Chapter 5, Problem 137AP

Both glucose and fructose are simple sugars with the same molecular formula of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$. Sucrose $\left({\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\right)$, or table sugar, consists of a glucose molecule bonded to a fructose molecule (a water molecule is eliminated in the formation of sucrose), (a) Calculate the energy released when a 2.0-g glucose tablet is burned in air. (b) To what height can a 65-kg person climb after ingesting such a tablet, assuming only 30 percent of the energy released is available for work? (See the hint foe Problem 5.123.) Repeat the calculations foe a 2.0-g sucrose tablet.

Interpretation Introduction

Interpretation:

The energy released in the burning of glucose tabletand the height at which the person can climb after eating this tabletareto be calculated.

Concept Introduction:

The standard enthalpy change for a reaction is the amount of enthalpy change thatoccurs at the standard conditions.

The standard enthalpy of reaction is determined using the equation as follows:

$\Delta H{°}_{rxn}={\displaystyle \sum n}\Delta H_f^{°}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_f^{°}\left(\text{reactants}\right).$

Here, the stoichiometric coefficients are represented by m for reactants and n for products and the enthalpies of formation at standard conditions are represented by $\Delta H_f^{°}$.

The standard enthalpy of formation is the amount of heat change when one mole of compound is formed from its integral elements that are present in their standard states.

The value of the enthalpy of formation of an element is zero at its most stable state.

The work done is calculated as follows:

$w=mgh,$

where m is the mass, g is the gravitational constant, h is the height, and w is the work done.

Answer to Problem 137AP

Solution:

a)

The energy released on burning $2\text{ g}$ of glucose is $31\text{ kJ}$ and on burning $2\text{ g}$ of sucrose is $33\text{ kJ}$.

b)

The height that a person can climb after ingesting $2\text{ g}$ of glucose is $\text{15 m}$, and $\text{16 m}$ after ingesting $2\text{ g}$ of sucrose.

Explanation of Solution

a)The energy released when 2.0 g glucose tablet is burned in air.

The reaction involved in photosynthesis is given as follows:

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

From appendix 2, the enthalpy of formation values of the given compounds are given as follows:

$\begin{array}{c}\Delta H{°}_f\left[{\text{CO}}_2\left(g\right)\right]=-393.5\text{ kJ}/\text{mol;}\\ \Delta H{°}_f\left[{\text{H}}_2\text{O}\left(l\right)\right]=-285.8\text{ kJ}/\text{mol;}\\ \Delta H{°}_f\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\left(s\right)\right]=-1274.5\text{ kJ}/\text{mol}\text{.}\end{array}$

The oxygen gas is in its most stable form, so its enthalpy of formation is zero.

The standard enthalpy of reaction is calculated by the expression, which isas follows:

$\Delta H{°}_{rxn}={\displaystyle \sum n}\Delta H_f^{°}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_f^{°}\left(\text{reactants}\right)$.

The standard enthalpy of the given reaction is calculated as follows:

$\begin{array}{c}\Delta H{°}_{rxn}=\left\{\Delta H{°}_f\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6\left(s\right)\right]+6\Delta H{°}_f\left[{\text{O}}_2\left(g\right)\right]\right\}-\left\{\Delta H{°}_f\left[{\text{CO}}_2\left(g\right)\right]+\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{O}\left(l\right) \right]\right\}\\ =\left\{\left(-1274.5\text{ kJ}/\text{mol}\right)+6\left(0\right)\right\}-\left[6\left(-393.5\text{ kJ}/\text{mol}\right)+6\left(-285.8\text{ kJ}/\text{mol}\right)\right]\\ =\left(-1274.5\text{ kJ}/\text{mol}\right)-\left(-2361\text{ kJ}/\text{mol}-1714.8\text{ kJ}/\text{mol}\right)\\ =2801.3\text{ kJ}/\text{mol}\text{.}\end{array}$

So, the heat of the combustion of glucose will be reverse in sign as compared to theenthalpy change of photosynthesis.

$\Delta H°=-2801.3\text{ KJ}/\text{mol}\text{.}$

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

The molar mass of glucose is $M_{\text{glucose}}=180.2\text{ g}/\text{mol}\text{.}$

So, the energy released for $2\text{ g}$ of glucose is calculated by the expression, which is as follows:

$q=\frac{m_{\text{glucose}}}{M_{\text{glucose}}}\times \Delta H{°}_{rxn}.$

Substitute $2\text{ g}$ for $m_{\text{glucose}}$ and $180.2\text{ g}/\text{mol}$ for $M_{\text{glucose}}$ in the above equation.

$\begin{array}{c}q=\frac{2\cancel{\text{g}}}{180.2\cancel{\text{g}}/\cancel{\text{mol}}}\times 2801.3\text{ kJ}/\cancel{\text{mol}}\\ =0.011\times 2801.3\text{ kJ}\\ \approx 31\text{ kJ}\text{.}\end{array}$

From appendix 2, the following enthalpy of formation values of the given compounds are given as:

$\Delta H{°}_f\left[{\text{C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(s\right)\right]=-5644\text{ kJ}/\text{mol}\text{.}$

The mass of sucrose given is $m_{\text{sucrose}}=2\text{ g}\text{.}$

The molar mass of sucrose is $M_{\text{sucrose}}=342.3\text{ g}/\text{mol}\text{.}$

So, the energy released for $2\text{ g}$ of sucrose is calculated by the expression, which is as follows:

$q=\frac{m_{\text{sucrose}}}{M_{\text{sucrose}}}\times \Delta H{°}_{rxn}.$

Substitute $2\text{ g}$ for $m_{\text{sucrose}}$ and $342.3\text{ g}/\text{mol}$ for $M_{\text{sucrose}}$ in the above equation.

$\begin{array}{c}q=\frac{2\text{ g}}{342.3\text{ g}/\text{mol}}\times 5644\text{ kJ}/\text{mol}\\ =0.00058\times 5644\text{ kJ}\\ =32.98\text{ kJ}\\ \approx 33\text{ kJ}\text{.}\end{array}$

b)The height that a 65 kg person can climb after ingesting given tablet.

The energy offered for work is 30% of the energy released. So, the work done forglucose is calculated as follows:

$\begin{array}{c}{w}_g=31\text{ kJ}\times \frac{30}{100}\\ =31\times 0.3\text{ kJ}\\ =9.3\text{ kJ}\\ =9.3\times {10}^3\text{ J}\text{.}\end{array}$

The work done is calculated by the expression, which is as follows:

$w=mgh.$

This can be rewritten as follows:

$h=\frac{w_g}{mg}.$ …… (1)

Substitute $9.8\text{ m}/{\text{s}}^2$ for $g$, $65\text{ Kg}$ for $m$ and $9.3\times {10}^3\text{ J}$ for $w_g$ in the equation (1).

$\begin{array}{c}h=\frac{9.3\times {10}^3\text{ J}}{\left(65\text{ kg}\right)\left(9.8\text{ m}/{\text{s}}^2\right)}\\ =\frac{9.3\times {10}^3\cancel{\text{ kg}\text{.}}{\text{m}}^2/\cancel{{\text{s}}^2}}{637\cancel{\text{kg}}\text{.m}/\cancel{{\text{s}}^2}}\\ =0.0146\times {10}^3\text{ m}\\ \approx \text{15 m}\text{.}\end{array}$

Similarly, the energy offered for work is 30% of the energy released. So, the work for sucrose is calculated as follows:

$\begin{array}{c}{w}_s=33\text{ kJ}\times \frac{30}{100}\\ =33\times 0.3\text{ kJ}\\ =9.9\text{ kJ}\\ =9.9\times {10}^3\text{ J}\text{.}\end{array}$

Substitute $9.8\text{ m}/{\text{s}}^2$ for $g$, $65\text{ kg}$ for $m$ and $9.9\times {10}^3\text{ J}$ for $w_g$ in the equation (1).

$\begin{array}{c}h=\frac{9.9\times {10}^3\text{ J}}{\left(65\text{ kg}\right)\left(9.8\text{ m}/{\text{s}}^2\right)}\\ =\frac{9.9\times {10}^3\cancel{\text{kg}}{\text{.m}}^2/\cancel{{\text{s}}^2}}{637\cancel{\text{kg}}\text{.m}/\cancel{{\text{s}}^2}}\\ =0.0156\times {10}^3\text{ m}\\ \approx \text{16 m}\text{.}\end{array}$