Trial 1
Table VI: Enthalpies of Reaction After Manipulating The Given Equations for MgO
Table VII: Enthalpy of Formation of MgO
Table VIII: Enthalpies of Reaction After Manipulating The Given Equations for CaO
Note: For CaCl2(aq) + H2O(l) → CaO(s) + 2HCl(aq) , the enthalpy of reaction from Trial 2 was used since Trial 1 data was not available. Trial 1 information was not included in the lab due to large human error during the experiment.
Table IX: Enthalpy of Formation of CaO
Trial 2
Table X: Enthalpies of Reaction After Manipulating The Given Equations for MgO
Table XI: Enthalpy of Formation of MgO
Table XII: Enthalpies of Reaction After Manipulating The Given Equations for CaO
Table XIII: Enthalpy of Formation of CaO
Notes:
- The enthalpy of formation was calculated by adding together the enthalpies of reaction
Ex:
Combustion of magnesium Trial 1:
∆Hf = (101.5 – 366.1 – 285.8)kJ/mol
= - 550.4 kJ/mol ± 0.9 kJ/mol
Error:
0.5 + 0.4 = 0.9
Table XIV: Enthalpy of Formation Averages
Notes:
- The average enthalpy of formation was calculated by adding the enthalpies of formation from the two trials and then dividing by two
Ex:
Combustion of magnesium:
(- 550.4 ± 0.9)kJ/mol + (- 526.2 ± 0.9)kJ/mol
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ = - 538.3 kJ/mol ± 1.8 kJ/mol
2
Error:
0.9 + 0.9 = 1.8
- There are two significant digits in this answer, as ∆T in the calculation of Q has two significant digits. Significant digits were only applied at the end, when calculating the average enthalpy of formation, so as not to lose accuracy through the multiple calculations.
Table XV: Percent Error
Notes:
- Percent error was calculated using the following formula:
% Error = (|accepted value – experimental value|)
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x 100%
accepted value
Ex: Mg(s) + ½O2(g) → MgO(s) :
% Error = (|accepted value – experimental value|)
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x 100%
accepted value
% Error = (|- 601.6 – (-538.3)|) kJ/mol
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x 100%
- 601.6 kJ/mol
% Error = 63.3 kJ/mol
¯¯¯¯¯¯¯¯¯¯¯¯¯ x 100%
601.6 kJ/mol
% Error = 10.5 %
Conclusion and Evaluation
Conclusion
The experiment conducted was meant to determine the enthalpy of formation of MgO(s) and CaO(s). Through experimentation, it was determined that the enthalpy of formation of MgO(s) is – 540 kJ/mol ± 1.8 kJ/mol, with 10.5% percent error when compared to the accepted value, and for CaO(s) it was -330 kJ/mol ± 2.9 kJ/mol, with 48.5% percent error.
Evaluation
Clearly, while determining the enthalpy of formation of CaO(s), there were many more errors present than when the enthalpy of formation of MgO(s) was determined experimentally. There were many sources of error that could have accounted for this.
When the enthalpy of reaction for magnesium was calculated, the magnesium was first sanded with sandpaper to remove and magnesium oxide that had formed on it. Similarly, calcium oxide forms on calcium, however no attempt was made to remove the calcium oxide. This would mean that more moles of calcium were accounted for than were actually present, and that calcium oxide was also reacting but this was not recorded and the amount of calcium oxide was not measured. As is visible in Table VIII, the enthalpy of reaction of calcium is greater than that of calcium oxide. Since it was not all calcium and some calcium oxide reacting in the trial that was meant to be only calcium, the calcium oxide could be responsible for the enthalpy of reaction being lower than the accepted value. This difference between the magnesium and calcium experiments could be one source of error accounting for the large difference in percent error between the two. The experiment could be improved by sanding the calcium as well.
