Chapter 5, Problem 48Q

(a)

Interpretation Introduction

Interpretation:

The merits of the calculation data has to be discussed.

Concept introduction:

Combustion reaction:

Combustion is an exothermic reaction between a fuel (the reductant) and an oxidant. Oxidant is usually atmospheric oxygen.  Generally organic molecule undergoes combustion reaction which produces carbon dioxide and water molecule.

${\text{Hydrocarbons + O}}_{\text{2}}\text{(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

Explanation of Solution

Generally, the number of carbon atoms is more (number of atoms is more), obviously it has higher molar masses in the fuels therefore it release more energy when burned.

Therefore, the number of atoms in octane is more so, it has more chemical bonds, hence it has a larger heat of combustion than butane.

The comparison of the molar heats of combustion of the molecule is based on the same amount of each substance and heat released per gram of each fuel.

$\begin{array}{l}\text{The}\text{heat}\text{released}\text{per}\text{gram}\text{octane}\text{burned is given below,}\\ {\text{Octane (C}}_{\text{8}}{\text{H}}_{\text{18}}\text{): }\frac{\text{5450 kJ}}{\text{1}\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}{\text{114}\text{.2 g}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\text{=}\frac{\text{47}\text{.72 kJ}}{\text{g}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\end{array}$

$\begin{array}{l}\text{The}\text{heat}\text{released}\text{per}\text{gram}\text{butane}\text{burned is given below,}\\ {\text{Butane (C}}_4{\text{H}}_{\text{10}}\text{): }\frac{\text{2859 kJ}}{\text{1}\text{mol}{\text{C}}_4{\text{H}}_{\text{10}}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_4{\text{H}}_{\text{10}}}{\text{58}\text{.1 g}{\text{C}}_4{\text{H}}_{\text{10}}}\text{=}\frac{\text{49}\text{.2 kJ}}{\text{g}{\text{C}}_4{\text{H}}_{\text{10}}}\end{array}$

Both the values are closer. It is not possible to found with just two data points.

Though, butane has less number of hydrocarbons than octane and it releases slightly more heat per gram than the octane.

Therefore, the result clearly tells that the heat comparisons should be based on the same mass of fuel.

(b)

Interpretation Introduction

Interpretation:

The heat of combustion per gram of candle wax (${\text{C}}_{\text{25}}{\text{H}}_{\text{52}}$) has to be compared with octane and explanation has to be given.

Concept introduction:

Combustion reaction:

Combustion is an exothermic reaction between a fuel (the reductant) and an oxidant. Oxidant is usually atmospheric oxygen.  Generally organic molecule undergoes combustion reaction which produces carbon dioxide and water molecule.

${\text{Hydrocarbons + O}}_{\text{2}}\text{(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

Explanation of Solution

Generally, the number of carbon atoms is more (number of atoms is more), obviously it has higher molar masses in the fuels therefore it release more energy when burned.

Therefore, the number of atoms in octane is more so, it has more chemical bonds, hence It has a larger heat of combustion than butane.

The comparison of the molar heats of combustion of the molecule is based on the same amount of each substance and heat released per gram of each fuel.

$\begin{array}{l}\text{The}\text{heat}\text{released}\text{per}\text{gram}\text{octane}\text{burned is given below,}\\ {\text{Octane (C}}_{\text{8}}{\text{H}}_{\text{18}}\text{): }\frac{\text{5450 kJ}}{\text{1}\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}{\text{114}\text{.2 g}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\text{=}\frac{\text{47}\text{.72 kJ}}{\text{g}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\end{array}$

$\begin{array}{l}\text{The}\text{heat}\text{released}\text{per}\text{gram}\text{butane}\text{burned is given below,}\\ {\text{Butane (C}}_4{\text{H}}_{\text{10}}\text{): }\frac{\text{2859 kJ}}{\text{1}\text{mol}{\text{C}}_4{\text{H}}_{\text{10}}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_4{\text{H}}_{\text{10}}}{\text{58}\text{.1 g}{\text{C}}_4{\text{H}}_{\text{10}}}\text{=}\frac{\text{49}\text{.2 kJ}}{\text{g}{\text{C}}_4{\text{H}}_{\text{10}}}\end{array}$

Both the values are closer.  It is not possible to found with just two data points.

Though, butane has less number of hydrocarbons than octane and it releases slightly more heat per gram than the octane.

Therefore, the result clearly tells that the heat comparisons should be based on the same mass of fuel.

The number of atoms is more in Candle wax, it has high molar mass.  When comparing the above data values, we can predict the heat of combustion per gram is expected to be slightly smaller and the heat of combustion per mole is expected to be larger than octane.