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 enthalpy change for each component in the gasoline and then sum them up.
The formula for octane, which represents gasoline, consists of several elements and bonds. Let's calculate the enthalpy change for each of these components:
1. For 0 = 0 (oxygen-oxygen bond), the enthalpy change is 0 kcal/mol.
2. For C = 0 (carbon-carbon triple bond), the enthalpy change is 173 kcal/mol.
3. For 0 h (oxygen-hydrogen bond), the enthalpy change is 111 kcal/mol.
4. For C-C (carbon-carbon single bond), the enthalpy change is 83 kcal/mol.
5. For C-H (carbon-hydrogen bond), the enthalpy change is 99 kcal/mol.
Now, we need to determine the number of moles of each component in 28.5 liters of gasoline. To do this, we'll use the bulk density of gasoline, which is given as 704 g/L. We'll assume that the density of gasoline is roughly equal to its molar mass.
First, we need to convert the volume of gasoline to grams:
28.5 liters x 704 g/L = 20,088 grams
Next, we need to convert grams to moles by dividing by the molar mass of gasoline. The molar mass of gasoline can vary, as it is a mixture of different hydrocarbons, but for this calculation, we'll assume the molar mass is roughly 114 g/mol.
Number of moles of gasoline = 20,088 grams / 114 g/mol ≈ 176.17 mol
Now we can calculate the enthalpy change by multiplying the number of moles of each component by their respective enthalpy values:
Enthalpy change = (Number of moles of 0 = 0) x (Enthalpy of 0 = 0)
+ (Number of moles of C = 0) x (Enthalpy of C = 0)
+ (Number of moles of 0 h) x (Enthalpy of 0 h)
+ (Number of moles of C-C) x (Enthalpy of C-C)
+ (Number of moles of C-H) x (Enthalpy of C-H)
Plugging in the values:
Enthalpy change = (0 mol) x (0 kcal/mol)
+ (0 mol) x (173 kcal/mol)
+ (0 mol) x (111 kcal/mol)
+ (176.17 mol) x (83 kcal/mol)
+ (176.17 mol) x (99 kcal/mol)
Simplifying:
Enthalpy change ≈ (176.17 mol) x (83 kcal/mol) + (176.17 mol) x (99 kcal/mol)
Calculating the sum:
Enthalpy change ≈ 14,602.11 kcal + 17,420.83 kcal ≈ 32,022.94 kcal
Therefore, the estimated variation in enthalpy when burning 28.5 liters of gasoline is approximately 32,022.94 kcal.