Results
Below is the equation for the reaction i carried out:
CH3OH + (11/2) O2 → (2) H2O + CO2
To make it easier to calculate the enthalpy of the bonds broken I doubled the above equation:
(2)CH3OH + (3) O2 → (4) H2O + (2)CO2
C-H BONDS = 3 X (+413) =1239 X 2 O=O BONDS = 3 X (+498) H-O BONDS = 2 X (+464) X 4
C-O BONDS = (1 X (+358) =358 X 2 C-O BONDS = 2 X (+805) X 2
O-H BONDS = (1 X (+464) =464 X 2
BONDS BROKEN - BONDS FORMED = Enthalpy
5616 – 6932 = -1316
The enthalpy of combustion of 1 mol methanol is (-1316 / 2 ) = ΔH-658 KJ/mol-1
Mass of 1 mole of methanol = 32g (C=12+H=(1x4)+O=16)
Mass of fuel burned= Average/Mass → 0.658 / 32 = 0.0205625
Energy Transferred by 1 Mole of Fuel = Mass of Water (g) x Change in Temperature x 4.18J (this is the energy required to raise the temperature of 1g of water by 1oC)
= 50g x 8.32 x 4.18J = 1738.88J
Heat given out by 1 mole of methanol = 1738.88 / 0.0205625= -84565.59J → -84.5656 KJ/mol-1
Describe the terms endothermic and exothermic in terms of bond breaking and bond forming
When a chemical reaction takes place bonds are broken in the reactants this is known as endothermic as energy is absorbed from the surroundings to break these bonds you can see this happening in the above ENDOTHERMIC REACTION the reactants are below the products meaning that the reactants need more energy from the surrounding the overcome the activation barrier and less energy is given out. The bonds of the products are weaker than those of the reactants and this is seen as a + (plus)
The new chemical bonds formed give out energy higher than the energy of the reactants this is known as EXOTHERMIC REACTION energy is given out to the surrounding you can see this happening in the above EXOTHERMIC REACTION the reactants are above the products meaning that less energy is needed to break the bonds and more energy is released from the products to the surroundings the bonds of the products are stronger than those of the reactants and this seen as a – (minus)
The reaction i carried out was an exothermic reaction as more energy was given out to the surroundings.
TASK 2
Procedural Errors throughout the experiment
I think my results are consistent but i know that this experiment was not as accurately done this could have been due to some procedural errors:-
- Heat could have escaped from the gaps that i were unable to cover due to the equipment this would have affected the results of mass and temperature given lower values as not all the heat would have heated the water.
- Also the heat may have not reached the calorimeter as the wind may have changed the direction of heat meaning that it couldn’t have heated the water directly
I could have setup my apparatus like this in a closed container to stop heat from escaping giving me more reliable and accurate results
- The main error which could have given me invalid results was the calorimeter as when it gets heated it also gets covered in soot & heat loss
- Also as the spirit burner is covered up there is another factor which could affect the experiment which is the oxygen levels as if they drop incomplete combustion.
To improve the practical i could use a different copper calorimeter each time or rinse off the black soot.
Procedural Errors:
A Weighing balance was used to measure the mass of the spirit burner before and after the experiment the balance was to 2 decimal places
Methanol before the experiment
Percentage error = 0.005 x 100 = 0.0225%
22.83g
Methanol after the experiment
Percentage error = 0.005 x 100 = 0.0225%
22.212g
I also used a measuring cylinder to measure the amount of water i used 2 x 100cm3 measuring cylinder to get 200cm3 i ensured lots of accuracy by looking at the bottom meniscus while measuring to produce more accurate reading i could have used pipette.
