Chapter 4, Problem 48Q

(a)

Interpretation Introduction

Interpretation:

The statement, liberation of more heat per gram by hydrocarbon fuel containing larger molecule than smaller molecule has to be validated using the given data.

Concept Introduction:

Molar heat of combustion: It is the amount of heat released when one mole of hydrocarbon is burnt with excess of oxygen.

Heat of combustion per gram: It is the amount of heat liberated when one gram of hydrocarbon is burnt with excess of oxygen.

Heat of combustion per gram can be calculated as follows,

$\begin{array}{l}\text{Heat of combustion per gram of hydrocarbon }\\ \text{= }\frac{\text{Molar heat of combustion of hydrocarbon}}{\text{Molecular mass of hydrocarbon}}\end{array}$

Explanation of Solution

Given,

    $\begin{array}{l}\begin{array}{l}{\text{Molar heat of combustion of octane,C}}_{\text{8}}{\text{H}}_{\text{18}}\text{ = 5070 kJ}\hfill \\ \hfill \end{array}\\ {\text{Molar heat of combustion of butane,C}}_{\text{4}}{\text{H}}_{\text{10}}\text{ = 2658 kJ}\end{array}$

Since, octane has more atoms and more bonds, molar heat of combustion is more for octane than butane. But for comparison same amount of substance should be taken. So, heat released from each gram of substance has to be calculated as follows,

$\begin{array}{l}\text{Amount of heat liberated when 1 gram octane burned=}\frac{\text{5070 kJ}}{{\text{1mol C}}_{\text{8}}{\text{H}}_{\text{18}}}\text{×}\frac{{\text{1 mol C}}_{\text{8}}{\text{H}}_{\text{18}}}{\text{114}{\text{.2 g C}}_{\text{8}}{\text{H}}_{\text{18}}}\\ \text{=}\frac{\text{44}\text{.39 kJ}}{{\text{g C}}_{\text{8}}{\text{H}}_{\text{18}}}\\ \\ \text{Amount of heat liberated when 1 gram butane burned=}\frac{\text{2658 kJ}}{{\text{1 mol C}}_{\text{4}}{\text{H}}_{\text{10}}}\text{×}\frac{{\text{1 mol C}}_{\text{8}}{\text{H}}_{\text{18}}}{\text{58}{\text{.1g C}}_{\text{4}}{\text{H}}_{\text{10}}}\\ \text{=}\frac{\text{45}\text{.74 kJ}}{{\text{g C}}_{\text{4}}{\text{H}}_{\text{10}}}\end{array}$

From the calculation, the amount of heat released by one gram of butane is slightly more than the amount of heat liberated from one gram of octane. Since, the amount of heat liberated from octane and butane have very small difference it is not possible to establish a trend using only two data. Even though from the calculation, butane the smaller hydrocarbon release more energy than the larger octane.

So the statement is wrong, because comparison should be done by taking same mass of fuel not same number of mole.

(b)

Interpretation Introduction

Interpretation:

Heat of combustion per gram and molar heat of combustion of candle wax and octane has to be compared.

Concept Introduction:

Molar heat of combustion: It is the amount of heat released when one mole of hydrocarbon is burnt with excess of oxygen.

Heat of combustion per gram: It is the amount of heat liberated when one gram of hydrocarbon is burnt with excess of oxygen.

Heat of combustion per gram can be calculated as follows,

$\begin{array}{l}\text{Heat of combustion per gram of hydrocarbon }\\ \text{= }\frac{\text{Molar heat of combustion of hydrocarbon}}{\text{Molecular mass of hydrocarbon}}\end{array}$

Explanation of Solution

Candle wax $\left({\text{C}}_{\text{25}}{\text{H}}_{\text{52}}\right)$ is a larger hydrocarbon compared to octane $\left({\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\right)\text{.}$ Analyzing the calculations from part a, the heat released from one gram of candle wax will be lower than the heat released from one gram of octane.

But, the molar heat of combustion of candle wax will be higher than the molar heat of combustion of octane.