Chapter 14, Problem 14.22QP

Using data from Appendix 2, calculate ΔS°_rxn and ΔS_surr for each of the reactions m Problem 14.10 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 14.22QP

The standard entropy of formation , ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{=188}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The entropy of surroundings, ${\text{ΔS}}_{\text{surr}}{\text{=-284JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The given reaction is non-spontaneous

Explanation of Solution

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}{\text{(KClO}}_{\text{3}}\text{)=142}\text{.9}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S}}^{\text{°}}{\text{(O}}_{\text{2}}\text{)=}\text{205}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S°(KClO}}_{\text{4}}\text{)=151}{\text{.0 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{Η}}^{\text{°}}{\text{(KClO}}_{\text{3}}\text{) = -391}{\text{.20 KJmol}}^{\text{-1}}\\ \\ {\text{Η}}^{\text{°}}{\text{(O}}_{\text{2}}\text{)=0}\\ \\ {\text{Η°(KClO}}_{\text{4}}\text{)=-433}{\text{.46 KJmol}}^{\text{-1}}\\ \\ \text{T=298K}\end{array}$

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 the given equation.

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}$ $\begin{array}{l}{\text{= 2S}}^{\text{°}}{\text{(KClO}}_{\text{3}}{\text{)+S}}^{\text{°}}{\text{(O}}_{\text{2}}{\text{)- 2S°(KClO}}_{\text{4}}\text{)}\\ \text{=(2)(142}\text{.9}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(1)(205}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(2)(151}{\text{.0 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ \text{=188}\text{.9}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}{\text{= 2Η}}^{\text{°}}{\text{(KClO}}_{\text{3}}{\text{) + Η}}^{\text{°}}{\text{(O}}_{\text{2}}{\text{)- 2Η°(KClO}}_{\text{4}}\text{)}\\ \text{=(2)(-391}{\text{.20 kJmol}}^{\text{-1}}\text{)+ 0 -}\left[\text{(2)(-433}{\text{.46 kJmol}}^{\text{-1}}\text{)}\right]\\ \text{= -84}{\text{.52 KJmol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}\text{= -}\frac{\text{(-84}{\text{.52 kJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= - 0}{\text{.284 kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ ={\text{- 284 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}\text{= 188}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{+(-284 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})\\ {\text{= -95 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since , ${\text{ΔS}}_{\text{univ}}\text{< 0}$ the given process is non-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 14.22QP

The standard entropy of formation , ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{=-118}{\text{.8 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The entropy of surroundings, ${\text{ΔS}}_{\text{surr}}{\text{=148 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The given reaction is spontaneous

Explanation of Solution

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}\left[{\text{H}}_{\text{2}}\text{O(l)}\right]\text{=69}\text{.9}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ \text{S°}\left[{\text{H}}_{\text{2}}\text{O(l)}\right]\text{=188}\text{.7}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}_{\text{f}}{}^{\text{°}}\left[{\text{H}}_{\text{2}}\text{O(l)}\right]\text{=-285}\text{.8}{\text{kJmol}}^{\text{-1}}\\ \\ {\text{ΔΗ}}_{\text{f}}{}^{\text{°}}\left[{\text{H}}_{\text{2}}\text{O(g)}\right]\text{=-241}{\text{.8 kJmol}}^{\text{-1}}\\ \\ \text{T = 298K}\end{array}$

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 the given equation.

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

$\begin{array}{l}{\text{= S}}^{\text{°}}\left[{\text{H}}_{\text{2}}\text{O(l)}\right]\text{- S°}\left[{\text{H}}_{\text{2}}\text{O(g)}\right]\\ \text{=69}\text{.9}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{- 188}\text{.7}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{=-118}\text{.8}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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 the given equation.

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

$\begin{array}{l}{\text{= ΔΗ}}_{\text{f}}{}^{\text{°}}\left[{\text{H}}_{\text{2}}\text{O(l)}\right]{\text{- ΔΗ}}_{\text{f}}{}^{\text{°}}\left[{\text{H}}_{\text{2}}\text{O(g)}\right]\\ \text{=-285}\text{.8}{\text{KJmol}}^{\text{-1}}\text{- (-241}{\text{.8 KJmol}}^{\text{-1}}\text{)}\\ \text{=- 44}\text{.0}{\text{KJmol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}\text{= -}\frac{\text{(-44}{\text{.0 kJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= 0}{\text{.148 kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ ={\text{148 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}\text{= 118}{\text{.8 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{+148 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{= 29 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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 14.22QP

The standard entropy of formation , ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{=-11}{\text{.4 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

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

The given reaction is spontaneous

Explanation of Solution

To record the given data

$\begin{array}{l}{\text{S}}^{\text{°}}\left[{\text{Na}}^{\text{+}}\text{(aq)}\right]\text{=60}\text{.25}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ \text{2S°}\left[{\text{OH}}^{\text{-}}\text{(aq)}\right]{\text{=-105JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S°(H}}_{\text{2}}\text{)=131}\text{.0}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{2S°(Na}}^{\text{+}}\text{)=51}{\text{.05 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{2S°(H}}_{\text{2}}\text{O(l))=69}\text{.9}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}^{\text{°}}{}_{\text{f}}\left[{\text{Na}}^{\text{+}}\text{(aq)}\right]=\text{-239}\text{.66}{\text{KJmol}}^{\text{-1}}\\ \\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}\left[{\text{OH}}^{\text{-}}\text{(aq)}\right]={\text{-229KJmol}}^{\text{-1}}\\ \\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(H}}_{\text{2}}\text{)}=0\\ \\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(H}}_{\text{2}}\text{O(l)}=\text{-229}{\text{.94KJmol}}^{\text{-1}}\end{array}$

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

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

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

$\begin{array}{l}{\text{= 2S}}^{\text{°}}\left[{\text{Na}}^{\text{+}}\text{(aq)}\right]\text{+2S°}\left[{\text{OH}}^{\text{-}}\text{(aq)}\right]{\text{+S°(H}}_{\text{2}}\text{)-}\left[{\text{2S°(Na}}^{\text{+}}{\text{)+2S°(H}}_{\text{2}}\text{O(l))}\right]\\ \text{=(2)(60}\text{.25}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{) + (2)(-105JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+131}\text{.0}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{-}\left[\text{(2)(51}{\text{.05 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(2)(69}\text{.9}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ \text{=-11}\text{.4}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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 the given equation.

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

$\begin{array}{l}{\text{= 2ΔΗ}}^{\text{°}}{}_{\text{f}}\left[{\text{Na}}^{\text{+}}\text{(aq)}\right]{\text{+2ΔΗ}}^{\text{°}}{}_{\text{f}}\left[{\text{OH}}^{\text{-}}\text{(aq)}\right]{\text{+ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(H}}_{\text{2}}\text{)-}\left[{\text{2ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(Na}}^{\text{+}}{\text{)+2ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(H}}_{\text{2}}\text{O(l))}\right]\\ \text{=(2)(-239}\text{.66}{\text{KJmol}}^{\text{-1}}{\text{) + (2)(-229KJmol}}^{\text{-1}}\text{)+0-}\left[\text{0+(2)(-229}{\text{.94KJmol}}^{\text{-1}}\text{)}\right]\\ \text{=-367}\text{.6}{\text{KJmol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}\text{= -}\frac{\text{(-376}{\text{.6 kJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= 1}{\text{.23 kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =\text{1}\text{.23}\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}\text{= -11}{\text{.4 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+1}\text{.23}\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{= 1}\text{.22}\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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 14.22QP

The standard entropy of formation , ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{=115}{\text{.1 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The entropy of surroundings, ${\text{ΔS}}_{\text{surr}}\text{=-3}\text{.16}\times {\text{10}}^{23}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The given reaction is non-spontaneous

Explanation of Solution

To record the given data

$\begin{array}{l}{\text{S}}^{\text{°}}\left[\text{N(g)}\right]=\text{153}\text{.3}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ \text{S°}\left[{\text{N}}_{\text{2}}\left(\text{g}\right)\right]=\text{191}{\text{.5JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\Delta {{\rm H}}_{\text{f}}{}^{\text{°}}\left[{\text{N}}_{\text{2}}\left(\text{g}\right)\right]=\text{941}{\text{.4 KJmol}}^{\text{-1}}$

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 the given equation.

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

$\begin{array}{l}{\text{= 2S}}^{\text{°}}\left[\text{N(g)}\right]\text{-S°}\left[{\text{N}}_{\text{2}}\left(\text{g}\right)\right]\\ \text{=(2)(153}\text{.3}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{) - (1)(191}{\text{.5JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\\ \text{=115}\text{.1}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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 the given equation.

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

$\begin{array}{l}{\text{= (2)ΔΗ}}_{\text{f}}{}^{\text{°}}\left[\text{N(g)}\right]{\text{-ΔΗ}}_{\text{f}}{}^{\text{°}}\left[{\text{N}}_{\text{2}}\left(\text{g}\right)\right]\\ \text{= (2)(941}{\text{.4 KJmol}}^{\text{-1}}\text{) - (1)(941}{\text{.4KJmol}}^{\text{-1}}\text{)}\\ \text{= 941}{\text{.4KJmol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{surr}$

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

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

$\begin{array}{l}\text{= -}\frac{\text{(941}{\text{.4 kJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= -3}{\text{.16 kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =\text{-3}\text{.16}\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}\text{= 115}{\text{.1JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+(-3}\text{.16}\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})\\ \text{= -}3.04\times {\text{10}}^3{\text{ JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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