Chapter 18, Problem 18.92QP

Tungsten is usually produced by the reduction of WO₃ with hydrogen:

${\text{WO}}_3(s)+3{\text{H}}_2(g)\stackrel{}{\to }\text{W}(s)+3{\text{H}}_2\text{O}(g)$

Consider the following data:

	WO3(s)		H2O(g)
$\Delta {\text{H}}_f^{°}(\text{kJ/mol})$	−839.9	−241.8
$\Delta {\text{G}}_f^{°}(\text{kJ/mol})$	−763–1	−228.6

1. a It K > 1 or < 1 at 25°C? Explain your answer.
2. b What is the value of ΔS° at 25°C?
3. c What is the temperature at which ΔG° equals zero for this reaction at 1 atm pressure?
4. d What is the driving force of this reaction?

Interpretation Introduction

Interpretation:

For the given reaction, the value of $\text{K,}{\text{ΔS}}^{\text{o}}\text{,}$ the temperature at which the free energy changes is equal to zero and the driving force for the reaction has to be given.

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.

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

Spontaneous process:

The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.

Answer to Problem 18.92QP

For the given reaction the sign of free energy change $\left({\text{ΔG}}^{\text{o}}\right)$ is positive.  Hence, $\text{K}$ will be less than 1.

Explanation of Solution

To calculate: The value of $\text{K}$

Given reaction and information,

$\begin{array}{l}{\text{WO}}_{\text{3}}{}_{\text{(s)}}\text{+ }{\text{3H}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ W}}_{\left(\text{s}\right)}\text{+}{\text{3H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{:}\text{-839}\text{.9}\text{3}\text{×}\text{-241}\text{.8}\text{kJ}\\ {\text{ΔG}}^{\text{o}}{}_{\text{f}}\text{:}\text{-763}\text{.1}\text{3}\text{×}\text{-228}\text{.6}\text{kJ}\\ {\text{Temperature is 25 }}^{\text{o}}\text{C}\text{.}\end{array}$

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

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔG}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔG}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\left(\text{3}\text{×}\text{-228}\text{.6}\right)\text{-}\left(\text{-763}\text{.1}\right)\right]\text{kJ}\\ \text{=}\text{77}\text{.3}\text{kJ}\end{array}$

The sign of free energy change $\left({\text{ΔG}}^{\text{o}}\right)$ is positive $\left(\text{ΔG}\text{>}\text{0}\right)$.  Hence, $\text{K}$ will be less than 1.

Interpretation Introduction

Interpretation:

For the given reaction, the value of $\text{K,}{\text{ΔS}}^{\text{o}}\text{,}$ the temperature at which the free energy changes is equal to zero and the driving force for the reaction has to be given.

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.

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

Spontaneous process:

The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.

Answer to Problem 18.92QP

The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ at $\text{25}{}^{\text{o}}\text{C}$ is $\text{125}\text{J/K}\text{.}$

Explanation of Solution

To calculate: The value of ${\text{ΔS}}^{\text{o}}$

Calculate the value of ${\text{ΔH}}^{\text{o}}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\left(\text{3}\text{×}\text{-241}\text{.8}\right)\text{-}\left(\text{-839}\text{.9}\right)\right]\text{kJ}\\ \text{=}\text{114}\text{.5}\text{kJ}\text{.}\end{array}$

Calculate the value of ${\text{ΔS}}^{\text{o}}$ by using the relationship between ${\text{ΔG}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{77}\text{.3}\text{kJ}\text{=}\text{114}\text{.5}\text{kJ}\text{-}\left(\text{298}\text{K}\right){\text{ΔS}}^{\text{o}}\\ {\text{ΔS}}^{\text{o}}\text{=}\text{-}\frac{\text{77}\text{.3}\text{kJ}\text{-}\text{114}\text{.5}\text{kJ}}{\text{298}\text{K}}\text{=}\text{0}\text{.1248}\text{kJ/K}\\ \text{=}\text{125}\text{J/K}\text{.}\end{array}$

The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ is $\text{125}\text{J/K}\text{.}$

Interpretation Introduction

Interpretation:

For the given reaction, the value of $\text{K,}{\text{ΔS}}^{\text{o}}\text{,}$ the temperature at which the free energy changes is equal to zero and the driving force for the reaction has to be given.

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.

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

Spontaneous process:

The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.

Answer to Problem 18.92QP

At temperature $\left(\text{T}\right)$ $\text{916}\text{K}$ the value of free energy change equals to zero.

Explanation of Solution

To calculate: At which temperature the free energy change equals to zero

Consider free energy change $\left({\text{ΔG}}^{\text{o}}\right)$ is zero.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{0}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{Substitute}\text{the}\text{values}\text{of}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}\\ \text{T}\text{=}\frac{\text{ΔH}}{\text{ΔS}}\text{=}\frac{\text{114}\text{.5}\text{×}{\text{10}}^{\text{3}}\text{J}}{\text{125}\text{J/K}}\\ \text{=}\text{916}\text{K}\text{.}\end{array}$

The value of $\text{T}$ is $\text{916}\text{K}\text{.}$

Interpretation Introduction

Interpretation:

For the given reaction, the value of $\text{K,}{\text{ΔS}}^{\text{o}}\text{,}$ the temperature at which the free energy changes is equal to zero and the driving force for the reaction has to be given.

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.

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

Spontaneous process:

The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.

Answer to Problem 18.92QP

The entropy change is the driving force.  The value of $\text{TΔS}$ is very important at higher temperatures.

Explanation of Solution

To identify: The driving force for the given reaction

The entropy change is the driving force.

The value of $\text{TΔS}$ is very important at higher temperatures.