Chapter 14, Problem 14.20QP

Calculate ΔS_surr for each of the reactions in Problem 14.14 and determine if each reaction is spontaneous at 25°C.

Interpretation Introduction

Interpretation: The entropy change of surroundings $(\Delta S_{surr})$ in the given set of reactions has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in system ${\text{ΔS}}_{sys}$ is the difference in entropy of the products and reactants.  ${\text{ΔS}}_{sys}$ can be calculated by the following equation.

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

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

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.

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

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

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

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

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

Answer to Problem 14.20QP

The given reaction is found to be a spontaneous process with   $\Delta {\text{S}}_{surr}\text{=9}\text{.95}\times {\text{10}}^2{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

To record the given data

Given,

${\text{S}}^{\text{°}}{\text{(SO}}_{\text{2}}\text{)}=\text{248}{\text{.5JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

${\text{S}}^{\text{°}}\text{(S)}=\text{31}{\text{.88 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

${\text{S}}^{\text{°}}\text{(}S{O}_2\text{) = 205}{\text{.0J K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

$\begin{array}{l}\Delta {{\rm H}}^{°}{}_f{\text{(SO}}_{\text{2}}\text{)}=\text{-296}{\text{.4KJmol}}^{\text{-1}}\\ \Delta {{\rm H}}^{°}{}_f\text{(}{O}_2\text{)}=0\\ \Delta {{\rm H}}^{°}{}_f\text{(S)}=0\\ \text{T = 298K}\end{array}$

To calculate ${\text{ΔH}}_{system}$

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

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

$\begin{array}{l}\text{= }\Delta {{\rm H}}^{°}{}_f{\text{(SO}}_{\text{2}}\text{)- }\left[\Delta {{\rm H}}^{°}{}_f\text{(}{O}_2\text{)+}\Delta {{\rm H}}^{°}{}_f\text{(S)}\right]\\ =\text{(1)(-296}{\text{.4KJmol}}^{\text{-1}}\text{)-(0+0)}\\ \text{= -296}{\text{.4KJmol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}\text{=-}\frac{\text{(-296}{\text{.4KJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{=0}{\text{.995KJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{sys}$

The ${\text{ΔS}}_{sys}$ 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.

${\text{ΔS}}_{sys}{\text{= S}}^{\text{°}}{}_{\text{produdcts}}{\text{- S}}^{\text{°}}{}_{\text{reacttants}}$

$\begin{array}{l}{\text{= S}}^{\text{°}}{\text{(SO}}_{\text{2}}\text{)- }\left[{\text{S}}^{\text{°}}\text{(}S{O}_2{\text{)+S}}^{\text{°}}\text{(S)}\right]\\ =\text{(1)(248}{\text{.5JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(1)(205}{\text{.0J K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(1)(31}{\text{.88 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ =11.6{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate $\Delta S_{univ}$

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

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

$\begin{array}{l}\text{=11}{\text{.6JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+9}{\text{.95×10}}^{\text{-2}}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{=1}{\text{.007×10}}^{\text{-3}}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$ Since , $\Delta S_{univ}>0$ the given process is spontaneous

Conclusion

The entropy change of surroundings $(\Delta S_{surr})$ in the given reaction has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Interpretation Introduction

Interpretation: The entropy change of surroundings $(\Delta S_{surr})$ in the given set of reactions has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in system ${\text{ΔS}}_{sys}$ is the difference in entropy of the products and reactants.  ${\text{ΔS}}_{sys}$ Can be calculated by the following equation.

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

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

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.

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

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

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

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

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

Answer to Problem 14.20QP

The given reaction is found to be a non-spontaneous process with    ${\text{ΔS}}_{\text{surr}}{\text{=-395JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

To record the given data

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}\text{(MgO)}=\text{26}{\text{.78 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}{\text{(CO}}_{\text{2}}\text{)}=\text{213}{\text{.6JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}{\text{(MgCO}}_{\text{3}}\text{) =65}{\text{.69 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}\Delta {{\rm H}}^{°}{}_f\text{(MgO) = -601}{\text{.8 KJmol}}^{\text{-1}}\\ \Delta {{\rm H}}^{°}{}_f{\text{(CO}}_2\text{) = -393}{\text{.5 KJmol}}^{\text{-1}}\\ \Delta {{\rm H}}^{°}{}_f\text{(MgC}{O}_3\text{)}=\text{-1112}{\text{.9KJmol}}^{\text{-1}}\\ \text{T=298K}\end{array}$

To calculate ${\text{ΔH}}_{system}$

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

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

$\begin{array}{l}\text{= }\Delta {{\rm H}}^{°}{}_f\text{(MgO)+}\Delta {{\rm H}}^{°}{}_f{\text{(CO}}_2\text{)- }\left[\Delta {{\rm H}}^{°}{}_f\text{(MgC}{O}_3\text{)}\right]\\ =\text{(1)(-601}{\text{.8 KJmol}}^{\text{-1}}\text{)+(1)(-393}{\text{.5 KJmol}}^{\text{-1}}\text{)-(1)(1112}{\text{.9KJmol}}^{\text{-1}})\\ \text{= 117}{\text{.6 KJmol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}\text{=-}\frac{\text{(117}{\text{.6KJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{=-0}{\text{.395 KJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{sys}$

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

${\text{ΔS}}^{}{}_{sys}{\text{ = S}}^{\text{°}}{}_{\text{produdcts}}{\text{- S}}^{\text{°}}{}_{\text{reacttants}}$

$\begin{array}{l}\text{= }\left[{\text{S}}^{\text{°}}{\text{(MgO)+S}}^{\text{°}}{\text{(CO}}_{\text{2}}\text{)}\right]{\text{-S}}^{\text{°}}{\text{(MgCO}}_{\text{3}}\text{)}\\ \text{= (1)(26}{\text{.78 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(1)(213}{\text{.6JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(1)(65}{\text{.69 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ \text{= 174}{\text{.7JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate $\Delta S_{univ}$

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

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

$\begin{array}{l}=\text{174}{\text{.7JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{+(-395JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\\ {\text{= -220JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

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

Conclusion

The entropy change of surroundings $(\Delta S_{surr})$ in the given reaction has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Interpretation Introduction

Interpretation: The entropy change of surroundings $(\Delta S_{surr})$ in the given set of reactions has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in system ${\text{ΔS}}_{sys}$ is the difference in entropy of the products and reactants.  ${\text{ΔS}}_{sys}$ can be calculated by the following equation.

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

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

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.

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

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

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

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

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

Answer to Problem 14.20QP

The given reaction is found to be a spontaneous process with   $\Delta {\text{S}}_{surr}\text{=9}\text{.9}\times {\text{10}}^3{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

To record the given data

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}{\text{(CO}}_{\text{2}}\text{)}=\text{213}{\text{.6 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{S}°{\text{(H}}_2\text{O(}l\text{)) = 69}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}{\text{(C}}_2{\text{H}}_6\text{)}=\text{229}{\text{.5J K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}{\text{(O}}_2\text{)}=\text{205}{\text{.0 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}\Delta {{\rm H}}^{°}{}_f{\text{(CO}}_2\text{)}=\text{-393}{\text{.5 KJmol}}^{\text{-1}}\\ \Delta {{\rm H}}^{°}{}_f\text{(}{H}_2\text{O)}=\text{-285}{\text{.8 KJmol}}^{\text{-1}}\\ \Delta {{\rm H}}^{°}{}_f\text{(}{C}_2{\text{H}}_6\text{)}=\text{-84}{\text{.7KJmol}}^{\text{-1}}\\ \Delta {{\rm H}}^{°}{}_f\text{(}{O}_2\text{)}=0\end{array}$

To calculate ${\text{ΔH}}_{system}$

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

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

$\begin{array}{l}\text{= 4}\Delta {{\rm H}}^{°}{}_f{\text{(CO}}_2\text{)+6}\Delta {{\rm H}}^{°}{}_f\text{(}{H}_2\text{O)- }\left[2\Delta {{\rm H}}^{°}{}_f\text{(}{C}_2{\text{H}}_6\text{)+7}\Delta {{\rm H}}^{°}{}_f\text{(}{O}_2\text{)}\right]\\ =\text{(4)(-393}{\text{.5 KJmol}}^{\text{-1}}\text{)+(6)(-285}{\text{.8 KJmol}}^{\text{-1}}\text{)-}\left[\text{(2)(-84}{\text{.7KJmol}}^{\text{-1}})+0\right]\\ \text{= -3}\text{.119}\times {\text{10}}^2{\text{ KJmol}}^{\text{-1}}\end{array}$

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.

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

$\begin{array}{l}\text{=-}\frac{{\text{(-3119KJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{=10}{\text{.5 KJ K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{sys}$

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

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

$\begin{array}{l}{\text{= 4S}}^{\text{°}}{\text{(CO}}_{\text{2}}\text{)+6S}°{\text{(H}}_2\text{O(}l\text{)) - }\left[{\text{2S}}^{\text{°}}{\text{(C}}_2{\text{H}}_6{\text{)+7S}}^{\text{°}}{\text{(O}}_2\text{)}\right]\\ =\text{(4)(213}{\text{.6 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(6)(69}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(2)(229}{\text{.5J K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(7)(205}{\text{.0 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ =-620.2{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate $\Delta S_{univ}$

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

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

$\begin{array}{l}=\text{-620}{\text{.2JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+(1}\text{.05}\times {\text{10}}^4{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\\ \text{= 9}\text{.9}\times {\text{10}}^3{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since, $\Delta S_{univ}>0$ the given process is non spontaneous