Using the balanced equation and the data in the table below, calculate the theoretical enthalpy of combustion.
Note: you will need to include the enthalpy of vaporisation for the liquid components which are also given.
CH3OH(l) + 1.5O2(g) → CO2(g) + 2H2O(l)
Average Bond Enthalpies KJmol-1
C-H 412
C-C 348
C-O 358
O=O 496
C=O 743
O-H 463
Enthalpy of vaporisation KJmol-1
Methanol 35
Water 41
This is a somewhat tedious problem. Tell me what you've done with it, what you don't understand and how I can help OTHER THAN WORKING THE PROBLEM COMPLETELY FOR YOU.
a) I think this is the way to find the average bond enthalpy:
3 x 412 = 1236 kJ
1.5 x 496 = 744 kJ
1 x 463 = 463 kJ
358 kJ
1486 kJ
Formed 2 x 743 = 1486 kJ
4 x 463 = 1852 kJ
I did not want to do anything else before checking this was right.
The note says to include the enthalpy of vaporisation for the liquid components which are also given. I did not. I do not know what to do.
See your other post above,
To calculate the theoretical enthalpy of combustion using the balanced equation and the given data, we need to calculate the energy required to break the bonds in the reactants and the energy released from the formation of bonds in the products.
Step 1: Calculate the energy required to break the bonds in the reactants.
In the reactants, we have bonds in CH3OH(l) and 1.5O2(g).
- CH3OH(l) has 3 C-H bonds and 1 C-O bond.
Energy required to break the C-H bonds = 3 * 412 kJ/mol = 1236 kJ/mol
Energy required to break the C-O bond = 1 * 358 kJ/mol = 358 kJ/mol
- 1.5O2(g) has 3 O=O bonds.
Energy required to break the O=O bonds = 3 * 496 kJ/mol = 1488 kJ/mol
Total energy required to break the bonds in the reactants = 1236 kJ/mol + 358 kJ/mol + 1488 kJ/mol = 3082 kJ/mol
Step 2: Calculate the energy released from the formation of bonds in the products.
In the products, we have bonds in CO2(g) and 2H2O(l).
- CO2(g) has 2 C=O bonds.
Energy released from the formation of C=O bonds = 2 * 743 kJ/mol = 1486 kJ/mol
- 2H2O(l) has 4 O-H bonds.
Energy released from the formation of O-H bonds = 4 * 463 kJ/mol = 1852 kJ/mol
Total energy released from the formation of bonds in the products = 1486 kJ/mol + 1852 kJ/mol = 3338 kJ/mol
Step 3: Calculate the enthalpy of vaporization for the liquid components.
For methanol, the enthalpy of vaporization is given as 35 kJ/mol.
For water, the enthalpy of vaporization is given as 41 kJ/mol.
Total enthalpy of vaporization = 35 kJ/mol (methanol) + 41 kJ/mol (water) = 76 kJ/mol
Step 4: Calculate the theoretical enthalpy of combustion.
Theoretical enthalpy of combustion = Total energy required to break the bonds in the reactants - Total energy released from the formation of bonds in the products + Total enthalpy of vaporization
= 3082 kJ/mol - 3338 kJ/mol + 76 kJ/mol
= -180 kJ/mol
Therefore, the theoretical enthalpy of combustion is -180 kJ/mol.