Chapter 20, Problem 20.95P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the ${\text{ΔS}}_{\text{rxn}}^{\text{o}}$ value has to be calculated at $298K$, using the given equilibrium $K$ value.

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.

  ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}{\text{S}}^{\text{°}}{}_{\text{Products}}-{\displaystyle \sum \text{n}}{\text{S}}^{\text{°}}{}_{\text{reactants}}$

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.

Answer to Problem 20.95P

The standard entropy changes of the given reaction value is ${\text{ΔS}}_{rxn}^o=\underset{\_}{121.5J/K}.$

Explanation of Solution

The given equilibrium reaction is,

  $2NOBr\left(g\right)\stackrel{}{\rightleftharpoons }2NO\left(g\right)+B{r}_2\left(g\right)K=0.42at373K$

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

$\Delta S_{rxn}^o$ Formations of values are,

  $\begin{array}{c}NO\left(g\right)=210.65J/K\cdot mol\\ \\ B{r}_2\left(g\right)=245.38J/K\cdot mol\\ \\ NOBr\left(g\right)=272.6J/K\cdot mol\end{array}$

Calculate the change in entropy for this reaction as follows,

  ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}{\text{S}}^{\text{°}}{}_{\text{Products}}-{\displaystyle \sum \text{n}}{\text{S}}^{\text{°}}{}_{\text{reactants}}$

Calculate the change in entropy for this reaction as follows

  $\begin{array}{l}{\text{ΔS}}_{rxn}^o\text{ =}\left[\left(2\text{mol}NO\right)\left(S^{o}ofNO\right)+\left(1\text{mol}{\text{Br}}_2\right)\left(S^{o}ofB{r}_2\right)\right]\\ -\left[\left(2\text{mol}ofNOBr\right)\left(S^{o}ofNOBr\right)\right]\end{array}$

Substituting the values of standard entropy $\left(S^o\right)$ of individual species in the above equation

  $\begin{array}{l}{\text{ΔS}}_{rxn}^o\text{ =}\left[\left(2\text{mol}\text{of NO}\right)\left(210.65\text{J/mol}\cdot K\right)+\left(1\text{mol}{\text{of Br}}_2\right)\left(245.38\text{J/mol}\cdot K\right)\right]-\\ \left[\left(2molofNOBr\right)\left(272.6\text{J/mol}\cdot K\right)\right]\\ \\ {\text{ΔS}}_{rxn}^o\text{ =}121.48=121.5J/K.\end{array}$

The standard entropy of the reaction ${\text{ΔS}}_{rxn}^o=\underset{\_}{121.5J/K}$ and the entropy change is positive.

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ value has to be calculated at $373K$, using the given data’s.

Concept introduction:

Free energy change$\text{ΔG}$: change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant $\text{K}$,

  $\begin{array}{l}\text{ΔG =}{\text{ΔG}}^{\text{o}}\text{+}\text{RT}\ln \left(\text{K}\right)\\ \\ {\text{ΔG}}^{\text{o}}\text{=- RT}\ln \left(\text{K}\right)\end{array}$

  Where,

  $\text{T}$ is the temperature

  $\text{ΔG}$ is the free energy

  ${\text{ΔG}}^{\text{o}}$ is standard free energy change.

Answer to Problem 20.95P

Given reaction the standard free ${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}$ energy value is $\text{2}{\text{.7×10}}^{\text{3}}\text{J/mol}\text{.}$

Explanation of Solution

The given equilibrium reaction is,

  $2NOBr\left(g\right)\stackrel{}{\rightleftharpoons }2NO\left(g\right)+B{r}_2\left(g\right)K=0.42at373K$

Solving for $\Delta G_{rxn}^o$ using following free energy equation,

  $\Delta G=\Delta G^o+RT\ln \left(Q\right)$

Calculate the $\Delta G_{rxn}^o$ of given equilibrium reaction is as follows,

     $\begin{array}{c}\text{ΔG=}0={\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{+RTlnK}\\ \\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=-\text{RTlnK}\\ \\ =-\left(8.314\text{J/mol}\times \text{K}\right)\left(373\text{K}\right)\text{In}\left(0.42\right)\\ \\ =-\left(8.314\text{J/mol}\times \text{K}\right)\left(373\text{K}\right)\times \left(-0.867500\right)\\ \\ =2690.223\text{J/mol}\\ \\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}2.7\times 10^{3}J/mol.\end{array}$

Therefore, the given reaction $\left({\text{ΔG}}_{\text{rxn}}^{\text{o}}\right)$ value is $2.7\times 10^{3}J/mol.$

(c)

Interpretation Introduction

Interpretation:

For the following reaction, the ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ value has to be calculated at $373K$, using the given data’s.

Concept introduction:

Free energy change is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ${\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}$

Where,

  $\Delta {{\rm H}}^{\text{o}}$ is the change in enthalpy of the system

  ${\text{ΔG}}^{\text{o}}$ is the standard change in free energy of the system

  $\text{T}$ is the absolute value of the temperature

Answer to Problem 20.95P

Given reaction enthalpy $\left(\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\right)$ changes is $4.80\times 10^{4}J/mol$

Explanation of Solution

The given equilibrium reaction is,

  $2NOBr\left(g\right)\stackrel{}{\rightleftharpoons }2NO\left(g\right)+B{r}_2\left(g\right)K=0.42at373K$

Calculation for enthalpy change  ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$

Standared Free energy change equation is,

  $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}{\text{ = ΔG}}_{\text{rxn}}^{\text{o}}\text{- T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$

Calcualted entropy $\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$ and  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ values are

  $\Delta {\text{S}}_{\text{rxn}}^{\text{o}}=121.48J/K$

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=2690.225\text{J/mol}$

These values are plugging following above equation,

  $\begin{array}{c}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=2690.225\text{J/mol}\text{+}\left(373K\right)\left(121.48\text{J/K}\right)\\ \\ =4.8002265\times {10}^4\\ \\ \Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=4.80\times 10^{4}J/mol.\end{array}$

Therefore, the given reaction $\left(\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\right)$ value is $4.80\times 10^{4}J/mol.$

(d)

Interpretation Introduction

Interpretation:

For the following reaction, the ${\text{ΔH}}_{\text{f}}^{\text{o}}$ value has to be calculated at $298K$.

Concept introduction:

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.

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

Where,

  ${\text{ΔH}}_{\text{f}}^{\text{o}}{}_{\left(\text{reactants}\right)}$ is the standard enthalpy of the reactants

  ${\text{ΔH}}_{\text{f}}^{\text{o}}{}_{\left(\text{produdcts}\right)}$ is the standard enthalpy of the products

Answer to Problem 20.95P

Given reaction standard enthalpy changes is ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}=\underset{\_}{81.7kJ/mol}.$

Explanation of Solution

Consider the equilibrium reaction is,

  $2NOBr\left(g\right)\stackrel{}{\rightleftharpoons }2NO\left(g\right)+B{r}_2\left(g\right)K=0.42at373K$

Standard enthalpy change is,

The equation for the standard enthalpy of the above reaction is,

  ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}\Delta {\text{H}}^{\text{°}}{}_{\text{f(Products)}}-{\displaystyle \sum \text{n}}\Delta {\text{H}}^{\text{°}}{}_{\text{f(reactants)}}$

Calculate the change in entropy for this reaction as follows

  $\begin{array}{l}{\text{ΔH}}_{rxn}^o\text{ =}\left[\left(2\text{mol}NO\right)\left({\text{ΔH}}_f^{o}ofNO\right)+\left(1\text{mol}B{r}_2\right)\left({\text{ΔH}}_f^{o}ofB{r}_2\right)\right]\\ \\ -\left[\left(2\text{mol}ofNOBr\right)\left({\text{ΔH}}_f^{o}ofNOBr\right)\right]\end{array}$

Substituting the values of standard enthalpies of individual species in the above equation

  $\begin{array}{l}{\text{ΔH}}_{rxn}^o\text{ =}\left[\left(2\text{mol}\text{of NO}\right)\left(90.29kJ/mol\right)+\left(1\text{mol}{\text{of Br}}_2\right)\left(30.91\text{kJ/mol}\right)\right]-\\ \left[\left(2molofNOBr\right)\left(4.8002265\times {10}^{4}J\right)\left(1kJ/{10}^{3}J\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o\text{of}\text{NOBr =}81.7438675=81.7kJ/mol.\end{array}$

Hence, the standard enthalpy of the reaction ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}=\underset{\_}{81.7kJ/mol}.$

(e)

Interpretation Introduction

Interpretation:

For the following reaction, the ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ value has to be calculated at $298K$.

Concept introduction:

Free energy (or) entropy change is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ${\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}$

Where,

  ${\text{ΔG}}^{\text{o}}$ is the standard change in free energy of the system

  $\Delta {{\rm H}}^{\text{o}}$ is the standard change in enthalpy of the system

  $\text{T}$ is the absolute value of the temperature

  $\Delta {\text{S}}^{\text{o}}$ is the change in entropy in the system

Answer to Problem 20.95P

The ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ value of the reaction is $\underset{\_}{1.18\times 10^{4}J/mol}.$

Explanation of Solution

Consider the equilibrium reaction is,

  $2NOBr\left(g\right)\stackrel{}{\rightleftharpoons }2NO\left(g\right)+B{r}_2\left(g\right)K=0.42at373K$

Calculate the Free energy change  $\Delta G_{rxn}^o$

Standared Free energy change equation is,

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{ = }\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{- T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$

Calcualted enthalpy and entropy values are

  $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=4.8002265\times 10^{4}J/mol$

  $\Delta {\text{S}}_{\text{rxn}}^{\text{o}}=121.48kJ/K$

These values are plugging following standard free energy equation,

  $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{ =4}{\text{.8002265×10}}^{\text{4}}\text{J/mol-}\left(298\text{K}\right)\left(121.48\text{J/mol}\cdot \text{K}\right)\\ \\ =1.1801225\times {10}^5\\ \\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=1.18\times 10^{4}J/mol.\end{array}$

Therefore, the ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ value of the reaction is $\underset{\_}{1.18\times 10^{4}J/mol}.$

(f)

Interpretation Introduction

Interpretation:

For the following reaction, the $NOBr$ ${\text{ΔG}}_{\text{f}}^{\text{o}}$ value has to be calculated at $298K$.

Concept introduction:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}{\displaystyle \sum {\text{mΔG}}_{\text{f}}{}^{\text{°}}}\text{(Products)-}{\displaystyle \sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}}\text{(Reactants)}$

Where,

  ${\text{nΔG}}_{\text{f}}{{}^{\text{°}}}_{\left(\text{Reactants}\right)}$ is the standard entropy of the reactants

  ${\text{mΔG}}_{\text{f}}{{}^{\text{°}}}_{\left(\text{products}\right)}$ is the standard free energy of the products

Answer to Problem 20.95P

The standard free energy value is $\Delta G_f^o=82.3kJ/mol.$

Explanation of Solution

Given equilibrium reaction is,

  $2NOBr\left(g\right)\stackrel{}{\rightleftharpoons }2NO\left(g\right)+B{r}_2\left(g\right)K=0.42at373K$

Free energy changes ${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}$

Gibbs free energy equation is,

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}{\displaystyle \sum {\text{mΔG}}_{\text{f}}{}^{\text{°}}}\text{(Products)-}{\displaystyle \sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}}\text{(Reactants)}$

Free energy change for the reaction is calculated as follows,

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(2\text{mol}\text{of NO}\right)\left({\text{ΔG}}_f^o\text{of NO}\right)+\left(1\text{mol}\text{of}{\text{Br}}_2\right)\left({\text{ΔG}}_f^o{\text{of Br}}_2\right)\right]\\ -\left[\left(2molofNOBr\right)\left({\text{ΔG}}_f^{o}ofNOBr\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(2\text{mol}\text{of NO}\right)\left(86.60kJ/mol\right)+\left(1\text{mol}{\text{of Br}}_2\right)\left(3.13kJ/mol\right)\right]-\\ \left[\left(2molofNOBr\right)\left(1.1801225\times {10}^{4}J\right)\left(1kJ/{10}^{3}J\right)\right]\\ \\ {\text{ΔG}}_f^o\text{ of}\text{NOBr}\text{=}82.264=82.3kJ/mol.\end{array}$

Hence, the standard free energy value is $\Delta G_f^o=82.3kJ/mol.$