Table 2 shows the different reaction that will be assigned to the groups. Each group are assigned to do 4 reaction. But due to time constraints only group 5 finished the experiment. That is why the instructor assigned groups where we will get the data for every reaction.
Table 3. The ▵T and their corresponding Ave ▵Hrxn and Percent Error
By repeating the same procedure done in the first step using the respective reactions assigned to each group, the students were able to gather temperature readings necessary for the calculation for the respective ▵Hrxn’s. Table 3 contains the observed ▵T and their corresponding average ▵Hrxn from two trials. Equation 1 was used to compute for the ▵Hrxn’s of the reactions
Table 3 shows that there’s a significant %error on the majority of the reactions which implies that there might be some errors caused by the students.
First, the students assumed that the reagent solutions, initially prepared by the students, are all accurate which led to a variety of errors since this is a quantitative analysis e.g. higher temperature reading, lower temperature reading.
Second, the students assumed in the calibration of the calorimeter that the calorimeter is a completely isolated system, which implies that the students assumed that only water and the calorimeter itself will absorb the heat caused by the reaction of HCl(aq) and NaOH(aq), thus totally neglecting the possibility of heat loss. Which in turn leads to a lower temperature reading since this is a quantitative analysis, therefore reverberating to the calculation of a higher Ccal.
Third, the students didn’t consider the volatility certain reagents e.g. NH3, HCl when preparing the desired amount and molarity of the solutions. Therefore if the transferring of liquids took too long, the liquid may have evaporated and therefore the measured quantity of the liquid may no longer be true, since this is a quantitative analysis it altered the results.
Fourth, some students exhibited carelessness in handling the laboratory material, for instance some students have neglected the fact that there should be a uniform pipette for every reagent in order to avoid further miscalculation from the repeated cleaning of the pipettes where there is a probability that minute amounts of water and chemicals previously measured could dilute and react with the solution being measured, thus being a quantitative analysis, it will alter the results.
Fifth, some students may have exhibited personal bias, for instance some students may have not recorded the initial temperature of the system before the other reactant was added and only started recording after the reaction started, therefore this will lead to a severe miscalculation since the value of ▵T will decrease this will therefore result to a higher value for the average Ccal therefore a higher value for ▵Hrxn if exhibited in the calibration of the calorimeter and a lower value for ▵Hrxn if exhibited upon the calculation of the ▵Hrxn’s of the reactions assigned to them.
BIBLIOGRAPHY
[1] “Calorimetry: Measurement of Enthalphy Change” < >
[2] Babbage, C. “ Error Analysis and Significant Figures” <>
[3] Petrucci, R.H. General Chemistry: Principles and Modern Applications, 10th edition. Pearson Education, Inc, 2010,250-251.
APPENDIX
ANSWERS TO QUESTIONS
1.)
First, the students assumed that the reagent solutions, initially prepared by the students, are all accurate which led to a variety of errors since this is a quantitative analysis e.g. higher temperature reading, lower temperature reading.
Second, the students assumed in the calibration of the calorimeter that the calorimeter is a completely isolated system, which implies that the students assumed that only water and the calorimeter itself will absorb the heat caused by the reaction of HCl(aq) and NaOH(aq), thus totally neglecting the possibility of heat loss. Which in turn leads to a lower temperature reading since this is a quantitative analysis, therefore reverberating to the calculation of a higher Ccal.
Third, the students didn’t consider the volatility certain reagents e.g. NH3, HCl when preparing the desired amount and molarity of the solutions. Therefore if the transferring of liquids took too long, the liquid may have evaporated and therefore the measured quantity of the liquid may no longer be true, since this is a quantitative analysis it altered the results.
Fourth, some students exhibited carelessness in handling the laboratory material, for instance some students have neglected the fact that there should be a uniform pipette for every reagent in order to avoid further miscalculation from the repeated cleaning of the pipettes where there is a probability that minute amounts of water and chemicals previously measured could dilute and react with the solution being measured, thus being a quantitative analysis, it will alter the results.
Sixth, some students may have exhibited personal bias, for instance some students may have not recorded the initial temperature of the system before the other reactant was added and only started recording after the reaction started, therefore this will lead to a severe miscalculation since the value of ▵T will decrease this will therefore result to a higher value for the average Ccal therefore a higher value for ▵Hrxn if exhibited in the calibration of the calorimeter and a lower value for ▵Hrxn if exhibited upon the calculation of the ▵Hrxn’s of the reactions assigned to them.[2][4]
2.)
a.1) The 6-inch test tube might overflow if a greater amount of the solution will be used.
a.2) Another is that, since the total volume of the reagents used in the calibration of the calorimeter is 15mL we added the specific heat of the 15mL solution with the specific heat of the set-up therefore it makes it important for us to know that the volume of the solutions are exactly 15mL because any variation will alter the solution
b.) It is important to know the exact concentrations of the reactions because they are essential for the computation of the ▵Hrxn.
c.) It is important to know the exact weight of the metal solids used because they are essential for the computation of the ▵Hrxn.
3.)
a.) 1mol HA× -6.0kJ200mL*1L1000mL*0.5moles1L = -60.kJ is the enthalpy change for the neutralization of 1 mol HA.
b) HA is a weak acid since a reaction between an strong acid and a strong bases would usually have a ▵Hrxn of -55.8 kJ/mol as seen in the reactions HCl + NaOH and HNO3 + NaOH because they have the same net ionic equation,
OH-(aq) + H+(aq) → H2O(l) + 55.8 kJ
In comparison to the reaction between CH3COOH (aq) and NaOH(aq),
OH(aq) + CH3COOH(aq) → CH3COO-(aq) + H2O(aq) + 56.09 kJ
Therefore based from the data derived from the experiment it can be concluded that HA is a weak acid.
c) HA(aq) + -OH-(aq) → A-(aq) + H2O(l)
4.)
a) H+(aq) + OH-(aq) → H2O(l)
b) Cu2+(aq) + Zn(s)→ Cu(s) + Zn2+(aq)
c)
(Ccal)=-(∆Hrxn)nLRcoeffLR(∆T)
(Ccal)=-(-55.8kJ/mol).001mol(5.60°C)
Ccal=9.964285714×10-3kJ/°C
Ccal=9.964285714 J/°C
d)
.264 g Zn ×1 mol Zn65.37 g×1 mol Cu1 mol Zn×63.54 g Cu1 mol Cu= .2566094539 g Cu=0.257g Cu
∆Hrxn=-∆TCcal+(msolid prod.(csolid prod.)nLRcoeffLR
∆Hrxn=-(8.83°C)9.964285714J°C+(0.2566094539 g Cu(0.3846J/g∙°C)4.038549793×10-3 Zn1 mol Zn
∆Hrxn=22001.98035Jmol=22.00198035kJmol=22.0kJ/mol
5.)
∆Hrxn= ΣnΔH°fproducts- ΣnΔH°freactants
-55.8 kJ=1mol-285kJmol-[1molH°f OH+1mol0kJmol]
SAMPLE CALCULATIONS
Calbration of the Calorimeter
Qrxn = (∆Hrxn)nLRcoeffLR
= (-55.8 kJ/mol) (.005 mol)
=-279
Qcal = -Qrxn
= - (-279 J/mol)
= 279 J/mol
(Ccal)=-(∆Hrxn)nLRcoeffLR(∆T)
(Ccal)=-(-55800 J/mol).005mol(3.0°C)
Ccal=93 J/°C
Calculating ∆Hrxn
∆Hrxn=Qrxn/molLR
Qrxn=-Qcal
Qcal=C∆T
Qcal=93J°C3.0 °C
Qcal=279 J
Qrxn =-279 J
∆Hrxn=-279J.005mol
∆Hrxn=-55800 J/mol
=(0.005 mol HCl)(
∆Hrxn=-(3.0°C)(55.67 J/g∙°C)0.005
∆Hrxn=-167.4 kJ/mol
% Error= [(Theoretical Yield- Actual Yield)/Theoretical] *100%
= [(-167.4--52.22) / -52.22] *100%
= 220.57 %
by
yuya990 (student)
