Enter your answer in the provided box. Ethanol (C 2 H 5 OH) and gasoline (assume

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 H

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

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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)

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Explanation / Answer

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.

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