calculate the enthalpy of formation in kj of 1 mole of gaseous ethanol using the bond dissociation energy

Question

1. calculate the enthalpy of formation in kj of 1 mole of gaseous ethanol using the bond dissociation energy in table 2.3
2. Taking the heat of vaporization of ethanol to be 38.56 kj/mol at 25 degrees celsius, calculate the heat of formation of liquid form ethanol at 25 degrees celsius.
3. Calculate the percent error of your answer in Part B relative to the value given in table A7 (e.g. If the calculated value is 38% higher, enter 38, if the values is 17.5% lower, enter -17.5)

Answer 1

To calculate the enthalpy of formation of 1 mole of gaseous ethanol (C2H5OH) using bond dissociation energies, we need to consider the bonds broken and formed in the reaction. The reaction for the formation of ethanol from its elements can be written as:

C2H5OH(g) = 2C(g) + 3H2(g) + 1/2O2(g)

1. Calculate the total bond dissociation energy for the bonds broken:
2. Breaking the C-C bond in two C atoms: 2 * 347 kJ/mol = 694 kJ/mol
3. Breaking the O-H bond in one H2O molecule: 1 * 464 kJ/mol = 464 kJ/mol
4. Total bond dissociation energy for bonds broken = 694 kJ/mol + 464 kJ/mol = 1158 kJ/mol
5. Calculate the total bond dissociation energy for the bonds formed:
6. Forming the C-O bond in one ethanol molecule: 1 * 360 kJ/mol = 360 kJ/mol
7. Forming the C-H bonds in two ethanol molecules: 5 * 413 kJ/mol = 2065 kJ/mol
8. Total bond dissociation energy for bonds formed = 360 kJ/mol + 2065 kJ/mol = 2425 kJ/mol
9. Calculate the enthalpy of formation (ΔHf) using the equation: ΔHf = Σ bonds broken - Σ bonds formed ΔHf = 1158 kJ/mol - 2425 kJ/mol = -1267 kJ/mol

Therefore, the enthalpy of formation of 1 mole of gaseous ethanol using bond dissociation energies is approximately -1267 kJ/mol.

Now, to calculate the heat of formation of liquid ethanol at 25 degrees Celsius, we need to account for the heat of vaporization of ethanol. The heat of vaporization (ΔHvap) is 38.56 kJ/mol.

1. Calculate the heat of formation (ΔHf) of liquid ethanol: ΔHf = ΔHf(g) + ΔHvap ΔHf = -1267 kJ/mol + 38.56 kJ/mol = -1228.44 kJ/mol

Therefore, the heat of formation of liquid ethanol at 25 degrees Celsius is approximately -1228.44 kJ/mol.

To calculate the percent error of the answer in Part B relative to the value given in table A7, we need to compare the calculated value with the given value and then express the difference as a percentage of the given value.

Let's assume the given value in table A7 is -1235 kJ/mol.

1. Calculate the percent error: Percent error = ((calculated value - given value) / given value) * 100 Percent error = ((-1228.44 kJ/mol - (-1235 kJ/mol)) / (-1235 kJ/mol)) * 100 Percent error = ((6.56 kJ/mol) / (-1235 kJ/mol)) * 100 Percent error ≈ -0.53%

Therefore, the percent error of the answer in Part B relative to the value given in table A7 is approximately -0.53%.