Table 2: Data observed for Reaction #2 between hydrochloric acid solution and magnesium oxide
CALCULATIONS (Processed Data)
Graph with extrapolated lines for experiment 1 A
Graph with extrapolated lines for experiment 1 B
Graph with extrapolated lines for experiment 1 C
Graph with extrapolated lines for experiment 2 A
Graph with extrapolated lines for experiment 2 B
Graph with extrapolated lines for experiment 2 C
Results Table:
Therefore,
First trial
For Mg
Energy change Q = mass x specific heat capacity x change in temperature.
→mass of water= mass of acid (diluted)
That is, we say, that the density is 1.0g/cm3 even though it is not water
Now, mass = density* volume
So when the density is 1 g/cm3, then mass is equal to volume.
→specific heat capacity of water= 4200 J/kg °C (data booklet)
→change in temperature= 22 kJ/mol
Q (energy)= m(mass of HCl solution)* c(specific heat capacity of water) *∆T (temperature change)
Q = 0.05kg x 4200 J/kg°C(from data booklet) x 17 °C
= 4620 J
= 4.62 kJ
I will carry out the following calculations to find the limiting reactant
The ratio between magnesium oxide and hydrochloric acid is 1:2 according to the balanced chemical equation.
magnesium is the limiting reactant.
This is the amount of energy which is given out for 0.25g.
If we convert this to mol, we will get = (mass/molar mass) = (0.25g/24gmol-1data book let) = 0.0104mol.
Therefore, for one mol the energy given out would be:
∆H = where Q is the heat energy and n is the number of moles.
∆H1= (negative sign because the reaction is exothermic)
For MgO
Energy change Q = mass x specific heat capacity x change in temperature.
→mass of water= mass of acid(diluted)
Now, mass = density* volume
So when the density is 1 g/cm3, then mass is equal to volume.
That is, we say, that the density is 1.0g/cm3 even though it is not water.
Now, mass = density* volume
So when the density is 1 g/cm3, then mass is equal to volume.
→specific heat capacity of water= 4200 J/kg °C
→change in temperature= 7 °C
Q (energy)= m(mass of HCl solution)* c(specific heat capacity of water) *∆T (temperature change)
Q = 0.05kg x 4200 J/kg°C x 7 °C
= 1680 J
= 1.68 kJ
it is necessary to
show by calculation, that Mg is the limiting reactant
The ratio between magnesium oxide and hydrochloric acid is 1:2
Which means that for (0.00992 mol) of MgO, only (0.01984 mol) of HCl is consumed.
Therefore magnesium oxide is the limiting reactant.
This is the amount of energy which is given out for 0.40g of MgO.
Therefore, for one mol the energy given out would be:
∆H 2 = -1.68 kJ/mol x (1/0.00992) = 169.3 = -1.69 * 102 kJ/mol.
( we have a negative sign because the reaction is exothermic)
Enthalpy change of a reaction is the same regardless of what pathway is taken to achieve the products.
Therefore, I will use the mathematical procedure to find the heat of combustion of magnesium.
We have determined the following values for the two reaction and one is given to us.
∆H 1 = Mg(s) + 2HCl (aq) → MgCl2(aq) + H2(g) -4.44* 102 kJ/mol
∆H 2 = MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) -1.69 * 102 kJ/mol
∆H 3 = H2(g) + ½ O2(g) → H2O(l) - 2.86*102 kJ/mol (given)
We have to find the heat of combustion of magnesium from the above 3 reactions.
Mg +1/2O2 → MgO ?
To find the ∆H of this reaction mathematical way, we should have MgO on the product side in ∆H 2, so we reverse ∆H 2 and thus change the sign to positive for the Heat of reaction which should be 1.27 * 102 kJ/mol . So
∆H combustion (Mg)= ∆H 1 +∆H 2 + ∆H 3
= (-4.44* 102 kJ/mol) + 1.69 * 102 kJ/mol + (- 2.86*102 kJ/mol)
∆H combustion (Mg) = - 5.61*102 kJ/mol + 76 kJ/mol
Heat of reaction= heat of formation of reactants - heat of formation of products.
Calculation of absolute uncertainty for the first trial
First of all we have to find the percentage error for both of the experiments.
Mg
As Q= mcΔT
So AU(ΔT) = +/-(0.5 +0.5)oC = ± 1 oC
RU(ΔT) = (AU(ΔT)/ ΔT)*100%
So RU(ΔT)= +/-(1/22 *100) = +/- 4.5 %
AU(V(HCl)) = +/- 1 cm3
RU(V(HCl)) = (AU(V(HCl))/ V(HCl))*100%
RU(V(HCl)) = +/- 1/50 * 100 = +/- 2%
As mass = volume * density
By Setting AU of density to zero because it has been as that of water which is a table value
Therefore
RU(m(HCl))= RU(V(HCl))
RU( C) = 0 ( table value )
So percentage uncertainty for
RU(Q)=RU(m)+ RU(ΔT)
RU(Q) = 2+ 4.5 = +/- 6.5 %
So percentage uncertainty for
AU(m(Mg))=+/- 0.01 g
Then
RU(m(Mg))= (AU(m(Mg))/ m(Mg))*100%
RU(m(Mg)) = (0.01/0.25)*100 = 1/0.25 = ± 4 %
As AU(M(Mg))=0 because this value is given and not calculated by us so,
RU(n) =RU(m)
So RU(n) = ± 4 %
RU(ΔH1)=RU(Q) + RU(n)
RU(ΔH1) = 4% + 6.5% = 10.5 %
Then
AU(ΔH1)=( ΔH1* RU(ΔH1)/ 100)
AU(ΔH1 )= (10.5/100)* 444KJ/mol = ±47 kJ/mol
MgO
As Q= mcΔT
So AU(ΔT) = +/-(0.5 +0.5)oC = ± 1 oC
RU(ΔT) = (AU(ΔT)/ ΔT)*100%
So RU(ΔT)= +/-(1/8 *100) = +/- 12.5 %
AU(V(HCl)) = +/- 1 cm3
RU(V(HCl)) = (AU(V(HCl))/ V(HCl))*100%
RU(V(HCl)) = +/- 1/50 * 100 = +/- 2%
As mass = volume * density
By Setting AU of density to zero because it has been as that of water which is a table value
Therefore
RU(m(HCl))= RU(V(HCl))
RU ( c) = 0 ( table value )
So percentage uncertainty for
RU(Q)=RU(m)+ RU(ΔT)
RU(Q) = 2+ 12.5 = +/- 14.5 %
Now, the absolute uncertainty for MgO’s mass is,
AU(m(MgO))=+/-0.01 g
Then
RU(m(MgO))= (AU(m(Mg))/ m(Mg))*100%
RU(m(MgO)) = (0.01/0.40)*100 = 1/0.40 = ± 2.5 %
As AU(M(Mg))=0 because this value is given and not calculated by us so,
RU(n) =RU(m)
So RU(n) = ± 2.5 %
RU(ΔH2)=RU(Q) + RU(n)
RU(ΔH2) = 2.5% + 14.5% = 17 %
Then
AU(ΔH2)=( ΔH1* RU(ΔH1)/ 100)
AU(ΔH2 )= (17/100)* 169KJ/mol = ± 29 kJ/mol
Absolute uncertainty for ΔH react 1:
As the uncertainty of ΔH3 is set to zero because of it being a given value, the
AU(ΔH react) =47 + 29 = ±76
Table for the results and uncertainty of 1st 2nd and 3rd trial
CONCLUSION
Theoretical value of heat of combustion of magnesium = - 602 kJ/mol
The heat of combustion for magnesium was calculated algebraically by adding the molar enthalpies of three different intermediate chemical reactions. The molar enthalpy of one reaction was given and the other molar enthalpies were determined experimentally. A temperature probe was used to calculate the temperature change in both of the reactions. Because of poor isolation, all the graphs were extrapolated according to their cooling rate after the maximum temperature, but only to the point where the reaction had started and this is the temperature values used in the calculations. This new value compensated for the heat loss during the reaction and this value was then used in the equation Q = mcΔT. The negative sign shows that both of the reactions are exothermic. Also the specific heat capacity value used in the equation was that of water (4.18 kJ/gºC) since the solution was excessively diluted and the amount of other chemical was negligible. The molar enthalpies for both reactions were calculated. Both of these reactions were carried out three times each. The last intermediate reaction was not done because it was almost impossible to carry out that experiment by using the available equipments but instead table values were given for those reactions. Hence, with the molar enthalpies of the three intermediate chemical reactions, the heat of combustion for magnesium was calculated algebraically. The values obtained experimentally are slightly different and inconsistent to those of table values. If these results are compared to the accepted value of - 602 kJ/mol it can be seen that the calculated uncertainty can account for the discrepancy in all the trials. The results obtained from all the trials lie relatively close to each other and also to the table value which means that my experiment was quite precise and accurate. And I can therefore conclude that the purpose has been achieved with a satisfactory accuracy and precision indicating that some decisive errors have not been encountered. Having said that there were reasons for some differences in precision and accuracy which can be avoided to improve the results. The average result for all the three results is 549 kJ/mol, which can be accounted by with the uncertainty we had in the experiment.
EVALUATION
There were several discrepancies which could have potentially affected the outcome of the experiment and which will translate into the inaccuracies I had in my final result. Although I almost canceled the effects of heat loss which is a systematic error by graph extrapolation, but still that there are some inaccuracies with the results in the first and final trial which might be explained better by some random errors during the experiment. The percentage error for the three trials is as under
Percentage error for the first trial:
= (table value - experimental value) = 602- 561 = 40
=
Percentage error for the second trial:
= ( table value - experimental value) = 602- 602 = 0
=
Percentage error for the third trial:
=( table value - experimental value) = 602- 562 = 40
=
It is very clear from the above calculations that the most accurate experiment I performed was the 2nd one. It is same as the table value taking into account the apparatus we had which were not very advanced.
As long as my first and third experiments are concerned, the values were 6.8% and 6.6% lower than the table values which is quite a small figure. But it is in the range of the uncertainty I have for the first trial.
My results are reasonably precise and they lie close to each other although the difference could be explained by some mistakes made during the experiment and by some random errors.
Accuracy can be improved further if use more precise equipments and a very well isolated system.
To take into account the heat loss because of poor isolation, I have extrapolated the graph to show that the maximum temperature would have been higher if there was no heat loss to the surroundings. The graphs were extrapolated using the gradient of the cooling rate of the liquid.
Having discussed all the trials individually, I will take the average of my results which will present me with a general picture of the results.
The average value from the trials =
The average of uncertainty from the trials =
The percentage difference between the table value and my average values
Which shows that my values are (100% - 99.5) = 4.5% lower than the table values. My values are lying close to each to other which means that I am achieving precise results to some extent and my results are fairly accurate because they are lying just 4.5% below the table values. These points towards the fact that there are some small problems with the system for my results not being more accurate and I am going to discuss it in the paragraph below.
Heat could have been lost to the surroundings through the Styrofoam cups because we were not provided with the proper equipment to cover the cup while the experiment was going on. This would affect the temperature change value and in turn, the molar enthalpy value for that reaction by giving us lower results in comparison to the Table value. There was also concern regarding the equal sharing of the reactants within the hydrochloric acid solution. It was much easier for the magnesium oxide to react with HCl because it was in powder form and the surface area was very large so it could react completely. In case of magnesium strip, I could not guarantee that a complete reaction took place and all the magnesium strip was used during the reaction because of the small surface area. I could also observe some residues of at the bottom of the cup which must have been some impurities. That amount was very small but still it could affect the outcome of the results by a significant degree. I used the thermometer to stir the reactants during this experiment and while recording the temperature, I lifted it out of the solution by mistake for a very short period of time, which might have played a role in the inaccuracies I have in my result. The thermometers could have been influenced by the temperature outside of the cup as the experiment was being carried out next to a window. The uncompleted reaction, use of thermometer as a stirrer and the heat loss means that the values obtained for heat loss are slightly lower than those of Table values. Finally, due to the time restrictions and chemical limitations implemented during this lab, further trials were unable to be completed to ensure the accuracy of this lab. However, these would have helped in determining a more accurate heat of combustion of magnesium, as an average of all results would be obtained that best reflected the actual result. Also, these values would be more accurate with more practices. There is a lot of room for improvement and all those errors can be prevented to obtain better results which will be much closer to the Table values. The following changes can be made in order to improve the experiment by a great deal. They are listed as under:
Improvements:
- It is extremely important to minimize the heat loss from the system to the surrounding during the experiment. We should have an advanced isolated system which will be able to prevent any heat loss to the surroundings. The easiest we could have done that was by using a proper lid and double walls with air between them. Although our results were not far shot from the Table values, but by using proper equipment we could have achieved a more accurate result.
- Thermometer should not be used as a stirrer, as the chances of getting inaccurate results are very high. A magnetic stirrer provided that we take it’s specific heat capacity into consideration.
- To ensure the even distribution of magnesium, magnesium powder should have been used instead of the magnesium strip, which would allow for a better reaction due to the larger surface area.
- During calculations, we could have measured the mass instead of the volume, because the scale has a smaller RU, and by doing that, we did not need to know the density.
- Through the benefit of more trials, the heat of combustion could have been calculated more accurately by reducing random errors.
- One of the major errors was the fact that we assumed that the hydrochloric acid was equal to the amount of water. We can’t really change this considering that 100% pure HCl is in gaseous state. It is a small error that we must take account of though. We could have determined the specific heat capacity for this solution in an experiment.
Discussion
After collection and organization the data was interpreted. Through the uses of Hess’s Law of heat summation, we were capable of deriving the change in enthalpy of a reaction from alternative reactions. Our final answers were fairly precise accurate. These errors may be accounted for by inaccurate measurements, mathematical mistakes, incomplete reactions, poor heat collection, incorrect recording of data, and poorly calibrated tools. To avoid such error one should label all materials, check each tool before use, take extra care in reading and recording of measurements, double check all calculations, and most of all be patient, labs take time and a rushed procedure leads to inaccurate data and incorrect analysis.