Chapter 20, Problem 20.111P

(a)

Interpretation Introduction

Interpretation:

For the steam forming and water-gas shift reaction, the ${\text{ΔG}}^{\text{o}}\text{and}{\text{ΔH}}^{\text{o}}$ values have to be calculated at $1000{}^{o}C$, using the equilibrium free energy and enthalpy values.

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 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 20.111P

The steam reformaing reaction the standared free energy $\left(\Delta G_{\text{rxn}}^{\text{o}}\right)$ changes  is $-66.84kJ.$

Water gas shift reaction the standared free energy $\left(\Delta G_{\text{rxn}}^{\text{o}}\right)$ changes  is  $12.2kJ.$

Explanation of Solution

Given,

The steam reformaing reaction is,

  $C{H}_4\left(g\right)+H_{2}O\left(g\right)\stackrel{}{\to }CO\left(g\right)+3H_2\left(g\right)steam\text{re-formaing}$

The equal amount of methane react with water to produce $COand{H}_2$ it is a steam re-forming reaction.

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

The enthalpy change for the reaction is calculated as follows,

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

  $\begin{array}{l}{\text{ΔH}}_{rxn}^o\text{ =}\left[\left(1\text{mol}CO\right)\left({\text{ΔH}}_f^{o}ofCO\right)+\left(3\text{mol}{H}_2\right)\left({\text{ΔH}}_f^{o}of{H}_2\right)\right]\\ -\left[\left(1\text{mol}C{H}_4\right)\left({\text{ΔH}}_f^{o}ofC{H}_4\right)+\left(1\text{mol}{H}_{2}O\right)\left({\text{ΔH}}_f^{o}of{H}_{2}O\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o\text{ =}\left[\left(1\text{mol}\right)\left(-110.5kJ/mol\right)+\left(3\text{mol}\right)\left(0kJ/mol\right)\right]-\\ \left[\left(1mol\right)\left(-74.87kJ/mol\right)+\left(\text{1}mol\right)\left(-241.826kJ/mol\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o=\left[\left(-110.5kJ/mol\right)\right]-\left[\left(-316.696kJ/mol\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o\text{ =}206.196\text{kJ}\\ \\ {\text{ΔH}}_{rxn}^o\text{ =}206.2kJ.\end{array}$

Hence, the standard enthalpy of the reaction ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}=\underset{\_}{206.2kJ}.$

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.

  $\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}CO\right)\left(S^{o}ofCO\right)+\left(3\text{mol}{H}_2\right)\left(S^{o}of{H}_2\right)\right]\\ -\left[\left(1molC{H}_4\right)\left(S^{o}ofC{H}_4\right)+\left(1mol{H}_{2}O\right)\left(S^{o}of{H}_{2}O\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(\text{1 mol}\right)\left(197.5J/mol\times K\right)+\left(\text{3}\text{mol}\right)\left(130.6J/mol\times K\right)\right]-\\ \left[\left(1mol\right)\left(186.1J/mol\times K\right)+\left(1mol\right)\left(188.72J/mol\times K\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}214.48\text{J/K}=214.5J/K.\end{array}$

The standard entropy of the reaction ${\text{ΔS}}_{rxn}^o=\underset{\_}{214.5J/K}.$

Calculation for free energy change  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$

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 entropy $\Delta {\text{H}}_{\text{rxn}}^{\text{o}}$ and  ${\text{ΔS}}_{\text{rxn}}^{\text{o}}$ values are

  $\Delta H_{\text{rxn}}^{\text{o}}=206.196\text{kJ}$

  ${\text{ΔS}}_{\text{rxn}}^{\text{o}}=214.48\text{J/K}$

These values are plugging above standard free energy equation,

  $\begin{array}{c}\Delta G_{\text{rxn}}^{\text{o}}=\left(206.196kJ\right)-\left(1273K\right)\left(214.48\text{J/K}\right)\left(1kJ/{10}^{3}J\right)\\ \\ =-66.83704\\ \\ \Delta G_{\text{rxn}}^{\text{o}}=-66.84kJ\end{array}$

Therefore, the given reaction $\left(\Delta G_{\text{rxn}}^{\text{o}}\right)$ value is $-66.84kJ.$

Next calculate the water-gas shift reactions

Given,

Water gas shift reaction is,

  $CO\left(g\right)+H_{2}O\left(g\right)\underset{}{\rightleftharpoons }C{O}_2\left(g\right)+H_2\left(g\right)\text{water-gas}\text{shift}\text{reaction}$

The equal amount of $CO$ react with water to produce $C{O}_{2}and{H}_2$ it is a water gas shift reaction.

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

The enthalpy change for the reaction is calculated as follows,

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

  $\begin{array}{l}{\text{ΔH}}_{rxn}^o\text{ =}\left[\left(1\text{mol}C{O}_2\right)\left({\text{ΔH}}_f^{o}ofC{O}_2\right)+\left(1\text{mol}{H}_2\right)\left({\text{ΔH}}_f^{o}of{H}_2\right)\right]\\ -\left[\left(1\text{mol}CO\right)\left({\text{ΔH}}_f^{o}ofCO\right)+\left(1\text{mol}{H}_{2}O\right)\left({\text{ΔH}}_f^{o}of{H}_{2}O\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o\text{ =}\left[\left(1\text{mol}\right)\left(-393.5kJ/mol\right)+\left(1\text{mol}\right)\left(0kJ/mol\right)\right]-\\ \left[\left(1mol\right)\left(-110.5kJ/mol\right)+\left(\text{1}mol\right)\left(-241.826kJ/mol\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o=\left[\left(-393.5kJ/mol\right)\right]-\left[\left(-352.326kJ/mol\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o\text{ =}-41.174\text{kJ}\\ \\ {\text{ΔH}}_{rxn}^o\text{ =}-41.2kJ.\end{array}$

Standard enthalpy of the given reaction is ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}=\underset{\_}{-41.2kJ}.$

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.

  $\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{O}_2\right)\left(S^{o}ofC{O}_2\right)+\left(1\text{mol}{H}_2\right)\left(S^{o}of{H}_2\right)\right]\\ -\left[\left(1molCO\right)\left(S^{o}ofCO\right)+\left(1mol{H}_{2}O\right)\left(S^{o}of{H}_{2}O\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(\text{1 mol}\right)\left(213.7J/mol\times K\right)+\left(1\text{mol}\right)\left(130.6J/mol\times K\right)\right]-\\ \left[\left(1mol\right)\left(197.7J/mol\times K\right)+\left(1mol\right)\left(188.72J/mol\times K\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}-41.92\text{J/K}\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}=-41.9J/K.\end{array}$

The standard entropy of the reaction ${\text{ΔS}}_{rxn}^o=\underset{\_}{-41.9J/K}.$

Calculation for free energy change  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$

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 entropy $\Delta {\text{H}}_{\text{rxn}}^{\text{o}}$ and  ${\text{ΔS}}_{\text{rxn}}^{\text{o}}$ values are

  $\Delta H_{\text{rxn}}^{\text{o}}=-41.174\text{kJ}$

  ${\text{ΔS}}_{\text{rxn}}^{\text{o}}=-41.92\text{J/K}$

These values are plugging above standard free energy equation,

  $\begin{array}{c}\Delta G_{\text{rxn}}^{\text{o}}=\left(-41.174kJ\right)-\left(1273K\right)\left(-41.92\text{J/K}\right)\left(1kJ/{10}^{3}J\right)\\ \\ =12.19016\\ \\ \Delta G_{\text{rxn}}^{\text{o}}=12.2kJ\end{array}$

Therefore, the given reaction $\left(\Delta G_{\text{rxn}}^{\text{o}}\right)$ value is $12.2kJ.$

(b)

Interpretation Introduction

Interpretation:

The given reaction is become a spontaneous at $1000{}^{o}C$ and standard $\Delta G^o$ state of this process has to determined.

Concept introduction:

Any natural process or a chemical reaction taking place in a laboratory can be classified into two categories, spontaneous or nonspontaneous. Spontaneous process occurs by itself, without the influence of external energy. In spontaneous process the free energy of the system decreases and entropy of the system increases.  Nonspontaneous process requires an external influence for initiation.  In nonspontaneous process the free energy of the system increases but entropy of the system decreases.

Answer to Problem 20.111P

The steam reformaing reaction the standared free energy $\left(\Delta G_{\text{rxn}}^{\text{o}}\right)$ changes  is $-121.5kJ$

Explanation of Solution

Given,

The steam reformaing reaction is,

  $2C{H}_4\left(g\right)+3H_{2}O\left(g\right)\underset{}{\rightleftharpoons }C{O}_2\left(g\right)+CO\left(g\right)+7H_2$

This reactions is two re-forming reactions and one water-gas shift reaction,

So,

$\Delta G^o=2\left(\Delta G^o\right)$ for the steam Re-forming reaction $+\Delta G^o$ for the water-gas shift reaction.

Hence,

  $\begin{array}{c}\Delta G^o=2\left(-66.84kJ\right)+12.1387kJ\\ \\ =-121.5413\\ \\ \Delta G^o=-121.5kJ\end{array}$

Thues, the $\Delta G^o$ is negative, the reaction is spontaneoues at this temprature.

(c)

Interpretation Introduction

Interpretation:

$\Delta G$ at $50\%$  conversion has to be determined where $50\%$ of starting material 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.

Answer to Problem 20.111P

The steam reformaing reaction the standared free energy $\left(\Delta G_{\text{rxn}}^{\text{o}}\right)$ changes is $55.4kJ/mol.$

Explanation of Solution

Consider the following steam re-forming reaction,

  $2C{H}_4\left(g\right)+3H_{2}O\left(g\right)\underset{}{\rightleftharpoons }C{O}_2\left(g\right)+CO\left(g\right)+7H_{2}steam\text{re-formaing}$

At $98atm$ and $50\%$ conversion, the partial pressure of the reactants and products are as follows,

The co-efficent are,

     $\begin{array}{l}C{H}_4=14atm;H_{2}O=21atm,\\ \\ C{O}_2=7.0atm;CO=7.0atm;H_2=49atm\end{array}$

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

The reaction quotient expression is,

  $\begin{array}{l}Q=\frac{\left(C{O}_2\right)\left(CO\right){\left(H_2\right)}^7}{{\left(C{H}_4\right)}^4{\left(H_{2}O\right)}^3}=\frac{\left(7.0\right)\left(7.0\right){\left(49\right)}^7}{{\left(14\right)}^2{\left(21\right)}^3}=1.8309\times {10}^7\\ \\ Q=1.8309\times 10^7\end{array}$

Calculate the Free enrgy change  $\Delta G$

Standared Free energy change equation is,

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

  $\begin{array}{c}\text{ΔG}=-121.5413kJ/mol+\left(1kJ/{10}^{3}J\right)\left(8.314J/mol\cdot K\right)\left(1273K\right)In\left(1.8309\times {10}^7\right)\\ \\ =-121.5413kJ/mol+\left(1kJ/{10}^{3}J\right)\left(8.314J/mol\cdot K\right)\left(1273K\right)\left(16.72290\right)\\ \\ \text{ΔG}=55.4476=55.4kJ/mol.\end{array}$

This reaction is not spontaneous at this point. Thus, the free energy value is $\text{ΔG}=\underset{\_}{55.4kJ/mol}$.

(d)

Interpretation Introduction

Interpretation:

$\Delta G$ at $90\%$  conversion has to be determined where $90\%$ of starting material 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.

Answer to Problem 20.111P

The steam reformaing reaction the free energy $\left(\Delta G\right)$ changes is $188kJ/mol.$

Explanation of Solution

Consider the following steam re-forming reaction,

  $2C{H}_4\left(g\right)+3H_{2}O\left(g\right)\underset{}{\rightleftharpoons }C{O}_2\left(g\right)+CO\left(g\right)+7H_{2}steam\text{re-formaing}$

At $98atm$ and $90\%$ conversion, the partial pressure of the reactants and products are as follows,

The co-efficent are,

     $\begin{array}{l}C{H}_4=2.3atm;H_{2}O=3.4atm,\\ \\ C{O}_2=10.3atm;CO=10.3atm;H_2=71.8atm\end{array}$

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

The reaction quotient expression is,

  $\begin{array}{l}Q=\frac{\left(C{O}_2\right)\left(CO\right){\left(H_2\right)}^7}{{\left(C{H}_4\right)}^4{\left(H_{2}O\right)}^3}=\frac{\left(10.3\right)\left(10.3\right){\left(71.8\right)}^7}{{\left(2.3\right)}^2{\left(3.4\right)}^3}=5.019415\times {10}^{12}\\ \\ Q=5.019415\times 10^{12}.\end{array}$

Calculate the Free enrgy change  $\Delta G$

Standared Free energy change equation is,

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

  $\begin{array}{c}\text{ΔG}=-121.5413kJ/mol+\left(1kJ/{10}^{3}J\right)\left(8.314J/mol\cdot K\right)\left(1273K\right)In\left(5.019\times {10}^{12}\right)\\ \\ =-121.5413kJ/mol+\left(1kJ/{10}^{3}J\right)\left(8.314J/mol\cdot K\right)\left(1273K\right)\left(29.24425\right)\\ \\ \text{ΔG}=187.9726=188kJ/mol.\end{array}$

This reaction is not spontaneous at this point. Thus, the free energy value is $\text{ΔG}=\underset{\_}{188kJ/mol}$.