O-H x1 = 463 1488 ---------- -----------
-------- 2972 2778
3231
4719 5750
4719 – 5750 = -1031
Practical
50g x 200 x 4.2 92000
------------------- x 46 = --------- = 92
- 1000
50g x 290 x 4.2 114811.4754
-------------------- x 46 = ----------------- = 114.8114754
- 1000
Butanol
C4H9OH + 6O2 4CO2 + 5H2O
H H H H O=O O=C=O O-H-O
| | | | O=O O=C=O O-H-O
H – C – C – C – C – C - O – H + O=O O=C=O + O-H-O
| | | | O=O O=C=O O-H-O
H H H H O=O O-H-O
O=O
Theoretical
H-C x9 (412 x9) O=O x6 O=C x8 O-H x10
C-C x3 (348 x3) + (496 x6) (743 x8) (463 x10)
O-C x1 360 ---------- ---------- ------------
O-H x1 463 2976 5944 4630
------------
5575
8551 10574
8551-10574 = -2023
Practical
50g x 520 x 4.2 191035.461
------------------- x 74 = --------------- = 191.035461
4.23 1000
Proponol
C3H7OH + 4.5O2 3CO2 + 4H2O
H H H O=O C=O H-O-H
| | | O=O || H-O-H
H – C – C – C – C - O – H + O=O C + H-O-H
| | | O=O || H-O-H
H H H (O=O%2) C=O
||
C
||
C=O
||
C
Theoretical
H – C x7 (412 x 7) O=O x (4.5) C=O x6 H – O x8
C – C x2 (348 x2) + (496 x 4.5 (743 x6) + (463 x8)
O – C x2 360 --------------- ---------- -------------
O – H x1 463 2232 4458 3704
----------------
4403
6635 8162
6635 – 8162 = -1527
Practical
50g x 200 x 4.2 280000
------------------- x 60 = --------------- = 280
- 1000
50g x 200 x 4.2 315000
------------------- x 60 = --------------- = 315
0.8 1000
Methanol
2CH3OH + 3O2 2CO2 + 4H2O
H O=O O H-O-H
| + O=O || H-O-H
H – C – O – H O=O C=O + H-O-H
| || H-O-H
H O
||
C=O
H
|
H – C – O – H
|
H
Theoretical
H – C x6 (412 x6) O=O x 3 O=C x4 H- O x8
C – O x2 (360 x2) + (496 x 3 (743 x4) + (463 x8)
O – H x2 (463 x2) -------------- ---------- ----------
----------- 1488 2972 3704
4118
5606 6676
5606 – 6676 = -1070 [/2 = -535]
Practical
50g x 200 x 4.2 82962.96296
------------------- x32 --------------- = 82.96296296
1.62 1000
50g x 200 x 4.2 101818.1818
------------------- x 32 = --------------- = 101.8181818
- 1000
50g x 200 x 4.2 76363.63636
------------------- x 32 = --------------- = 76.36363636
1.76 1000
Results for all alcohols
The results show that different levels of energy are released from the substances at different heat levels.
In each case the results show that a change in chemical energy has taken place and the end mass is less than the start mass.
The chemical formula for the change in bond after the combustion shows that oxidisation has taken place. Oxidisation is where oxygen is added to a substance [in this case through combustion or burning the substance in air] or where hydrogen is lost.
The energy lost during the combustion has a)turned to water or steam and evaporated into the air. For example with Ethanol, when this is burned in air or oxygen the products are carbon dioxide and water. The water will evaporate into the air – this is where some of the energy goes as well as b)being taken up by breaking of bonds during the burning of the substance. When bonds are formed energy is released. If bonds that hold particles are to be broken or changed then energy has to be supplied to enable this to happen. The change is the difference between the energy liberated to form the bonds and that required to break the bonds.
The experiment also proves that Exothermic process is followed because energy is liberated during the reaction proven by the end masses of the substances being lighter than the start masses. The quantities of energy produced depends upon the quantity of the chemical burnt as shown by the results. I.e. the higher the level of mass burnt or combusted the higher the level of energy released.
