Nathan Stokes Enthalpy Coursework
Determining the Enthalpy Change of the Decomposition of Calcium Carbonate.
The objective of this experiment was to determine the enthalpy change for the decomposition of calcium carbonate with heat, using an indirect method based on Hess’ law.
The formulae for calcium carbonate, CaCO3, decomposing with heat:
CaCO3 (s) heat CaO (s) + CO2 (g)
Hess’ Law states that ‘ the total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place provided the initial and the final conditions are the same.
The equation for the experiment not just for the decomposition:
H3 HCl H2 HCl
Both Calcium oxide and Calcium carbonate react readily with 2 mol dm-3 Hydrochloric acid.
The temperature changes during the reaction will be measured and the enthalpy changes H1 and H2 calculated. From these enthalpy changes the value for H3 can be calculated.
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1. Weigh out a beaker containing between 2.4g and 2.6g of calcium carbonate. Record the results.
2. Using the measuring cylinder provided place 50cm³ of 2 mol dm³ hydrochloric acid in a 250cm³ beaker.
3. Measure the temperature of the acid using the thermometer provided and record this value. The thermometer should not be left unsupported or it may break.
4. Add the calcium carbonate to the acid. Take the temperature, then again when the reaction is complete. Record this value.
5. Weigh the beaker afterwards.
6. Repeat steps 1 to 5 using an accurately weighed mass of calcium oxide, in the range 1.3g to 1.5g, instead of calcium carbonate.
CaCo3 reacting with 2 mol dm-3 HCl.
CaO reacting with 2 mol dm-3 hydrochloric acid.
Analysis of the results
The data that is in the tables can be combined, in the formula Enthalpy Change = mass of liquid x specific heat capacity of liquid x temperature change or ∆H = MC∆T, with the specific heat capacity of HCl, which is 4.2 jg -1k -1 and its density, which is 10 g.cm-3. This formula will be used to calculate the values needed to find the H3.
∆H = MC∆T
CaCO3 = 50g x 4.2 jg -1k -1 x 1.66 C
2.5g of CaCO3 = 348.6j
so therefore: 1g = 348.6 / 2.5 = 139.44j
Mol mass of CaCO3 = 100
139.44 j x 100 = 13944 mol -1
H1 = -13.9 Kj mol -1
∆H = MC∆T
CaO = 50g x 4.2 jg -1k -1 x 8.33 C
1.4g = 1749.3
1g = 1749.3 / 1.4 = 1249.5j
Mol mass of CaO = 56
1249.5 x 56 = 69972 mol -1
H2 = - 69.9 Kj mol -1
Using Hess’ Law.
H2 = H2 + H3
H3 + H2 = H2
H3 + (- 69.9) = - 13.9
H3 = -13.9 + 69.9
H3 = 56 Kj mol -1
This result shows that this reaction is endothermic.
The enthalpy change for the reaction
CaCO3 (s) heat CaO (s) + CO2 (g)
Should have been +178 Kj mol –1 , this figure was according to the text book ‘understanding chemistry for advanced level ’
Errors must have occurred in the experiment because my result for the H3 was over 100 Kj mol –1 out.
There are numerous ways in which errors have occurred in my experiment, the main reason however is greatly due to heat loss into the atmosphere and the convection of heat into the glass beaker as glass is a poor insulator.
A modification that could have been made to the experiment would be to use a polystyrene cup and a lid to carry out the reaction, instead of a glass beaker. The polystyrene is an insulator and would keep maximum amount of heat in and the lid would stop heat from escaping out of the top as this is where most of the heat is lost due to the fact heat rises.
Another error that affected the results was that the thermometer is only accurate to 1 C this affects the accuracy of the results. An improvement that could be made to this is to use a more accurate thermometer i.e. an electric thermometer. A smaller beaker should also have been used instead of a large beaker, so the acid was at a higher level so that the bulb of the thermometer was totally covered by the solution.
An error, which also caused problems with accuracy, was that two different weighing scales were used due to availability. One scale measured to two decimal places where as the other only measured to one, the results would have been more accurate if the two decimal scales were used.
An improvement that could have been made to my experiment is that the CaCO3 should have been used as a powdered form instead of a lump form as it would have a higher surface area, so the reaction would take place quicker. If the reaction is faster then it is less likely that so much heat will be lost.