Chapter 6, Problem 6.87P

(a)

Interpretation Introduction

Interpretation:

A balanced equation for the combustion of stearic acid is to be written.

Concept introduction:

The standard enthalpy of formation $\left(\Delta H_{\text{f}}^{°}\right)$ is the enthalpy change of the formation of $1\text{mol}$ of compound from its elements and the reactants and products, in this case, is also present in their standard state.

Standard state includes $1\text{atm}$ pressure for gases, $1M$ concentration for substances in the aqueous phase and the most stable form of substance at $1\text{atm}$ pressure and $298\text{K}$ temperature. For example, metal like sodium exists in solid state and molecular elements like halogen exist in a gaseous state.

Answer to Problem 6.87P

A balanced equation for the combustion of stearic acid is as follows:

${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}\left(s\right)+26{\text{O}}_2\left(g\right)\to 18{\text{CO}}_2\left(g\right)+18{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

Combustion reaction involves the reaction of stearic acid $\left({\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}\right)$ with an oxygen molecule to form carbon dioxide and the water molecule

One mole of stearic acid $\left({\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}\right)$ reacts with twenty-four moles of ${\text{O}}_2$ to form eighteen moles of ${\text{CO}}_2$ and eighteen moles of ${\text{H}}_{\text{2}}\text{O}$.

A balanced equation for the combustion of stearic acid is as follows:

${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}\left(s\right)+26{\text{O}}_2\left(g\right)\to 18{\text{CO}}_2\left(g\right)+18{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Conclusion

A balanced equation for the combustion of stearic acid is as follows:

${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}\left(s\right)+26{\text{O}}_2\left(g\right)\to 18{\text{CO}}_2\left(g\right)+18{\text{H}}_{\text{2}}\text{O}\left(g\right)$

(b)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ for the combustion of stearic acid 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 6.87P

$\Delta H_{\text{rxn}}^{°}$ for the combustion of stearic acid is $-10488\text{kJ}$.

Explanation of Solution

The given balanced chemical equation is as follows:

${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}\left(s\right)+26{\text{O}}_2\left(g\right)\to 18{\text{CO}}_2\left(g\right)+18{\text{H}}_{\text{2}}\text{O}\left(g\right)$

One mole of stearic acid $\left({\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}\right)$ reacts with twenty-four moles of ${\text{O}}_2$ to form eighteen moles of ${\text{CO}}_2$ and eighteen moles of ${\text{H}}_{\text{2}}\text{O}$.

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

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{18\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(l\right)\right]+18\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right\}-\\ \left\{\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}\right]+26\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right\}\end{array}\right]$ (1)

Substitute $-393.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(l\right)\right]$, $-241.826\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]$, $-948\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]$ in the equation (1).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left(18\text{mol}\right)\left(-393.5\text{kJ/mol}\right)+\left(18\text{mol}\right)\left(-241.826\text{kJ/mol}\right)-\\ \left\{\left(1\text{mol}\right)\left(-948\text{kJ/mol}\right)+\left(26\text{mol}\right)\left(0\right)\right\}\end{array}\right]\\ =-10478.868\text{kJ}\\ \approx -10488\text{kJ}\end{array}$

Conclusion

$\Delta H_{\text{rxn}}^{°}$ for the combustion of stearic acid is $-10488\text{kJ}$.

(c)

Interpretation Introduction

Interpretation:

The heat that is released in $\text{kJ}$ and $\text{kcal}$ when $1.00\text{ g}$ of stearic acid is burned completely is to be calculated.

Concept introduction:

Heat $\left(q\right)$ is the energy transferred as a consequence of the difference of the temperature between the system and surrounding.

The heat released by the system is negative and the heat taken by the system is positive.

The amount of heat released by the system is equal to the amount of amount absorbed by the surrounding.

The expression to convert $\text{kcal}$ into $\text{kJ}$ is:

$\text{1}\text{kcal}=4.184\text{kJ}$

Answer to Problem 6.87P

The heat released in $\text{kJ}$ and $\text{kcal}$ is $-36.9\text{kJ}$ and $-8.81\text{kcal}$ respectively.

Explanation of Solution

The formula to calculate the moles of stearic acid $\left({\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}\right)$ is as follows:

${\text{Moles of C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}=\left(\frac{{\text{ given mass of C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}}{{\text{molar mass of C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}}\right)$ (2)

Substitute $1.00\text{ g}$ for given mass of ${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$ and $284.47\text{g/mol}$ for molar mass of ${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$ in the equation (2).

$\begin{array}{c}{\text{Moles of C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}=\left(\frac{1.00\text{ g}}{284.47\text{g/mol}}\right)\\ =0.00351530\text{mol}\end{array}$

The heat released by combustion of $1\text{mol}$ of ${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$ is $-10488\text{kJ}$.

Therefore, the heat released by combustion of $0.00351530\text{mol}$ of ${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$ is as follows:

$\begin{array}{c}\text{Heat}\text{released}=\left(0.00351530\text{moll}\right)\left(\frac{-10488\text{kJ}}{\text{1}\text{mol}}\right)\\ =-36.8631\text{kJ}\\ \approx -36.9\text{kJ}\end{array}$

The heat released by combustion of $0.00351530\text{mol}$ of ${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$ in $\text{kcal}$ is as follows:

$\begin{array}{c}\text{Heat}\text{released}=-36.8631\text{kJ}\left(\frac{\text{1}\text{kcal}}{4.184\text{kJ}}\right)\\ =-8.81168\text{kcal}\\ \approx -8.81\text{kcal}\end{array}$

Conclusion

The heat released in $\text{kJ}$ and $\text{kcal}$ is $-36.9\text{kJ}$ and $-8.81\text{kcal}$ respectively.

(d)

Interpretation Introduction

Interpretation:

Whether the data that a candy containing $11.0\text{ g}$ fat and provides $100\text{ Cal}$ energy is consistent with the answer calculated in part (c) is to be determined.

Concept introduction:

Heat $\left(q\right)$ is the energy transferred as a consequence of the difference of the temperature between the system and surrounding.

The heat released by the system is negative and the heat taken by the system is positive.

The amount of heat released by the system is equal to the amount of amount absorbed by the surrounding.

The conversion factor to convert nutritional calorie $\left(\text{Calorie}\right)$ into $\text{kcal}$ is:

$\text{1}\text{Calorie}=1\text{kcal}$

Answer to Problem 6.87P

Yes, the data is consistent with the given package information.

Explanation of Solution

The heat released by $1\text{g}$ of fat is $-8.81168\text{kcal}$.

The formula to calculate heat released by $11\text{}\text{g}$ of fat is as follows:

$\text{Heat released}=\left(\text{mass of fat}\right)\left(\frac{-8.81168\text{kcal}}{1\text{g fat}}\right)$ (3)

Substitute $11\text{}\text{g}$ for the mass of fat in the equation (3).

$\begin{array}{c}\text{Heat released}=\left(11\text{}\text{g}\right)\left(\frac{-8.81168\text{kcal}}{1\text{g fat}}\right)\\ =-96.9286\text{kcal}\\ \approx -96.9\text{kcal}\end{array}$

The heat released in $\text{Cal}$ is,

$\begin{array}{c}\text{Heat released}=96.9\text{kcal}\left(\frac{1\text{Cal}}{\text{1}\text{kcal}}\right)\\ =-96.9\text{Cal}\end{array}$

The calculated value is close to $\text{100}\text{Calorie}$. Therefore, the calculated value is consistent with the package information.

Conclusion

The heat released in $\text{Cal}$ is $-96.9\text{Cal}$. Therefore, the calculated value is consistent with the package information.