Chapter 2, Problem 2C.11P

(a)

Interpretation Introduction

Interpretation: The values of the standard molar enthalpy of combustion, the standard internal energy of combustion and the standard enthalpy of formation of glucose have to be calculated.

Concept Introduction: Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.  The measure of amount of heat supplied to the system gives the enthalpy change of the system.

In the process of combustion, the enthalpy change under standard conditions refers to the enthalpy change which occurs when one mole of a compound is burned completely in air.

Answer to Problem 2C.11P

The standard molar enthalpy for the combustion of glucose is $\underset{\_}{-2806.17kJmo{l}^{-1}}$.

The standard internal energy of combustion is $\underset{\_}{-2806.10kJmo{l}^{-1}}$.

The standard enthalpy of formation of glucose is $\underset{\_}{-1269.81kJmo{l}^{-1}}$.

Explanation of Solution

(i) The chemical equation for the combustion reaction of glucose is,

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

The given mass of glucose is $0.321\text{2 g}$.

The calorimeter constant is $64{\text{1 JK}}^{-1}$.

The temperature of bomb calorimeter is $\text{298 K}$.

The raise in temperature in bomb calorimeter is $\text{7}\text{.793 K}$.

First, the number of moles of glucose in the sample is calculated by the formula,

    $\text{Moles}=\frac{\text{Mass}}{\text{Molar mass}}$                                                                                    (1)

The molar mass of glucose is $180.15{\text{6 g mol}}^{-1}$.

Substitute the values of mass and molar mass in equation (1) to calculate the moles of glucose.

    $\begin{array}{c}\text{Moles}=\frac{0.321\text{2 g}}{180.15{\text{6 g mol}}^{-1}}\\ =1.78\times {10}^{-3}\text{ mol}\end{array}$

The enthalpy of combustion for the sample is calculated by the formula,

    ${\Delta }_{c}H=\text{Calorimeter constant}\times \Delta T$                                                                 (2)

Substitute the value of calorimeter constant and change in temperature in equation (1) to calculate the enthalpy of combustion for the sample.

    $\begin{array}{c}{\Delta }_{c}H=64{\text{1 JK}}^{-1}\times \text{7}\text{.793 K}\\ =4995.\text{3 J}\end{array}$

The conversion of J to kJ is done as,

    $\text{1 J}={10}^{-3}\text{ kJ}$

Therefore, the conversion of $4995.\text{3 J}$ into kJ is done as,

    $\begin{array}{c}4995.\text{3 J}=4995.\text{3}\times {10}^{-3}\text{ kJ}\\ =4.99\text{5 kJ}\end{array}$

Since, the heat lost during the combustion of glucose is equal to the heat gained by the calorimeter.  Therefore,

    $\begin{array}{l}{q}_c=q_{\text{gained}}\\ q_c=-4.99\text{5 kJ}\end{array}$

Now, the standard molar enthalpy for the combustion of glucose is calculated as,

    $\begin{array}{c}{\Delta }_{c}H=\frac{q_c}{\text{moles}}\\ =\frac{-4.99\text{5 kJ}}{1.78\times {10}^{-3}\text{ mol}}\\ =\underset{\_}{-2806.17kJmo{l}^{-1}}\end{array}$

Hence, the standard molar enthalpy for the combustion of glucose is $\underset{\_}{-2806.17kJmo{l}^{-1}}$.

(ii) The reaction takes place at constant volume, the change in volume is $\Delta V=0$.

The relation between internal energy and enthalpy is given by the expression,

    $H=U+pV$

For $n$ mol of gas, the ideal gas equation is,

    $pV=nRT$

Substitute the value of $pV$ in the above equation.

    $H=U+nRT$

The enthalpy of combustion is given by the expression,

    ${\Delta }_{c}H={\Delta }_{c}U+nR{\Delta }_{c}T$

Therefore, the internal energy of combustion is given by the expression,

    ${\Delta }_{c}U={\Delta }_{c}H-nR{\Delta }_{c}T$

The standard internal energy of combustion is calculated by the formula,

    ${\Delta }_c{U}^{\text{ο}}={\Delta }_c{H}^{\text{ο}}-R{\Delta }_{c}T$                                                                                    (3)

Substitute the values of ${\Delta }_c{H}^{\text{ο}}$, $R$ and ${\Delta }_{c}T$ in equation (3) to calculate the standard internal energy of combustion.

    $\begin{array}{c}{\Delta }_c{U}^{\text{ο}}=-\text{2806}{\text{.17 kJ mol}}^{-1}-8.31\text{4}\times {\text{10}}^{-3}{\text{ KJ K}}^{-1}{\text{ mol}}^{-1}\times -7.79\text{3 K}\\ =-\text{2806}{\text{.17 kJ mol}}^{-1}+0.06479\\ =\underset{\_}{-2806.10kJmo{l}^{-1}}\end{array}$

Hence, standard internal energy of combustion is $\underset{\_}{-2806.10kJmo{l}^{-1}}$.

(iii) The standard enthalpy of combustion is given by the expression,

    ${\Delta }_{\text{c}}{H}^{\text{ο}}={\displaystyle \sum m\cdot \Delta H_{\text{f}}^{\text{ο}}}\left(\text{products}\right)-{\displaystyle \sum n\cdot \Delta H_{\text{f}}^{\text{ο}}}\left(\text{reactants}\right)$                                (4)

Where,

• • $\Delta H_{\text{f}}^{\text{ο}}\left(\text{products}\right)$ is the standard enthalpy of formation of products.
• • $\Delta H_{\text{f}}^{\text{ο}}\left(\text{reactants}\right)$ is the standard enthalpy of formation of reactants.
• • $m$ is the total moles of products.
• • $n$ is the total moles of reactants.

The chemical equation for the combustion reaction of glucose is,

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

For standard enthalpy of combustion of glucose, equation (1) is written as,

    ${\Delta }_{\text{c}}{H}^{\text{ο}}=\left(6\times \Delta H_{\text{f}}^{\text{ο}}\left({\text{CO}}_{\text{2}}\right)+6\times \Delta H_{\text{f}}^{\text{ο}}\left({\text{H}}_{\text{2}}\text{O}\right)-\left(1\times \Delta H_{\text{f}}^{\text{ο}}{\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6\left(\text{s}\right)+6\times \Delta H_{\text{f}}^{\text{ο}}\left({\text{O}}_{\text{2}}\right)\right)\right)$

The standard enthalpy of formation of ${\text{CO}}_{\text{2}}$ is $-393.5{\text{ kJ mol}}^{-1}$.

The standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}$ is $-285.83{\text{ kJ mol}}^{-1}$.

The standard enthalpy of formation of ${\text{O}}_2$ is ${\text{0 kJ mol}}^{-1}$.

The standard enthalpy for the combustion of glucose is $-\text{2806}{\text{.17 kJ mol}}^{-1}$.

Substitute the values of standard enthalpy of formation of reactants and products in the above equation to calculate the standard enthalpy of formation of glucose.

    $\begin{array}{c}-\text{2806}{\text{.17 kJ mol}}^{-1}=\left(\begin{array}{l}6\times -393.5{\text{ kJ mol}}^{-1}+6\times -285.83{\text{ kJ mol}}^{-1}-\\ \left(1\times \Delta H_{\text{f}}^{\text{ο}}{\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6\left(\text{s}\right)+0{\text{ kJ mol}}^{-1}\right)\end{array}\right)\\ -\text{2806}{\text{.17 kJ mol}}^{-1}=\left(-2361-1714.98\right){\text{kJ mol}}^{-1}-\Delta H_{\text{f}}^{\text{ο}}{\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6\left(\text{s}\right)\\ -\text{2806}{\text{.17 kJ mol}}^{-1}=-4075.9{\text{8 kJ mol}}^{-1}-\Delta H_{\text{f}}^{\text{ο}}{\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6\left(\text{s}\right)\\ \Delta H_{\text{f}}^{\text{ο}}{\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6\left(\text{s}\right)=-4075.9{\text{8 kJ mol}}^{-1}+\text{2806}{\text{.17 kJ mol}}^{-1}\end{array}$

Simplify the above equation,

    $\Delta H_{\text{f}}^{\text{ο}}{\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6\left(\text{s}\right)=\underset{\_}{-1269.81kJmo{l}^{-1}}$

Hence, the standard enthalpy of formation of glucose is $\underset{\_}{-1269.81kJmo{l}^{-1}}$.

(b)

Interpretation Introduction

Interpretation: The biological advantage of complete aerobic oxidation than the anaerobic glycolysis to lactic acid has to be stated.

Concept Introduction: Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.  The measure of amount of heat supplied to the system gives the enthalpy change of the system.

Answer to Problem 2C.11P

The complete aerobic oxidation has biological advantage as compared to the anaerobic glycolysis to lactic acid as more amount of energy is released in complete oxidation of glucose as compared to the energy released in glycolysis.

The biological advantage is $\underset{\_}{-2687.98kJmo{l}^{-1}}$.

Explanation of Solution

The complete aerobic oxidation has biological advantage as compared to the anaerobic glycolysis to lactic acid.  The anaerobic glycolysis reaction of glucose takes place in the muscles.  The formation of lactic acid is shown by the reaction as,

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\to {\text{2CH}}_{\text{3}}\text{C}\left(\text{OH}\right)\text{COOH}$

The molar enthalpy of formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is $-\text{1269}{\text{.81 kJ mol}}^{-1}$.

The molar enthalpy of formation of ${\text{2CH}}_{\text{3}}\text{C}\left(\text{OH}\right)\text{COOH}$ $-694{\text{ kJ mol}}^{-1}$

Therefore, the energy amount of released per mole due to combustion is calculated as follows,

    $\begin{array}{c}{\Delta }_{\text{c}}{H}^{\text{ο}}={\displaystyle \sum m\cdot \Delta H_{\text{f}}^{\text{ο}}}\left(\text{products}\right)-{\displaystyle \sum n\cdot \Delta H_{\text{f}}^{\text{ο}}}\left(\text{reactants}\right)\\ =2\times -694{\text{ kJ mol}}^{-1}-\left(-\text{1269}{\text{.81 kJ mol}}^{-1}\right)\\ =\left(-1388+\text{1269}\text{.81}\right){\text{kJ mol}}^{-1}\\ =-118.1{\text{9 kJ mol}}^{-1}\end{array}$

Hence, the biological advantage of complete aerobic oxidation over aerobic glycolysis is,

    $\begin{array}{l}=-\text{2806}{\text{.17 kJ mol}}^{-1}-\left(-118.1{\text{9 kJ mol}}^{-1}\right)\\ =\underset{\_}{-2687.98kJmo{l}^{-1}}\end{array}$

Thus, the biological advantage is $\underset{\_}{-2687.98kJmo{l}^{-1}}$.

The complete aerobic oxidation is a preferred method of pyruvate breakdown from glycolysis because it requires oxygen to generate energy.  It also requires that pyruvate enter the mitochondria to be fully oxidized by Krebs cycle.  On the other hand, in the absence of oxygen, pyruvate is not metabolized by the process of cellular respiration but undergoes fermentation process.  The pyruvate is not transported into the mitochondria and remains in the cytoplasm.  In the cytoplasm, it is converted into waste products that may be removed from the cell.  Hence, the complete aerobic oxidation has biological advantage as compared to the anaerobic glycolysis to lactic acid as more amount of energy is released in complete oxidation of glucose as compared to the energy released in glycolysis.