Chapter 5, Problem 5.108QE

Interpretation Introduction

Interpretation:

Amount of $C{H}_4$ required to burn to heat the air has to be determined.

Concept Introduction:

The temperature change $\left(\Delta T\right)$ taking place when heat (q) is given to the system depends on amount of the substance (n)and its specific heat(C_s).

  $q{\text{ = nC}}_s\Delta T$

Enthalpy$\left(H\right)$: It is the total amount of heat in a particular system.

The value of standard enthalpy change ${\text{ΔH}}^{\text{ο}}$ of the given reaction is calculated by the formula,

  ${\text{ΔH}}_{rxn}^{\circ }\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

${\text{ΔH}}_{\text{f}}^{\text{ο}}$ is the standard enthalpies of formation.

n is the number of moles.

Answer to Problem 5.108QE

Amount of $C{H}_4$ required to burn to heat the air is $\text{118 g}$.

Explanation of Solution

Change in temperature is obtained as follows,

  $\begin{array}{c}\Delta T{\text{ = T}}_{final}-T_{initial}\\ \\ =\left(\text{22}\text{.0}°\text{C}\right)-\left(\text{15}\text{.0}°\text{C}\right)\\ \\ =\text{7}°\text{C}\end{array}$

Volume of air is determined as follows,

  $\begin{array}{c}Volume\text{ = Area}\times \text{Height}\\ \\ \text{= }\left({\text{275 m}}^{\text{2}}\right)\left(\text{2}\text{.50 m}\right)\text{ = 687}{\text{.5 m}}^3\\ \\ \text{ = 687}{\text{.5 m}}^3\times \frac{\text{1000 L}}{1{\text{ m}}^3}\\ \\ =\text{ 6}\text{.875}\times {\text{10}}^5\text{ L}\end{array}$

Number of moles of air inside the house can be determined from the density and average molar mass of air values as follows,

  $\begin{array}{c}Mass\text{ = }Density\times Volume\\ \\ \text{= }\left(\text{1}\text{.22 g/L}\right)\left(\text{6}\text{.875}\times {\text{10}}^5\text{ L}\right)\text{ = 838750 g}\\ \\ No.\text{ of moles = }\frac{Mass}{Molar\text{ mass}}\\ \\ \text{ = }\frac{\text{838750 g}}{\text{28}\text{.9 g/mol}}\\ \\ =\text{ 2}\text{.902}\times {\text{10}}^4\text{ mol}\end{array}$

Heat required to increase the temperature can be determined as follows,

  $\begin{array}{c}q{\text{ = nC}}_s\Delta T\\ \\ =\left(\text{2}\text{.902}\times {\text{10}}^4\text{ mol}\right)\left(\text{29}\text{.1 J/mol}\text{.}°\text{C}\right)\left(\text{7}°\text{C}\right)\\ \\ =5.91\times {10}^6\text{ J }=5.91\times {10}^3\text{ kJ }\end{array}$

The balanced equation for the combustion of natural gas is given as:

  ${\text{CH}}_4\left(g\right){\text{ + 2 O}}_2\left(g\right)\underset{}{\to }{\text{CO}}_{\text{2}}\left(g\right){\text{+ 2 H}}_{2}O\left(g\right)$

Standard enthalpy of formation values are given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of CO}}_{\text{2}}\left(g\right)\text{=}-\text{393}\text{.51}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of H}}_{2}O\left(l\right)\text{=}-\text{241}\text{.82}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of }{\text{CH}}_4\left(g\right)\text{=}-\text{74}\text{.81}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of}{\text{O}}_2\left(g\right)\text{=}0\text{kJ}/\text{mol}\end{array}$

Change in enthalpy can be calculated by the equation:

  ${\text{ΔH}}_{rxn}^{\circ }\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Substitute the values as follows,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{1 mol}\times -\text{393}\text{.51}\text{kJ}/\text{mol}\right)\text{+}\left(\text{2 mol×}-\text{241}\text{.82}\text{kJ}/\text{mol}\right)\right]\\ -\left[\left(\text{1 mol}\times -\text{74}\text{.81}\text{kJ}/\text{mol}\right)\text{+}\left(\text{2 mol×}0\text{kJ}/\text{mol}\right)\right]\\ \\ \text{=}-802.34\text{kJ}\end{array}$

$\text{ΔH}°$ for the combustion of one mole of natural gas is $-802.34\text{kJ}$. Hence, energy per gram is determined as follows,

  $\begin{array}{c}\text{ΔH}°\text{ =}\frac{-802.34\text{kJ}}{1\text{ mol}}\times \frac{1\text{ mol}}{16.04\text{ g}}\\ \\ =-50.02\text{ kJ/g}\end{array}$

Amount of $C{H}_4$ required to burn to heat the air is determined as follows,

  $\begin{array}{c}Mass\text{ =}\frac{5.91\times {10}^3\text{ kJ }}{50.02\text{ kJ/g}}\\ \\ ={\text{118 g CH}}_4\end{array}$