Chapter 15, Problem 112A

sample of natural gas is analyzed and found to be88.4% methane $(C{H}_4)$ and 11.6% ethane $(C_2{H}_6)$ bymass. The standard enthalpy of combustion of methaneto gaseous carbon dioxide $(C{O}_2)$ and liquid water $(H_{2}O)$ is -891 kJ/mol. Write the equation for the combustionof gaseous ethane to carbon dioxide and water.Calculate the standard enthalpy of combustion of ethaneusing standard enthalpies of formation from Table R-11on page 975. Using that result and the standard enthalpyof combustion of methane in Table 15.3, calculate theenergy released by the combustion of 1 kg of natural gas.

Interpretation Introduction

Interpretation:

The energy released by burning of 1 kg of natural gas is to be calculated.

Concept introduction:

The energy of combustion of a substance can be found out using its formation enthalpy.

Answer to Problem 112A

The energy of 1 kg natural gas is -55269.03kJ.

Explanation of Solution

Chemical equation for combustion of ethane:

$2C_{2\text{}}{H}_{6\text{}}\left(g\right)+7O_2\text{}\left(g\right)=4C{O}_2\text{}\left(g\right)+6H_{2\text{}}O\left(l\right);\Delta H=?$

The equation involves 2 moles of  C2 ? H6 ?; heat of combustion of ethane will be  2?H $\begin{array}{l}\Delta H=\Delta H{°}_f\left(products\right)-\Delta H{°}_f\left(reactants\right)\text{}\\ \left[4\times \Delta H\left(C{O}_2\text{}\right)\text{}+6\Delta H\left(H_2\text{}O\right)\text{}\right]-[2\Delta H\left(C_2\text{}{H}_6\text{}\right)\text{}+7\Delta H\left(O_2\text{}\right)\text{}]\\ =\left[4\times \left(-94.1\right)+6\times \left(-68.3\right)\right]-\left[2\times \left(-21.1\right)+7\times 0\right]\\ =-744.0kcal\end{array}$$\begin{array}{l}\begin{array}{l}Heatofcombustionofethane\hfill \\ =-\frac{744}{2}\text{}=-372.0 kcal\hfill \end{array}\\ =-372\times 4.2\\ =-1562.4kJ\end{array}$

Out of 1 kg natural gas, 884 g is methane and 116g is ethane.

$\begin{array}{l}Energyof1mol\left(16g\right)methane=-891kJ\hfill \\ \begin{array}{l}Energyof884gmethane=\frac{-891\times 884}{16}=-49227.75kJ\\ \begin{array}{l}Energyof1mol\left(30g\right)ethane=-1562.4kJ\hfill \\ \begin{array}{l}Energyof\text{ 116}gethane=\frac{-1562.4\times 116}{30}=-6041.28kJ\\ Totalenergy=-49227.75-6041.28=-55269.03kJ\end{array}\hfill \end{array}\end{array}\hfill \end{array}$