/*! 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 40 \(\Delta G\) cannot generally be... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\Delta G\) cannot generally be equated with \(\Delta G^{\circ}\). To a very good first approximation \(\Delta H\) can be equated with \(\Delta H^{\circ}\). Explain.

Short Answer

Expert verified
\( \Delta G \neq \Delta G^{\circ} \) due to varying conditions; \( \Delta H \approx \Delta H^{\circ} \) because enthalpy is less condition-sensitive.

Step by step solution

01

Understanding Gibbs Free Energy Change

Gibbs free energy change (\( \Delta G \)) varies with the reaction conditions like temperature and pressure, while \( \Delta G^{\circ} \), the standard Gibbs free energy change, is measured under standard conditions (1 bar pressure, 298 K temperature). Therefore, \( \Delta G \) is not equivalent to \( \Delta G^{\circ} \) because they refer to different sets of conditions.
02

Enthalpy Change and Standard Conditions

Enthalpy change (\( \Delta H \)) is less sensitive to changes in pressure and temperature for many reactions, especially when considering chemical bonds or phases involved are consistent. This enables \( \Delta H \) to often approximate \( \Delta H^{\circ} \) closely, as they share minimal variance in the conditions affecting enthalpy.

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 91Ó°ÊÓ!

Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physical science focused on heat, work, temperature, energy, and how they are interrelated. One significant aspect of thermodynamics is the study of energy changes during chemical reactions. Gibbs Free Energy, denoted as \( \Delta G \), is a pivotal part of this study. It combines enthalpy (\( \Delta H \)), entropy (\( \Delta S \)), and temperature (\( T \)) into one equation: \[\Delta G = \Delta H - T\Delta S\]This equation helps predict whether a reaction will occur spontaneously. For a reaction to be spontaneous at constant temperature and pressure, \( \Delta G \) must be negative.
There are several key aspects of thermodynamics to remember:
  • Systems can exchange energy, often in the form of heat, with their surroundings.
  • The First Law of Thermodynamics states that energy cannot be created or destroyed, only transformed.
  • The Second Law of Thermodynamics introduces the concept of entropy, often associated with disorder.
Integrating these principles, Gibbs Free Energy offers a way to understand and predict the energy changes and feasibility of reactions.
Standard Conditions
Standard conditions, often represented by the superscript "°" (as in \( \Delta G^{\circ} \)), are a set of baseline parameters used by scientists to report measurements consistently. For chemistry, these typically include a temperature of 298 K (25°C) and a pressure of 1 bar. Under these conditions, values such as Gibbs Free Energy and Enthalpy are reported as \( \Delta G^{\circ} \) and \( \Delta H^{\circ} \), respectively.
Why use standard conditions?
  • They provide a common reference point, making it easier to compare results across different studies.
  • They simplify calculations, as constant parameters reduce variability.
However, real-world reactions seldom occur under these idealized conditions, resulting in a potential difference between values calculated under standard conditions and those in actual scenarios. \( \Delta G \) often differs from \( \Delta G^{\circ} \) due to factors like variable temperature and pressure, crucial for accurate assessments of reaction dynamics outside the lab.
Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), measures the total heat content change within a system during a chemical reaction. In many cases, \( \Delta H \) tends to approximate \( \Delta H^{\circ} \) closely under varying conditions because enthalpy is less temperature-sensitive compared to other thermodynamic quantities. This makes enthalpy change a more stable measure across different scenarios.
Key characteristics of Enthalpy Change:
  • If \( \Delta H \) is negative, the reaction is exothermic (releases heat).
  • If \( \Delta H \) is positive, the reaction is endothermic (absorbs heat).
Enthalpy change helps in understanding broader reaction energetics and is pivotal in processes like calorimetry, where precise heat measurements are crucial. Despite its minor sensitivity to pressure and temperature shifts, always consider the reaction environment, as these factors may affect other conditions of the reaction, contributing to the final outcome of \( \Delta G \).

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

When a photon in the visible range is absorbed in the retina by rhodopsin, the photoreceptor in rod cells, 11-cis-retinal is converted to the all-trans isomer. Light energy is transformed into molecular motion. The efficiency of photons to initiate the reaction is about \(20 \%\) at \(500 \mathrm{~nm}\left(57 \mathrm{kcal} \mathrm{mol}^{-1}\right)\). About \(50 \%\) of the absorbed energy is available for the next signaling step. This process takes about \(10 \mathrm{~ms}\). In the absence of light, spontaneous isomerization of 11 -cis- retinal is very slow, on the order of \(0.001 \mathrm{yr}^{-1}\) ! Experimental studies have shown that the equilibrium energetics of retinal isomerization are \(\Delta S^{\circ}=4.4 \mathrm{cal}\) \(\mathrm{mol}^{-1} \mathrm{~K}^{-1}\) and \(\Delta \mathrm{H}^{\circ}=150 \mathrm{cal} \mathrm{mol}^{-1}\). Calculate the equilibrium constant for the reaction.

Cytochromes are redox-active proteins that occur in all organisms except a few types of obligate anaerobes. These proteins contain heme groups, the iron atom of which reversibly alternates between the \(\mathrm{Fe}(\mathrm{II})\) and \(\mathrm{Fe}(\mathrm{III})\) oxidation states during electron transport. Consider the reaction cytc \(\left(\mathrm{Fe}^{2+}\right)+\operatorname{cyt}\left(\mathrm{Fe}^{3+}\right) \Leftrightarrow \operatorname{cytc}\left(\mathrm{Fe}^{3+}\right)+\operatorname{cyt}\left(\mathrm{Fe}^{2+}\right)\) involving cytochromes \(c\) and \(f\). If \(V^{01}=0.365 \mathrm{~V}\) for electron transfer to cytf \(\left(\mathrm{Fe}^{3+}\right)\), and \(V^{\prime \prime}=0.254 \mathrm{~V}\) for electron transfer to cytc \(\left(\mathrm{Fe}^{3+}\right)\), can ferrocytochrome \(c(2+\) oxidation state) reduce ferricytochrome \(f\) ( \(3+\) oxidation state) spontaneously?

What is the \(\mathrm{pH}\) value of \(0.001 \mathrm{M} \mathrm{HCl}\) solution?

State whether the following phrases pertain to (A) spontaneity, (B) reversibility, (C) both spontaneity and reversibility, or (D) neither spontaneity nor reversibility. (1) Established for \(\Delta G<0\) at constant \(T\). (2) Established for \(\Delta S<0\). (3) Established for a process in which the work done is a maximum. (4) Illustrated by the migration of a solute from a region of high concentration to low concentration. (5) Required for determination of \(\Delta S\) by the heat transferred. (6) Implies that the coupling of coupled reaction is very efficient.

Which one of the following equations is used to evaluate free energy changes in cells under physiological conditions? What makes it appropriate? (a) \(\Delta G=R T \ln K_{\mathrm{eq}}\). (b) \(\Delta G=\Delta G^{\prime \prime}+R T \ln [\) products \(][\) [reactants]. (c) \(\Delta G=R T \ln [\) products \(] /[\) reactants \(]\). (d) \(\Delta G=\Delta H-\mathrm{T} \Delta S\). (e) \(\Delta G=\Delta G^{\circ \prime}+R T\) [products] [ [reactants].

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.