Calculate the energy in kJ that is released as heat (70%) by the combustion of exactly 1 gallon of gasoline.
The combustion of gasoline releases and only about 30% of the energy from gasoline is used as work with the rest expelled as heat.
The chemical reaction of combustion is:
2 C8H18(l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g) H = –10110 kJ
The density of octane is 0.702 g/mL. Assume gasoline is pure octane (1 gallon = 3.7854 L).
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When gasoline, which is taken to be pure octane, undergoes combustion in an internal combustion engine 30% of the energy released is converted to mechanical work and the rest is expelled as heat. 1 gallon of octane is 3.7854 L and the density of octane is 0.702 kg/L
From the chemical equation:2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) `Delta H`=–10110 kJ
The molar mass of octane is 114 g/mole. The mass of 1 gallon of gasoline is 2.657 kg. This is approximately 2.657/114 = 23.31 moles of octane. Each mole of octane releases (10110/2)*0.7 = 5055*0.7 = 3538.5 kJ of heat.
23.31 moles releases 3538.5*23.31 = 82482.43 kJ of heat.
The amount of heat released when one gallon of gasoline is burnt is approximately 82482.43 kJ
1 gallon = 3.7854 L = 3785.4 mL
so the volume of ocatne is 3785.4mL and the density is 0.702 g/mL
density = mass/volume
mass= density * volume
Mass of octane = 0.702 g/mL * 3785.4 mL = 2657.35 g
molar mass of octane = 114.23 g/mol and mass is 2657.35 g
moles = mass/molar mass
moles of octane = 2657.35 g/(114.23 g/mol) = 23.26 mol
now the effiency is 70 % that is out of 100 moles, only 70 mol is combusted. so the number of moles combusted would be
23.26 mol * (70/100)
16.28 mol of octane.
now the reaction is
2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g) H = –10110 kJ
[dividing bothside by 2]
C8H18 (l) + 25/2 O2 (g) → 8 CO2 (g) + 9 H2O (g) H = –5055 kJ/mol
so 1 mol of octane gives -5055 kj of heat, but we have 16.28 mol(70% efficient).
so the amount of heat released would be
16.28 mol * (-5055 kj/1 mol)
82925.4 Kj -- this is the amount of heat released.
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