Chapter 6, Problem 6.61P

(a)

Interpretation Introduction

Interpretation:The value of $\Delta H^0{}_{rxn}$ of fermentation and respiration needs to be determined.

Concept Introduction : The reactions of fermentation, respiration and combustion of glucose are as follows:

1. Fermentation of glucose

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}(s)\to 2{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH(}l{\text{)+2CO}}_2(g)$

2. Respiration

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}(s)+6{\text{O}}_{\text{2}}(g)\to 6{\text{CO}}_2(g)+6{\text{H}}_{\text{2}}\text{O(}g\text{)}$

The amount of energy released on combustion of an organic compound is known as enthalpy of combustion. The formula of determination of $\Delta H_{\text{rxn}}$ is as follows:

  $\Delta H^0{}_{\text{rxn}}={\displaystyle \sum _{}^{}m\Delta H^0{}_{\text{f}}(\text{products})-}{\displaystyle \sum n\Delta H^0{}_{\text{f}}}(\text{reactants})$

Here, ${\displaystyle \sum }$ represents the sum of terms, m and n are number of moles of products and reactants respectively. The term $H^0{}_{\text{f}}$ is standard heat of formation.

Answer to Problem 6.61P

  $\Delta H^0{}_{rxn}$ of fermentation reaction is $-68.96\text{kJ}$

  $\Delta H^0{}_{rxn}$ of respiration reaction is $-2538.656\text{kJ}$ .

Explanation of Solution

The value of standard heat of formation of compounds is as follows:

  $\begin{array}{c}{H}^0{}_{\text{f}}({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}(s))=-1273.3\text{kJ}{\text{mol}}^{-1}\\ H^0{}_{\text{f}}({\text{CO}}_2)=-393.5\text{kJ}{\text{mol}}^{-1}\\ H^0{}_{\text{f}}({\text{H}}_2\text{O(g)})=-241.826\text{kJ}{\text{mol}}^{-1}\\ H^0{}_{\text{f}}({\text{O}}_2)=0\text{kJ}{\text{mol}}^{-1}\end{array}$

Now calculate $\Delta H^0{}_{rxn}$ of fermentation reaction.

Fermentation of glucose

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}(s)\to 2{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH(}l{\text{)+2CO}}_2(g)$

  $\begin{array}{c}\Delta H^0{}_{rxn}=\left[2\text{mol}\Delta H^0{}_{\text{f}}({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH(}l\text{)})+2\text{mol}\Delta H^0{}_{\text{f}}({\text{CO}}_2\text{(}g\text{)})\right]-\left[1\text{mol}\Delta H^0{}_{\text{f}}({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\text{(}s\text{)})\right]\\ =\left[2\text{mol}\times \left(-277.63\frac{\text{kJ}}{\text{mol}}\right)+2\text{mol}\times \left(-393.5\frac{\text{kJ}}{\text{mol}}\right)\right]-\left[1\text{mol}\times \left(-1273.3\frac{\text{kJ}}{\text{mol}}\right)\right]\\ =-555.26\text{kJ}-787\text{kJ+1273}\text{.3}\text{kJ}\\ \text{=}-68.96\text{kJ}\end{array}$

So, $\Delta H^0{}_{rxn}$ of fermentation reaction is $-68.96\text{kJ}$ .

Now calculate $\Delta H^0{}_{rxn}$ of respiration reaction.

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}(s)+6{\text{O}}_{\text{2}}(g)\to 6{\text{CO}}_2(g)+6{\text{H}}_{\text{2}}\text{O(}g\text{)}$

  $\begin{array}{c}\Delta H^0{}_{rxn}=\left[6\text{mol}\Delta H^0{}_{\text{f}}({\text{CO}}_2\text{(}g\text{)})+6\text{mol}\Delta H^0{}_{\text{f}}({\text{H}}_2\text{O(}g\text{)})\right]-\left[\begin{array}{l}1\text{mol}\Delta H^0{}_{\text{f}}({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\text{(}s\text{)})+\\ 6\text{mol}\Delta H^0{}_{\text{f}}({\text{O}}_2\text{(}g\text{)})\end{array}\right]\\ =\left[6\text{mol}\times \left(-393.5\frac{\text{kJ}}{\text{mol}}\right)+6\text{mol}\times \left(-241.826\frac{\text{kJ}}{\text{mol}}\right)\right]-\left[\begin{array}{l}1\text{mol}\times \left(-1273.3\frac{\text{kJ}}{\text{mol}}\right)\\ +6\text{mol}\times 0\end{array}\right]\\ =-2361\text{kJ}-1450.956\text{kJ+1273}\text{.3}\text{kJ}\\ \text{=}-2538.656\text{kJ}\end{array}$

Hence, $\Delta H^0{}_{rxn}$ of respiration reaction is $-2538.656\text{kJ}$ .

(b)

Interpretation Introduction

Interpretation:A combustion reaction for ethanol needs to be determined and whether ethanol or sugar has the higher $\Delta H^0{}_{rxn}$ for combustion per mole of C needs to be determined.

Concept Introduction : The amount of energy released on combustion of an organic compound is known as enthalpy of combustion. The formula of determination of $\Delta H_{\text{rxn}}$ is as follows:

  $\Delta H^0{}_{\text{rxn}}={\displaystyle \sum _{}^{}m\Delta H^0{}_{\text{f}}(\text{products})-}{\displaystyle \sum n\Delta H^0{}_{\text{f}}}(\text{reactants})$

Here, ${\displaystyle \sum }$ represents the sum of terms, m and n are number of moles of products and reactants respectively. The term $H^0{}_{\text{f}}$ is standard heat of formation.

Answer to Problem 6.61P

The balanced chemical reaction for combustion of ethanol is,

  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH(}l{\text{)+3O}}_2(g)\to 2{\text{CO}}_{\text{2}}(g)+3{\text{H}}_{\text{2}}\text{O}$

Ethanol has higher $\Delta H^0{}_{rxn}$ for the combustion of per mole of carbon than sugar.

Explanation of Solution

The balanced chemical reaction for combustion of ethanol is,

  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH(}l{\text{)+3O}}_2(g)\to 2{\text{CO}}_{\text{2}}(g)+3{\text{H}}_{\text{2}}\text{O}$

Use appendix B to get the value of standard heat of formation of compound.

  $\begin{array}{c}{H}^0{}_{\text{f}}({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH(}l\text{)})=-277.63\text{kJ}{\text{mol}}^{-1}\\ H^0{}_{\text{f}}({\text{CO}}_2)=-393.5\text{kJ}{\text{mol}}^{-1}\\ H^0{}_{\text{f}}({\text{H}}_2\text{O(g)})=-241.826\text{kJ}{\text{mol}}^{-1}\\ H^0{}_{\text{f}}({\text{O}}_2)=0\text{kJ}{\text{mol}}^{-1}\end{array}$

Putting the values,

  $\begin{array}{c}\Delta H^0{}_{rxn}=\left[2\text{mol}\Delta H^0{}_{\text{f}}({\text{CO}}_2\text{(}g\text{)})+3\text{mol}\Delta H^0{}_{\text{f}}({\text{H}}_2\text{O(}g\text{)})\right]-\left[\begin{array}{l}1\text{mol}\Delta H^0{}_{\text{f}}({\text{CH}}_3{\text{CH}}_2\text{OH(}l\text{)})+\\ \text{3}\text{mol}\Delta H^0{}_{\text{f}}({\text{O}}_2\text{(}g\text{)})\end{array}\right]\\ =\left[2\text{mol}\times \left(-393.5\frac{\text{kJ}}{\text{mol}}\right)+3\text{mol}\times \left(-241.826\frac{\text{kJ}}{\text{mol}}\right)\right]-\left[\begin{array}{l}1\text{mol}\times \left(-277.63\frac{\text{kJ}}{\text{mol}}\right)\\ +3\text{mol}\times 0\end{array}\right]\\ =-787\text{kJ}-725.478\text{kJ+277}\text{.63}\text{kJ}\\ \text{=}-1234.848\text{kJ}\end{array}$

Thus, standard heat of reaction for combustion of ethanol per mole will be $-1234.848\text{kJ}$

The molar relationship of carbon in sugar and ethanol are as follows:

  $\begin{array}{l}1\text{mol}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}=6\text{mol}\text{C}\\ \text{1}\text{mol}{\text{CH}}_3{\text{CH}}_{\text{2}}\text{OH=2}\text{mol}\text{C}\end{array}$

Use the molar relationships and heats of reaction of sugar and ethanol to determine the standard heat of reaction for the combustion of per mole of carbon.

  $\begin{array}{c}\Delta H^0{}_{\text{rxn}}=1.00\text{mol}\text{C}\times \frac{1\text{mol}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}{6\text{mol C}}\times \frac{-2538.6\text{kJ}}{1\text{mol}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}\\ =-423.1\text{kJ}\text{(From}{\text{C}}_6{\text{H}}_{12}{\text{O}}_6\text{)}\\ \Delta H^0{}_{\text{rxn}}=1.00\text{mol}\text{C}\times \frac{1\text{mol}{\text{CH}}_3{\text{CH}}_2\text{OH}}{2\text{mol C}}\times \frac{-1234.8\text{kJ}}{1\text{mol}{\text{CH}}_3{\text{CH}}_2\text{OH}}\\ =-617.4\text{kJ}\text{(From}{\text{CH}}_3{\text{CH}}_2\text{OH)}\end{array}$

Hence, ethanol has higher $\Delta H^0{}_{rxn}$ for the combustion of per mole of carbon than sugar.