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

What is the correct relationship between free energy change and equilibrium constant of a reaction : (a) \(\Delta G^{\circ}=R T \ln K\) (b) \(\Delta G^{\circ}=-R T \ln K\) (c) \(\Lambda G=B T \ln K\) (d) \(\Delta G=-R T \ln K\)

Short Answer

Expert verified
\(\Delta G^\circ=-RT \ln K\)

Step by step solution

01

Understanding the Relationship Between Free Energy Change and Equilibrium Constant

The free energy change \(\Delta G^\circ\) at standard conditions and equilibrium constant \(K\) of a reaction have a well-defined relationship given by the Gibbs free energy equation, which states that \(\Delta G^\circ = -RT \ln K\), where \(R\) is the universal gas constant and \(T\) is the temperature in Kelvin.
02

Checking the Given Options

Evaluate each option to see which one correctly matches the Gibbs free energy equation. Remember that the sign before \(RT\) must be negative as per the actual relationship.
03

Identifying the Correct Relationship

Comparing the given options with the correct Gibbs free energy equation, we find that option (b) \(\Delta G^\circ=-RT \ln K\) correctly reflects the relationship between the free energy change and equilibrium constant.

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.

Free Energy Change
The concept of free energy change lies at the heart of chemical thermodynamics, providing vital insights into whether a reaction can proceed spontaneously. Fundamentally, the free energy change, denoted as \(\Delta G\), represents the amount of energy available to do work during a chemical process at constant temperature and pressure. When \(\Delta G\) is negative, a reaction is regarded as spontaneous, meaning it has the inherent potential to occur without external energy input. Conversely, a positive \(\Delta G\) indicates that a reaction requires an input of energy to proceed.

In the context of a chemical reaction reaching equilibrium, \(\Delta G\) becomes zero because the system's free energy is at its minimum, and no further net change is possible. At this juncture, the forward and reverse reactions occur at the same rate, leading to the important relationship between \(\Delta G^\circ\) and the equilibrium constant \(K\). This is encapsulated in the equation \(\Delta G^\circ = -RT \ln K\), linking thermodynamic potentials to chemical kinetics.
Equilibrium Constant
The equilibrium constant, represented as \(K\), is a quantitative measure reflecting the ratio of concentrations of the products to reactants at the state of chemical equilibrium. Each concentration is raised to the power of its stoichiometric coefficient in the balanced chemical equation. The higher the value of \(K\), the greater the concentration of products over reactants at equilibrium, suggesting a reaction that strongly favors the formation of products.

Understanding \(K\) in the Context of Free Energy

The linkage between \(K\) and \(\Delta G^\circ\) outlines the inherent directionality of a reaction. If \(K\) is greater than 1, implying a reaction favoring products, \(\Delta G^\circ\) is negative, echoing a spontaneous process. However, if \(K\) is less than 1, showing a reactant-favored reaction, \(\Delta G^\circ\) is positive, indicating non-spontaneity. When \(K = 1\), the system is at equilibrium, and \(\Delta G^\circ\) is zero. This pivotal relationship allows for the prediction of reaction spontaneity and direction from the magnitude of the equilibrium constant.
Thermodynamics
Thermodynamics is a broad and fundamental branch of physics, which also underpins much of chemistry, concerned with heat, work, and the energy transformations that accompany physical and chemical changes. At its core, it revolves around the concepts of energy conservation, entropy, and the flow of thermal energy.

Within the context of chemical reactions, thermodynamics focuses particularly on changes to the internal energy of systems (\(\Delta U\)), enthalpy (\(\Delta H\)), entropy (\(\Delta S\)), and Gibbs free energy (\(\Delta G\)). These quantities speak to the feasibility and extent to which reactions proceed. The equation \(\Delta G^\circ = -RT \ln K\) is a prime example of thermodynamic principles applied to chemical kinetics, expressing how the potential free energy change at standard conditions influences the system's equilibrium state.
Chemical Kinetics
Chemical kinetics is the study of reaction rates and the factors that affect them. This branch of chemistry is concerned with understanding how quickly reactions occur and determining the pathways they follow. Unlike thermodynamics—which tells us the potential for a reaction to occur and the energy change involved—kinetics provides insights into the speed of the reaction and the mechanism through which it proceeds.

While Gibbs free energy provides an understanding of whether a reaction is thermodynamically favored, it does not provide direct information about the rate of the reaction. Both the free energy change (\(\Delta G\)) and the equilibrium constant (\(K\)) are crucial for determining a reaction's spontaneity but tell us nothing about how fast it will reach equilibrium. This is where chemical kinetics plays a complementary role to thermodynamic principles, emphasizing the practical aspects of chemical 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

What is the unit of \(K_{p}\) for the reaction? $$ \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) $$ (a) atm (b) \(\mathrm{atm}^{-2}\) (c) \(\mathrm{atm}^{2}\) (d) \(\mathrm{atm}^{-1}\)

Consider the following reactions at equilibrium and determine which of the indicated changes will cause the reaction to proceed to the right. (1) \(\mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\left(\right.\) add \(\left.\mathrm{CH}_{4}\right)\) (2) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (remove \(\mathrm{NH}_{3}\) ) (3) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) (add \(\mathrm{F}_{2}\) ) (4) \(\mathrm{BaO}(s)+\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{BaSO}_{4}(s)\) (add BaO) (a) \(2.3\) (b) \(1.4\) (c) \(2.4\) (d) \(2,3,4\)

In a chemical reaction equilibrium is established when : (a) Opposing reaction ceases (b) Concentrations of reactants and product are equal (c) Velocity of opposing reaction is the same as that of forward reaction (d) Reaction ceases to generate heat

If the pressure in a reaction vessel for the following reaction is increased by decreasing the volume, what will happen to the concentrations of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) ? $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)+\text { Heat } $$ (a) both the [CO] and [CO \(\left._{2}\right]\) will decrease (b) neither the [CO] nor the \(\left[\mathrm{CO}_{2}\right]\) will change (c) the [CO] will decrease and the [CO \(_{2}\) ] will increase (d) both the [CO] and [CO \(\left._{2}\right]\) will increase

Using molar concentrations, what is the unit of \(K_{c}\) for the reaction? $$ \mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) $$ (a) \(M^{-2}\) (b) \(M^{2}\) (c) \(M^{-1}\) (d) \(M\)

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