Chapter 6, Problem 6.102P

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the combustion of kerosene is to be written.

Concept introduction:

A combustion reaction is the reaction in which reactant is reacted with molecular oxygen to form the product. Heat is released and the energy is produced in the reaction. Molecular oxygen is employed as an oxidizing agent in these reactions.

(b)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{f}}^{°}$ of kerosene when $\Delta H_{\text{rxn}}^{°}$ is $-1.5\times {10}^4\text{kJ}$ is to be calculated.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum m\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(c)

Interpretation Introduction

Interpretation:

The heat released by the combustion of $0.50\text{ gal}$ of kerosene $\left({\text{C}}_{\text{12}}{\text{H}}_{26}\right)$ is to be determined.

Concept introduction:

Heat $\left(q\right)$ is the energy transferred as a consequence of the difference of the temperature between the system and surrounding.

The heat released by the system is negative and the heat taken by the system is positive.

The amount of heat released by the system is equal to the amount of heat absorbed by the surrounding.

The conversion factor to convert $\text{gal}$ into $\text{qt}$ is:

$\text{1}\text{gal}=4\text{qt}$

The conversion factor to convert $\text{L}$ into $\text{qt}$ is:

$\text{1}\text{L}=1.057\text{qt}$

(d)

Interpretation Introduction

Interpretation:

The volume of kerosene in $\text{gal}$ that must be burned for a kerosene furnace to produce $1250\text{Btu}$ is to be calculated.

Concept introduction:

Heat $\left(q\right)$ is the energy transferred as a consequence of the difference of the temperature between the system and surrounding.

The heat released by the system is negative and the heat taken by the system is positive.

The amount of heat released by the system is equal to the amount of heat absorbed by the surrounding.

The conversion factor to convert $\text{Btu}$ into $\text{kJ}$$\text{qt}$ is:

$\text{1}\text{Btu}=1.055\text{kJ}$