Question

Ethanol (C2H5OH) and gasoline (C8H18) are both used as automobile fuel. If gasoline is selling for $2.50/gal, what would the price of ethanol have to be in order to provide the same amount of heat per dollar? Necessary ΔH°f values are found in the table below. The density of octane is 0.7025 g/mL and the density of ethanol is 0.7894 g/mL. 1 gal = 3.785 L.

C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(g)

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Substance	ΔH°f (kJ/mol)
H2O(g)	−241.8
H2O(l)	−285.8
CO2(g)	−393.5
O2(g)	0
C2H5OH	−277.0
C8H18	−249.9

Answer 1

Sol:-

Calculation of heat produce by ethanol :-

given

C2H5OH (l) + 3 O2 --------> 2 CO2 (g) + 3 H2O (l) , ΔH°_c = ΔH°_rxn = ?

we know

ΔH°_c = ΔH°_rxn = [sum of ΔH°_f of products] - [sum of ΔH°_f of reactants]

ΔH°_c =[ 2xΔH°_f of CO2 + 3x ΔH°_f of H2O] -[ 1 x ΔH°_f of C2H5OH + 3x ΔH°_f of O2]

ΔH°_c =[2 (−393.5) + 3(−285.8)] - [ 1(−277.0) + 3 x 0]

ΔH°_c = [ -787 - 857.4 ] - [ - 277.0 ]

ΔH°_c = -1644.4 + 277.0

ΔH°_c = - 1367.4 KJ/mol

Now ΔH°_c per gram = - 1367.4 KJmol^-1 / 46.08 gmol^-1 = - 29.67 KJ/g

[ Gram molar mass of ethanol = 46.08 g/mol]

ΔH°_c per mL = - 29.67 KJg^-1 x 0.7894 g mL^-1 = - 23.4215 KJ/mL

ΔH°_c per L = - 23.42 x 10³ KJ/ L = - 23421.5 KJ/L

ΔH°_c per gallon =  - 23421.5 KJ / 0.2642 gal = - 88650.6 KJ/gal

[ because 1 gal = 3.785 L so 1 L = 0.2642 gal ]

Hence heat produce by ethanol = - 88650.6 KJ/gal

similarly

Calculation of heat produce by gasoline :-

given

C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(g), ΔH°_c = ΔH°_rxn = ?

ΔH°_c =[ 8xΔH°_f of CO2 + 9x ΔH°_f of H2O] -[ 1 x ΔH°_f of C8H18 + 25/2x ΔH°_f of O2]

ΔH°_c =[8 (−393.5) + 9(−241.8 )] - [ 1(−249.9) + 25/2 x 0]

ΔH°_c = [ -3148 - 2176.2 ] - [ - 249.9 ]

ΔH°_c = -5324.2 + 249.9

ΔH°_c = - 5074.3 KJ/mol

Now ΔH°_c per gram = - 5074.3 KJ mol^-1 / 114 gmol^-1 = - 44.5 KJ/g

[ Gram molar mass of gasoline = 114 g/mol ]

ΔH°_c per mL = - 44.5KJg^-1 x 0.7025 g mL^-1 = - 31.26125 KJ/mL

ΔH°_c per L = - 31.26125 x 10³ KJ/ L = - 31261.25 KJ/L

ΔH°_c per gallon =  - 31261.25 KJ / 0.2642 gal = - 118324.2 KJ/gal

[ because 1 gal = 3.785 L so 1 L = 0.2642 gal ]

Hence heat produce by gasoline =  - 118324.2 KJ/gal

So

Ratio of price = heat produce by ethanol / heat produce by gasoline

Ratio of price = - 88650.6 KJ/gal / - 118324.2 KJ/gal

Ratio of price = 0.7492

So

Cost of ethanol = 0.7492 x $2.50/gal = 1.87 $/ gal