Q21.56CP - Chemistry: Molecular Nature Of Matter And Change

Question

Gasoline is a mixture of hydrocarbons, but the heat released when it burns is close to that of octane, ${\mathbf{C}}_{\mathbf{8}}{\mathbf{H}}_{\mathbf{18}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{250}\mathbf{.}\mathbf{1}\mathbf{ }\mathbf{KJ}\mathbf{/}\mathbf{mol}$ As an alternative to gasoline, research is underway to use   from the electrolysis of water in fuel cells to power cars. 

(a) Calculate $\mathbf{∆}{\mathit{H}}^{\mathbf{o}}$when $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{gal}$ of gasoline $\mathbf{(}\mathbf{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7028}\mathbf{ }\mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{)}$ burns to produce carbon dioxide gas and water vapour.

(b) How many litres of ${\mathbf{H}}_{\mathbf{2}}$ at $\mathbf{25}\mathbf{°}\mathbf{C}$and $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{atm}$must burn to produce this quantity of energy?

(c) How long would it take to produce this amount of ${\mathbf{H}}_{\mathbf{2}}$ by electrolysis with a current of $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{ }\mathbf{A}$ at $\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{V}$.

(d) How much power in kilowatt hours $\mathbf{(}\mathbf{kW}\mathbf{\times }\mathbf{h}\mathbf{)}$ is required to generate this amount of ${\mathbf{H}}_{\mathbf{2}}$? $\mathbf{(}\mathbf{1}\mathbf{ }\mathbf{W}\mathbf{=}\mathbf{1}\mathbf{ }\mathbf{J}\mathbf{/}\mathbf{s}\mathbf{,}\mathbf{1}\mathbf{ }\mathbf{J}\mathbf{=}\mathbf{1}\mathbf{C}\mathbf{\times }\mathbf{V}\mathbf{,}\mathbf{and}\mathbf{1}\mathbf{ }\mathbf{kW}\mathbf{\times }\mathbf{h}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{ }\mathbf{J}\mathbf{.}\mathbf{)}$

(e) If the cell is $\mathbf{88}\mathbf{.}\mathbf{0}\mathbf{\%}$ efficient and electricity costs $\mathbf{0}\mathbf{.}\mathbf{950}\mathbf{C}$ per $\mathbf{kW}\mathbf{\times }\mathbf{h}$, what is the cost of producing the amount of ${\mathbf{H}}_{\mathbf{2}}$ equivalent to $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{gal}$ of gasoline?

Step-by-Step Solution

Answer

(a) The value is $-1.18\times {10}^5\mathrm{kJ}$.

(b) The amount is $1.195\times {10}^4 \mathrm{L}$.

(c) The time is $9.43\times {10}^4\mathrm{s}$.

(d) The value is $131\mathrm{kW}\times \mathrm{h}$.

(e) The cost is $141.42\mathrm{dollars}$.

Step 1: Definition of energy

Change from liquid to vapour phase involves heat of vaporisation, which requires energy to be consumed or released. Chemical energy is the energy associated with a molecule's chemical bonds.

Step 2: Subpart (a)

Equation given, $2{\mathrm{C}}_8{\mathrm{H}}_8+25{\mathrm{O}}_2\to 16{\mathrm{CO}}_2+18{\mathrm{H}}_2\mathrm{O}$

Calculate the ${\mathrm{\Delta H}}^{\mathrm{o}}$using formation heats.

$\begin{array}{c}{\mathrm{\Delta H}}^{\mathrm{o}}{=\mathrm{\Delta H}}_{\mathrm{products}}^{\mathrm{o}}{-\mathrm{\Delta H}}_{\mathrm{reactants}}^{\mathrm{o}}\\ =\left({16\mathrm{\Delta H}}_{{\mathrm{CO}}_2}^{\mathrm{o}}{+18\mathrm{\Delta H}}_{{\mathrm{H}}_2\mathrm{O}}^{\mathrm{o}}\right)-\left({2\mathrm{\Delta H}}_{{\mathrm{C}}_8{\mathrm{H}}_8}^{\mathrm{o}}{+25\mathrm{\Delta H}}_{{\mathrm{O}}_2}^{\mathrm{o}}\right)\\ =(16\times -393.5+18\times -241.826)-(2\times -250.1+25\times 0)\\ =10,149\mathrm{kJ}\end{array}$

Convert the volume of gasoline to grammes by multiplying by the density, then dividing by the molar mass to get moles of octane.

$1.00\mathrm{gal}\times \frac{3785 \mathrm{mL}}{1\mathrm{gal}}\times \frac{0.7028 \mathrm{g}}{1 \mathrm{mL}}\times \frac{1 {\mathrm{molC}}_8{\mathrm{H}}_8}{114.24 \mathrm{g}}=23.29\mathrm{mol}$

To calculate the heat emitted, multiply by the ${\mathrm{\Delta H}}_{\mathrm{f}}$

$23.29 \mathrm{mol}\times \frac{-10,149\mathrm{kJ}}{2 {\mathrm{molC}}_8{\mathrm{H}}_{18}}=-1.18\times {10}^5\mathrm{kJ}$

Therefore, the value is $-1.18\times {10}^5\mathrm{kJ}$.

Step 3: Subpart (b)

Equation given, ${\mathrm{H}}_{2\left(\mathrm{g}\right)}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}$has ${\mathrm{\Delta H}}^{\mathrm{o}}=-241.826 \mathrm{kJ}$, Divide the ${\mathrm{\Delta H}}^{\mathrm{o}}$ from (a) by the energy released by hydrogen combustion.

$-1.18\times {10}^5\mathrm{kJ}\times \frac{1 {\mathrm{molH}}_2}{-241.826 \mathrm{kJ}}=488.71 {\mathrm{molH}}_2$

To calculate the amount of hydrogen in litres, use the ideal gas equation 

$\begin{array}{c}\mathrm{PV}=\mathrm{nRT}\\ \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}\\ =\frac{488.71\times 0.08206\times 298.15}{1 \mathrm{atm}}\\ =1.195\times {10}^4 \mathrm{L}\end{array}$

Therefore, the amount is $1.195\times {10}^4 \mathrm{L}$.

Step 4: Subpart (c)

The time required to make $488.71$mol of hydrogen (from b) can be computed using the equation

$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

Convert the moles of hydrogen to electrons, then multiply by Faraday's constant to get the charge. The generation of hydrogen gas from   requires two electrons.

$488.71 {\mathrm{molH}}_2\times \frac{2{\mathrm{mole}}^{-}}{{\mathrm{molH}}_2}\times \frac{96,500\mathrm{C}}{{\mathrm{mole}}^{-}}=9.43\times {10}^7\mathrm{C}$

To calculate the required time, divide by the current time.

$9.43\times {10}^7\times \frac{1\mathrm{s}}{1000\mathrm{C}}=9.43\times {10}^4\mathrm{s}$

Therefore, the time is $9.43\times {10}^4\mathrm{s}$.

Step 5: Subpart (d)

Calculate kWh by multiplying the charge from (c) by the voltage.

$9.43\times {10}^7\mathrm{C}\times 5.00\mathrm{V}\times \frac{1\mathrm{J}}{1\mathrm{CV}}\times \frac{1\mathrm{kW}\times \mathrm{h}}{3.6\times {10}^6\mathrm{J}}=131\mathrm{kW}\times \mathrm{h}$

Therefore, the value is $131\mathrm{kW}\times \mathrm{h}$.

Step 6: Subpart (e)

Take the cost and multiply it by the power (d).

$131\mathrm{kw}\times \mathrm{H}\times \frac{0.950\mathrm{cents}}{1\mathrm{kW}\times \mathrm{h}}=124.45\mathrm{dollars}$

Calculate the cost of achieving $100$ percent efficiency.

$1.24\mathrm{dollars}\times \frac{100}{88}=141.42\mathrm{dollars}$

Therefore, the cost is $141.42\mathrm{dollars}$.