Q20.62 P - Chemistry: Molecular Nature Of Matter And Change

Question

As a fuel, ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{( g)}}$ produces only nonpolluting   ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(g)}}$when it burns. Moreover, it combines with  ${\mathbf{\text{O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$in a fuel cell (Chapter 21) to provide electrical energy.

(a) Calculate ${\mathbf{\text{ΔH}}}^{\mathbf{\text{°}}}{\mathbf{\text{, ΔS}}}^{\mathbf{\text{°}}}$ , and ${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}$ per mole of  ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}$at 298 K.

(b) Is the spontaneity of this reaction dependent on T? Explain.

(c) At what temperature does the reaction become spontaneous?

Step-by-Step Solution

Answer

1. The given reaction standard free energy value is$\text{-228.6kJ}$  .
2. This reaction is not dependent on temperature T and enthalpy and entropy values are negative.
3. Given the reaction the temperature value is $\text{7281.45 K}$  and this reaction is non-spontaneous.

Step 1: Definition of Concept

 Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

Step 2: Calculate ΔH ° , ΔS ° , and ΔG ° per mole

(a)

The reaction of ${\text{H}}_{\text{2}}$  with ${\text{O}}_{\text{2}}$  to give ${\text{H}}_{\text{2}}\text{O}$  formation is, 

 ${\text{H}}_{\text{2}}\text{( g)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{( g)}\to {\text{H}}_{\text{2}}\text{O(g)}$

The coefficient are written this way instead of

 ${\text{2H}}_{\text{2}}{\text{( g)+O}}_{\text{2}}\text{( g)}\to {\text{2H}}_{\text{2}}\text{O(g)}$

The specific thermodynamic values (per ${\text{1 molH}}_{\text{2}}$) not per ${\text{2molH}}_{\text{2}}$ .

Standard enthalpy change is

${\text{ΔH}}^{\text{o}}$ Formation of values,

 $\begin{array}{l}{\text{H}}_{\text{2}}\text{(g)=0kJ/mol}\\ {\text{O}}_{\text{2}}\text{(g)=0kJ/mol}\\ {\text{H}}_{\text{2}}\text{O(g)=-241.826 kJ/mol}\end{array}$

The enthalpy change for the reaction is calculated as follows,

 $\begin{array}{l}{\text{ΔH}}_{\text{rxn }}^{\text{°}}\text{=}\sum \text{m}{\text{ΔH}}_{\text{f( Products )}}^{\text{°}}\text{-}\sum \text{n}{\text{ΔH}}^{\text{°}}{\text{f}}_{\text{(reactants )}}\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{1mol}}_{\text{2}}\text{O}\right)\left({{\text{ΔH}}_{\text{f}}^{\text{°}}}_{\text{of}}{\text{H}}_{\text{2}}\text{O}\right)\right]\end{array}$

$\begin{array}{c}\text{-}\left[\left({\text{1molH}}_{\text{2}}\right)\left({{\text{ΔH}}_{\text{f}}^{\text{°}}}_{\text{f}}{\text{ of H}}_{\text{2}}\right)\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\left({\text{ΔH}}_{\text{f}}^{\text{°}}{\text{ of O}}_{\text{2}}\right)\right]\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{1 molH}}_{\text{2}}\text{O}\right)\text{(-241.826 kJ/mol)}\right]\\ \text{-}[\left({\text{1molH}}_{\text{2}}\right)\text{(0 kJ/mol)}\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\text{(0 kJ/mol)}]\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=-241.826 kJ}\end{array}$

 

The enthalpy change is negative. 

Hence, the enthalpy $\left({\text{ΔH}}_{\text{rxn }}^{\text{°}}\right)\text{ changes is -241.826kJ}$.

Entropy change  ${{\text{ΔS}}^{\text{°}}}_{\text{system }}$

Calculate the change in entropy for this reaction as follows,

 $\begin{array}{c}{\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}\sum {\text{mS}}_{\text{Products }}^{\text{0}}\text{-}\sum {\text{nS}}_{\text{reactants }}^{\text{0}}\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{1 molH}}_{\text{2}}\text{O}\right)\left({\text{S}}^{\text{°}}{\text{ of H}}_{\text{2}}\text{O}\right)\right]\\ \text{-}\left[\text{(1 molH2)}\left({\text{S}}^{\text{o}}{\text{ of N}}_{\text{2}}\right)\text{+}\left({\text{1/2 molO}}_{\text{2}}\right)\left({\text{S}}^{\text{o}}{\text{ of O}}_{\text{2}}\right)\right]\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{1 molH}}_{\text{2}}\text{O}\right)\text{(188.72 J/mol×K)}\right]\\ \text{-}\left[\text{(1 molH2)(130.6 J/mol×K)+}\left({\text{1/2 molO}}_{\text{2}}\right)\text{(205.0 J/mol×K)}\right]\\ {{\text{ΔS}}^{\text{°}}}_{\text{rxn}}\text{=-44.38}\\ \text{ =-44.4 J/K}\\ {\text{ΔS}}_{\text{rxn }}^{\text{°}}\text{=-0.0444 kJ/K}\end{array}$

Hence, the  $\left({\text{ΔS}}_{\text{rxn}}^{\text{0}}\right)$ of the reaction is  $\text{-44.4 J/K}$

Calculate the Free energy change  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ is,

Standard Free energy change equation is,

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

Free energy change  ${\text{ΔG}}_{\text{f}}^{\text{o}}$

Enthalpy and entropy values are

 $\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=-241.826}\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-0.0444 kJ/K}\end{array}$

These values are plugging above standard free energy equation,

 $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=(-241.826 kJ)-[(298 K)(-0.04438 kJ/K)]}\\ \text{ =-228.6008 k J/mol}\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-228.6kJ}\end{array}$

Therefore, the given reaction standard free energy value is  $\text{-228.6kJ}$.

Step 3: Explain the spontaneity of this reaction is dependent on T

 (b)

Given reaction,

 ${\text{2H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{2H}}_{\text{2}}\text{O(g)}$

This reaction is does not dependent on temperature T, because enthalpy   and  values are negative.

This reaction will become non-spontaneous at higher temperature because the positive$\text{(-TΔS)}$  term becomes larger than negative  $\text{ΔH}$ term. Further in this reaction.

Step 4: Find at what temperature does the reaction become spontaneous

(c)

Given reaction,

 ${\text{2H}}_{\text{2}}{\text{( g)+O}}_{\text{2}}\text{( g)}\to {\text{2H}}_{\text{2}}\text{O(g)}$

The reaction becomes spontaneous below the temperature where  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=0}$

Consider the following free energy equation,

 $\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}{\text{=0=ΔH}}_{\text{rxn}}^{\text{o}}{\text{-TΔS}}_{\text{rxn}}^{\text{o}}\text{-----[1]}\\ {\text{ΔH}}_{\text{rxn}}^{\text{o}}{\text{=TΔS}}_{\text{rxn}}^{\text{°}}\text{------[2]}\end{array}$

Rearrange equation (2) to calculate temperature T,

 $\text{T=}\frac{{\text{ΔH}}_{\text{rnn}}^{\text{o}}}{{\text{ΔS}}_{\text{ran}}^{\text{o}}}$

Hence,

 $\begin{array}{c}\text{T=}\frac{\text{-241.826 kJ}}{\text{-0.04438 kJ/K}}\\ \text{ =5448.986}\\ {\text{T=5.45×10}}^{\text{3}}\text{K}\end{array}$

At temperature become above  $\text{7281.45 K}$, this reaction is non-spontaneous. Because both enthalpy ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$and entropy  ${\text{ΔS}}_{\text{rxn}}^{\text{o}}$values are negative, the reaction becomes non-spontaneous calculated temperature T.