Using Hess’ cycle we can calculate a value for ∆H3
We can calculate ∆H3 by following the path from CaCO3 down the ΔH1 arrow to the ΔH2. This pathway shows us to do ΔH1 – ΔH2. We know it is a minus as the arrows pointing downwards means plus and an arrow pointing down but going up to the CaO (s) + CO2.
The formula to find the enthalpy change is:
∆H = M x C x ∆T
In word form this is:
Enthalpy change= Mass in grams of solution and reactant x heat capacity of the solution x temperature change
After doing the equation above, we can then divide the answer by the number of moles of the reactant to find the enthalpy change per mole.
The formula to find the moles is:
Moles = M/Ram
So to combine these 2 formula’s into one simple formula, it would be:
MC∆T/ moles
I will now calculate the values for ΔH1 + ΔH2 and will assume that the specific heat capacity of Hcl(aq) is 4.2 J g-1 K-1 and that the density of Hcl is 1.0g cm-3.
I will now find out the ∆H for the reaction between CaCO3 + Hcl:
The enthalpy change is as follows:
∆H = M x C x ∆T
= 52.48 x 4.2 x -2
= -440.83 J
ΔH1 = -0.44 kj mol-1
Moles of CaCO3 = 2.48/(40 + 12 + 48) = 0.0248 = 0.025 moles
Therefore the enthalpy change per mole = -440.83/0.025
=-17633.20 J
ΔH1 = -17.63 kj mol-1
This shows the reaction was exothermic as the enthalpy change was negative and heat was absorbed therefore the temperature increased. Therefore the enthalpy change would be a negative figure.
Now we will do the same for CaO:
∆H = M x C x ∆T
= 51.41 x 4.2 x -12
= -2591.06 J
ΔH2= -2.59 kj mol-1
Moles of CaO = 1.41/(40 + 16) = 0.0251 moles
Therefore the enthalpy change per mole = -2591.06/0.025
= -103642.4 J
ΔH2 = - 103.64 kj mol-1
This shows the reaction was again exothermic as the enthalpy change was negative and heat was absorbed therefore the temperature increased. Therefore the enthalpy change must be a negative figure.
Looking at the enthalpy changes to both reactions, we can see that a heat reaction will take place. For the first reaction with the calcium carbonate, we can see that ΔH1 = -0.44 kj which is ΔH1 = -17.63 kj mol-1 and this shows that 17.63 kj mol-1 of heat has been given out in the exothermic reaction which has taken place in the first reaction. We know this as the temperature has increased and heat has been given out. For the second reaction we see that the enthalpy change is ΔH2= -2.59 kj and it shows that ΔH2 = - 103.64 kj mol-1 is given out per mole. Looking at these figures we can see that both these reactions are exothermic as the temperature has increased and the figures are negative which indicates that heat has been given out.
We can now find the enthalpy change for when calcium carbonate combusts to give calcium oxide and oxygen ΔH3, we can do this by using Hess’ law. As explained above. We do:
ΔH1 - ΔH2 = ΔH3
Looking at the cycle above, we already have the values for the combustion product (CaCl2) and the element (CaCO3) we can work out the enthalpy change of the compound (CaO + CO2) by using Hess’ law.
As worked out before, CaCO3 enthalpy change of -17.63 kj mol-1 per mole and CaCl2 has a enthalpy change of 103.64 kj mol-1
These were already worked out before. Therefore following the calculation we made before the enthalpy change of ΔH3 CaO and CO2 will be:
(-17.63 kj mol-1) - (-103.64 kj mol-1) = 86.01 kj mol-1
Therefore this shows that the enthalpy change of ΔH3 is:
CaCO3(s) CaO(aq) + CO2(g) is 86.01 kj mol-1.
This equation is a endothermic reaction as the reaction absorbs energy. We can tell this by the positive ΔH. In this reaction the temperature often falls.
All answers are given to 2 decimal places
The correct theoretical value for ΔH3 should be 178 kj mol-1
As quoted from
As we can see from the results I obtained and the theoretical value, it shows that my answer is quite far off to the theoretical value of calcium carbonate. We will look at reasons why this may be later on.
Level of accuracy in apparatus
Equipment errors = error / how much u used/measured x 100
- 250cm3 measuring cylinder- this was correct to every 0.5 cm3 +/-, therefore the percentage error will be worked out by: 0.5 / 50 x 100 =1.00% (exp 1+2)
- 0-100ºc thermometer 1ºc +/- % error = 1/24 x 100 = 4.17% (exp. 1+2)
- 0-100ºc thermometer 1ºc +/- % error = 1/26 x 100 = 3.85% (exp 1)
- 0-100ºc thermometer 1ºc +/- % error = 1/36 x 100 = 2.78% (exp 2)
- Balance 0.005+/- % error = 0.005/2.48 x 100 = 0.20% (exp 1)
- Balance 0.005+/- % error = 0.005/1.41 x 100 = 0.35% (exp 2)
My Overall percentage error will be used using the formula MCΔT
0.005 X 4.2 X 12 /0.025 = 10.8%
As the worst case scenario was 24-36ºc
This shows that the overall percentage error was 10.8%
Evaluation
There were many errors which may have occurred during the experiment. These can be split into 2 kinds of errors. Experimental and procedure errors, My experimental errors were that the equipment used in the experiment were not very accurate, as seen from above, the thermometer has quite a high error in accuracy which can affect my experiment as my measurements will not be accurate. Therefore to rectify this matter, we should use better and more accurate thermometers to increase the accuracy of the experiment. Another experimental error would be the balance, as I was using quite a small amount to be weighed, this would lead to a higher experiment error than it would if I used a larger amount for my experiment. Another error would be the measuring cylinder as it was a 1% error and as it read to each cm3, the error of it would be 0.5+/-. My procedure errors were that the calcium had impurities and that and this would have affected the reaction rate and how much would have actually reacted with the Hcl acid as some of the mass obtained will be from the impurities. So the calcium carbonate used should be either powdered or cut into the most same sized pieces. Also calcium oxide may have been formed as it was kept in the atmosphere for quite a while. Also a calorimeter was not used, this would have decreased the efficiency of the experiment as heat would have been lost through the beaker whereas using a calorimeter, less heat would have been lost as the only way heat would have been lost would be through the top. Whereas in a glass beaker heat energy can escape everywhere.
Looking at my percentage errors in my equipment, it shows that my total percentage error for my first experiment with CaCo3 is 9.22% and for my second experiment with CaO is 8.39%. As my % errors show, it shows that the second experiment was more accurately done than my first experiment. This is because it has a lower percentage error.
Overall we can see that the experiment was quite unreliable as the theoretical value for calcium carbonate is 178kj mol-1 and the answer I obtained was 86.01 kj mol-1 . This shows a 91.99kj mol-1 difference.
To find the experimental percentage difference I will find the difference between my value and the theoretical value, divide by the theoretical value times 100:
Difference = 91.99 → 91.99/178 x 100 = 51.68% difference
The most significant reason for this error would be because we did not use a calorimeter therefore a lot of heat energy would have been lost through the beaker glass and was unable to be measured.
There are 6 possible ways in which I could have improved with the experiment, these are:
- I should have used a calorimeter instead of a beaker as the beaker lost sufficient heat, and the experiment was based on temperature change, and as a beaker can lose heat from all around the glass. The calorimeter will only lose heat from the top as the rest of the body is insulated.
- Also we should have used a burette instead of using a measuring cylinder as a measuring cylinder is measured to the nearest 2cm (therefore the results obtatined will be +/- 1) and the burette is measured to the nearest 0.025cm and this would have improved the experiment reliability quite largely.
- Impurities on the calcium oxide: this will have affected the experiment as when we were weighing out the amount of grams we were using, some people got bits of impurities in their reactant (due to manufacturers fault) and some people didn’t realize it and put it into their reaction to find out that their was a stone impurity which was not react able and would have affected the amount of ACTUAL calcium oxide which was in the reactor. To increase the reliability of this, it would have been better to use powdered calcium oxide/carbonate.
- Surface area on the calcium oxides and calcium carbonate: different sizes of calcium oxide/carbonate would have affected the rate of the reaction, so to improve this factor I would suggest to either use powdered calcium oxide or relatively large pieces of CaO which are very similar to each other size wise.
- Thermometer: this went up in graduations of 1ºc. This means that the results we collected could have varied by +/- 0.5cm3. To improve the accuracy and reliability of the experiment, a digital thermometer to 2dp would have been the better choice to ensure a reliable and accurate reading for the experiment.
- Measuring balance: These were accurate to 2 decimal places and the balance error was to the nearest +/- 0.005. I have made calculations above to show the percentage error and although the percentage error is not relatively large, to increase the experiment, we should use balances accurate to at least 3 decimal places.
Therefore it is obvious that these significant errors may have made my results less accurate and the improvements which have been suggested e.g. use a digital thermometer may increase the reliability of the experiment. As we have several variances of results which were used to obtain the correct enthalpy change. The variance in calculations has caused the ending results for ΔH3 to be inaccurate.
