Chapter U5, Problem C19.4RE

a)

Interpretation Introduction

Interpretation:

Balanced equation for the combustion of propane must be given.

Concept Introduction:

Hydrocarbons undergo combustion reaction to produce carbon dioxide and water as shown below.

  ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{(g)+(x+}\frac{\text{y}}{\text{4}}{\text{)O}}_{\text{2}}{\text{(g)=xCO}}_{\text{2}}\text{(g)+}\frac{\text{y}}{\text{2}}{\text{H}}_{\text{2}}\text{O(g)}\text{.}$

Here x and y represents number of carbon and hydrogen atoms in the hydrocarbon.

Answer to Problem C19.4RE

  ${\text{C}}_3{\text{H}}_8{\text{(g)+5O}}_{\text{2}}{\text{(g)=3CO}}_{\text{2}}{\text{(g)+4H}}_{\text{2}}\text{O(g)}\text{.}$

Explanation of Solution

  $\begin{array}{l}{\text{C}}_3{\text{H}}_8\text{(g)+(3+}\frac{8}{\text{4}}{\text{)O}}_{\text{2}}{\text{(g)=3CO}}_{\text{2}}\text{(g)+}\frac{8}{\text{2}}{\text{H}}_{\text{2}}\text{O(g)}\text{.}\\ {\text{Or, C}}_3{\text{H}}_8{\text{(g)+5O}}_{\text{2}}{\text{(g)=3CO}}_{\text{2}}{\text{(g)+4H}}_{\text{2}}\text{O(g)}\text{.}\end{array}$

There are 3 carbons and 8 hydrogens. So from 1 mole propane, 3 moles of carbon dioxide and 4 moles of water were obtained after complete combustion.

b)

Interpretation Introduction

Interpretation:

Heat of combustion of propane must be converted from kJ/mol unit to kJ/g unit.

Concept Introduction:

Molar mass of propane (C₃H₈) is 44 g/mol.

Heat of combustion in kJ/g $=\frac{\text{Heat of combustion in kJ/mol}}{\text{Molar mass in g}}.$

Answer to Problem C19.4RE

Heat of combustion of propane is -50.45 kJ/g.

Explanation of Solution

Heat of combustion of propane is given as -2220 kJ/mol.

To convert this to kJ/g unit, the energy value must be divided by the molar mass of propane which is 44 g/mol.

Heat of combustion in kJ/g

  $\begin{array}{l}\text{=}\frac{\text{Heat of combustion in kJ/mol}}{\text{Molar mass in g}}\\ \text{=}\frac{\text{-2220 kJ/mol}}{\text{44 g/mol}}\\ \text{=-50}\text{.45 kJ/g}\text{.}\end{array}$

c)

Interpretation Introduction

Interpretation:

Whether heat of combustion of propane in kJ/g is greater than that of methane must be explained. Also, whether heat of combustion of propane in kJ/mol is greater than that of methane must be explained.

Concept Introduction:

Heat of combustion is the heat energy released when 1 mol of fuel is combusted with oxygen completely. It can be expressed generally in kJ/mol or in kJ/g unit.

Answer to Problem C19.4RE

Heat of combustion of propane is lesser than that of methane in kJ/g unit.

Heat of combustion of propane is greater than that of methane in kJ/ mol unit.

Explanation of Solution

Heat of combustion of propane is given as -2220 kJ/mol.

To convert this to kJ/g unit, the energy value must be divided by the molar mass of propane which is 44 g/mol.

Heat of combustion in kJ/g

  $\begin{array}{l}\text{=}\frac{\text{Heat of combustion in kJ/mol}}{\text{Molar mass in g}}\\ \text{=}\frac{\text{-2220 kJ/mol}}{\text{44 g/mol}}\\ \text{=-50}\text{.45 kJ/g}\text{.}\end{array}$

Heat of combustion of methane is given as -889 kJ/mol.

To convert this to kJ/g unit, the energy value must be divided by the molar mass of methane which is 16 g/mol.

Heat of combustion in kJ/g

  $\begin{array}{l}\text{=}\frac{\text{Heat of combustion in kJ/mol}}{\text{Molar mass in g}}\\ \text{=}\frac{\text{-889 kJ/mol}}{\text{16 g/mol}}\\ \text{=-55}\text{.56 kJ/g}\text{.}\end{array}$

Thus heat of combustion of methane is greater than that of propane in kJ/g unit. But heat of combustion of propane is greater than that of methane in kJ/mol unit.