the activation energy will result in an increase in the reaction rate constant, making the reaction faster. This is because higher temperatures provide more energy to the reactants, while lower activation energies make it easier for the reaction to proceed. Understanding this relationship is crucial in fields like chemistry and biology, where controlling reaction rates can be pivotal.
enzyme catalysis
Enzyme catalysis is a fascinating process through which biological catalysts, known as enzymes, accelerate chemical reactions. Enzymes are proteins that facilitate cellular reactions by lowering the activation energy needed, allowing reactions to proceed faster even at lower temperatures.

In the context of enzyme catalysis, enzymes offer several advantages:
  • They provide an alternative reaction pathway with a lower activation energy.
  • They enhance reaction specificity by binding to particular substrates.

  • They remain unchanged after the reaction, able to catalyze multiple cycles.
The mechanism usually involves the enzyme binding to reactants (substrates) to form an enzyme-substrate complex. This complex provides a more stable transition state, reducing the energy barrier required for the reaction.

Through these processes, enzyme catalysis plays a vital role in sustaining life by ensuring that biochemical reactions proceed efficiently and swiftly within the gentle confines of living cells.
reaction rate
The reaction rate is a critical concept in chemistry that measures how fast a chemical reaction occurs. It is governed by several factors, including temperature, concentration of reactants, and the presence of catalysts such as enzymes. Let鈥檚 explore these factors:
  • Temperature: As per the Arrhenius equation, an increase in temperature generally increases the reaction rate. More thermal energy results in more frequent and energetic molecular collisions, facilitating faster reactions.

  • Concentration of Reactants: Higher concentrations lead to more frequent collisions between reactants, thus increasing the reaction rate.

  • Catalysts: Catalysts, including enzymes, speed up reactions by lowering the activation energy. They enable the reaction to proceed quicker without being consumed in the process.
Understanding these factors helps us manipulate and control reaction rates in various applications, from industrial processes to biological pathways. This knowledge is essential for optimizing conditions to achieve desired reaction speeds, ensuring productivity and efficiency.

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) Can an intermediate appear as a reactant in the first step of a reaction mechanism? (b) On a reaction energy profile diagram, is an intermediate represented as a peak or a valley? (c) If a molecule like \(\mathrm{Cl}_{2}\) falls apart in an elementary reaction, what is the molecularity of the reaction?

Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carboncontaining fragment such as \(\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2},\) and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a C atom for "R"). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

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-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 \(0.100 ?\)

(a) The reaction \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\) is first order with in \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)\) and zero-order in \(\mathrm{H}_{2} \mathrm{O}\). At \(300 \mathrm{~K}\) the rate constant equals \(3.30 \times 10^{-2} \mathrm{~min}^{-1} .\) Calculate the half- life at this temperature. \((\mathbf{b})\) If the activation energy for this reaction is \(80.0 \mathrm{~kJ} / \mathrm{mol}\), at what temperature would the reaction rate be doubled?

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21^{\circ} \mathrm{C}\). (a) Write out the balanced equation for the reaction catalyzed by urease. \((\mathbf{b})\) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C}\), what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

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.