Chapter 20, Problem 20.98P

(a)

Interpretation Introduction

Interpretation:

The molar entropy $S^o$ of given standard formation reaction has to be founded at $\text{267}\text{.1}\text{J/mol}\cdot \text{K}$ 

Concept Introduction:

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 of one mole of substance at a given standard state is called standard molar entropy ($S^o$).

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.

Explanation of Solution

The formation reaction for propylene is,

  $3C\left(s\right)+3H_2\left(g\right)\stackrel{}{\to }C{H}_{3}CH=C{H}_2\left(g\right)$

Equal molar of carbon $C$ and hydrogen $\left(H\right)$ produced to one mole of propylene molecule.

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.

$S_f^o$ Formation of values,

  $\begin{array}{c}C\left(s\right)=5.686J/mol\cdot K\\ H_2\left(g\right)=130.6J/mol\cdot K\\ C{H}_{3}CHC{H}_2\left(g\right)=267.1J/mol\cdot K\end{array}$

Calculate the molar entropy for given reaction,

  $\begin{array}{c}{S}_f^o=\left[\left(1molC{H}_{3}CH=C{H}_2\right)\left(S^{o}ofC{H}_{3}CH=C{H}_2\right)\right]\\ -\left[\left(3molC\right)\left(S^{o}ofC\right)+\left(3mol{H}_2\right)\left(S^{o}of{H}_2\right)\right]\\ =\left[\left(1mol\right)\left(267.1J/mol\cdot K\right)\right]\\ -\left[\left(3molC\right)\left(5.686J/mol\cdot K\right)+\left(3mol{H}_2\right)\left(130.6J/mol\cdot K\right)\right]\\ =\left[\left(1mol\right)\left(267.1J/mol\cdot K\right)\right]-\left[\left(17.058J/mol\cdot K\right)+\left(391.8J/mol\cdot K\right)\right]\\ =\left[\left(267.1J/mol\cdot K\right)\right]-\left[\left(408.858J/mol\cdot K\right)\right]\\ =-\text{141}\text{.758}\text{J/K}\\ S_f^o=-141.8J/K\end{array}$

The ${\text{(S}}^{\text{°}}{}_{\text{f}}\text{)}$ of the reaction is $\underset{\_}{-141.8J/K}$

The molar entropy change is negative sign for ${\text{(S}}^{\text{°}}{}_{\text{f}}\text{)}$ indicates the formation of propylene is disfavored.

(b)

Interpretation Introduction

Interpretation:

For the propylene formation reaction $\Delta G_f^o$ value should be found.

Concept introduction:

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 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.

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

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

  ${\text{ΔG}}^{\text{o}}$, ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$ is standard free energy, enthalpy and entropy values.

Explanation of Solution

The formation reaction for propylene is,

  $3C\left(s\right)+3H_2\left(g\right)\stackrel{}{\to }C{H}_{3}CH=C{H}_2\left(g\right)$

Standared Free energy change equation is,

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

Free energy change $\Delta G_f^o$

Calcualted enthalpy and entropy  values are

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=20.4kJ/mol\\ \Delta {\text{G}}_{\text{rxn}}^{\text{o}}=-\text{141}\text{.758}\text{J/K}\end{array}$

These values are plugging above standard free energy equation,

   $\begin{array}{l}{\text{ΔG}}_f^{\text{o}}=\left(20.4kJ/mol\right)-\left(298K\right)\left(-\text{141}\text{.758}\text{J/K}\right)\left({10}^{3}J/1kJ\right)\\ {\text{ΔG}}_f^{\text{o}}=62.643884kJ/mol\\ {\text{ΔG}}_f^{\text{o}}=62.6kJ/mol\end{array}$

Hence the free energy $\left({\text{ΔG}}_{\text{f}}^{\text{o}}\right)$ of the reaction is $\underset{\_}{62.6kJ/mol}$.

(c)

Interpretation Introduction

Interpretation:

For the dehydrogenation reaction the enthalpy $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}$ and free energy $\Delta {\text{G}}_{\text{rxn}}^{\text{o}}$ values have to be calculated.

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

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 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.

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

Explanation of Solution

The dehydrogantion  reaction is as follows,

  $C{H}_{3}C{H}_{2}C{H}_3\left(g\right)\underset{}{\to }C{H}_{3}CH=C{H}_2\left(g\right)+H_2\left(g\right)$

Standard enthalpy change is,

Let us find enthalpy change for the reaction,

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

  $\begin{array}{l}{\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{3}CH=C{H}_2\right)\left({\text{ΔH}}_f^{o}ofC{H}_{3}CH=C{H}_2\right)+\left(2\text{mol}{H}_2\right)\left({\text{ΔH}}_f^{o}of{H}_2\right)\right]\\ -\left[\left(1\text{mol}C{H}_{3}C{H}_{2}C{H}_3\right)\left({\text{ΔH}}_f^{o}ofC{H}_{3}C{H}_{2}C{H}_3\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}\right)\left(20.4kJ/mol\right)+\left(1\text{mol}\right)\left(0kJ/mol\right)\right]-\\ \left[\left(1\text{mol}\right)\left(-105kJ/mol\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}125.4=125kJ\end{array}$

Hence, the enthalpy ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ value is $\underset{\_}{125kJ}$

Standared Free energy change equation iss,

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

Free energy change $\Delta G_f^o$

Calculated enthalpy and entropy  values are

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=125.4kJ/mol\\ \Delta {\text{G}}_{\text{rxn}}^{\text{o}}=62.643884kJ/mol\end{array}$

These values are plugging above standard free energy equation,

$\begin{array}{l}{\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{3}CH=C{H}_2\right)\left({\text{ΔG}}_f^{o}ofC{H}_{3}CH=C{H}_2\right)+\left(1\text{mol}{H}_2\right)\left({\text{ΔG}}_f^{o}of{H}_2\right)\right]\\ -\left[\left(1\text{mol}C{H}_{3}C{H}_{2}C{H}_3\right)\left({\text{ΔG}}_f^{o}ofC{H}_{3}C{H}_{2}C{H}_3\right)\right]\\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}\right)\left(62.643884kJ/mol\right)+\left(1\text{mol}\right)\left(0kJ/mol\right)\right]-\\ \left[\left(1\text{mol}\right)\left(-24.5kJ/mol\right)\right]\\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}87.1438874=87.1kJ\end{array}$

The free energy $\left({\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\right)$ of the reaction is $\underset{\_}{87.1kJ}$.

(d)

Interpretation Introduction

Interpretation:

For the formation of propylene theoretical yield has to be calculated at ${580}^{o}C$.

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

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 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.

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

Explanation of Solution

The formation reaction for propylene is,

  $C{H}_{3}C{H}_{2}C{H}_3\left(g\right)\stackrel{}{\to }C{H}_{3}CH=C{H}_2\left(g\right)+H_2\left(g\right)$

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

Calculate the change in entropy for this reaction as follows,

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

  $\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{3}CH=C{H}_2\right)\left(S^{o}ofC{H}_{3}CH=C{H}_2\right)+\left(1\text{mol}{H}_2\right)\left(S^{o}of{H}_2\right)\right]\\ -\left[\left(1\text{mol}C{H}_{3}C{H}_{2}C{H}_3\right)\left(S^{o}ofC{H}_{3}C{H}_{2}C{H}_3\right)\right]\\ {\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}\right)\left(267.1\text{J/mol}\cdot \text{K}\right)+\left(1\text{mol}\right)\left(\text{130}\text{.6}\text{J/mol}\cdot \text{K}\right)\right]-\\ \left[\left(1\text{mol}\right)\left(\text{269}\text{.9}\text{J/mol}\cdot \text{K}\right)\right]\\ {\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}127.8kJ\end{array}$

Standared Free energy change equation is,

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

Free energy change $\Delta G_f^o$

Calcualted enthalpy and entropy  values are

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=1125.4kJ\\ \Delta {\text{G}}_{\text{rxn}}^{\text{o}}=\text{127}\text{.8}\text{kJ}\end{array}$

These values are plugging above standard free energy equation,

   $\begin{array}{l}{\text{ΔG}}_f^{\text{o}}=\left(125.4kJ/mol\right)-\left(273+580K\right)\left(127.8\text{J/K}\right)\left({10}^{3}J/1kJ\right)\\ {\text{ΔG}}_f^{\text{o}}=16.3866kJ/mol\end{array}$

Calculation for equilibrium pressure${\text{K}}_{\text{p}}$

The equilibrium equation is,

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

Rearrange the above equation,

  $\begin{array}{c}\ln \text{K}=\frac{{\text{ΔG}}^{\text{o}}}{\text{-RT}}=\left(\frac{16.3866k\text{J/mol}}{-\left(8.314\text{J/mol}\cdot \text{K}\right)\left(853\text{K}\right)}\right)\left(\frac{{10}^{3}J}{1kJ}\right)\\ \ln \text{K}=-2.3106267\\ \text{K=}{\text{e}}^{-2.3106267}\\ \text{K=}0.099199\\ \text{K=}0.0992\end{array}$

The equilibrium reaction is,

  $C{H}_{3}C{H}_{2}C{H}_3\left(g\right)\stackrel{}{\to }C{H}_{3}CH=C{H}_2\left(g\right)+H_2\left(g\right)$

  $K=\frac{\left(P_{C{H}_{3}CH=C{H}_2}\right)\left(P_{H_2}\right)}{\left(P_{C{H}_{3}C{H}_{2}C{H}_3}\right)}$

Here,

  $\begin{array}{l}x=pressureofC{H}_{3}CH=C{H}_{2}and{H}_2\\ 1.00-xispressureofC{H}_{3}C{H}_{2}C{H}_3\end{array}$

So,

  $\begin{array}{l}0.099199=\frac{\left(x\right)\left(x\right)}{\left(1.00-x\right)}\\ x^2+0.099199x-0.099199=0\end{array}$

Solve the above quadratic equation,

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-\left(0.099199\right)\pm \sqrt{{\left(0.099199\right)}^2-4\left(1\right)\left(0.099199\right)}}{2\left(1\right)}\\ x=0.2692408\\ x=0.269atm\end{array}$

Hence, the theoretical yield of propylene is $27\%$.

(e)

Interpretation Introduction

Interpretation:

Identify whether there is any yield change if the reactor wall were preamble to hydrogen $H_2$.

Concept introduction:

Theoretical yield: The amount of product formed, assuming complete reaction of the limiting reagent.

Actual yield: The amount of product actually formed in a reaction.

Percent yield: The percentage of the theoretical yield actually obtained from a chemical reaction.

$\text{Percent yield =}\frac{\text{Actual}\text{yield}}{\text{Theoretical yield}}\times 100$

Explanation of Solution

The formation reaction for propylene,

  $C{H}_{3}C{H}_{2}C{H}_3\left(g\right)\stackrel{}{\to }C{H}_{3}CH=C{H}_2\left(g\right)+H_2\left(g\right)$

If hydrogen could escape through the reactor walls, the reaction would be shifted to the right side and improving the yield.

(f)

Interpretation Introduction

Interpretation:

The temp at which the dehydrogenation spontaneous has to be identified, provided all substances in the standard state.

Concept introduction:

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.

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

Where,

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

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

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

Explanation of Solution

The formation reaction for propylene,

  $C{H}_{3}C{H}_{2}C{H}_3\left(g\right)\stackrel{}{\to }C{H}_{3}CH=C{H}_2\left(g\right)+H_2\left(g\right)$

Standared Free energy change equation is,

  $\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}\\ {\text{ΔG}}^{\text{o}}=0=\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}-------\left[1\right]\\ \Delta {{\rm H}}^{\text{o}}=\text{T}\Delta {\text{S}}^{\text{o}}------------\left[2\right]\end{array}$

Rearrange the equation (2) to calculate temprature T,

$\text{T}=\frac{\Delta {{\rm H}}^{\text{o}}}{{\text{ΔS}}^{\text{o}}}------\left[3\right]$

Hence,

Enthalpy and entropy values are

  $\begin{array}{l}\Delta {{\rm H}}_f^{\text{o}}=125.4kJ\\ \Delta {\text{G}}_{\text{rxn}}^{\text{o}}=0.1278k\text{J/K}\end{array}$

  $\begin{array}{c}T=\frac{125.4kJ}{0.1278\text{J/K}}\left(\frac{{10}^3}{1kJ}\right)=981.22066\\ T=981K\Rightarrow 707.85{}^{o}C=\underset{\_}{708{}^{o}C}\end{array}$

Hence, the propylene formation founded temperature value is $T=708{}^{o}C$.