Chapter 14, Problem 14.23QP

Using data from Appendix 2, calculate ΔS°_rxn and ΔS_surr for each of the reactions in Problem 14.11 and determine if each reaction is spontaneous at 25°C.

Interpretation Introduction

Interpretation:

The standard entropy change of the reaction ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ and entropy change in the surroundings ${\text{(ΔS}}^{\text{°}}{}_{\text{surr}}\text{)}$ has to be calculated for the given set of reactions and the reactions has to be predicted to be spontaneous on non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$  can be calculated by the following equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same. Enthalpy is the amount energy absorbed or released in a process.

 The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

${\text{ΔH}}_{\text{rxn}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}$ is the standard entropy of the products

Standard enthalpy change in a reaction and entropy change in the system are same.

Using the value for the change in enthalpy in a system and the temperature, we can calculate ${\text{ΔS}}_{\text{surr}}$ for an isothermal reaction.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe( ${\text{ΔS}}_{\text{univ}}$). If the ${\text{ΔS}}_{\text{univ}}$ is positive, then the reaction will be spontaneous.  If the ${\text{ΔS}}_{\text{univ}}$ is negative then the reaction will be non-spontaneous.

Answer to Problem 14.23QP

The standard entropy of formation, ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{=-75}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The entropy of surroundings, ${\text{ΔS}}_{\text{surr}}\text{=185}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The given reaction is spontaneous

Explanation of Solution

Given,

$\begin{array}{l}{\text{S°(PCl}}_{\text{5}}\text{)=364}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S°(PCl}}_{\text{3}}{}_{\text{(g)}}\text{)=217}{\text{.1 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S°(Cl}}_{\text{2}}\text{)=223}\text{.0}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(PCl}}_{\text{5}}\text{)=-374}{\text{.9 kJmol}}^{\text{-1}}\\ \\ {\text{ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(PCl}}_{\text{3}}{}_{\text{(g)}}\text{)=-319}\text{.7}{\text{kJmol}}^{\text{-1}}\\ \\ {\text{ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(Cl}}_{\text{2}}\text{)=0}\\ \\ \text{T=298K}\end{array}$

To calculate ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  This is calculated by plugging in the values of standard entropy of the products and reactants in the given equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= S°(PCl}}_{\text{5}}\text{) -}\left[{\text{S°(PCl}}_{\text{3}}{}_{\text{(g)}}{\text{)+S°(Cl}}_{\text{2}}\text{)}\right]\\ \text{= 364}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{-}\left[\text{217}{\text{.1 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+223}\text{.0}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right]\\ \text{= -75}\text{.6}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$

The ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$ is the difference in enthalpy of the reactants and products.  This is calculated by plugging in the values of standard enthalpy of the products and reactants in the given equation.

${\text{ΔH}}_{\text{rxn}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reacttants}}$

$\begin{array}{l}{\text{= ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(PCl}}_{\text{5}}\text{) -}\left[{\text{ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(PCl}}_{\text{3}}{}_{\text{(g)}}{\text{) +ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(Cl}}_{\text{2}}\text{)}\right]\\ \text{= -374}{\text{.9 kJmol}}^{\text{-1}}\text{ -}\left[\text{-319}\text{.7}{\text{kJmol}}^{\text{-1}}\text{+0}\right]\\ \text{= -55}{\text{.2 KJmol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{surr}}$

${\text{ΔS}}_{\text{surr}}$ is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

$\begin{array}{l}\text{= -}\frac{\text{(-55}{\text{.2 kJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= 0}{\text{.185 kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{=185 JK}}^{\text{-1}}\text{mol}\end{array}$

To calculate ${\text{ΔS}}_{\text{surr}}$

${\text{ΔS}}_{\text{univ}}$ is calculated by plugging in the values of ${\text{ΔS}}_{\text{sys}}$ and ${\text{ΔS}}_{\text{surr}}$ in the given equation.

${\text{ΔS}}_{\text{univ}}{\text{=ΔS}}_{\text{sys}}{\text{+ΔS}}_{\text{surr}}$

$\begin{array}{l}\text{= -75}{\text{.6 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{+185 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{= 109}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since , ${\text{ΔS}}_{\text{univ}}\text{> 0}$ the given process is spontaneous

Interpretation Introduction

Interpretation:

The standard entropy change of the reaction ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ and entropy change in the surroundings ${\text{(ΔS}}^{\text{°}}{}_{\text{surr}}\text{)}$ has to be calculated for the given set of reactions and the reactions has to be predicted to be spontaneous on non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$  can be calculated by the following equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same. Enthalpy is the amount energy absorbed or released in a process.

 The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

${\text{ΔH}}_{\text{rxn}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}$ is the standard entropy of the products

Standard enthalpy change in a reaction and entropy change in the system are same.

Using the value for the change in enthalpy in a system and the temperature, we can calculate ${\text{ΔS}}_{\text{surr}}$ for an isothermal reaction.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe( ${\text{ΔS}}_{\text{univ}}$). If the ${\text{ΔS}}_{\text{univ}}$ is positive, then the reaction will be spontaneous.  If the ${\text{ΔS}}_{\text{univ}}$ is negative then the reaction will be non-spontaneous.

Answer to Problem 14.23QP

The standard entropy of formation, ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{= 215}{\text{.8 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The entropy of surroundings, ${\text{ΔS}}_{\text{surr}}{\text{=-609 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The given reaction is non-spontaneous

Explanation of Solution

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}\left[\text{Hg}\right]\text{=77}\text{.4}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ \text{S°}\left[{\text{O}}_{\text{2}}\right]\text{=205}{\text{.0 JK}}^{\text{-1}}{\text{mol}}^{-1}\\ \\ \text{S°(HgO)=72}\text{.0}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}_{\text{f}}{}^{\text{°}}\text{(Hg)=0}\\ \\ {\text{ΔΗ}}_{\text{f}}{}^{\text{°}}\left[{\text{O}}_{\text{2}}\right]\text{=0}\\ \\ {\text{ΔΗ}}_{\text{f}}{}^{\text{°}}\text{(HgO)=-90}{\text{.7kJmol}}^{\text{-1}}\\ \\ \text{T=298K}\end{array}$

To calculate ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$

The ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$ is the difference in entropy of the reactants and products.  This is calculated by plugging in the values of standard entropy of the products and reactants in the given equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{products}}{\text{- S}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= 2S}}^{\text{°}}\left[\text{Hg}\right]\text{+ S°}\left[{\text{O}}_{\text{2}}\right]\text{-2S°(HgO)}\\ \text{= (2)(77}\text{.4}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+205}{\text{.0 JK}}^{\text{-1}}\text{mol-}\left[\text{(2)(72}\text{.0}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right]\\ \text{= 215}{\text{.8JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$

The ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$ is the difference in enthalpy of the reactants and products.  This is calculated by plugging in the values of standard enthalpy of the products and reactants in the given equation.

${\text{ΔH}}_{\text{rxn}}{\text{ = ΔH}}^{\text{°}}{}_{\text{products}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= 2ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(Hg)+ ΔΗ}}_{\text{f}}{}^{\text{°}}\left[{\text{O}}_{\text{2}}\right]{\text{-2ΔΗ}}_{\text{f}}{}^{\text{°}}\text{(HgO)}\\ \text{= 0+0 - }\left[\text{2(-90}{\text{.7kJmol}}^{\text{-1}}\text{)}\right]\\ \text{=181}\text{.4}{\text{kJmol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{surr}}$

${\text{ΔS}}_{\text{surr}}$ is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

$\begin{array}{l}\text{= -}\frac{\text{(181}{\text{.4 kJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= -0}{\text{.609 kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ ={\text{-609 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate $\Delta S_{\text{univ}}$

${\text{ΔS}}_{\text{univ}}$ is calculated by plugging in the values of $\Delta S_{\text{sys}}$ and $\Delta S_{\text{surr}}$ in the given equation.

$\Delta S_{\text{univ}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}$

$\begin{array}{l}\text{= 215}{\text{.8 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+(-609}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})\\ \text{= -393}{\text{.2 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since, ${\text{ΔS}}_{\text{univ}}\text{< 0}$ the given process is spontaneous.

Interpretation Introduction

Interpretation:

The standard entropy change of the reaction ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ and entropy change in the surroundings ${\text{(ΔS}}^{\text{°}}{}_{\text{surr}}\text{)}$ has to be calculated for the given set of reactions and the reactions has to be predicted to be spontaneous on non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$  can be calculated by the following equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same. Enthalpy is the amount energy absorbed or released in a process.

 The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

${\text{ΔH}}_{\text{rxn}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}$ is the standard entropy of the products

Standard enthalpy change in a reaction and entropy change in the system are same.

Using the value for the change in enthalpy in a system and the temperature, we can calculate ${\text{ΔS}}_{\text{surr}}$ for an isothermal reaction.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe( ${\text{ΔS}}_{\text{univ}}$). If the ${\text{ΔS}}_{\text{univ}}$ is positive, then the reaction will be spontaneous.  If the ${\text{ΔS}}_{\text{univ}}$ is negative then the reaction will be non-spontaneous.

Answer to Problem 14.23QP

The standard entropy of formation, ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{=98}\text{.2}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The entropy of surroundings, ${\text{ΔS}}_{\text{surr}}\text{=-1}\text{.46}\times {\text{10}}^{23}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The given reaction is nonspontaneous

Explanation of Solution

To record the given data

$\begin{array}{l}{\text{ΔΗ}}^{\text{°}}{}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{(g)}\right]\text{=436}{\text{.4kJmol}}^{\text{-1}}\\ \\ \text{T=298K}\end{array}$

To calculate ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$

Explanation:

The ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$ is the difference in entropy of the reactants and products.  This is calculated by plugging in the values of standard entropy of the products and reactants in the given equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{products}}{\text{- S}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= 2S}}^{\text{°}}\left[\text{H(g)}\right]\text{-S°}\left[{\text{H}}_2\text{(g)}\right]\\ \text{=(2)(114}\text{.6}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{) - (1)(131}{\text{.0JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\\ \text{=98}\text{.2}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$

The ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$ is the difference in enthalpy of the reactants and products.  This is calculated by plugging in the values of standard enthalpy of the products and reactants in the given equation.

${\text{ΔH}}_{\text{rxn}}{\text{ = ΔH}}^{\text{°}}{}_{\text{products}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= 2ΔΗ}}^{\text{°}}{}_{\text{f}}\left[\text{H(g)}\right]{\text{-ΔΗ}}^{\text{°}}{}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{(g)}\right]\\ \text{=(2)(436}\text{.4}{\text{kJmol}}^{\text{-1}}\text{) - }\left[\text{(1)(436}{\text{.4kJmol}}^{\text{-1}}\text{)}\right]\\ \text{=436}\text{.6}{\text{kJmol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{surr}}$

${\text{ΔS}}_{\text{surr}}$ is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

$\begin{array}{l}\text{= -}\frac{\text{(436}{\text{.4 kJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= -1}{\text{.46 kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =\text{-1}\text{.46}\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate $\Delta S_{\text{univ}}$

${\text{ΔS}}_{\text{univ}}$ is calculated by plugging in the values of $\Delta S_{\text{sys}}$ and $\Delta S_{\text{surr}}$ in the given equation.

$\Delta S_{\text{univ}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}$

$\begin{array}{l}\text{= 98}{\text{.2 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+(-1}\text{.46}\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})\\ \text{= -1}\text{.36}\times {\text{10}}^3\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since, $\Delta S_{\text{univ}}<0$ the given process is spontaneous.

Interpretation Introduction

Interpretation:

The standard entropy change of the reaction ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ and entropy change in the surroundings ${\text{(ΔS}}^{\text{°}}{}_{\text{surr}}\text{)}$ has to be calculated for the given set of reactions and the reactions has to be predicted to be spontaneous on non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$  can be calculated by the following equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same. Enthalpy is the amount energy absorbed or released in a process.

 The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

${\text{ΔH}}_{\text{rxn}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}$ is the standard entropy of the products

Standard enthalpy change in a reaction and entropy change in the system are same.

Using the value for the change in enthalpy in a system and the temperature, we can calculate ${\text{ΔS}}_{\text{surr}}$ for an isothermal reaction.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe( ${\text{ΔS}}_{\text{univ}}$). If the ${\text{ΔS}}_{\text{univ}}$ is positive, then the reaction will be spontaneous.  If the ${\text{ΔS}}_{\text{univ}}$ is negative then the reaction will be non-spontaneous.

Answer to Problem 14.23QP

The standard entropy of formation, ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{=-282 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The entropy of surroundings, ${\text{ΔS}}_{\text{surr}}\text{=7}\text{.20}\times {\text{10}}^{23}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The given reaction is spontaneous

Explanation of Solution

To record the given data

$\begin{array}{l}{\text{S}}^{\text{°}}{\text{(UF}}_{\text{6}}\text{)=378}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ \text{S°(U)=50}{\text{.21JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S°(F}}_{\text{2}}\text{)=203}{\text{.34JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(UF}}_{\text{6}}{\text{)=-2147 kJmol}}^{\text{-1}}\\ \\ {\text{ΔΗ}}_{\text{f}}{}^{\text{°}}\text{(U)=0}\\ \\ {\text{ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(F}}_{\text{2}}\text{)=0}\\ \\ \text{T=298K}\end{array}$

To calculate ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$

The ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$ is the difference in entropy of the reactants and products.  This is calculated by plugging in the values of standard entropy of the products and reactants in the given equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{produdcts}}{\text{- S}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= 2S}}^{\text{°}}{\text{(UF}}_6\text{)-}\left[S°(U)+3S°(F_2)\right]\\ \text{= 378}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{-}\left[50.21{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}+\text{(3)(203}{\text{.34JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ \text{=-282}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$

The ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$ is the difference in enthalpy of the reactants and products.  This is calculated by plugging in the values of standard enthalpy of the products and reactants in the given equation.

${\text{ΔH}}_{\text{rxn}}{\text{ = ΔH}}^{\text{°}}{}_{\text{products}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(UF}}_{\text{6}}{\text{)-[ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(U)+3ΔΗ}}_{\text{f}}{}^{\text{°}}{\text{(F}}_{\text{2}}\text{)}\\ {\text{= -2147 kJmol}}^{\text{-1}}\text{ - (0+0)}\\ {\text{= -2147 kJmol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{surr}}$

${\text{ΔS}}_{surr}$ is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

$\begin{array}{l}\text{= -}\frac{{\text{(-2147 kJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= 7}{\text{.20 kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =\text{7}\text{.20}\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{univ}}$

${\text{ΔS}}_{\text{univ}}$ is calculated by plugging in the values of ${\text{ΔS}}_{\text{sys}}$ and ${\text{ΔS}}_{\text{surr}}$ in the given equation.

${\text{ΔS}}_{\text{univ}}{\text{= ΔS}}_{\text{sys}}{\text{+ΔS}}_{\text{surr}}$

$\begin{array}{l}\text{= -282}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+7}\text{.20}\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{= }6.92\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since, ${\text{ΔS}}_{\text{univ}}\text{> 0}$ the given process is spontaneous.