Chapter 6, Problem 6.88P

(a)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ for the three reactions involved in the combustion of candle $\left({\text{C}}_{\text{21}}{\text{H}}_{44}\right)$ 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 6.88P

$\Delta H_{\text{rxn}}^{°}$ for the three reactions are $–13108\text{kJ}$, $–7165\text{kJ}$ and $–4844\text{kJ}$ respectively.

Explanation of Solution

The balanced chemical equations for the combustion of paraffin $\left({\text{C}}_{\text{21}}{\text{H}}_{44}\right)$ are as follows:

$\begin{array}{c}{\text{C}}_{\text{21}}{\text{H}}_{44}\left(s\right)+32{\text{O}}_2\left(g\right)\to {\text{21CO}}_2\left(g\right)+22{\text{H}}_2\text{O}\left(g\right)\\ {\text{C}}_{\text{21}}{\text{H}}_{44}\left(s\right)+\frac{43}{2}{\text{O}}_2\left(g\right)\to \text{21CO}\left(g\right)+22{\text{H}}_2\text{O}\left(g\right)\\ {\text{C}}_{\text{21}}{\text{H}}_{44}\left(s\right)+11{\text{O}}_2\left(g\right)\to \text{21C}\left(s\right)+22{\text{H}}_2\text{O}\left(g\right)\end{array}$

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 for first reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{21\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]+22\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right\}\\ -\left\{1\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{21}}{\text{H}}_{44}\left(s\right)\right]+32\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right\}\end{array}\right]$ (1)

Substitute $-393.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]$, $-241.826\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]$, $-476\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{21}}{\text{H}}_{44}\left(s\right)\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]$ in the equation (1).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(21\text{mol}\right)\left(-393.5\text{kJ/mol}\right)+\left(22\text{mol}\right)\left(-241.826\text{kJ/mol}\right)\right\}\\ -\left\{\left(1\text{mol}\right)\left(-476\text{kJ/mol}\right)+\left(22\text{mol}\right)\left(0\right)\right\}\end{array}\right]\\ =–13107.672\text{kJ}\\ \approx –13108\text{kJ}\end{array}$

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

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{21\Delta H_{\text{f}}^{°}\left[\text{CO}\left(g\right)\right]+22\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right\}\\ -\left\{1\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{21}}{\text{H}}_{44}\left(s\right)\right]+\frac{43}{2}\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right\}\end{array}\right]$ (2)

Substitute $-110.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[\text{CO}\left(g\right)\right]$, $-241.826\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]$, $-476\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{21}}{\text{H}}_{44}\left(s\right)\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]$ in the equation (2).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(21\text{mol}\right)\left(-110.5\text{kJ/mol}\right)+\left(22\text{mol}\right)\left(-241.826\text{kJ/mol}\right)\right\}\\ -\left\{\left(1\text{mol}\right)\left(-476\text{kJ/mol}\right)+\left(\frac{43}{2}\text{mol}\right)\left(0\right)\right\}\end{array}\right]\\ =–7164.672\text{kJ}\\ \approx –7165\text{kJ}\end{array}$

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

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{21\Delta H_{\text{f}}^{°}\left[\text{C}\left(s\right)\right]+22\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right\}\\ -\left\{1\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{21}}{\text{H}}_{44}\left(s\right)\right]+11\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right\}\end{array}\right]$ (3)

Substitute $0$ for $\Delta H_{\text{f}}^{°}\left[\text{C}\left(s\right)\right]$, $-241.826\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]$, $-476\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{21}}{\text{H}}_{44}\left(s\right)\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]$ in the equation (3).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(21\text{mol}\right)\left(0\right)+\left(22\text{mol}\right)\left(-241.826\text{kJ/mol}\right)\right\}\\ -\left\{\left(1\text{mol}\right)\left(-476\text{kJ/mol}\right)+\left(11\text{mol}\right)\left(0\right)\right\}\end{array}\right]\\ =–4844.172\text{kJ}\\ \approx –4844\text{kJ}\end{array}$

Conclusion

$\Delta H_{\text{rxn}}^{°}$ for the three reactions are $–13108\text{kJ}$, $–7165\text{kJ}$ and $–4844\text{kJ}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The heat released when a $254\text{g}$ candle burns 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 heat absorbed by the surrounding.

Answer to Problem 6.88P

The heat released when a $254\text{g}$ candle burns is $-1.12\times {10}^4\text{kJ}$.

Explanation of Solution

The formula to calculate heat released when $254\text{g}$ candle burns is as follows:

$\text{Heat released}=\left(\frac{{\text{mass of C}}_{\text{21}}{\text{H}}_{44}}{{\text{molar mass of C}}_{\text{21}}{\text{H}}_{44}}\right)\left(\text{heat released per mole}\right)$ (4)

Substitute $254\text{g}$ for mass of ${\text{C}}_{\text{21}}{\text{H}}_{44}$, $296.56\text{g/mol}$ for molar mass of ${\text{C}}_{\text{21}}{\text{H}}_{44}$ and $–13107.672\text{kJ/mol}$ for heat released per mole in the equation (4).

$\begin{array}{c}\text{Heat released}=\left(\frac{254\text{g}}{296.56\text{g/mol}}\right)\left(–13107.672\text{kJ/mol}\right)\\ =-1.12266\times {10}^4\text{kJ}\\ \approx -1.12\times {10}^4\text{kJ}\end{array}$

Conclusion

The heat released when a $254\text{g}$ candle burns is $-1.12\times {10}^4\text{kJ}$.

(c)

Interpretation Introduction

Interpretation:

The heat released when $8.00\%$ by mass of the candle burns incompletely and $5.00\%$ by mass of it undergoes soot formation 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 heat absorbed by the surrounding.

Answer to Problem 6.88P

The heat released is $-1.05\times {10}^4\mathrm{kJ}$.

Explanation of Solution

Consider $87.00\%$ by mass of the candle undergoes complete combustion.

The formula to calculate moles of candle $\left({\text{C}}_{\text{21}}{\text{H}}_{44}\right)$ is as follows:

${\text{Moles of C}}_{\text{21}}{\text{H}}_{44}=\left(\frac{{\text{mass of C}}_{\text{21}}{\text{H}}_{44}}{{\text{molar mass of C}}_{\text{21}}{\text{H}}_{44}}\right)$ (5)

Substitute $254\text{g}$ for mass of ${\text{C}}_{\text{21}}{\text{H}}_{44}$ and $296.56\text{g/mol}$ for molar mass of ${\text{C}}_{\text{21}}{\text{H}}_{44}$ in the equation (5).

$\begin{array}{c}{\text{Moles of C}}_{\text{21}}{\text{H}}_{44}=\left(\frac{254\text{g}}{296.56\text{g/mol}}\right)\\ =0.856488\text{mol}\end{array}$

The formula to calculate heat released when candle burns is as follows:

$\text{Heat released}=\left[\begin{array}{l}\left(\frac{\text{mass \% of complete combustion}}{100\%}\right)\\ \left({\text{moles of C}}_{\text{21}}{\text{H}}_{44}\right)\left(\Delta H_{\text{rxn}}^{°}\text{of complete combustion}\right)\\ \left(\frac{\text{mass \% of incomplete combustion}}{100\%}\right)\\ \left({\text{moles of C}}_{\text{21}}{\text{H}}_{44}\right)\left(\Delta H_{\text{rxn}}^{°}\text{of incomplete combustion}\right)\\ \left(\frac{\text{mass \% of soot formation}}{100\%}\right)\\ \left({\text{moles of C}}_{\text{21}}{\text{H}}_{44}\right)\left(\Delta H_{\text{rxn}}^{°}\text{of soot formation}\right)\end{array}\right]$ (6)

Substitute $87.00\%$ for $\text{mass \%}$ of complete combustion, $0.856488\text{mol}$ for moles of ${\text{C}}_{\text{21}}{\text{H}}_{44}$, $–13107.672\text{kJ/mol}$ for $\Delta H_{\text{rxn}}^{°}$ of complete combustion, $8.00\%$ for $\text{mass \%}$ of incomplete combustion, $–7164.672\text{kJ/mol}$ for $\Delta H_{\text{rxn}}^{°}$ of incomplete combustion, $5.00\%$ for $\text{mass \%}$ of soot formation and $–4844.172\text{kJ/mol}$ for $\Delta H_{\text{rxn}}^{°}$ of soot formation in the equation (6).

$\begin{array}{c}\text{Heat released}=\left[\begin{array}{l}\left(\frac{87.00\%}{100\%}\right)\left(0.856488\text{mol}\right)\left(–13107.672\text{kJ/mol}\right)\\ \left(\frac{8.00\%}{100\%}\right)\left(0.856488\text{mol}\right)\left(–7164.672\text{kJ/moln}\right)\\ \left(\frac{5.00\%}{100\%}\right)\left(0.856488\text{mol}\right)\left(–4844.172\text{kJ/mol}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(0.87\right)\left(0.856488\text{mol}\right)\left(–13107.672\text{kJ/mol}\right)\\ \left(0.080\right)\left(0.856488\text{mol}\right)\left(–7164.672\text{kJ/moln}\right)\\ \left(0.050\right)\left(0.856488\text{mol}\right)\left(–4844.172\text{kJ/mol}\right)\end{array}\right]\\ =-1.04655\times {10}^4\mathrm{kJ}\\ \approx -1.05\times {10}^4\mathrm{kJ}\end{array}$

Conclusion

The heat released is $-1.05\times {10}^4\mathrm{kJ}$.