This document provides instructions on calculating uncertainty in enthalpy change calculations for a chemical reaction. It discusses two main methods for uncertainty calculation - using percentage uncertainty and using maximum and minimum values. It emphasizes the importance of performing multiple trials of the experiment and taking the average or standard deviation. It also highlights properly accounting for uncertainty in temperature change, mass, volume and moles of reactants when determining the overall uncertainty in enthalpy change.

1. http://lawrencekok.blogspot.com Prepared by Lawrence Kok Video Tutorial on Uncertainty Calculation, standard deviation and Error analysis in Enthalpy Change

3. 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

5. More Accurate way by taking Uncertainty in Moles Cu 1 st Method Using %Uncertainty %Uncertainty moles of Cu used Conc CuSO 4 = (1M±0) , Vol = (20.0 ± 0.3) ml Moles 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

6. 2 nd Method Using Max/Min Conc CuSO 4 = (1M±0) , Vol = (20.0 ± 0.3)ml Moles 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= 42.5 m = 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 = 0.0203 Min 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----------0.02mol Min Q = 3503.0J-----------0.02mol Uncertainty moles of Cu = 0.02 ± (0.0203--0.0197) Max moles Cu = 0.0203 Min moles Cu = 0.0197

7. 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 Cu Max Q = 3677.7/0.0197------1 mol Min Q = 3503.0/0.023--------1mol Max Q = 186.7kJ/mol Min Q = 152.3kJ/mol Q = 179.5 ± ( 186.7---152.3 )kJ/mol % Uncertainty Method Max/Min Method Q = ( 186 ----172 )kJ/mol Q = ( 186.7---152.3 )kJ/mol

9. Two Methods used 1 st Method Average 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

10. 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

11. 2 nd Method Taking average Temp change Click HERE for data Average Temp data is taken. Displacement for Zinc + CuSO 4 Zinc excess, Copper limiting, Conc CuSO 4 = (1M± 0) Vol = 20.0 ± 0.3 ml Mol of Cu ions= M x V = 0.02 mol △ t = (25.7 ± 0.2) - (21.6 ± 0.2)= (4.1 ± 0.4)

12. 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 4.1 Q = 343.1J ---------0.02 mol Q = 343.1/0.02------1 mol Q = 17.155kJ/mol

13. More accurate way is to consider Uncertainty for moles Cu Conc CuSO 4 = (1M± 0), Vol = (20.0 ± 0.3) ml Mol 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

14. 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

16. Acknowledgements Thanks to source of pictures and video used in this presentation Thanks to Creative Commons for excellent contribution on licenses http://creativecommons.org/licenses/ Prepared by Lawrence Kok Check out more video tutorials from my site and hope you enjoy this tutorial http://lawrencekok.blogspot.com