Chapter 6, Problem 6.90P

(a)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ for the fermentation and of respiration of sugar $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\right)$ is to be calculated.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum m\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

Answer to Problem 6.90P

$\Delta H_{\text{rxn}}^{°}$ for the fermentation and of respiration of sugar $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\right)$ is $-69.0\text{kJ}$ and $–2538.7\text{kJ}$ respectively.

Explanation of Solution

The balanced chemical equation for the respiration of sugar $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\right)$ is as follows:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\left(s\right)+6{\text{O}}_2\left(g\right)\to {\text{6CO}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)$

The standard state of oxygen is ${\text{O}}_2\left(g\right)$ so the standard enthalpy of formation for ${\text{O}}_2$ is zero.

The formula to calculate the standard enthalpy of a given reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{6\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]+6\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right\}\\ -\left\{1\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\left(s\right)\right]+6\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right\}\end{array}\right]$ (1)

Substitute $-393.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]$, $-241.8\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]$, $-1273.3\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\left(s\right)\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]$ in the equation (1).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(6\text{mol}\right)\left(-393.5\text{kJ/mol}\right)+\left(6\text{mol}\right)\left(-241.8\text{kJ/mol}\right)\right\}\\ -\left\{\left(1\text{mol}\right)\left(-1273.3\text{kJ/mol}\right)+\left(6\text{mol}\right)\left(0\right)\right\}\end{array}\right]\\ =–2538.656\text{kJ}\\ \approx –2538.7\text{kJ}\end{array}$

The balanced chemical equation for the fermentation of sugar $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\right)$ is as follows:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\left(s\right)\to {\text{2CO}}_2\left(g\right)+2{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right)$

The formula to calculate the standard enthalpy of a given reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{2\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]+2\Delta H_{\text{f}}^{°}\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right)\right]\right\}\\ -\left\{1\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\left(s\right)\right]\right\}\end{array}\right]$ (2)

Substitute $-393.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]$, $-277.63\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right)\right]$ and $-1273.3\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\left(s\right)\right]$ in the equation (2).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(2\text{mol}\right)\left(-393.5\text{kJ/mol}\right)+\left(2\text{mol}\right)\left(-277.63\text{kJ/mol}\right)\right\}\\ -\left\{\left(1\text{mol}\right)\left(-1273.3\text{kJ/mol}\right)\right\}\end{array}\right]\\ =-68.96\text{kJ}\\ \approx -69.0\text{kJ}\end{array}$

Conclusion

$\Delta H_{\text{rxn}}^{°}$ for the fermentation and of respiration of sugar $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\right)$ is $-69.0\text{kJ}$ and $–2538.7\text{kJ}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The combustion reaction for ethanol is to be written. Whether sugar or ethanol has higher value heat of combustion per mole of $\text{C}$ is to be determined.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum m\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

Answer to Problem 6.90P

The balanced chemical equation for the respiration of enthanol $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\right)$ is as follows:

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

Enthanol has higher value than sugar for heat of combustion per mole of $\text{C}$.

Explanation of Solution

The balanced chemical equation for the respiration of enthanol $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\right)$ is as follows:

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

The formula to calculate the standard enthalpy of a given reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{2\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]+3\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right\}\\ -\left\{1\Delta H_{\text{f}}^{°}\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right)\right]+3\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right\}\end{array}\right]$ (3)

Substitute $-393.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]$, $-241.8\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]$, $-277.63\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right)\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]$ in the equation (3).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(\text{2}\text{mol}\right)\left(-393.5\text{kJ/mol}\right)+\left(3\text{mol}\right)\left(-241.8\text{kJ/mol}\right)\right\}\\ -\left\{\left(\text{1}\text{mol}\right)\left(-277.63\text{kJ/mol}\right)+\left(\text{3}\text{mol}\right)\left(0\right)\right\}\end{array}\right]\\ =-1234.848\text{kJ}\\ \approx -1234.8\text{kJ}\end{array}$

The formula to calculate the heat of combustion per mole of $\text{C}$ in sugar is,

$\text{Heat of combustion/C}=\left(\frac{\text{heat of combustion}}{\text{mole of carbon atom in sugar}}\right)$ (4)

Sustitute $–2538.7\text{kJ}$ for heat of combustion and $6\text{mol}$ for mole of carbon atom in sugar in equation (4).

$\begin{array}{c}\text{Heat of combustion/mol}\text{C}=\left(\frac{–2538.7\text{kJ}}{6\text{mol}\text{C}}\right)\\ =–423.1093\text{kJ/mol}\text{C}\\ \approx –423.11\text{}\text{kJ/mol}\text{C}\end{array}$

The formula to calculate the heat of combustion per mole of $\text{C}$ in ethanol is,

$\text{Heat of combustion/C}=\left(\frac{\text{heat of combustion}}{\text{mole of carbon atom in ethanol}}\right)$ (5)

Sustitute $-1234.8\text{kJ}$ for heat of combustion and $2\text{mol}$ for mole of carbon atom in ethanol in equation (5).

$\begin{array}{c}\text{Heat of combustion/mol}\text{C}=\left(\frac{-1234.8\text{kJ}}{2\text{mol}\text{C}}\right)\\ =–617.424\text{kJ/mol}\text{C}\\ \approx –617.42\text{}\text{kJ/mol}\text{C}\end{array}$

Hence, enthanol has higher value.

Conclusion

The balanced chemical equation for the respiration of enthanol $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\right)$ is as follows:

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

Enthanol has higher value than sugar for heat of combustion per mole of $\text{C}$.