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
2) SUGGEST SOME REASONS AS TO WHY THIS VALUE IS DIFFERENT TO THE ONE OBRTAINED IN THE PRACTICAL (practical- a student investigated the enthalpy of combustion (deltaHc) of methanol under standard conditions using the apparatus shown in the diagram. The measurements the student recorded are shown in the table. Use this information to answer the questions below.).
Practical information:
Alcohol mass before burning: 80.6g
Alcohol mass after burning: 75.9g
Water heated: 100g
Methanol Mr: 32
Initial temperature of water: 21.5 C
Final temperature of water: 32.4 C
Q = 4556.2 J
To calculate the theoretical enthalpy of combustion, we need to use average bond enthalpies of the reactants and products.
Step 1: Calculate the energy required to break the bonds in the reactants: CH3OH(l) + 1.5O2(g)
- We need to break the bonds in methane (CH4) and oxygen (O2) in order to form methanol (CH3OH).
- CH4(g) --> C(g) + 4H(g)
- O2(g) --> 2O(g)
The total energy required to break the bonds in CH3OH is:
(1 x C-H bond energy) + (3 x C-H bond energy) + (1.5 x O=O bond energy) = (412 + 3 x 412) + (1.5 x 496) = 3816 kJ
Step 2: Calculate the energy released when new bonds are formed in the products: CO2(g) + 2H2O(l)
- We need to form new bonds in carbon dioxide (CO2) and water (H2O) from the atoms released in the reactants.
- C(g) + 2O(g) --> CO2(g)
- 4H(g) + O(g) --> 2H2O(l)
The total energy released when new bonds are formed in the products is:
(1 x C=O bond energy) + (2 x O-H bond energy) + (2 x O-H bond energy) = (1 x 743) + (2 x 463) + (2 x 463) = 2585 kJ
Step 3: Calculate the change in enthalpy (ΔH) by subtracting the energy required to break the bonds from the energy released when new bonds are formed:
ΔH = Energy released - Energy required = 2585 kJ - 3816 kJ = -1231 kJ
So, the theoretical enthalpy of combustion is -1231 kJ.
Now, let's address the second part of the question regarding the difference between the theoretical and practical values of the enthalpy of combustion.
Potential reasons for the difference could include:
1. Incomplete combustion in the practical experiment: The practical experiment may not have achieved complete combustion, resulting in the release of less energy than expected.
2. Heat losses to the surroundings: The heat generated during the combustion reaction could have been lost to the environment, causing a decrease in the observed enthalpy change.
3. Experimental errors in measurements: Errors in measuring the masses of alcohol, water, and the temperature change could have led to inaccuracies in calculating the enthalpy change.
4. Assumption of standard conditions: The theoretical calculation assumes standard conditions (25°C and 1 atm), whereas the practical experiment may not have exactly matched these conditions.
These factors, among others, can contribute to the discrepancy between the theoretical and practical values of the enthalpy of combustion.