Chapter 5, Problem 5.11PS

(a)

Summary Introduction

To show: At equilibrium, $\Delta G\text{'}$ is 0 for the given reaction, using the provided equations:

$a\text{A}+b\text{B}\rightleftharpoons cC+d\text{D}$

$\begin{array}{c}\Delta G°\text{'}=-592\ln K{\text{'}}_{eq}\\ \Delta G\text{'}=\Delta G°\text{'}+592\ln \frac{{\left[\text{C}\right]}_{\text{pr}}^c{\left[\text{D}\right]}_{\text{pr}}^d}{{\left[\text{A}\right]}_{\text{pr}}^a{\left[\text{B}\right]}_{\text{pr}}^b}\end{array}$

Introduction: Free energy of a system is that amount of energy that can be changed into work in a system provided that the temperature and pressure of the system remain the same. It is denoted by $\Delta G\text{'}$. Free energy is informative of the extent to which a reaction can proceed so that all the reactants and products can reach a concentration value of 1.0 M when the conditions are maintained the standard.

(b)

Summary Introduction

To determine: The amount of work that can be done by the system.

Introduction: Free energy of a system is that amount of energy that can be changed into work in a system provided that the temperature and pressure of the system remain the same. It is denoted by $\Delta G\text{'}$. Free energy is informative of the extent to which a reaction can proceed so that all the reactants and products can reach a concentration value of 1.0 M when the conditions are maintained the standard.