Chapter 21, Problem 21.153P

a)

Interpretation Introduction

Interpretation:

Burning of $\text{1}\text{.00}\text{gal}$ gasoline to produce carbon-dioxide gas and water vapour, value of $\Delta {\text{H}}^{\circ }$ has to be calculated.

Concept Introduction:

The standard enthalpy of reaction is quantity of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elemental form. It is denoted as ${\text{ΔH}}_{\text{rxn}}^{\text{°}}$.

$\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{°}}={\text{ ΔH}}_{\text{p}}^{\text{°}}-{\text{ ΔH}}_{\text{r}}^{\text{°}}\\ {\text{ΔH}}_{\text{p}}^{\text{°}}\text{ = standard enthalpy of product}\\ {\text{ΔH}}_{\text{r}}^{\text{°}}\text{ = standard enthalpy of reactant}\end{array}$

Explanation of Solution

The given chemical equation:

${\text{2C}}_{\text{8}}{\text{H}}_{\text{8}}\left(\text{l}\right)\text{+ 25}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{16CO}}_{\text{2}}\left(\text{g}\right)\text{+18}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

As known, heat of reaction is calculated using heat of formation as follows,

${\text{ΔH}}_{\text{rxn}}^{\text{°}}=\sum \text{m }\Delta {\text{H}}_{f\left(poducts\right)}^{\text{°}}-\sum \text{m }\Delta {\text{H}}_{f\left(reac\tan ts\right)}^{\text{°}}$

$\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left(\text{16}{\text{mol CO}}_{\text{2}}\right)\left({\text{ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{CO}}_{\text{2}}\right)\text{+}\left(\text{18}{\text{mol H}}_{\text{2}}\text{O}\right)\left({\text{ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{H}}_{\text{2}}\text{O}\right)\right]\text{-}\\ \left[\left(\text{2}\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\right)\left({\text{ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\right)\text{+}\left(\text{25}{\text{mol O}}_{\text{2}}\right)\left({\text{ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{O}}_{\text{2}}\right)\right]\\ \\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left(\text{16}{\text{mol CO}}_{\text{2}}\right)\left(\text{-393}\text{.5}\text{kJ/mol}\right)\text{+}\left(\text{18}{\text{mol H}}_{\text{2}}\text{O}\right)\left(\text{-241}\text{.826}\text{kJ/mol}\right)\right]\text{-}\\ \left[\left(\text{2}\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\right)\left(\text{-250}\text{.1}\text{kJ/mol}\right)\text{+}\left(\text{25}{\text{mol O}}_{\text{2}}\right)\left(\text{0}\text{kJ/mol}\right)\right]\\ \\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\text{-10148}\text{.868}\text{=}\text{-10148}\text{.9}\text{kJ}\text{.}\end{array}$

The energy from $\text{1}\text{.00}\text{gal}$ of gasoline as follows,

$\begin{array}{l}\text{Energy }\left(kJ\right)\text{=}\left(\text{1}\text{.00}\text{gal}\right)\left(\frac{4qt}{1gal}\right)\left(\frac{1L}{1.057qt}\right)\left(\frac{1mL}{{10}^{-3}L}\right)\left(\frac{0.7028g}{mL}\right)\left(\frac{1mol{C}_8{H}_{18}}{114.22g{C}_8{H}_{18}}\right)\left(\frac{-10148.9kJ}{2mol{C}_8{H}_{18}}\right)\\ \\ \text{=}\text{-1}\text{.18158}\times {\text{10}}^5\text{=}-1.18\times 10^{5}kJ\text{.}\end{array}$

Hence, the calculated value of $\Delta {\text{H}}^{\circ }$ is $-1.18\times 10^{5}kJ.$

b)

Interpretation Introduction

Interpretation:

Liters of ${\text{H}}_2$ required to burn the given quantity of energy has to be calculated.

Concept introduction:

Ideal gas: A hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas law exactly.

Ideal gas equation: An equation describes the relationship among the four variables P, V, T, and n.

$PV=nRT$

The unit of each term in the equation is,

Where, R is proportionality constant called gas constant $\left(atm.L/mol.K\right)$

  V is volume $\left(L\right)$

  N is number of moles $\left(moles\right)$

  P is Pressure $\left(atm\right)$

  T is temperature $\left(K\right)$

Standard temperature and pressure: The conditions ${\text{0}}^{\text{o}}\text{C}$ and $\text{1}\text{atm}$ are called standard atmospheric conditions.

Explanation of Solution

The reaction as follows,

${\text{H}}_2\left(\text{g}\right)\text{+ }\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The heat of formation of water vapour is ${\text{ΔH}}^{\text{o}}\text{=}\text{-241}\text{.826}\text{kJ}\text{.}$

The moles of hydrogen needed to produce the energy from given part a) is,

${\text{Moles of H}}_{\text{2}}\text{=}\left(\text{-1}\text{.18158}\text{×}{\text{10}}^{\text{5}}\text{kJ}\right)\left(\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}}{\text{-241}\text{.826}\text{kJ}}\right)\text{=}\text{488}\text{.6}\text{mol}\text{.}$

In order to determine the volume, the ideal gas equation is used as shown below,

$\text{V}\text{=}\frac{\text{nRT}}{\text{P}}\text{=}\left(\frac{\left(\text{488}\text{.6}\text{mol}{\text{H}}_{\text{2}}\right)\left(\text{0}\text{.0821}\frac{\text{L}\text{.atm}}{\text{mol}\text{.K}}\right)\left(\text{298}\text{K}\right)}{\text{1}\text{.00}\text{atm}}\right)\text{=}\text{1}\text{.195}\times {\text{10}}^4=1.20\times 10^{4}L\text{.}$

Therefore, the volume required calculated as $1.20\times 10^{4}L.$

c)

Interpretation Introduction

Interpretation:

For the amount of ${\text{H}}_2$ to produce the time required by electrolysis with a current of $\text{1}\text{.00}\text{×}{\text{10}}^{\text{3}}\text{A}\text{at 6}\text{.00}\text{V}$ has to be calculated.

Explanation of Solution

The reaction as follows,

${\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+ 2}{\text{e}}^{\text{-}}\to {\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2OH}}^{\text{-}}\left(\text{aq}\right)$

Using the known unit conversion; $\text{1A}\text{ =}\text{1}\text{C/s}$

$\text{Time }\left(\text{s}\right)\text{=}\left(\text{488}\text{.6}\text{mol}{\text{H}}_{\text{2}}\right)\left(\frac{\text{2}\text{mol}{\text{e}}^{\text{-}}}{\text{2}\text{mol}{\text{H}}_{\text{2}}}\right)\left(\frac{\text{96485}\text{C}}{\text{1}\text{mol}{\text{e}}^{\text{-}}}\right)\left(\frac{\text{s}}{\text{1}\text{.00}\text{×}{\text{10}}^{\text{3}}\text{C}}\right)\text{=}9.43\times 10^4\sec onds\text{.}$

Therefore, the required time calculated as $9.43\times 10^4\sec onds.$

d)

Interpretation Introduction

Interpretation:

Power required generating the amount of ${\text{H}}_{\text{2}}$ in kilowatt-hours has to be calculated.

Explanation of Solution

The coulombs involved in electrolysis of $\text{488}\text{.6}\text{mol}\text{of}{\text{H}}_{\text{2}}.$

$\text{Coulombs}\text{=}\left(\text{488}\text{.6}\text{mol}{\text{H}}_2\right)\left(\frac{2\text{mol}{\text{e}}^{-}}{1\text{mol}{\text{H}}_{\text{2}}}\right)\left(\frac{96485C}{1\text{mol}{\text{e}}^{-}}\right)\text{=}\text{94,286,589}\text{C}\text{.}$

Conversion of Coulomb into Joules:

$\text{Joules}\text{=}\text{C}\text{×}\text{V=}\text{94,286,589}\text{C}\text{×}\text{6}\text{.00V}\text{=}565,719,534J\text{.}$

Conversion of Joules into Power:

$\text{Power}\left(\text{kW}\text{.h}\right)\text{=}\left(\text{565,719,534}\text{J}\right)\left[\frac{\text{1}\text{kW}\text{.h}}{\text{3}\text{.6}{\text{×10}}^{\text{6}}\text{J}}\right]\text{=}\text{157}\text{.144}\text{=}157kW.h\text{.}$

Therefore, the Power required in order of generating the amount of ${\text{H}}_{\text{2}}$ in kilowatt-hours calculated as $157kW.h.$

e)

Interpretation Introduction

Interpretation:

The cost of producing the amount of ${\text{H}}_{\text{2}}$ equivalent to $\text{1}\text{.00}\text{gal}$ of gasoline has to be calculated.

Explanation of Solution

Given:

Cell has efficiency of $88.0\%$; electricity costs of $\text{Per}\text{kW}\text{.h}$ is $\$\mathrm{0.123.}$

Additional electricity is necessary to produce sufficient hydrogen; the purpose of $\left(\frac{100\%}{80\%}\right)$ factor.

$\text{Cost}\text{=}\left(157kW.h\right)\left(\frac{100\%}{80\%}\right)\left[\frac{\text{0}\text{.123}\text{cents}}{\text{1}\text{kW}\text{.h}}\right]\text{=}21.96\text{=}22.0cents\text{.}$

Therefore, the cost of producing the amount of ${\text{H}}_{\text{2}}$ equivalent to $\text{1}\text{.00}\text{gal}$ of gasoline calculated as $22.0cents.$