Energetics Practical

Energetics Practical 2: To Determine the Enthalpy Change of a Reaction
Background
Calcium Carbonate, CaCO₃, decomposes with heat to form CaO and CO₂. The objective of this practical was to determine the enthalpy change for this reaction by using an indirect method based upon the foundation of Hess’ Law. Both calcium oxide and calcium carbonate react readily with 2 mol/dm3. The reaction can be demonstrated as so:
CaCO₃ (s) -> CaO(s) + CO₂ (g)
Results: Raw Data
Calcium Carbonate
Mass of CaCO₃ + Weighing Bottle (in grams) 3.62g±0.10
Mass of empty weighing bottle (in grams) 1.02g±0.05
Mass of CaCO₃ used (in grams) 2.60g±0.05
Initial temperature of the acid (in degrees centigrade) 16°C±0.05
Final Temperature of the solution (in degrees centigrade) 19.5°C±0.05
Temperature Change of the Reaction (in degrees centigrade) 3.5°C±0.05

Calcium Oxide
Mass of CaO + Weighing Bottle (in grams) 3.58g±0.10
Mass of empty weighing bottle (in grams) 1.02g±0.05
Mass of CaO used 2.56g±0.05
Initial temperature of the acid (in degrees centigrade) 18°C±0.05
Final Temperature of the solution (in degrees centigrade) 26.5°C±0.05
Temperature Change of the Reaction (in degrees centigrade) 8.5°C±0.05

Calculations
The results that I have recorded allowed me to calculate the required enthalpy changes. To measure enthalpy change of a reaction, the following formula must be used:
ΔH=mcΔt where:
-‘m’ is the mass of the substance
-‘c’ is the specific heat capacity of the substance
-And ΔT is the temperature change
For this experiment, we made one major assumption – the specific heat capacity of the solution that we measured was 4.2 g-1 k-1.
 Enthalpy Change between Calcium Carbonate (CaCO3) and the Hydrochloric Acid (HCl)
ΔH=mcΔt
ΔH1 = 52.6 x 4.2 x 3.5 = 773.22J = 0.77322kJ
As standard enthalpy change is the standard enthalpy change per mole, the number of moles in the substance will need to be calculated.
Moles of a substance = Mass of a substance / Relative Molecular Mass of a Substance = 2.60 / (40 + 12 + (16.00 x 3)) = 2.60 / 100 = 0.026 mol
Simply to find the enthalpy change per mole, the value calculated from the enthalpy change is divided the moles of the substance, CaCO3.
0.77322 / 0.026 = 29.739 kJ mol-1 (TO 3DP)
 Enthalpy Change between Calcium Oxide (CaO) and the Hydrochloric Acid
ΔH=mcΔt
ΔH2 = 52.56 x 4.2 x 8.5 = 1876.392 J = 1.876392 kJ
From the point, we now continue with the same procedure as before – calculating the moles, and then using that figure to divide the first value of the energy change.
Moles of a substance = Mass of a substance / Relative Molecular Mass of a Substance = 2.56 / (40 + 16) = 0.046 mol (TO 3DP)
Enthalpy Change per mole = 1.876392 / 0.046 = 41.046075 kJ mol-1

Now that I have calculated ΔH1 and ΔH2, I am now able to calculate the enthalpy change is this decomposition, using Hess’ Law.
ΔH3 = ΔH1- ΔH2 = -29.739 kJ mol-1 – (-41.046075 kJ mol-1) = 11.307075 kJ mol-1

Conclusion
From the result above, we can see that the reaction of heating calcium carbonate (CaCO3) is an endothermic reaction, because the enthalpy change is a positive one, meaning that it has taken in energy from its’ surroundings.
Evaluation
I feel that the experiment went relatively well, and went without any major hiccups. However, I feel as if there could have some aspects in where it could have been improved significantly. Perhaps if I had done the experiment again, I wouldn’t have received the results that I did. On the other side of this, if I had have done it again, it would have been beneficial for me, so that I would be able to see whether there was a correlation between all the results I collected, and in that way, checking the accuracy of the results. Another potential source of error could have been the actual reaction itself. Potentially, not all of the acid could have reacted with the Calcium Carbonate and the Calcium Oxide, meaning that wrong mass could have been calculated for both of the enthalpies. A solution for this could be to do the experiment in a fume cupboard, and in this way, pressure and temperature could be regulated to make sure that the results would be both accurate and precise. Another problem that we encountered was the fact that when the experiment was being done, there was no lid on top of the beaker. This means that heat could have been to its’ surroundings at quicker rate, which almost definitely changed the course of the experiment and the results. A possible solution to this problem would be to place a cover on the top of the beaker to regulate the heat given to the surroundings.