Chapter 18, Problem 18.104QP

(a)

Interpretation Introduction

Interpretation:

For the formation of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\text{,}$ the value of ${\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{,}{\text{ΔS}}^{\text{o}}{}_{\text{f}}\text{,}{\text{ΔG}}^{\text{o}}{}_{\text{f}}$ have to be calculated and also the result obtained by the student has to be checked.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Answer to Problem 18.104QP

The value of standard enthalpy change of formation $\left({\text{ΔH}}^{\text{o}}{}_{\text{f}}\right)$ of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ is $\text{-241}\text{.8}\text{kJ/mol}\text{.}$

Explanation of Solution

To calculate: The value of ${\text{ΔH}}^{\text{o}}{}_{\text{f}}$

The value of standard enthalpy change of formation $\left({\text{ΔH}}^{\text{o}}{}_{\text{f}}\right)$ of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ is $\text{-241}\text{.8}\text{kJ/mol}\text{.}$

(b)

Interpretation Introduction

Interpretation:

For the formation of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\text{,}$ the value of ${\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{,}{\text{ΔS}}^{\text{o}}{}_{\text{f}}\text{,}{\text{ΔG}}^{\text{o}}{}_{\text{f}}$ have to be calculated and also the result obtained by the student has to be checked.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Answer to Problem 18.104QP

The value of standard entropy of formation of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ is $\text{188}\text{.7}\text{J/mol}\cdot \text{K}\text{.}$

Explanation of Solution

To calculate: The value of standard entropy of formation of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$

The value of standard entropy of formation of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ is $\text{188}\text{.7}\text{J/mol}\cdot \text{K}\text{.}$

(c)

Interpretation Introduction

Interpretation:

For the formation of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\text{,}$ the value of ${\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{,}{\text{ΔS}}^{\text{o}}{}_{\text{f}}\text{,}{\text{ΔG}}^{\text{o}}{}_{\text{f}}$ have to be calculated and also the result obtained by the student has to be checked.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Answer to Problem 18.104QP

The calculated value of ${\text{ΔG}}^{\text{o}}{}_{\text{f}}$ for ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ is the same as the value given in Appendix C.

Explanation of Solution

To compare: The calculated value of ${\text{ΔG}}^{\text{o}}{}_{\text{f}}$ for ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ with the value given in Appendix C

Given reaction and information

$\begin{array}{l}{\text{H}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{1/2O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{:}\text{0}\text{0}\text{-241}\text{.8}\text{kJ}\\ {\text{S}}^{\text{o}}\text{:}\text{130}\text{.6}\text{1/2}\text{×}\text{205}\text{188}\text{.7}\text{J/K}\\ {\text{Temperature is 25 }}^{\text{o}}\text{C}\text{.}\end{array}$

Calculate the change in enthalpy and entropy

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{f}}^{\text{o}}\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{g}}\right]\\ \text{=}\text{-241}\text{.8}\text{kJ}\text{.}\\ {\text{ΔS}}^{\text{o}}\text{=}\left[\left(\text{188}\text{.7}\right)\text{-}\left(\text{130}\text{.6}\right)\text{-}\left(\text{1/2}\text{×}\text{205}\right)\right]\text{J/K}\\ \text{=}\text{-0}\text{.0444}\text{kJ/K}\end{array}$

Calculate the value of ${\text{ΔG}}_{\text{f}}^{\text{o}}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{=}\text{-241}\text{kJ}\text{-}\left(\text{298}\text{.15}\text{K}\right)\left(\text{-0}\text{.0444}\text{kJ/K}\right)\\ \text{=}\left(\text{-241}\text{+}\text{13}\text{.2378}\right)\text{kJ}\\ \text{=}\text{-228}\text{.5621}\text{kJ/mol}\\ \text{=}\text{-228}\text{.6}\text{kJ/mol}\text{.}\end{array}$

The calculated value of ${\text{ΔG}}^{\text{o}}{}_{\text{f}}$ for ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ is the same as the value given in Appendix C

(d)

Interpretation Introduction

Interpretation:

For the formation of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\text{,}$ the value of ${\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{,}{\text{ΔS}}^{\text{o}}{}_{\text{f}}\text{,}{\text{ΔG}}^{\text{o}}{}_{\text{f}}$ have to be calculated and also the result obtained by the student has to be checked.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Answer to Problem 18.104QP

The result obtained by the student is not correct because the value of ${\text{ΔS}}^{\text{o}}$ is not equal to ${\text{S}}^{\text{o}}$ for ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\text{.}$

Explanation of Solution

To check: The result obtained by the student

The correct equation is given by

${\text{ΔG}}_{\text{f}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}$

The equation used by the student is given below

${\text{ΔG}}_{\text{f}}^{\text{o}}\text{=}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{-}{\text{TS}}^{\text{o}}$

The result obtained by the student is not correct because the value of ${\text{ΔS}}^{\text{o}}$ is not equal to ${\text{S}}^{\text{o}}$ for ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\text{.}$