Chapter 6, Problem 6.161QP

(a)

Interpretation Introduction

Interpretation:

By using the given data, high heating value (HHV) vs molar mass of hydrocarbon graph has to be drawn and also the HHV for propane has to be calculated by using that graph.

Explanation of Solution

Figure 1

Calculate the HHV for propane by using the trend line equation,

$\begin{array}{l}\text{Molar mass for propane is 44}\text{.097}\text{g/mol}\text{.}\\ \text{high heating value }\left(\text{HHV}\right)\text{=}\text{30376}{\left(\text{44}\text{.097}\text{g/mol}\right)}^{\text{-0}\text{.088}}\\ \text{=}\text{21768}\text{BTU/lb}\text{.}\end{array}$

(b)

Interpretation Introduction

Interpretation:

By using the given data, High heating value (HHV) - low heating value (LHV) vs molar mass of hydrocarbon graph has to be drawn and also the $\left|\text{HHV - LHV}\right|$ for propane has to be calculated by using that graph.

Explanation of Solution

Figure 2

Calculate the $\left|\text{HHV - LHV}\right|$ for propane by using the trend line equation,

$\begin{array}{l}\text{Molar mass for propane is 44}\text{.097}\text{g/mol}\text{.}\\ \left|\text{HHV - LHV}\right|\left(\text{BTU/lb}\right)\text{=}\text{5159}\text{.3}{\left(\text{44}\text{.097}\text{g/mol}\right)}^{\text{-0}\text{.287}}\\ \text{=}\text{1740}\text{BTU/lb}\text{.}\end{array}$

(c)

Interpretation Introduction

Interpretation:

By using the given and obtained data in part (a) and part (b), enthalpies of reaction for the given reaction has to be calculated.

Explanation of Solution

Given reaction

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{}_{\left(\text{g}\right)}\text{+}{\text{5O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{3CO}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{;}{\text{ΔH}}_{\text{rxn}}\text{=}\text{-}\text{21768}\text{BTU/lb}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$

Covert $\text{21768}\text{BTU/lb}\text{to}\text{kJ/mol}$ unit

$\begin{array}{l}\frac{\text{-}\text{21768}\text{BTU}}{\text{lb}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\left(\frac{\text{1}\text{lb}}{\text{453}\text{.59}\text{g}}\right)\left(\frac{\text{44}\text{.097}\text{g}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}{\text{1}\text{mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\right)\left(\frac{\text{1}\text{.055}\text{kJ}}{\text{1}\text{.000}\text{BTU}}\right)\text{=}\text{-}\text{2233}\text{kJ/mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\\ \text{HHV}\text{based}\text{thermochemical}\text{equation is given by}\\ {\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{}_{\left(\text{g}\right)}\text{+}{\text{5O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{3CO}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{;}{\text{ΔH}}_{\text{rxn}}\text{=}\text{-}\text{2233}\text{kJ}\text{.}\end{array}$

The second equation corresponds to propane’s LHV.  Using $\text{HHV}\text{-}\text{LHV}$ difference the value of LHV was calculated.

$\begin{array}{l}\text{LHV}\text{=}\text{HHV}\text{-}\left|\text{HHV}\text{-}\text{LHV}\right|\\ \text{=}\text{21768}\text{-}\text{1740}\text{=}\text{20028}\text{BTU}\text{/}\text{lb}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\end{array}$

Given reaction

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{}_{\left(\text{g}\right)}\text{+}{\text{5O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{3CO}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\text{;}{\text{ΔH}}_{\text{rxn}}\text{=}\text{-}\text{20028}\text{BTU/lb}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$

Covert $\text{20028}\text{BTU/lb}\text{to}\text{kJ/mol}$ unit

$\begin{array}{l}\frac{\text{-}\text{20028}\text{BTU}}{\text{lb}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\left(\frac{\text{1}\text{lb}}{\text{453}\text{.59}\text{g}}\right)\left(\frac{\text{44}\text{.097}\text{g}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}{\text{1}\text{mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\right)\left(\frac{\text{1}\text{.055}\text{kJ}}{\text{1}\text{.000}\text{BTU}}\right)\text{=}\text{-}\text{2054}\text{kJ/mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\\ \text{LHV}\text{based}\text{thermochemical}\text{equation is given by}\\ {\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{}_{\left(\text{g}\right)}\text{+}{\text{5O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{3CO}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{;}{\text{ΔH}}_{\text{rxn}}\text{=}\text{-}\text{2054}\text{kJ}\text{.}\end{array}$

The estimated enthalpies are $\text{-}\text{2233}\text{kJ}\text{and}\text{-}\text{2054}\text{kJ,}$ respectively.

(d)

Interpretation Introduction

Interpretation:

By using the given data and results obtained in part (c), the amount of heat energy released when $\text{1700}\text{kg}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{}_{\left(\text{g}\right)}$ combustion has to be calculated.

Answer to Problem 6.161QP

The amount of heat energy released from the combustion of $\text{1700}\text{kg}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{}_{\left(\text{g}\right)}$ is $\text{7}\text{.9}\text{×}{\text{10}}^{\text{7}}\text{kJ}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{mass}\text{of}\text{propane}\text{burned}\text{is}\text{1700}\text{kg}\\ \text{water}\text{vapor}\text{is}\text{the}\text{only}\text{product}\end{array}$

Use the enthalpy change associated with the LHV based thermochemical equation to calculate the heat energy,

$\text{1700}\text{kg}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\frac{\text{1000}\text{g}}{\text{1}\text{kg}}\right)\left(\frac{\text{1}\text{mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}{\text{44}\text{.097}\text{g}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\right)\left(\frac{\text{-}\text{2054}\text{kJ}}{\text{mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\right)\text{=}\text{-}\text{7}\text{.9}\text{×}{\text{10}}^{\text{7}}\text{kJ}$

The amount of heat energy released when $\text{1700}\text{kg}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{}_{\left(\text{g}\right)}$ combustion reaction is $\text{7}\text{.9}\text{×}{\text{10}}^{\text{7}}\text{kJ}\text{.}$

(e)

Interpretation Introduction

Interpretation:

Estimated results in part (d) has to be compared with accepted thermochemical heats of formation tabulated in appendix C.

Concept introduction:

Hess’s law:

This law is also known as Hess’s law of heat of summation.

This law states that, the sum of enthalpy changes for the individual steps will give the enthalpy change for the overall reaction.

The enthalpy change for the overall chemical change is the same.

Explanation of Solution

Calculate the enthalpy change using thermochemical heats of formation values

$\begin{array}{l}{\text{ΔH}}_{\text{rxn}}\text{=}\left[\text{4}\text{mol}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\left(\text{-}\text{241}\text{.8}\text{kJ/mol}\right)\text{+}\text{3}\text{mol}{\text{CO}}_{\text{2}}{}_{\left(\text{g}\right)}\left(\text{-393}\text{.5}\text{kJ/mol}\right)\right]\\ \text{-}\left[\text{1}\text{mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{-}\text{104}\text{.7}\text{kJ/mol}\right)\text{+}\text{5}\text{mol}{\text{O}}_{\text{2}}\left(\text{0}\text{kJ/mol}\right)\right]\\ \text{=}\text{-}\text{2043}\text{kJ/mol}\text{.}\end{array}$

The accepted LHV- based thermochemical equation is

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{}_{\left(\text{g}\right)}\text{+}{\text{5O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{3CO}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\text{;}{\text{ΔH}}_{\text{rxn}}\text{=}\text{-}\text{2043}\text{kJ/mol}$

Calculate the heat energy when combustion of $\text{1700}\text{kg}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$,

$\text{1700}\text{kg}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\frac{\text{1000}\text{g}}{\text{1}\text{kg}}\right)\left(\frac{\text{1}\text{mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}{\text{44}\text{.097}\text{g}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\right)\left(\frac{\text{-}\text{2043}\text{kJ}}{\text{mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\right)\text{=}\text{-}\text{7}\text{.9}\text{×}{\text{10}}^{\text{7}}\text{kJ}$

To two significant figures, the obtained value is the same as that calculated in part (d) so the percentage difference is negligible.