Gasoline, which can be represented by the formula octane, bulk density is 704 g/l. To estimate the variation in enthalpy when burning 28.5 liters
gasoline, based on the binding energies: 0 = 0 (117 kcal); C = 0 (173 kcal); 0 h (111 kcal); C-C (83 kcal); C-H (99 kcal)
To estimate the variation in enthalpy when burning 28.5 liters of gasoline, we need to calculate the total energy released during the combustion process.
First, we need to determine the chemical composition of gasoline, which is predominantly octane (C8H18). Octane is an alkane with eight carbon atoms and 18 hydrogen atoms.
The balanced chemical equation for the combustion of octane is:
2C8H18 + 25O2 -> 16CO2 + 18H2O
From the given binding energies, we can determine the energy released for each bond broken and formed during the combustion process.
1. Breaking the O=O bond: 0 kcal/mol
2. Breaking the C-C bonds (8 in total): 8 x 83 kcal/mol = 664 kcal/mol
3. Breaking the C-H bonds (18 in total): 18 x 99 kcal/mol = 1782 kcal/mol
4. Forming the CO2 bonds (16 in total): 16 x 0 kcal/mol = 0 kcal/mol
5. Forming the H2O bonds (18 in total): 18 x 111 kcal/mol = 1998 kcal/mol
Now, we can calculate the total energy released during the combustion:
Energy released = Energy of bond breaking - Energy of bond formation
= (664 + 1782) - (0 + 1998)
= 2448 - 1998
= 450 kcal/mol
Next, we need to convert the energy in kcal/mol to energy per liter of gasoline.
The bulk density of gasoline is given as 704 g/l, which means that the density of gasoline is 0.704 g/ml.
To find the number of moles in 28.5 liters of gasoline, we use the formula:
moles of gasoline = volume / molar volume = (28.5 L) / (0.704 g/mL x 1000 mL/L x molar mass of octane)
The molar mass of octane (C8H18) is (12.01 g/mol x 8) + (1.01 g/mol x 18) = 114.23 g/mol.
moles of gasoline = (28.5 L) / (0.704 g/mL x 1000 mL/L x 114.23 g/mol) = 355.94 moles
Finally, we can calculate the total energy released when burning 28.5 liters of gasoline:
Total energy released = energy released per mole x moles of gasoline
= 450 kcal/mol x 355.94 moles
= 160,676.26 kcal
Therefore, approximately 160,676.26 kcal of energy would be released when burning 28.5 liters of gasoline.
To estimate the variation in enthalpy when burning 28.5 liters of gasoline, we need to calculate the total energy released in the combustion reaction of the components in gasoline.
First, let's determine the components present in gasoline. The formula octane represents gasoline, which consists mainly of hydrocarbons. Octane (C8H18) is a major component of gasoline. Therefore, we can assume that gasoline is mostly octane.
Now, let's calculate the total energy released in the combustion of 28.5 liters of octane.
1. Calculate the mass of 28.5 liters of octane:
The bulk density of gasoline is given as 704 g/l.
Mass = Volume x Density
Mass = 28.5 liters x 704 g/l
Mass = 20040 grams or 20.04 kg
2. Determine the number of moles of octane:
The molar mass of octane (C8H18) is 114.22 g/mol (8 x 12.01 + 18 x 1.01).
Moles = Mass / Molar mass
Moles = 20040 g / 114.22 g/mol
Moles ā 175.4 mol
3. Calculate the energy released from the combustion of octane:
The enthalpy change for the combustion of different bonds is given:
C-C (83 kcal/mol), C-H (99 kcal/mol), 0 = 0 (117 kcal/mol), C = 0 (173 kcal/mol), 0 h (111 kcal/mol)
Octane can be represented as a sum of C-C and C-H bonds:
Octane = 8 x C-C + 18 x C-H
Energy released = (8 x 83 kcal/mol) + (18 x 99 kcal/mol)
Energy released = 664 kcal + 1782 kcal
Energy released ā 2446 kcal/mol
4. Calculate the total energy released in the combustion of 28.5 liters of octane:
Total energy released = Energy released x Moles
Total energy released = 2446 kcal/mol x 175.4 mol
Total energy released ā 428,993 kcal
Therefore, the estimated variation in enthalpy when burning 28.5 liters of gasoline (mainly octane) is approximately 428,993 kcal.