Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 81

When most biologic enzymes are heated, they lose their catalytic activity. The change Original enzyme \(\longrightarrow\) new form that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain.

Short Answer

Expert verified
The structure of the original enzyme is more ordered (has smaller positional probability) compared to its new form after heating. This is because the process of heating leads to an increase in entropy, which implies a more disordered state with higher positional probability.

Step by step solution

01

What happens to enzymes upon heating

When most biologic enzymes are heated, they lose their catalytic activity. This occurs because heat causes the enzyme to denature, which means its molecular structure changes. The original enzyme transitions to a new form with altered molecular structure and reduced enzyme activity. This process is endothermic (heat is absorbed) and spontaneous (occurs without external input).
02

Discuss spontaneity

A spontaneous process is one that occurs naturally without any external input of energy. In thermodynamics, spontaneity is governed by the concept of Gibbs free energy (G). A process is spontaneous when \(\Delta G < 0\). Gibbs free energy is a function of enthalpy (H), entropy (S), and temperature (T), and can be calculated using the formula \(\Delta G = \Delta H - T\Delta S\). In this problem, since the process is endothermic and spontaneous, it means that \(\Delta H > 0\) and \(\Delta S > 0\).
03

Entropy and positional probability

Entropy (S) is a measure of the disorder or randomness in a system. A system with higher entropy has a higher positional probability, which means its molecules are more disordered and less constrained in their movement. Conversely, a system with lower entropy has lower positional probability and is more ordered. In this problem, when the enzyme undergoes a change due to heating, the entropy of the system increases (\(\Delta S > 0\)).
04

Determine which enzyme form is more ordered

Since the increase in entropy (\(\Delta S > 0\)) suggests that the final state has higher positional probability (more disordered) than the initial state, we can conclude that the structure of the original enzyme is more ordered (has smaller positional probability) compared to its new form after heating.

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

Consider the following system at equilibrium at \(25^{\circ} \mathrm{C}\) : $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \quad \Delta G^{\circ}=-92.50 \mathrm{~kJ}$$ What will happen to the ratio of partial pressure of \(\mathrm{PCl}_{5}\) to partial pressure of \(\mathrm{PCl}_{3}\) if the temperature is raised? Explain completely.

Consider the following reaction at \(800 . \mathrm{K}\) : $$\mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{~atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{~atm}, P_{\mathrm{NF}_{3}}=0.48 \mathrm{~atm} .\) Calculate \(\Delta G^{\circ}\) for the reaction at \(800 . \mathrm{K}\).

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\). Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\). Consider the process \(\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\) a. Determine the sign of \(\Delta S, \Delta S_{\text {surr }}\), and \(\Delta S_{\text {univ }}\) for the process in vessel 1 . b. Determine the sign of \(\Delta S, \Delta S_{\text {surt }}\), and \(\Delta S_{\text {univ }}\) for the process in vessel 2 . (Hint: Think about the effect that a salt has on the freezing point of a solvent.)

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\). b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Making Measurements

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