Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Enzymes are remarkably efficient catalysts that can increase reaction rates by as many as 20 orders of magnitude. (a) How does an enzyme affect the transition state of a reaction, and how does this effect increase the reaction rate? (b) What characteristics of enzymes give them this effectiveness as catalysts?

Short Answer

Expert verified
Enzymes lower the activation energy by stabilizing the transition state, increasing the reaction rate. Their specificity, efficiency, and regulatory capabilities make them effective catalysts.

Step by step solution

01

Understand the role of enzymes in reactions

Enzymes act as catalysts, substances that increase the rate of chemical reactions without being consumed. By lowering the activation energy required to reach the transition state, enzymes make it easier for the reaction to proceed.
02

Impact on the transition state

Enzymes stabilize the transition state of a reaction. This stabilization lowers the activation energy and provides an alternative reaction pathway that requires less energy, thereby increasing the reaction rate.
03

Characteristics of enzymes

Enzymes possess unique features: specificity (each enzyme catalyzes a specific reaction), efficiency (operating under mild conditions such as low temperature and neutral pH), and the ability to be regulated as per the cellular requirements. These attributes contribute to their effectiveness as catalysts.

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 Vaia!

Key Concepts

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

Transition State Stabilization
Enzymes work their magic by stabilizing the transition state of a reaction. The transition state is a high-energy, unstable state that reactants must pass through to become products. By providing a more stable pathway, enzymes essentially lower the 'hill' that reactants need to climb. Enzymes achieve this by binding tightly to the transition state, making it easier for the reactants to transform into products. This stabilization decreases the activation energy required for the reaction, speeding up the process immensely. Imagine enzymes as skilled professionals who know the shortcuts within complicated tasks, making it far easier and quicker to get to the endpoint.
Activation Energy
Activation energy is the initial energy investment required for a reaction to occur. It's the barrier that reactants need to overcome. Enzymes lower this energy barrier, akin to constructing a lower fence that requires less effort to jump over. By lowering the activation energy, enzymes enable reactions to proceed at a quicker pace and more efficiently.
Without enzymes, reactions would be significantly slower or require much higher temperatures to reach completion—conditions that are often incompatible with life. With enzymes, these reactions can happen under mild, life-sustaining conditions.
In summary, by reducing the activation energy, enzymes make it easier for reactions to start and proceed, ensuring that essential biochemical processes occur swiftly and efficiently.
Enzyme Specificity
One of the most fascinating aspects of enzymes is their specificity. Each enzyme is designed to catalyze a specific reaction or type of reaction. This specificity arises from the unique shape and structure of the enzyme's active site which matches only specific substrates, much like a lock and key.
This 'fit' ensures that enzymes are extremely selective, often recognizing only one particular substrate. This specificity is crucial for numerous metabolic pathways, as it ensures that reactions occur precisely and without unwanted side reactions. Enzyme specificity not only contributes to efficiency but also to the regulation and coordination of complex biochemical networks.
Reaction Rate Enhancement
Enzymes dramatically enhance reaction rates. By lowering the activation energy and stabilizing transitional states, they make reactions happen significantly faster—sometimes up to millions of times faster.
The quickening of these reactions is vital for life since many biochemical reactions are necessary for basic cellular functions like energy production, DNA replication, and nutrient metabolism.
Without enzymes, these reactions would proceed at a snail's pace, proving insufficient for the demands of living cells. Enzyme-driven speed ensures that cells can respond rapidly to environmental changes and maintain homeostasis.
Catalytic Efficiency
Catalytic efficiency refers to an enzyme's ability to convert a substrate into a product as quickly and effectively as possible. This efficiency is measured by two main factors: how well the enzyme binds to the substrate and how quickly it converts the bound substrate into the product.
Enzymes with high catalytic efficiency operate close to the diffusion limit, meaning they are so efficient that the rate of product formation is limited only by how quickly substrates can diffuse to the enzyme. High catalytic efficiency ensures that metabolic processes are conducted rapidly, supporting the fast pace of cellular activities.
Thus, the incredible efficiency of enzymes is essential for maintaining the speed and balance of life-sustaining reactions.

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

In a study of nitrosyl halides, a chemist proposes the following mechanism for the synthesis of nitrosyl bromide: \(\mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(g) \rightleftharpoons \operatorname{NOBr}_{2}(g)\) [fast] \(\operatorname{NOBr}_{2}(g)+\mathrm{NO}(g) \longrightarrow 2 \mathrm{NOBr}(g)\) [slow] If the rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{Br}_{2}\right]\), is the proposed mechanism valid? If so, show that it satisfies the three criteria for validity.

Aqua regia, a mixture of \(\mathrm{HCl}\) and \(\mathrm{HNO}_{3},\) has been used since alchemical times to dissolve many metals, including gold. Its orange color is due to the presence of nitrosyl chloride. Consider this one-step reaction for the formation of this compound: \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \operatorname{NOCl}(g)+\mathrm{Cl}(g) \quad \Delta H^{\circ}=83 \mathrm{~kJ}\) (a) Draw a reaction energy diagram, given \(E_{\text {affwd }}=86 \mathrm{~kJ} / \mathrm{mol}\). (b) Calculate \(E_{\text {arrev }}\). (c) Sketch a possible transition state for the reaction. (Note: The atom sequence of nitrosyl chloride is \(\mathrm{Cl}-\mathrm{N}-\mathrm{O} .)\)

(a) For a reaction with a given \(E_{a},\) how does an increase in \(T\) affect the rate? (b) For a reaction at a given \(T,\) how does a decrease in \(E_{\mathrm{a}}\) affect the rate?

(a) What is the difference between an average rate and an instantancous rate? (b) What is the difference between an initial rate and an instantaneous rate?

The compound \(\mathrm{AX}_{2}\) decomposes according to the equation \(2 \mathrm{AX}_{2}(g) \rightarrow 2 \mathrm{AX}(g)+\mathrm{X}_{2}(g) .\) In one experiment, \(\left[\mathrm{AX}_{2}\right]\) was measured at various times and these data were obtained: $$ \begin{array}{cc} \text { Time (s) } & {\left[A X_{2}\right](\mathrm{mol} / \mathrm{L})} \\ \hline 0.0 & 0.0500 \\ 2.0 & 0.0448 \\ 6.0 & 0.0300 \\ 8.0 & 0.0249 \\ 10.0 & 0.0209 \\ 20.0 & 0.0088 \end{array} $$ (a) Find the average rate over the entire experiment. (b) Is the initial rate higher or lower than the rate in part (a)? Use graphical methods to estimate the initial rate.

See all solutions

Recommended explanations on Chemistry Textbooks

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