Question

Enter your answer in the provided box.

Ethanol (C₂H₅OH) and gasoline (assumed to be all octane, C₈H₁₈) are both used as automobile fuel. If gasoline is selling for $3.19/gal, what would the price of ethanol have to be in order to provide the same amount of heat per dollar? The density and ΔH

o

f

of octane are 0.7025 g/mL and −249.9 kJ/mol and the density and ΔH

o

f

of ethanol are 0.7894 g/mL and −277.0 kJ/mol, respectively. Assume that the products of combustion are CO₂(g) and H₂O(l). (1 gal = 3.785 L)

Answer 1

To compare this, you need to calculate the heat of combustion for both ethanol and gasoline.

the balanced chemical equation for combustion of gasoline : C8H18(l) + (25/2)O2(g) --> 8CO2(g) + 9H20(g)

Calculate heat of combustion of Octane. Refer to your text book fot the standard values.

delta H of Combustion = delta H formation of products - delta H formation of reactants
= 8*delta H CO2(g) + 9*delta H H20(g) - [1*delta H C8H18(l) + 25*delta H O2]
= 8*-393.52 + 9*-241.8 - [1*-249.9 + 25*0]
= -3148.16 - 2176.2 + 249.9
= -5074.46 kJ/mol

Change delta H combustion of gasoline in terms of one gallon,

molar mass of gasoline = 8*12.01g/mol + 18*1.01g/mol
= 114.09 g/mol

delta H combustion of gasoline per gram= -5074.46kJ/mol / 114.09g/mol
= -44.48 kJ/g

delta H combustion of gasoline per mL= -44.48kJ/g / 0.7025g/mL
= -63.31 kJ/mL
delta H combustion of gasoline per L = -63.31kJ/mL *1000mL/1L
= -63313.44 kJ/L

delta H combustion of gasoline per gallon = -63313.44kJ/L * 3.785 L / 1 gallon
= -2.40 x10^5 kJ / gallon
This is how much energy there is released in one gallon of gasoline.

heat per dollar = 2.40x10^5 kJ/gallon / $ 3.19 / gallon = 7.656 *10^5kJ/dollar for gasoline


Now find the delta H combustion of ethanol
The balanced chemical equation for combustion of ethanol :C2H5OH (l) + 3O2(g) ---> 2CO2(g) + 3H20(g)

delta H of Combustion = delta H formation of products - delta H formation of reactants
= 2*delta H CO2(g) + 3*delta H H20(g) - [1*delta H C2H5OH(l) + 3*delta H O2]
= 2*-393.52 + 3*-241.8 - [1*-277 + 3*0]
= -1235.44 kJ/mol

Change delta H combustion of ethanol in terms of one gallon,

molar mass of ethanol = 2*12.01g/mol + 6*1.01g/mol + 1*16g/mol
= 46.08 g/mol

delta H combustion of ethanol per gram= -1235.44 kJ/mol
46.08g/mol
= -26.81 kJ/g

delta H combustion of ethanol per mL= -26.81kJ/g / 0.7894g/mL
= -33.96 kJ/mL

delta H combustion of ethanol per L = -33.96kJ/mL *1000mL/1L
= -33963.47 kJ/L

delta H combustion of ethanol per gallon = -33963.47kJ/L * 3.785 L / 1 gallon
= 1.29 x10^5 kJ / gallon
This is how much energy is released in one gallon of ethanol.

heat per dollar ethanol = heat per dollar gasoline = 7.656 *10^5kJ/dollar

7.656 *10^5kJ/dollar = heat per gallon ethanol / dollar per gallon ethanol

Rearrange to get

dollar per gallon ethanol = heat per gallon ethanol / 1.54 x10^5kJ / dollar
= 1.29 x10^5 kJ / gallon / 7.656 *10^5kJ/dollar
= $0.17/gallon

So ethanol must be at $0.84/gallon to provide the same amount of heat per dollar as gasoline at $1.56/gallon.