Chapter 10, Problem 61E

Interpretation Introduction

Interpretation:

The value of $\text{ΔG}°$ , $\text{ΔH}°$ and $\text{ΔS}°$ should be calculated for the following reaction which produces acetic acid. The reaction should be identified which can be used as a commercial method for producing acetic acid. The value of temperature should be determined for the given reaction.

  $\begin{array}{l}{\text{CH}}_{\text{4}}{\text{(g)+CO}}_{\text{2}}\text{(g)}\to {\text{CH}}_{\text{3}}\text{COOH(l)}\\ {\text{CH}}_3\text{OH(g)+CO(g)}\to {\text{CH}}_{\text{3}}\text{COOH(l)}\end{array}$

Concept Introduction:

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{\text{f}}^{\text{o}}$.

If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$ or,

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

Answer to Problem 61E

For first reaction:

  $\Delta G{°}_{298}=56\text{ kJ/mol}$

  $\Delta S{°}_{298}=+240\text{ J/Kmol}$

  $\Delta H{°}_{298}=-16\text{ kJ/mol}$

For second reaction:

  $\Delta G{°}_{298}=-89\text{ kJ/mol}$

  $\Delta S{°}_{298}=-278\text{ J/Kmol}$

  $\Delta H{°}_{298}=-173\text{ kJ/mol}$

Second reaction is preferred over first reaction.

Temperatures below 622 K, reaction will be spontaneous

Explanation of Solution

The given reactions are:

  $\begin{array}{l}{\text{CH}}_{\text{4}}{\text{(g)+CO}}_{\text{2}}\text{(g)}\to {\text{CH}}_{\text{3}}\text{COOH(l)}\\ {\text{CH}}_3\text{OH(g)+CO(g)}\to {\text{CH}}_{\text{3}}\text{COOH(l)}\end{array}$

For first reaction:

The mathematical expression for the standard free energy at room temperature is:

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

The value of standard free energy for ${\text{CH}}_{\text{3}}\text{COOH(l)}$ is $-389\text{ kJ/mol}$

The value of standard free energy for ${\text{CH}}_4\text{(g)}$ is $-51\text{ kJ/mol}$

The value of standard free energy for ${\text{CO}}_2(g)$ is $\text{-394 kJ/mol}$

Put the values,

  $\Delta G{°}_{298}=(1\times G{\text{°}}_{\text{298}}{\text{(CH}}_{\text{3}}\text{COOH(l)}))-(1\times G{\text{°}}_{\text{298}}{\text{(CH}}_4\text{(g)})+1\times G{\text{°}}_{\text{298}}{\text{(CO}}_2\text{(g)}))$

  $\Delta G{°}_{298}=[(-389)-(-51-394)]\text{ kJ/mol}$

  $\Delta G{°}_{298}=56\text{ kJ/mol}$

The mathematical expression for the standard entropy at room temperature is:

  $\Delta S°=\Delta S{°}_{298}={\displaystyle \sum \text{n}}{\text{S°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{S°}}_{\text{298}}\text{(reactants)}$

The value of standard entropy for ${\text{CH}}_{\text{3}}\text{COOH(l)}$ is $160\text{ J/Kmol}$

The value of standard entropy for ${\text{CH}}_4\text{(g)}$ is $186\text{ J/Kmol}$

The value of standard entropy for ${\text{CO}}_2(g)$ is $214\text{ J/Kmol}$

Put the values,

  $\Delta S{°}_{298}=(1\times S{\text{°}}_{\text{298}}{\text{(CH}}_{\text{3}}\text{COOH(l)}))-(1\times S{\text{°}}_{\text{298}}{\text{(CH}}_4\text{(g)})+1\times S{\text{°}}_{\text{298}}{\text{(CO}}_2(g\text{)}))$

  $\Delta S{°}_{298}=[(160)-(186\text{ +214})]\text{ J/Kmol}$

  $\Delta S{°}_{298}=+240\text{ J/Kmol}$

The mathematical expression for the standard enthalpy at room temperature is:

  $\Delta H°=\Delta H{°}_{298}={\displaystyle \sum \text{n}}{\text{H°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{H°}}_{\text{298}}\text{(reactants)}$

The value of standard enthalpy for ${\text{CH}}_{\text{3}}\text{COOH(l)}$ is $-484\text{ kJ/mol}$

The value of standard enthalpy for ${\text{CH}}_4\text{(g)}$ is $-75\text{ kJ/mol}$

The value of standard enthalpy for ${\text{CO}}_2(g)$ is $-393.5\text{ kJ/mol}$

Put the values,

  $\Delta H{°}_{298}=(1\times H{\text{°}}_{\text{298}}{\text{(CH}}_{\text{3}}\text{COOH(l)}))-(1\times H{\text{°}}_{\text{298}}{\text{(CH}}_4\text{(g)})+1\times H{\text{°}}_{\text{298}}{\text{(CO}}_2(g)))$

  $\Delta H{°}_{298}=[(-484\text{)}-((-75)+(-393.5))]\text{ kJ/mol}$

  $\Delta H{°}_{298}=-16\text{ kJ/mol}$

For second reaction:

The mathematical expression for the standard free energy at room temperature is:

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

The value of standard free energy for ${\text{CH}}_{\text{3}}\text{COOH(l)}$ is $-389\text{ kJ/mol}$

The value of standard free energy for ${\text{CH}}_3\text{OH(g)}$ is $-163\text{ kJ/mol}$

The value of standard free energy for $\text{CO}(g)$ is $\text{-137 kJ/mol}$

Put the values,

  $\Delta G{°}_{298}=(1\times G{\text{°}}_{\text{298}}{\text{(CH}}_{\text{3}}\text{COOH(l)}))-(1\times G{\text{°}}_{\text{298}}{\text{(CH}}_3\text{OH(g)})+1\times G{\text{°}}_{\text{298}}\text{(CO(g)}))$

  $\Delta G{°}_{298}=[(-389)-(-163-137)]\text{ kJ/mol}$

  $\Delta G{°}_{298}=-89\text{ kJ/mol}$

The mathematical expression for the standard entropy at room temperature is:

  $\Delta S°=\Delta S{°}_{298}={\displaystyle \sum \text{n}}{\text{S°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{S°}}_{\text{298}}\text{(reactants)}$

The value of standard entropy for ${\text{CH}}_{\text{3}}\text{COOH(l)}$ is $160\text{ J/Kmol}$

The value of standard entropy for ${\text{CH}}_3\text{OH(g)}$ is $240\text{ J/Kmol}$

The value of standard entropy for $\text{CO}(g)$ is $198\text{ J/Kmol}$

Put the values,

  $\Delta S{°}_{298}=(1\times S{\text{°}}_{\text{298}}{\text{(CH}}_{\text{3}}\text{COOH(l)}))-(1\times S{\text{°}}_{\text{298}}{\text{(CH}}_3\text{OH(g)})+1\times S{\text{°}}_{\text{298}}\text{(CO}(g\text{)}))$

  $\Delta S{°}_{298}=[(160)-(240\text{ +198})]\text{ J/Kmol}$

  $\Delta S{°}_{298}=-278\text{ J/Kmol}$

The mathematical expression for the standard enthalpy at room temperature is:

  $\Delta H°=\Delta H{°}_{298}={\displaystyle \sum \text{n}}{\text{H°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{H°}}_{\text{298}}\text{(reactants)}$

The value of standard enthalpy for ${\text{CH}}_{\text{3}}\text{COOH(l)}$ is $-484\text{ kJ/mol}$

The value of standard enthalpy for ${\text{CH}}_3\text{OH(g)}$ is $-201\text{ kJ/mol}$

The value of standard enthalpy for $\text{CO}(g)$ is $-110.5\text{ kJ/mol}$

Put the values,

  $\Delta H{°}_{298}=(1\times H{\text{°}}_{\text{298}}{\text{(CH}}_{\text{3}}\text{COOH(l)}))-(1\times H{\text{°}}_{\text{298}}{\text{(CH}}_3\text{OH(g)})+1\times H{\text{°}}_{\text{298}}\text{(CO}(g)))$

  $\Delta H{°}_{298}=[(-484\text{)}-((-201)+(-110.5))]\text{ kJ/mol}$

  $\Delta H{°}_{298}=-173\text{ kJ/mol}$

On the basis of above calculation, second reaction is preferred over first reaction as the value of change in Gibbs free energy is negative that is $\Delta G{°}_{298}\text{<0}$ implies that the reaction is spontaneous. Thus, second reaction can be used commercially for the production of acetic acid.

Now, for calculating temperature:

Reaction is spontaneous at low temperatures and change in enthalpy favors the production of acetic acid which is exothermic implies $\Delta H°\text{<0}$ and thus, change in entropy favors the opposite process as positional entropy decreases. Thus, the temperature of a reaction at equilibrium is calculated by taking the value of change in Gibbs free energy which equals to zero.

Therefore,

  ${\text{ΔG}}_{}^{\text{o}}\text{=ΔH°-TΔS°}$

  $\text{0 =ΔH°-TΔS° }$

  $\text{T =}\frac{\text{ΔH}°}{\text{ΔS° }}$

Put the values from second reaction,

  $\text{T =}\frac{-173\text{ kJ}}{\text{-278 J/K }}\times \frac{1000\text{ J}}{1\text{ kJ}}$

= $622\text{ K}$

Therefore, at temperatures lower than 622 K, the magnitude of $\text{ΔH°}$ is smaller than $\text{TΔS°}$ , value of ${\text{ΔG}}_{}^{\text{o}}$ is negative. Hence, below 622 K, reaction will be spontaneous.