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

Most reactions, including enzyme-catalyzed reactions, proceed faster at higher temperatures. However, for a given enzyme, the rate drops off abruptly at a certain temperature. Account for this behavior.

Short Answer

Expert verified
Enzymes increase reaction rates with higher temperatures until denaturation occurs, reducing their activity.

Step by step solution

01

Understand Enzyme Activity and Temperature

Enzymes are biological catalysts that speed up chemical reactions in cells. As temperature increases, the kinetic energy of molecules also increases, leading to more frequent collisions between enzymes and substrates, thus increasing the reaction rate.
02

Identify the Optimal Temperature

For each enzyme, there is an optimal temperature at which the enzyme activity is at its peak. This is the temperature at which the enzyme structure is most conducive to catalyzing the reaction efficiently.
03

Explain Denaturation of Enzymes

Beyond the optimal temperature, the enzyme may begin to denature, losing its specific three-dimensional structure. Denaturation disrupts the active site, where the substrate binds to the enzyme, leading to a decrease in enzyme activity.
04

Connect Temperature Increase and Rate Decrease

As the temperature continues to rise past the optimal point, denaturation becomes more significant and abrupt, causing the enzyme to become inactive. The decrease in the number of functioning enzyme molecules leads to a rapid decline 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 Vaia!

Key Concepts

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

Enzyme Denaturation
Enzymes are delicate molecules that catalyze chemical reactions in cells efficiently. However, they can be quite sensitive to changes in environmental conditions, such as temperature. At elevated temperatures, enzymes can undergo a process called denaturation. This means that their intricate three-dimensional structures—essential for their function—begin to unravel and change. When an enzyme denatures, its active site, where the chemical reactions occur, gets distorted.
  • Denaturation is often irreversible.
  • The enzyme's ability to catalyze reactions diminishes or is completely lost.
Importantly, denaturation doesn't break the peptide bonds within the enzyme but rather disrupts the bonds that maintain its shape and function. Once an enzyme denatures, even reducing the temperature back to optimal levels won't restore its original structure and activity.
Optimal Temperature
Every enzyme has a specific optimal temperature range where it catalyzes reactions most efficiently. When enzymes function within this range, they adopt the most favorable conformation to facilitate reactions. The optimal temperature aligns with the environment the enzyme naturally operates in. For example:
  • Human enzymes typically have an optimal temperature near 37°C, which is body temperature.
  • Thermophilic bacteria, which live in hot springs, have enzymes optimal at much higher temperatures.
When an enzyme is at its optimal temperature, the increased kinetic energy of molecules enhances collisions between enzyme and substrate. This perfect balance leads to a peak in enzyme activity.
Reaction Rate
The reaction rate, in the context of enzyme activity, is how fast a substrate is converted into a product. As temperature increases, so does the reaction rate—up to a point. This increase in rate occurs because higher temperatures raise molecular motion, increasing the chances of enzyme and substrate interactions. But there’s a catch:
  • Beyond the optimal temperature, the reaction rate declines due to enzyme denaturation.
  • The abrupt slowdown is often because enzymes, losing their shape from heat, no longer bind to substrates effectively.
Therefore, while higher temperatures initially boost the reaction rate, once the temperature exceeds the optimal level, the efficiency declines sharply due to damaged enzyme structures.

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 protein molecule \(\mathrm{P}\) of molar mass \(\mathscr{M}\) dimerizes when it is allowed to stand in solution at room temperature. A plausible mechanism is that the protein molecule is first denatured (i.e., loses its activity due to a change in overall structure) before it dimerizes: \(\mathrm{P} \stackrel{k}{\longrightarrow} \mathrm{P}^{*}(\) denatured \() \quad\) (slow) $$ 2 \mathrm{P}^{*} \longrightarrow \mathrm{P}_{2} $$ (fast) where the asterisk denotes a denatured protein molecule. Derive an expression for the average molar mass (of \(\mathrm{P}\) and \(\left.\mathrm{P}_{2}\right), \bar{U},\) in terms of the initial protein concentration \([\mathrm{P}]_{0}\) and the concentration at time \(t,[\mathrm{P}]_{t},\) and \(\mathscr{M} .\) Describe how you would determine \(k\) from molar mass measurements.

Write an equation relating the concentration of a reactant A at \(t=0\) to that at \(t=t\) for a first-order reaction. Define all the terms, and give their units. Do the same for a second-order reaction.

Classify the following elementary reactions as unimolecular, bimolecular, or termolecular: (a) \(2 \mathrm{NO}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{NOBr}\) (b) \(\mathrm{CH}_{3} \mathrm{NC} \longrightarrow \mathrm{CH}_{3} \mathrm{CN}\) (c) \(\mathrm{SO}+\mathrm{O}_{2} \longrightarrow \mathrm{SO}_{2}+\mathrm{O}\)

When a mixture of methane and bromine is exposed to light, the following reaction occurs slowly: $$ \mathrm{CH}_{4}(g)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Br}(g)+\mathrm{HBr}(g) $$ Suggest a reasonable mechanism for this reaction. (Hint: Bromine vapor is deep red; methane is colorless.)

A certain reaction is known to proceed slowly at room temperature. Is it possible to make the reaction proceed at a faster rate without changing the temperature?

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