For any reaction that is carried out directly at a constant pressure, the heat flow is exactly equal to the difference between enthalpy of products and that of the reactant. The symbol delta H ( H ) is given to symbolise enthalpy change.
H Reaction = H Products - H Reactant
Combustion of methanol.
To calculate the H we need know the bond enthalpy of the bonds present.
C=O = + 805 kJ mol-1
C-H = + 413 kJ mol-1
C-O = + 360 kJ mol-1
O=O = + 498 kJ mol-1
O-H = + 464 kJ mol-1
Bonds broken( H Reactant )
3 x C-H = 3 x (+ 413 ) = + 1239 kJ mol-1
1 x C-O = 1 x (+ 360) = + 360 kJ mol-1
1 x O-H = 1 x (+ 464) = + 464 kJ mol-1
1.5 x O=O = 1.5 x (+ 498) = + 747 kJ mol-
Total H Reactant = +1239 + 360 + 464 + 747 = + 2810 kJ mol-1
Bond formed ( H Products)
4 x O-H = 4 x (+ 464) = + 1856 kJ mol-1
2 x C=O = 2 x (+ 805) = + 1610 kJ mol-1
Total H Products = +1856 + 1610 = + 3556 kJ mol-1
The total enthalpy change of formation
:
H Reaction = H Products - H Reactant
H Reaction = -(+3556 - 2810)
H Reaction = -( + 746)
H Reaction = - 746 kJ mol-1
Combustion of Ethanol
To calculate the H we need know the bond enthalpy of the bonds present.
C=O = + 805 kJ mol-1
C-H = + 413 kJ mol-1
C-O = + 360 kJ mol-1
O=O = + 498 kJ mol-1
O-H = + 464 kJ mol-1
C-C = + 347 kJ mol-1
Bonds broken( H Reactant )
5 x C-H = 5 x (+ 413 ) = + 2065 kJ mol-1
1 x C-O = 1 x (+ 360) = + 360 kJ mol-1
1 x O-H = 1 x (+ 464) = + 464 kJ mol-1
3 x O=O = 3 x (+ 498) = + 1494 kJ mol-1
1 x C-C = 1 x (+ 347) = +347 kJ mol-1
Total H Reactant = +2065 + 360 + 464 + 1494 + 347 = + 4730 kJ mol-1
Bond formation ( H Products)
6 x O-H = 6 x (+ 464) = + 2784 kJ mol-1
4 x C=O = 4 x (+ 805) = + 3220 kJ mol-1
Total H Products = +2784 + 3220 = +6004 kJ mol-1
The total enthalpy change of formation :
H Reaction = H Products - H Reactant
H Reaction = -(+ 6004 - 4730)
H Reaction = -( + 1274)
H Reaction = - 1274 kJ mol-1
To separate C-H bond you need to apply 413 kJ of energy. There are five such bonds in ethanol so you multiply 413 by five to get 2065 kJ. All of the other alcohols can be broken up in this way.
Aim: to calculate the different enthalpy changes of different alcohol
I predict that the more bonds there are holding the carbon, oxygen and hydrogen atoms together, more energy will be required to break them apart. For example Ethanol has the formula C2 H5OH. In this formula you have five C-H bonds, one C-C bond, one C-O bond and one O-H bond. To separate these types of bonds you require a certain amount of energy.
C=O = + 805 kJ mol-1
C-H = + 413 kJ mol-1
C-O = + 360 kJ mol-1
O=O = + 498 kJ mol-1
O-H = + 464 kJ mol-1
C-C = + 347 kJ mol-1
These are the alcohols I will be using.
- methanol
- ethanol
- butan-1-ol
- propan-1-ol
- propan-2-ol
The structures /display formulas of the alcohol are as follows:
1. Methanol
2. Ethanol
3. Propan-2-ol
4. propan-1-ol
5. Butan-1-ol
As you can see a longer molecule takes more energy to break its bonds, in this case Propan-2-ol. Compared to a smaller molecule, methanol that requires less energy to do so. I predict that the longer the molecular structure in the alcohol the more energy it will take to remove the bonds. So when I come to predicting results I can safely say that Propan-2-ol will need more input of energy than methanol simply because it has more bonds to break.
How to calculate energy transfer of alcohol after combustion.
Heat transfer = mc T x M
y
- m = mass of water
- c = specific heat capacity
- T= rise in temperature
- M = relative molecular mass of alcohol
- y = (initial mass of alcohol – mass of alcohol after combustion)
y = Difference
-
mass of water = 200cm3
- Specific heat capacity of water is 4.2 J
-
Rise in temperature is 20 oc
Example
Ar (H) = 1, (c) = 12 (O) = 16
RMM of (CH3OH) = 12 + (4X 1) + 16 = 32g
Heat transfer = 200cm3 x 4.2 J x 20 oc x (32g)
0.21(rough estimate)
Heat transfer = 16800 J x (32g)
0.21
Heat transfer = 16800 J x 15.23
Heat transfer = 25586400 J
Heat transfer = -(2558) kJ mol-1
Heat transfer = - 2558 kJ mol-1
Through this method I will calculate my energy transfer of different alcohol.
Predicted table