ib chemistry, ib biology on uncertainty calculation, error analysis and standard deviation on...
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IB Chemistry, IB Biology on Uncertainty calculation, error analysis and standard deviation on enthalpy changeTRANSCRIPT
http://lawrencekok.blogspot.com
Prepared by Lawrence Kok
Video Tutorial on Uncertainty Calculation, standard deviation and Error analysis in Enthalpy Change
IB Chemistry, Uncertainty Calculation, Error analysis, Standard Deviation
Enthalpy change for displacement Zn + CuSO4
2 methods for Uncertainty Calculation• 1st Method using % Uncertainty
Zinc excess, Copper limitingConc CuSO4 = (1M ± 0)Vol = (20.0 ± 0.3) mlMol of Cu ions= M x V= 0.02 mol△t = (68.0±0.2) - (25.1±0.2) = (42.9±0.4)m = 20.0 ± 0.3ml
Q = mc t△Q = 20.00 x 4.184 x t△Q = 20.00 x 4.184 x 42.9Q = 3589.8J ----------0.02 molQ = 3589.9/0.02------ 1molQ = 179.5kJ/mol
Extrapolation for Temp increase, 68.0 - 25.1= 42.9
%Uncertainty Q = % Uncertainty m + % Uncertainty t△%Uncertainty m = (0.3/20) x 100% = 1.5%%Uncertainty t = (0.2+0.2) / 42.9 x 100% = 0.93%△
%Total Uncertainty = (0.93 + 1.5) = 2.43%
Q = (179.5 ± 2.43%)Q = (179.5 ± 4.36) = (180 ± 4)kJ/mol
•2nd Method Using Max/Min
Q = mc t△Q = 20.00 x 4.184 x 42.9Q = 3589.8J ----------0.02 molQ = 3589.9/0.02------1molQ = 179.5kJ/mol
Q = mc t△△t = 42.9 ± (0.2+02) = (42.9±0.4) Max t = 42.9+ 0.4 = 43.3 Min t = 42.9- 0.4 = 42.5m = 20.0 ± 0.3ml Max m = 20.0+ 0.3=20.3 Min m = 20.0- 0.3 = 19.7
Max Q = Maximum m and Maximum t△ △ △Max Q = m c t = 20.3 x 4.184 x 43.3 = 3677.7J△ △Max Q = 3677.7J -----------0.02molMax Q = 3677.7/0.02-------1molMax Q = 183.3kJ/mol
Min Q = Minimum m and Minimum t△ △ △Min Q = mc t = 19.7 x 4.184 x 42.5 = 3503.0J△ △Min Q = 3503.0J------------0.02molMin Q = 3503.0J/0.02-------1molMin Q = 175.1kJ/molUsing Max/Min Method Using %Uncertainty MethodQ = 179.5 ± ( 183.3--175.1)kJ/mol Q = (179.5 ± 2.43%) = (180 ± 4)kJ/molQ = (183.3---175.1) Q = (184---176)
More Accurate way by taking Uncertainty in Moles Cu1st Method Using %Uncertainty
%Uncertainty moles of Cu usedConc CuSO4 = (1M±0) , Vol = (20.0 ± 0.3) mlMoles of Cu ions = M x V= 0.02 mol
%Uncertainty moles Cu = %Uncertainty in M + %Uncertainty in Vol %Uncertainty Cu = 0% + (0.3/ 20) x 100% = 0% +1.5% = 1.5%
Total %Uncertainty Q = %Uncertainty m + %Uncertainty t + %Uncertainty mol Cu△Total %Uncertainty Q = 1.5% + 0.93% + 1.5% = 3.93%
Q = (179.5 ± 3.93%)= (179.5 ± 7.05) = (179 ± 7)Q = ( 186 ----172 )kJ/mol
2nd Method Using Max/Min
Conc CuSO4 = (1M±0) , Vol = (20.0 ± 0.3)mlMoles of Cu ions = M x V = 0.02 mol
△t = 42.9 ± (0.2+02) = (42.9±0.4) Max t= 42.9+0.4 = 43.3 Min t= 42.9- 0.4= 42.5m = 20.0 ± 0.3ml Max m= 20.0+ 0.3=20.3 Min m= 20.0- 0.3= 19.7
Max Vol = 20.3ml, Min Vol = 19.7ml Molarity = (1M±0)
Max moles Cu = Max M x Max Vol = 1 x 20.3 = 0.0203Min moles Cu = Min M x Min Vol = 1 x 19.7 = 0.0197
Uncertainty moles of Cu = 0.020 ± (0.0203--0.0197)
Max Uncertainty Q = mc t = 20.3 x 4.184 x 43.3 = 3677.7J△Min Uncertainty Q = mc t = 19.7 x 4.184 x 42.5 = 3503.0J△
Max Q = 3677.7J----------0.02molMin Q = 3503.0J-----------0.02mol
Uncertainty moles of Cu = 0.02 ± (0.0203--0.0197)Max moles Cu = 0.0203Min moles Cu = 0.0197
Max Q = Max Uncertainty Q + Min Uncertainty mol Cu ( will give greatest error )Min Q = Min Uncertainty Q + Max Uncertainty mol Cu ( will give least error )
Max Q = 3677.7---------------0.0197 mol Cu Min Q = 3503.0-----------------0.0023mol CuMax Q = 3677.7/0.0197------1 mol Min Q = 3503.0/0.023--------1molMax Q = 186.7kJ/mol Min Q = 152.3kJ/mol
Q = 179.5 ± ( 186.7---152.3 )kJ/mol
% Uncertainty Method Max/Min MethodQ = ( 186 ----172 )kJ/mol Q = ( 186.7---152.3 )kJ/mol
IA must be done with minimum 3 trials, for valid conclusion.
Displacement of Zinc + CuSO4
Displacement done with 3 trials.
• Zinc excess, Copper limiting
• Conc CuSO4 = (1M±0)
• Vol = (20.0 ± 0.3) ml
• Mol of Cu ions = M x V = 0.02 mol
Click HERE for Data
Two Methods used1st MethodAverage Q for 3 trials and using Standard deviation as uncertainty
Temp change for 3 trials△t1 =( 26.2 ± 0.2)- (21.7± 0.2)=(4.5± 0.4)
△t2 = (26.1 ± 0.2)- (21.6± 0.2)=(4.5± 0.4)
△t3 = (25.8 ± 0.2)- (21.7± 0.2)=(4.1± 0.4)
Q1 = mc t1 = 20.0 x 4.184 x 4.5 = 376.5△
Q2 = mc t2 = 20.0 x 4.184 x 4.5 = 376.5△
Q3 = mc t3 = 20.0 x 4.184 x 4.1= 343.1△
Average Q = (376.5 + 376.5 + 343.1)/3 = 365.4J Average Q = 365.4J ----------0.02 mol Average Q = 365.4/0.02------1 mol Average Q = 18.3kJ/mol
Uncertainty Q = Standard Deviation Q= (19.28)
Uncertainty Q = 19.28-----------0.02 mol
Uncertainty Q = 19.28/0.02-----1 mol
Uncertainty Q = 0.96kJ/mol
Average Q = (18.3 ± 0.96)kJ/mol
2nd Method
Taking average Temp changeClick HERE for data
Average Temp data is taken.Displacement for Zinc + CuSO4
Zinc excess, Copper limiting,Conc CuSO4 = (1M± 0)Vol = 20.0 ± 0.3 mlMol of Cu ions= M x V = 0.02 mol△t = (25.7 ± 0.2) - (21.6 ± 0.2)= (4.1 ± 0.4)
By extrapolation using average temp△t = (25.7 ± 0.2) - (21.6 ± 0.2)= (4.1 ± 0.4)
Q = mc t△Q = 20.0 x 4.184 x t△Q = 20.0 x 4.184 x 4.1Q = 343.1J ---------0.02 molQ = 343.1/0.02------1 mol
Q = 17.155kJ/mol
More accurate way is to consider Uncertainty for moles Cu
Conc CuSO4 = (1M± 0), Vol = (20.0 ± 0.3) mlMol of Cu ions = M x V
%Uncertainty moles Cu = %Uncertainty in M + %Uncertainty in Vol %Uncertainty Cu = 0% + (0.3/ 20) x 100% = 0% +1.5% = 1.5%
%Total Uncertainty Q = %Uncertainty m + %Uncertainty t + %Uncertainty mol Cu△
%Total Uncertainty Q = 1.5% + 9.75% + 1.5% = 12.75%
Q = (17.15± 12.75%) = (17.15± 2.18)kJ/mol
Using %Uncertainty Method
Temp increase = (25.7 ± 0.2) - (21.6 ± 0.2)= (4.1 ± 0.4)%Uncertainty Q = % Uncertainty m + % Uncertainty t△%Uncertainty m = (0.3/20) x 100% = 1.5%%Uncertainty t = (0.4 / 4.1) x 100% = 9.75%△
%Total Uncertainty = (1.5 + 9.75)% = 11.25%
Q = (17.155± 11.25%)= (17.15±1.92)kJ/mol
In short,
• Make sure you perform 3 trials and compute the average or use std deviation
• Use %Uncertainty or Max/min Method of your choice
• Use lots of common sense for error treatment
• The only certainty in life is continual uncertainty
• Have fun with uncertainty
Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/
Prepared by Lawrence Kok
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