Chapter 20, Problem 35PS

(a)

Interpretation Introduction

Interpretation: The balanced equation for the burning of methyl $myristate$ has to be written.

Concept introduction:

• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation.

Answer to Problem 35PS

The balanced equation for the given reaction is,

 ${\text{C}}_{\text{13}}{\text{H}}_{\text{27}}{\text{CO}}_{\text{2}}{\text{CH}}_{\text{3}}\text{(l)}\text{+}\frac{\text{43}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to \text{15}{\text{CO}}_{\text{2}}\text{(g)}\text{+}\text{15}{\text{H}}_{\text{2}}\text{O}\text{(g)}$

Explanation of Solution

When methyl $myristate$ is burned the gas ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ are produced as the products.

The balanced equation for the combustion of methyl myristate is given as:

${\text{C}}_{\text{13}}{\text{H}}_{\text{27}}{\text{CO}}_{\text{2}}{\text{CH}}_{\text{3}}\text{(l)}\text{+}\frac{\text{43}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to \text{15}{\text{CO}}_{\text{2}}\text{(g)}\text{+}\text{15}{\text{H}}_{\text{2}}\text{O}\text{(g)}$.

(b)

Interpretation Introduction

Interpretation: Enthalpy change for the oxidation of methyl myristate should be determined using the given data.

Concept introduction:

Enthalpy of combustion: At standard conditions when one mole of substance is burnt completely in presence of oxygen, the enthalpy change involved is called enthalpy of combustion.

${\text{ΔH}}^{\text{0}}{}_{\text{comb}}\text{=}\sum \text{n}{\text{ΔH}}^{\text{0}}{\text{f}}_{\text{products}}\text{-}\sum \text{n}{\text{ΔH}}^{\text{0}}{\text{f}}_{\text{reactants}}$

Answer to Problem 35PS

Enthalpy change is calculated to be $\text{-8759}\text{.1}\text{kJ}/\text{mol}$

Explanation of Solution

Given:

$\begin{array}{c}{\text{ΔH}}^{\text{0}}\text{f}{\text{of CO}}_{\text{2}}\text{=}\text{-393}\text{.509}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}{\text{of H}}_{\text{2}}\text{O}\text{=}\text{-241}\text{.83}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}\text{of}{\text{C}}_{\text{13}}{\text{H}}_{\text{27}}{\text{CO}}_{\text{2}}{\text{CH}}_{\text{3}}\text{=}\text{-771}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}\text{of}{\text{O}}_{\text{2}}\text{=}\text{0}\text{kJ}/\text{mol}\end{array}$

Change in enthalpy can be calculated by the equation:

${\text{ΔH}}^{\text{0}}{}_{\text{comb}}\text{=}\sum \text{n}{\text{ΔH}}^{\text{0}}{\text{f}}_{\text{products}}\text{-}\sum \text{n}{\text{ΔH}}^{\text{0}}{\text{f}}_{\text{reactants}}$

$\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{comb}}\text{=}\left[\text{15×-393}\text{.509}\text{+}15\text{×-241}\text{.83}\right]\text{-}\left[\text{-771}\right]\\ \text{=}\text{-8759}\text{.1}\text{kJ}/\text{mol}\end{array}$

(c)

Interpretation Introduction

Interpretation: The fuel which provides more energy (Hexadecane or methyl $myristate$) should be determined.

Concept introduction:

Enthalpy of combustion: At standard conditions when one mole of substance is burnt completely in presence of oxygen, the enthalpy change involved is called enthalpy of combustion.

${\text{ΔH}}^{\text{0}}{}_{\text{comb}}\text{=}\sum \text{n}{\text{ΔH}}^{\text{0}}{\text{f}}_{\text{products}}\text{-}\sum \text{n}{\text{ΔH}}^{\text{0}}{\text{f}}_{\text{reactants}}$

Answer to Problem 35PS

Hexadecane provide more energy than methyl $myristate$.

Explanation of Solution

For methyl myristate:

$\begin{array}{c}{\text{ΔH}}^{\text{0}}\text{f}{\text{of CO}}_{\text{2}}\text{=}\text{-393}\text{.509}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}{\text{of H}}_{\text{2}}\text{O}\text{=}\text{-241}\text{.83}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}\text{of}{\text{C}}_{\text{13}}{\text{H}}_{\text{27}}{\text{CO}}_{\text{2}}{\text{CH}}_{\text{3}}\text{=}\text{-771}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}\text{of}{\text{O}}_{\text{2}}\text{=}\text{0}\text{kJ}/\text{mol}\end{array}$

Change in enthalpy can be calculated by the equation:

${\text{ΔH}}^{\text{0}}{}_{\text{comb}}\text{=}\sum \text{n}{\text{ΔH}}^{\text{0}}{\text{f}}_{\text{products}}\text{-}\sum \text{n}{\text{ΔH}}^{\text{0}}{\text{f}}_{\text{reactants}}$

$\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{comb}}\text{=}\left[\text{15×-393}\text{.509}\text{+}15\text{×-241}\text{.83}\right]\text{-}\left[\text{-771}\right]\\ \text{=}\text{-8759}\text{.1}\text{kJ}/\text{mol}\end{array}$

For hexadecane:

$\begin{array}{c}{\text{ΔH}}^{\text{0}}\text{f}{\text{of CO}}_{\text{2}}\text{=}\text{-393}\text{.509}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}{\text{of H}}_{\text{2}}\text{O}\text{=}\text{-241}\text{.83}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}\text{of}{\text{C}}_{\text{16}}{\text{H}}_{34}\text{=}\text{-456}\text{.1}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}\text{of}{\text{O}}_{\text{2}}\text{=}\text{0}\text{kJ}/\text{mol}\end{array}$

${\text{2C}}_6{\text{H}}_{34}\text{(l)}\text{+}49{\text{O}}_{\text{2}}\text{(g)}\to 32{\text{CO}}_{\text{2}}\text{(g)}\text{+}34{\text{H}}_{\text{2}}\text{O}\text{(g)}$

Change in enthalpy can be calculated by the equation:

${\text{ΔH}}^{\text{0}}{}_{\text{comb}}\text{=}\sum \text{n}{\text{ΔH}}^{\text{0}}{\text{f}}_{\text{products}}\text{-}\sum \text{n}{\text{ΔH}}^{\text{0}}{\text{f}}_{\text{reactants}}$

$\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{comb}}\text{=}\left[\text{32×-393}\text{.509}\text{+}34\text{×-241}\text{.83}\right]\text{-}\left[\text{-2}\times 456.1\right]\\ \text{=}\text{-9951}\text{.2}\text{kJ}/\text{mol}\end{array}$

Hexadecane provide more energy than methyl myristate.