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If 2.51 g of CaCO3 then: moles of CaCO3 = 2.51 g ÷ 100.00 = 0.0251 moles. Seeing as the molar mass of CaCO3 = 100.00
ΔH1 = 630 x (1 x 0.0251) = - 25.09 kJmol-1
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If 2.53 g of CaCO3 then: moles of CaCO3 = 2.53 g ÷ 100 = 0.0253 moles. Seeing as the molar mass of CaCO3 = 100.00
ΔH1 = 630 x (1 x 0.0253) = - 24.90 kJmol-1
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If 2.50 g of CaCO3 then: moles of CaCO3 = 2.50 g ÷ 100 = 0.0250 moles. Seeing as the molar mass of CaCO3 = 100.00
ΔH1 = 420 x (1 x 0.0250) = - 16.80 kJmol-1
I will not include the last result in my average for ΔH1, which is - 16.80 kJmol-1. This is because it’s way off the other results and would significantly affect my average results, it’s an anomaly.
AVERAGE for the ΔH1 for the reaction between CaCO3 + HCl:
(- 25.09 kJmol-1) + (- 24.90 kJmol-1)
2
ΔH1 = - 25.00 kJmol-1
This value for ΔH1 is negative because heat is lost to the surroundings. It’s an exothermic reaction.
Calculations for ΔH2 for the reactions between CaO (s) + HCl:
- E = 50 x 4.2 x 15 = 3150 joules
- E = 50 x 4.2 x 12 = 2520 joules
- E = 50 x 4.2 x 13 = 2730 joules
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If 1.38 g of CaO then: moles of CaO = 1.38 g ÷ 56.00 = 0.0246 moles Seeing as the molar mass of CaO = 56.00
ΔH2 = 3150 x (1 x 0.0246) = - 128.05 kJmol-1
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If 1.37 g of CaO then: moles of CaO = 1.37 g ÷ 56 = 0.0245 moles Seeing as the molar mass of CaO = 56.00
ΔH2 = 2520 x (1 x 0.0245) = - 102.86 kJmol-1
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If 1.39 g of CaO then: moles of CaO = 1.39 g ÷ 56 = 0.0245 moles Seeing as the molar mass of CaO = 56.00
ΔH2 = 2730 x (1 x 0.0245) = - 111.43 kJmol-1
I will not include the - 102.86 kJmol-1 result in my average for ΔH2. This is because it’s way off the other results and would significantly affect my average results, it’s an anomaly.
AVERAGE for the ΔH2 for the reaction between CaO + HCl:
(-128.05 kJmol-1) + (- 111.43 kJmol-1)
2
ΔH2 = - 119.74 kJmol-1
This value for ΔH1 is negative because heat is lost to the surroundings. It’s an exothermic reaction.
Using Hess’ cycle I will use the values that I have calculated for ΔH1 and ΔH2 to work out the value for ΔH3.
ΔH3= ΔH1 − ΔH2
= (- 25.00 kJmol-1) – (- 111.43 kJmol-1)
= 86.43 kJmol-1
This value is positive because heat is absorbed from the surroundings. It’s an endothermic reaction.
I have been told the actual value for ΔH3, which is 178.00. So I will calculate the percentage by which my value is out by the actual value.
178.00 −86.43 = 91.57
(91.57 ÷ 178.00) x 100 = 51%
Evaluation
Errors in procedure:
When the CaO and CaCO3 were put into the cup there was a delay before the lid was put on. This could have caused heat to escape out of the cup and the temperature change would not have been as great compared to if there was no delay.
Also when the lid was put on there was a hole made in it so that the thermometer could be put into the acid and measure the change in temperature. But the hole had a larger circumference than the circumference of the thermometer and heat could have escaped through this gap.
The CaO and CaCO3 were not stirred after being emptied into the cup so not all of the reactants could have reacted with the HCl. This means that the temperature change during the reaction could have been lower than it should have been.
When the thermometer was inserted through the lid, the bulb of the thermometer wasn’t checked to see if was completely submerged in the acid. Therefore the temperature change could have been higher if the bulb was fully submerged.
Errors in measurement:
The thermometer used in the experiment was of an analogue type and it was hard to read in between each degree. E.g. in between 3 ºc and 4 ºc. So if 0.1 of a degree was misread then E = 50 x 4.2 x 0.1 = 21 Jmol-1, there would be a difference of 21 Jmol-1 and could have effected my final results.
The measurement scales I used only went to 2.d.p and weren’t very accurate.
I used two different measuring scales to measure out the masses of CaO and CaCO3 during the experiment. This could have affected the accuracy of my results as each measurement scale was calibrated differently.
When measuring the CaO and CaCO3 bits of the CaO and CaCO3 were scattered on the top of the measuring scales and were weighted as being in the weighing bottle. But when the weighing bottle was taken away there wasn’t the amount of the CaO and CaCO3 that was thought to be in the weighing bottle.
Errors that were significant:
As mentioned in the error in measurement section, if there was 0.1 of a degree misread then there would be a 21 Jmol-1 difference in the final result. To resolve this problem a digital thermometer could have been used. This would improve the accuracy of the results that I would collect as a digital thermometer can read to the point of a degree automatically.
The CaCO3/ HCl reaction was more accurate compared to the CaO/HCl reaction. This is because the CaO used was damp as it reacted with moisture in the air and became more Ca(OH)2. So the reaction between the HCl was less exothermic and affected the final result significantly. Therefore the CaO should have been baked in the oven before it was used so all of the moisture was removed.
Because the HCl was used in excess the heat from the reaction was dissipated through the acid so it would have taken more energy to increase the temperature during the reaction.
Improvements to the procedure:
I could use a tube with a bung at the end of it, which could be built into the side of the insulating cup. Therefore no lid will need to be used and no heat would be lost when the lid was open as the CaO and CaCO3 was about to be put into the cup.
The digital thermometer that I have proposed to use could be inserted into the built in tube on the side of the cup with a piece of transparent material that was built into the bottom of the cup so I could see if the bulb of the digital thermometer was fully submerged in the acid.
I used a cup which was made of polystyrene as the vessel for the reaction to take place in. The polystyrene cup was adequate to make sure that most of the heat produced during the reaction was contained as polystyrene is an insulator. But not all of the heat produced was contained within the reaction vessel. Therefore I could have used another layer of polystyrene which surrounded this piece of polystyrene with an air gap in between. This would reduce the amount of heat loss during the reaction.
Here is a diagram to show the improvement that I have suggested:
The CaCO3 and CaO weren’t stirred after it was put into the cup. So lumps of the CaCO3 and CaO gathered at the bottom of the cup. Therefore not all of the reactants reacted with the HCl and when they did the reaction would have occurred slower and the temperature change wouldn’t have been as great. If I were to make an improvement here it would be to use a mechanical shaker for a fixed period of time so that all of the reactants reacted with the HCl.
Improvements to the method of taking measurements:
As I have already mentioned and discussed I will use a digital thermometer.
I will use a measurement scale that can read beyond 2.d.p so my results will be more accurate.
I will use the same measurement scales so there will be no variation in my results which would occur if I used two different scales due to them being calibrated differently.
Also because bits of the CaO and CaCO3 were scattered on the top of the weighing scale and were recorded as being in the weighing bottle. I could have blown the off the scattered bits of CaO and CaCO3 on the measuring scales away and measured the weighing bottle again to see if any more CaO and CaCO3 would need to be added.
