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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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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