Chapter 20, Problem 47GQ

(a)

Interpretation Introduction

Interpretation: The enthalpy of combustion of ethanol and isooctane, the fuel which release more ${\text{CO}}_{\text{2}}$ and better fuel in terms of energy production should be given.

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 47GQ

Isooctane release more energy than ethanol.

Explanation of Solution

Given the combustion of ethanol and isooctane:

$\begin{array}{l}\text{Ethanol}\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\text{(l)}\text{+3}{\text{O}}_{\text{2}}\text{(g)}\to {\text{2CO}}_{\text{2}}\text{(g)}\text{+}{\text{3H}}_{\text{2}}\text{O}\text{(l)}\end{array}$

$\begin{array}{c}{\text{ΔH}}^{\text{0}}\text{f}{\text{of CO}}_{\text{2}}\text{=}\text{-393}\text{.509}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}{\text{of H}}_{\text{2}}\text{O}\text{=}\text{-285}\text{.83}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\text{=}\text{-277}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}\text{of}{\text{O}}_{\text{2}}\text{=}\text{0}\text{kJ}/\text{mol}\end{array}$

Enthalpy of combustion can be calculated by the equation:

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

$\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{comb}}\text{=}\left[\text{2×-393}\text{.509}\text{+}\text{3×-285}\text{.83}\right]\text{-}\left[\text{-277}\right]\\ \text{=}\text{-1367}\text{.5}\text{kJ}/\text{mol}\end{array}$

$\begin{array}{l}\text{Isooctane}\\ {\text{C}}_8{\text{H}}_{18}\text{(l)}\text{+}\frac{25}{2}{\text{O}}_{\text{2}}\text{(g)}\to 8{\text{CO}}_{\text{2}}\text{(g)}\text{+}9{\text{H}}_{\text{2}}\text{O}\text{(l)}\end{array}$

$\begin{array}{c}{\text{ΔH}}^{\text{0}}\text{f}{\text{of CO}}_{\text{2}}\text{=}\text{-393}\text{.509}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}{\text{of H}}_{\text{2}}\text{O}\text{=}\text{-285}\text{.83}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}\text{of}{\text{C}}_8{\text{H}}_{18}\text{=}\text{-259}\text{.3}\text{kJ}/\text{mol}\\ {\text{ΔH}}^{\text{0}}\text{f}\text{of}{\text{O}}_{\text{2}}\text{=}\text{0}\text{kJ}/\text{mol}\end{array}$

Enthalpy of combustion can be calculated by the equation:

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

$\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{comb}}\text{=}\left[\text{8×-393}\text{.509}\text{+}9\text{×-285}\text{.83}\right]\text{-}\left[\text{-259}\text{.3}\right]\\ \text{=}\text{-5461}\text{.2}\text{kJ}/\text{mol}\end{array}$

Isooctane releases more energy per kilogram.

(b)

Interpretation Introduction

Interpretation: compare the enthalpy of combustion of ethanol and isooctane and the fuel which release more ${\text{CO}}_{\text{2}}$ per kilogram 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 47GQ

Ethanol produces less ${\text{CO}}_{\text{2}}$ per kilogram.

Explanation of Solution

${\text{ΔH}}^{\text{0}}{}_{\text{comb}}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is found to be $\text{-1367}\text{.5}\text{kJ}/\text{mol}$ and ${\text{ΔH}}^{\text{0}}{}_{\text{comb}}$ of ${\text{C}}_8{\text{H}}_{18}$ is found to be $\text{-5461}\text{.2}\text{kJ}/\text{mol}$.

$\begin{array}{l}\text{Ethanol}\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\text{(l)}\text{+3}{\text{O}}_{\text{2}}\text{(g)}\to {\text{2CO}}_{\text{2}}\text{(g)}\text{+}{\text{3H}}_{\text{2}}\text{O}\text{(l)}\end{array}$

One mole of ethanol produces two moles of ${\text{CO}}_{\text{2}}$.

The number of moles ${\text{CO}}_{\text{2}}$ released by ethanol will be $\text{43}\text{.4}\text{mol}{\text{CO}}_{\text{2}}$

$\begin{array}{l}\text{Isooctane}\\ {\text{C}}_8{\text{H}}_{18}\text{(l)}\text{+}\frac{25}{2}{\text{O}}_{\text{2}}\text{(g)}\to 8{\text{CO}}_{\text{2}}\text{(g)}\text{+}9{\text{H}}_{\text{2}}\text{O}\text{(l)}\end{array}$

One mole of isooctane produces eight moles of ${\text{CO}}_{\text{2}}$.

The number of moles ${\text{CO}}_{\text{2}}$ released by the number of moles ${\text{CO}}_{\text{2}}$ released by isooctane will be $\text{70}\text{mol}{\text{CO}}_{\text{2}}$

Isooctane release more amount of ${\text{CO}}_{\text{2}}$.

(c)

Interpretation Introduction

Interpretation: The better fuel in terms of energy production and greenhouse gases 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 47GQ

Isooctane releases more energy, but it releases more ${\text{CO}}_{\text{2}}$ too.

Explanation of Solution

In-order to say whether ethanol or isooctane is a better fuel, we have to compare its energy release.

Comparing ethanol and isooctane, isooctane releases more energy.

But the problem is that when we compare the ${\text{CO}}_{\text{2}}$ release by both ethanol and isooctane, isooctane releases more ${\text{CO}}_{\text{2}}$.${\text{CO}}_{\text{2}}$ is green-house gas and if the amount is increased, it will pollute the atmosphere. Isooctane releases more amount of ${\text{CO}}_{\text{2}}$ than ethanol.

On the basis of energy release isooctane can be considered as a better fuel, but when the release of ${\text{CO}}_{\text{2}}$ is considered it will not be a better fuel.