Chapter 20, Problem 20.97P

(a)

Interpretation Introduction

Interpretation:

For the following dehydrogenation reaction, the $\Delta {{\rm H}}^{\text{o}},\Delta {\text{G}}_{\text{rxn}}^{\text{o}}\text{and}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$ values have to be founded at $298K$, using the given data’s.

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

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.

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.

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

Answer to Problem 20.97P

The standard enthalpy changes of the given reaction is ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}=\underset{\_}{116.3kJ}.$

The standard free energy value is $\Delta G_{rxn}^o=82.8kJ.$

Standard entropy of the reaction ${\text{ΔS}}_{rxn}^o=\underset{\_}{114J/K}.$

Explanation of Solution

Given,

The dehydrogenation of ethylbenzen is,

  $C_6{H}_5-C{H}_2-C{H}_3\left(g\right)\underset{dehydrogantion}{\to }{C}_6{H}_5-CH=C{H}_2\left(g\right)+H_2\left(g\right)$

Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presence of superheated steam. The dehydration reaction is given above.

Standard enthalpy change is ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$,

The enthalpy change for the reaction is calculated as follows,

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

  $\begin{array}{l}{\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}{H}_2\right)\left({\text{ΔH}}^{\text{°}}{}_{f}of{H}_2\right)+\left(1\text{mol}{C}_6{H}_{5}CHC{H}_2\right)\left({\text{ΔH}}^{\text{°}}{}_{f}of{C}_6{H}_{5}CHC{H}_2\right)\right]\\ -\left[\left(1mol{C}_6{H}_{5}C{H}_{2}C{H}_3\right)\left({\text{ΔH}}^{\text{°}}{}_{f}of{C}_6{H}_{5}C{H}_{2}C{H}_3\right)\right]\\ \\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(\text{1 mol}\right)\left(0kJ/mol\right)+\left(\text{1 mol}\right)\left(103.8kJ/mol\right)\right]-\\ \left[\left(1mol\right)\left(-12.5kJ/mol\right)\right]\\ \\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}116.3kJ.\end{array}$

Hence, the standard enthalpy of the reaction ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}=\underset{\_}{116.3kJ}.$

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,

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}{H}_2\right)\left({\text{ΔG}}_f^{o}of{H}_2\right)+\left(1\text{mol}{C}_6{H}_{5}CHC{H}_2\right)\left({\text{ΔG}}_f^{o}of{C}_6{H}_{5}CHC{H}_2\right)\right]\\ -\left[\left(1mol{C}_6{H}_{5}C{H}_{2}C{H}_3\right)\left({\text{ΔG}}_f^{o}of{C}_6{H}_{5}C{H}_{2}C{H}_3\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(\text{1 mol}\right)\left(0kJ/mol\right)+\left(\text{1 mol}\right)\left(202.5kJ/mol\right)\right]-\\ \left[\left(1mol\right)\left(119.7kJ/mol\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}82.8kJ.\end{array}$

Hence, the standard free energy value is $\Delta G_{rxn}^o=82.8kJ$

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

Calculate the change in entropy for this reaction as follows,

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

Where, $\left(m\right)and\left(n\right)$ are the stoichiometric co-efficient.

$\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}{H}_2\right)\left(S^{o}of{H}_2\right)+\left(1\text{mol}{C}_6{H}_{5}CHC{H}_2\right)\left(S^{o}of{C}_6{H}_{5}CHC{H}_2\right)\right]\\ -\left[\left(1mol{C}_6{H}_{5}C{H}_{2}C{H}_3\right)\left(S^{o}of{C}_6{H}_{5}C{H}_{2}C{H}_3\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(\text{1 mol}\right)\left(130.6J/mol\times K\right)+\left(\text{1 mol}\right)\left(238J/mol\times K\right)\right]-\\ \left[\left(1mol\right)\left(255J/mol\times K\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}113.6=114J/K\left(or\right)0.1136kJ/K.\end{array}$

The standard entropy of the reaction ${\text{ΔS}}_{rxn}^o=\underset{\_}{114J/K}.$

(b)

Interpretation Introduction

Interpretation:

For the dehydration reaction, in which temperature become a spontaneous and temperature should be calculated.

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.

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

Entropy is a thermodynamic quantity, which is the measure of randomness in a system.  The term entropy is useful in explaining the spontaneity of a process. For all spontaneous process in an isolated system there will be an increase in entropy. Entropy is represented by the letter ‘S’.  It is a state function.  The change in entropy gives information about the magnitude and direction of a process.  The entropy changes associated with a phase transition reaction can be found by the following equation.

  $\text{Τ }\text{}\text{=}\text{}\text{}\frac{{\text{ΔΗ}}^{\text{o}}}{{\text{ΔS}}^{\text{o}}}$

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 20.97P

Given reaction the founded temperature value is $1020K.$

Explanation of Solution

The dehydrogenation of ethylbenzen is,

  $C_6{H}_5-C{H}_2-C{H}_3\left(g\right)\underset{dehydrogantion}{\to }{C}_6{H}_5-CH=C{H}_2\left(g\right)+H_2\left(g\right)$

Determination for temperature (T)

The reaction will become spontaneous when $\Delta G$ change from being positve to being negative. This point occuers when ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=0$.

Standared Free energy change equation is,

  $\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}\\ \\ {\text{ΔG}}^{\text{o}}=0=\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}\\ \\ \Delta {{\rm H}}^{\text{o}}=\text{T}\Delta {\text{S}}^{\text{o}}\end{array}$

Rearrange the above equation to calculate temprature T,

$\text{T}\text{=}\frac{\text{ΔΗ}}{\text{ΔS}}------\left[3\right]$

Hence,

Calculated enthalpy $\Delta H$ and entropy $\Delta S$ values are,

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=116.3\text{kJ}\\ \\ \Delta S_{\text{rxn}}^{\text{o}}=0.1136k\text{J/K}\end{array}$

  $\begin{array}{c}T=\frac{116.3kJ}{0.1136k\text{J/K}}\left(\frac{{10}^3}{1kJ}\right)=1023.7676\\ \\ T=1020K\end{array}$

At temprature will become above $1020K$, this also reaction spontaneous. Because both enthalpy $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}$ and entropy $\Delta S_{\text{rxn}}^{\text{o}}$ values are positve, the reaction becomes spontaneous above this temprature.

(c)

Interpretation Introduction

Interpretation:

For the dehydration reaction, the $\Delta G_{\text{rxn}}^{\text{o}}$ and equilibrium $K$ values have to be calculated at ${600}^{o}C$.

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,

  $\Delta {{\rm H}}^{\text{o}}$ 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 20.97P

The given reaction equilibrium constant value is $\text{K=}\underset{\_}{9.9\times 10^{-2}}.$

Explanation of Solution

The dehydrogenation of ethylbenzen is,

  $C_6{H}_5-C{H}_2-C{H}_3\left(g\right)\underset{dehydrogantion}{\to }{C}_6{H}_5-CH=C{H}_2\left(g\right)+H_2\left(g\right)$

Standared Free energy change 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

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=116.3\text{kJ}\\ \\ \Delta S_{\text{rxn}}^{\text{o}}=114\text{J/K}\\ \\ T=273K+{600}^{o}C=873K\end{array}$

These values are plugging above standard free energy equation,

   $\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}=\left(116.3\text{kJ}\right)\text{-}\left[\left(873\text{K}\right)\times \left(\text{114}\text{J/K}\right)\left(1kJ/{10}^3\right)\right]\\ \\ =\left(116.3\text{kJ}\right)-99.552J/mol\\ \\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=16.778=16.8kJ/mol.\end{array}$

Calculation for equilibrium constant$\text{K}$

We know that equilibrium equation,

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=-\text{RT}\ln \left({\text{K}}_{\text{p}}\right)$

Rearrange the above equation,

  $\begin{array}{c}\ln \text{K}=\frac{{\text{ΔG}}^{\text{o}}}{\text{-RT}}=\left(\frac{16.778\text{k}\text{J/mol}}{-\left(8.314\text{J/mol}\cdot \text{K}\right)\left(873\text{K}\right)}\right)\left(\frac{{10}^{3}J}{1kJ}\right)\\ \\ \ln \text{K}=-2.311617\\ \\ \text{K=}{\text{e}}^{-2.311617}=0.0991008\text{=}9.9100\times {10}^{-2}\text{=}9.9\times 10^{-2}.\end{array}$

Therefore, the given reaction equilibrium constant is $\text{K=}\underset{\_}{9.9\times 10^{-2}}$.

(d)

Interpretation Introduction

Interpretation:

$\Delta G$ at $50\%$  conversion has to be determined where $50\%$ of ethylbenzene is reacted.

Concept introduction:

Reaction quotient$\left(Q\right)$: Reaction quotient and equilibrium constant has same expression.  Reaction quotient is also the ratio between the concentrations of the reactant to product, but these concentrations are not necessarily the equilibrium concentrations.

        $\text{Q=}\frac{\left[\text{Product}\right]}{\left[\text{Reactant}\right]}$

The reaction quotient $\left(Q\right)$ is helpful in predicting the direction of the reaction.

• When $\text{Q}\text{>}{\text{K}}_{\text{c}}$, the reaction proceeds towards left to increase the concentration of the reactants.
• When $\text{Q}<{\text{K}}_{\text{c}}$, the reaction proceeds towards right to increase the concentration of the products.
• When $\text{Q}={\text{K}}_{\text{c}}$, the reaction is at equilibrium.

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 $\text{K}$

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

  Where,

  $\text{T}$ is the temperature

  $\text{ΔG}$ is the free energy

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

Mole fraction: Mole fraction ($X$) of any compound in a solution is the number of moles of the component divided by total number of moles making up a solution. It is denoted by $X$

  $Molefraction(X)=\frac{Molesofcomponent}{\begin{array}{l}Totalnumberofmoles\\ makingupthesolutions\end{array}}$

Answer to Problem 20.97P

Following reaction free energy value is $\text{ΔG}=\underset{\_}{0.07kJ/mol}.$

Explanation of Solution

Find the concentration of given reaction is as fallows,

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{-CH}}_{\text{2}}{\text{-CH}}_{\text{3}}\left(\text{g}\right)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{-CH=CH}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\left(\text{g}\right)\\ \text{Initial concentration: }1.0\text{0}\text{0}\\ \text{Reactant}\text{mixture :}\text{-0}\text{.50}+\text{0}\text{.50}\text{+0}\text{.50}\text{(50\%}\text{conversion)}\\ \text{-----------------------------------------------------------------------------------------------}\\ \text{Remaining: }0.50\text{0}\text{.5}0\text{0}\text{.50}\\ \text{------------------------------------------------------------------------------------------------}\end{array}$

Mole fraction:

Mole fraction of each gas,

  $\begin{array}{l}=\frac{amountofgas}{H_2+styrene+ethylbenzene+steam}\\ \\ =\frac{0.5}{0.5+0.5+0.5+5}=\frac{0.5}{6.5}=0.076923\end{array}$

Partial pressure of each gas

  $\begin{array}{l}=Molefraction\times Totalpressure\\ \\ =0.076923\times 1.3atm\\ \\ =0.10atm.\end{array}$

Next we calculate the reaction quotient and then use to find $\Delta G$

The reaction quotient expression is,

  $\begin{array}{l}Q=\frac{\left[H_2\right]\left[C_6{H}_{5}CHC{H}_2\right]}{\left[C_6{H}_{5}C{H}_{2}C{H}_3\right]}=\frac{\left[0.10atm\right]\left[0.10atm\right]}{\left[0.10atm\right]}=0.10\\ \\ Q=0.10\end{array}$

Calculate the Free enrgy change  $\Delta G$

Standared Free energy change equation is,

$\text{ΔG =}{\text{ΔG}}^{\text{o}}\text{+}\text{RT}\ln \left(\text{Q}\right)$

  $\begin{array}{c}\text{ΔG}=16.778kJ/mol+\left(1kJ/{10}^{3}J\right)\left(8.314J/mol\cdot K\right)\left(873K\right)In\left(0.10\right)\\ \\ =16.778kJ/mol+\left(1kJ/{10}^{3}J\right)\left(8.314J/mol\cdot K\right)\left(873K\right)\left(-2.302585\right)\\ \\ =16.778kJ/mol-16.71244kJ/mol\\ \\ \text{ΔG}=0.065565=0.07kJ/mol.\end{array}$

Therefore, the given reaction free energy value is $\text{ΔG}=\underset{\_}{0.07kJ/mol}.$