Chapter 17, Problem 17.38QP

Calculate ΔG for the reaction

${\text{H}}_{\text{2}}\text{O}(l)\stackrel{}{\rightleftharpoons }{\text{H}}^{+}(aq)+{\text{OH}}^{-}(aq)$

at 25°C for the following conditions:

1. (a) $\left[{\text{H}}^{+}\right]=1.0\times {10}^{-7}M,\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-7}M$
2. (b) $\left[{\text{H}}^{+}\right]=1.0\times {10}^{-3}M,\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-4}M$
3. (c) $\left[{\text{H}}^{+}\right]=1.0\times {10}^{-12}M,\left[{\text{OH}}^{-}\right]=2.0\times {10}^{-8}M$
4. (d) $\left[{\text{H}}^{+}\right]=3.5M,\left[{\text{OH}}^{-}\right]=4.8\times {10}^{-4}M$

Interpretation Introduction

Interpretation:

To calculate the free energy $(\text{ΔG})$ values for given aqueous phase equilibrium reaction at ${\text{25}}^{°}\text{C}$.

Concept Information:

Thermodynamics is the branch of science that relates heat and energy in a system.  The laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy. Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

$\text{ΔG = }\Delta {\rm H}\text{- T}\Delta \text{S}$

Where,

$\text{ΔG }$ is the change in free energy of the system

$\Delta {\rm H}$ is the change in enthalpy of the system

$\text{T}$ is the absolute value of the temperature

$\Delta \text{S}$ is the change in entropy in the system

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$

    $\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.0826}\text{l}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Explanation of Solution

In each part of this problem we can use the following equation to calculate $\Delta G$

$\begin{array}{l}\text{ΔG}={\text{ΔG}}^{\text{0}}+\text{RTlnQ}\\ {\text{ΔG=ΔG}}^{\text{0}}{\text{+RTln([H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])}\end{array}$

Given the concentrations are equilibrium concentrations at ${25}^{0}C$, Since the reaction is at equilibrium $\Delta G=0$. This is advantage, because it allows us to calculate free energy $\Delta G°$, the equilibrium $Q=K$, so we can write fallowing equation.  

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=}{\text{-RTlnK}}_{\text{w}}\\ \text{Given}\text{values}\text{(R,}\text{T)}\text{are}\text{substituted}\text{above}\text{equation}\\ {\text{ΔG}}^{\text{0}}\text{=}\text{-(8}\text{.314}\text{J/K}\cdot \text{mol)}\text{(298}\text{K)}\text{ln(1}\text{.0}{\text{×10}}^{\text{-14}}\text{)}\\ \text{=}\text{-2477}\text{.572}\times \text{ln(1}\text{.0}{\text{×10}}^{\text{-14}}\text{)}\text{[Q}\text{Here}\text{In}\text{1}\text{.0}{\text{×10}}^{\text{-14}}\text{=-29}\text{.9336]}\\ {\text{ΔG}}^{\text{0}}\text{=}\text{8}\text{.0}{\text{×10}}^{\text{4}}\text{J/mol}\end{array}$

Interpretation Introduction

Interpretation:

To calculate the free energy $(\text{ΔG})$ values for given aqueous phase equilibrium reaction at ${\text{25}}^{°}\text{C}$.

Concept Information:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

Thermodynamics is the branch of science that relates heat and energy in a system.  The four laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy. Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

$\text{ΔG = }\Delta {\rm H}\text{- T}\Delta \text{S}$

Where,

$\text{ΔG }$ is the change in free energy of the system

$\Delta {\rm H}$ is the change in enthalpy of the system

$\text{T}$ is the absolute value of the temperature

$\Delta \text{S}$ is the change in entropy in the system

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$

    $\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.0826}\text{l}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Explanation of Solution

$\begin{array}{l}\text{Let}\text{us consider following reaction (b)}\\ \text{b)}\text{.}{\text{[H}}^{\text{+}}\text{]=}\text{1}\text{.0}\times {\text{10}}^{\text{-3}}M,[O{H}^{-}]=\text{1}\text{.0}\times {\text{10}}^{\text{-4}}M\\ {\text{ΔG}}^{\text{0}}\text{=}\text{8}\text{.0}{\text{×10}}^{\text{4}}\text{J/mol}\\ {\text{ΔG}}^{\text{0}}\text{=}\text{-RTlnQ}\\ {\text{ΔG=ΔG}}^{\text{0}}{\text{+RTln([H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])}\\ \text{Given values are substituted above}\text{equatuion}\\ \text{ΔG=8}\text{.0}{\text{×10}}^{\text{4}}\text{J/mol+(8}\text{.314}\text{J/K}\cdot \text{mol)}\text{(298}\text{K)}\text{ln(1}\text{.0}{\text{×10}}^{\text{-3}}\text{)}\text{(1}\text{.0}{\text{×10}}^{\text{-4}}\text{)}\\ \text{ΔG=}\text{4}\text{.0}{\text{×10}}^{\text{4}}\text{J/mol}\end{array}$

Interpretation Introduction

Interpretation:

To calculate the free energy $(\text{ΔG})$ values for given aqueous phase equilibrium reaction at ${\text{25}}^{°}\text{C}$.

Concept Information:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$

    $\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.0826}\text{l}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Explanation of Solution

$\begin{array}{l}\text{Let us consider the following reaction (c)}\\ {\text{H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}^{\text{+}}\text{(aq)}\text{+}{\text{OH}}^{\text{-}}\text{(aq)}\\ {\text{ΔG}}^{\text{0}}\text{=}\text{-RTlnQ}\\ {\text{[H}}^{\text{+}}\text{]=}\text{1}{\text{.0×10}}^{\text{-12}}\text{M,}{\text{[OH}}^{\text{-}}\text{]=}\text{2}{\text{.0×10}}^{\text{-8}}\text{M}\\ {\text{ΔG=ΔG}}^{\text{0}}{\text{+RTln([H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])}\\ Givenvalues(R,\text{T})aresubstitutedequation\\ \Delta \text{G}=(8.0\times {10}^{4}J/mol)+(8.314J/\cancel{K}\cdot mol)(298\cancel{\text{K}})ln(\text{1}\text{.0}\times {\text{10}}^{\text{-12}})(2.\text{0}\times {\text{10}}^{\text{-8}})\\ \Delta \text{G}=-3.2\times {10}^{4}J/mol\end{array}$

Interpretation Introduction

Interpretation:

To calculate the free energy $(\text{ΔG})$ values for given aqueous phase equilibrium reaction at ${\text{25}}^{°}\text{C}$.

Concept Information:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$

    $\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.0826}\text{l}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Explanation of Solution

$\begin{array}{l}\text{Let us consider the following reaction (d)}\\ {\text{[H}}^{\text{+}}\text{]=}3.5\text{M},[O{H}^{-}]=4.\text{8}\times {\text{10}}^{\text{-4}}M\\ {\text{ΔG}}^{\text{0}}\text{=}\text{-RTlnQ}\\ {\text{ΔG=ΔG}}^{\text{0}}{\text{+RTln([H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])}\\ \text{Given values are substituted equatuion}\\ \Delta \text{G}=8.0\times {10}^{4}J/mol+(8.314J/\cancel{K}\cdot mol)(298\cancel{\text{K}})ln(3.5)(4.\text{8}\times {\text{10}}^{\text{-4}})\\ \Delta \text{G}=6.4\times {10}^{4}J/mol\end{array}$