1) Three compounds that have been used as fuel in cars and racing cars are ethan
ID: 542890 • Letter: 1
Question
1) Three compounds that have been used as fuel in cars and racing cars are ethanol, nitromethane (CH3NO2, H°f = -105.65 kJ/mole), and gasoline. Using octane (H°f = - 208.24 kJ/mole) as representative of gasoline, calculate the following quantities for each of the three compounds:
H° of combustion per mole of the fuel,
moles of gas produced per mole of fuel at 25°C and 1 atm pressure, heat (q) produced per gram of fuel,
liters of gas produced per gram of fuel at 25°C and 1 atm,
the q×V value.
Assume that all components are in the gas phase.
Which fuel would be the most efficient as a heat source based on these considerations? (Make
all of the simplifying assumptions regarding heat flow, ideal behavior, etc.)
Explanation / Answer
a) For ethanol H°f =-277.7 KJ / mole
CH3NO2, H°f = -105.65 kJ/mole
octane H°f = - 208.24 kJ/mole
CO2 H f = -393.5 kJ/mol
H2O (l) H f = -285.8 kJ/mol.
The combustion equation for ethanol will be
C2H5OH + 3O2 ---> 2CO2 (g) + 3H2O (l)
moles of gas produced per mole of Ethanol = 2moles of carbondioxide
DeltaH rxn = Heat of combustion of ethanol
= sum of enthalpy of formation of products- sum of enthalpy of formation of reactants
=[ 2 X Hf CO2 + 3 X H fH2O ] - [H f C2H5OH + H f O2]
= [2 X (-393.5) + 3 X (-285.8)] - [-277.7] = -1366.7 KJ / mole
Molecular weight of ethanol = 46 g / mole
So heat released per gram = -1366.7 / 46 = -29.71 KJ
2) for nitromethane
CH3NO2 + 3/4 O2 CO2 + 1.5H2O(l) + 0.5N2(g)
moles of gas produced per mole of nitromethane = 1 mole of carbondioxide + 0.5 moles of N2 = 1.5 moles of gas
DeltaH rxn = Heat of combustion of ethanol
= sum of enthalpy of formation of products- sum of enthalpy of formation of reactants
=[ Hf CO2 + 1.5 X H fH2O ] - [H f CH3NO2 + H f N2]
= [ (-393.5) + 1.5 X (-285.8)] - [-105.65] = -716.55 KJ / mole
Molecular weight of nitromethane = 61g / mole
So heat produced per gram = -176.55 / 61 = -11.75 KJ
3) for octane
C8H18 + 25/2 O2 ---> 8CO2 (g) + 9H2O(l)
moles of gas produced per mole of octane = 10 moles of carbondioxide
DeltaH rxn = Heat of combustion of ethanol
= sum of enthalpy of formation of products- sum of enthalpy of formation of reactants
=[ 8Hf CO2 + 9 X H fH2O ] - [H f C8H18 + H f O2]
= [ 8(-393.5) + 9 X (-285.8)] - [-208.24] = -5511.96 KJ / mole
Molecular weight of octane = 114g/ mole
heat evolved per gram = -5511.96 / 114 = -48.35 KJ
The heat released per gram for octane is 48.35 which is greater than heat released per gram of ethanola or nitromethane so due to high calorific value it can be considered as good fuel. Ethanol is also corrosive in nature.
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