Chemistry Experiment.
Dr. Watson.
Evaluating An Enthalpy Change That Cannot
Be Measured Directly.
Introduction.
We were told that sodium hydrogencarbonate decomposes on heating to give sodium
carbonate, water and carbon dioxide as shown in the equation below:-
2NaHCO3(s)--------> Na2CO3 (s) + H2O (l) + CO2 (g) = DeltaH1
This was given as deltaH1 and we had to calculate as part of the experiment.
This however cannot be measured directly, but can be found using the enthalpy
changes from two other reactions. These being that of sodium hydrogencarbonate and
hydrochloric acid and also sodium carbonate and hydrochloric acid.
We were given a list of instructions in how to carry out the experiment, which are
given later.
List of Apparatus Used.
1 x 500ml Beaker.
1 x Thermometer(-10 to 50oC).
1 x Polystyrene Cup.
1 x Weighing Balance.
1 x Weighing Bottle.
10 grams of Sodium Hydrogencarbonate.
10 grams of Sodium Carbonate.
A bottle of 2 molar HCL.
Diagram.
Method.
Three grams of sodium hydrogen carbonate was weighted out accurately using a
weighting bottle and a balance. Then thirty centimetres cubed of 2 molar HCL was
measured using a measuring cylinder. The acid was then placed into the polystyrene
cup and its temperature was taken and recorded using the thermometer. The pre-
weighted sodium hydrogencarbonate was then added to the solution, and the final
temperature was recorded.
The contents of the cup were then emptied out and the cup was washed out with
water and then thoroughly dried. This was done three times for the sodium hydrogen
carbonate so that I could remove any anomalies that were obtained.
The experiment was then repeated in exactly the same manner except sodium
carbonate was used instead of sodium hydrogen carbonate.
The results were then tabulated, this table is shown below.
Results Table.
Results Table for Sodium Hydrogencarbonate.
Results Table for Sodium Carbonate.
Calculations.
From these results I had to calculate deltaH2 and deltaH3. DeltaH2 refers to the
enthalpy change when sodium hydrogencarbonate reacts with hydrochloric acid, and
deltaH3 is the enthalpy change when the sodium carbonate reacts with the acid.
Firstly however it is necessary to show the equations for the two reactions:-
DeltaH2= 2NaHCO3 (s) + 2HCl (aq)----> 2NaCl (aq) + 2H2O (l) + 2CO2 (g).
DeltaH3= Na2CO3 + 2HCl (aq) ----> 2NaCl (aq) + H2O (l) + CO2 (g)
The enthalpy changes of the two reactions can be worked out using the formula
shown below :-
Energy Exchanged between = Specific Heat Capacity x Mass of the x Temperature
Reactants and Surroundings of the Solution Solution Change.
Therefore the DeltaH2 of the reaction when fitted into the formula is :-
Energy Exchanged between = 4.18 x (84 x 2) x -11.1
Reactants and Surroundings.
This gives the enthalpy change for DeltaH2 to be = -7794.9 Joules per mole.
The same formula is used for DeltaH3:-
Energy Exchanged Between = 4.18 x 106 x 21.8
Reactants and Surroundings.
This gives the Enthalpy change for DeltaH3 to be = 9659.1 Joules per mole.
From these two results we are able to work out what DeltaH1 is likely to be even
though we have not done the experiment. This is done using the formula :-
DeltaH1 = DeltaH3 + DeltaH2 =>
DeltaH1 = 9659.1 + (-7794.9) =>
DeltaH1 = 1864.2 Joules per mole.
Conclusions.
The result obtained will not be a very accurate due to the means by which the
experiment was done. The equipment used was not the most efficient for measuring
enthalpy changes, however it does give a rough estimate to work from. Some errors
of the equipment would have been heat lost through conduction from the reaction
vessel. Also heat may well have been lost through the open top of the container, even
though there was a lid this was not very secure some heat will have escaped through
here.
In summation the experiment was very difficult to undertake as the enthalpy change
for DeltaH1 is hard to determine due to the fact that it thermally decomposes in the
air, causing great problems in calculating its enthalpy change with its surroundings.
I believe you have made an enormous error in your calculations. ∆H is NOT equal to m*c*∆T, that is ∆Q, the change in heat energy for ONLY THE NUMBER OF MOL OF SUBSTANCE YOU USED. To find ∆H, one divides ∆Q by the # of mol. (units of enthalpy change are kJ/mol, not kJ).
ReplyDeleteCorrection:
∆Q of NaHCO3= -7.7949kJ
number of mol=10.00/Mr(NaHCO2)=10.00/84.007=0.1190 mol
∆H= -65.48 kJ
There so many erros. I think you need to erase the document. It is possible to measure directly this enthaply by DSC, the calculation is in error because you use incorrectelly masses values; you need calculate the energy using the total mass of the system, including mass of the cup, after that the energy need to be calculated for 1 mol of substances. Everythin is wrong. I am sorry so much.
ReplyDelete