Experiment13DeterminationofEnthalpyChanges：焓的变化测定实验.ppt

1、Determination of Enthalpy Changes Associated with a Reaction & Hesss Law,Jae Lee Period 2 Experiment 13,http:/,/distance/sci122/Programs/p21/image.gif,Background Information,Thermochemistry study of heat chances & transfers associated with chemical reactions Hesss law states

2、 that at constant pressure, the enthalpy change for a process is not dependent on the reaction pathway, but is dependent only upon the initial and final states of the system. H is used to write the change in heat,Equations you should know and will use,NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) NH4Cl(aq)+N

3、aOH(aq) NH3(aq)+NaCl(aq)+H2O(l) NH3(aq) + HCl(aq) NH4Cl(aq,Purpose of Lab,To determine the enthalpy change that occurs when a strong base, NaOH (Sodium Hydroxide), reacts with a strong acid, HCL (Hydrochloric acid,Materials,A calibrated temperature probe Thermometer Calorimeter Water 2.00 M HCl 2.00

4、 M NaOH 2.00 M NH4CL 2.00 M NH3,http:/,http:/www.hansatech-,More materials,Styrofoam cover Polystyrene cups 400 ml beaker Glass Stirring Rod,http:/,http:/www.slsonline.co.uk/asps/uploads/big/2148-1.jpg,http:/www.chem.ualberta.ca/ngee/ExptG1.gif,Procedure,Place 50.0 mL of room temp. distilled water i

5、n calorimeter Use calibrated temperature probe to determine/ record temperature Add 50.0 mL of warm distilled water to calorimeter Record temperature every 30 seconds for 3 min. Clean beaker for next experiment Measure temp chance water 50.0 ml of 2.00 M HCl is added to 50.0 ml of 2.00 M NaOH Record

6、 temperature change every 30 sec. for 3 min. Repeat step 7 but with 2.00 M NH4Cl with 2.00 M NaOH Repeat step 7 but with 2.00 M NH3 with 2.00 M HCl,A graph of the heat capacity of the calorimeter,From the graph, determine the heat capacity of the calorimeter. The specific gravity of water at 23.0 C

7、is 0.998 and at 61.0 C,Analysis,Chart created from Experiment 13: Determination of Enthalpy Changes Associated with a Reaction and Hesss law,Extrapolating the regression line to the Y axis (0 seconds) gives a temperature of 41.4 C at the moment the room temperature and warm water were mixed Average

8、temperature of room temperature and warm water: 23.0 C + 61.0 C = 42.0 C 2,Ccalorimeter = qcalorimeter (Tmix Tinitial) qcalorimeter = -qwater Qwater = (mass water)*(specific heat)*(Tmix Tavg) At 23.0C: 50.0 mL H2O x 0.983 g*mL-1 1 = 49.9 g H2O,At 61.0C: 50.0 mL H2O x 0.983 g*mL-1 1 = 49.1 g H2O Tota

9、l mass = 49.9 g + 49.1 g = 99.0 g H2O =(99.0 g)*(4.18 J/g * C)*(41.4C 42.0C) = -2.5x102 J Heat gained by calorimeter = -qwater = 2.5x 102 J Ccalormeter = qcalorimeter 2.5 x 102 J = 14 J* C (Tmix-Tinitial) (41.4 C 23.0 C,2.)Temperature changes for each of the 3 reactionsCalculate the heat evolved in

10、the reaction (kJ/mol of product). Assume the density of each solution = 1.00gxmL-1,Chart created from Experiment 13: Determination of Enthalpy Changes Associated with a Reaction and Hesss law,HCl + NaOH qrxn = -(masssoln) x (specific heat soln) x (Tsoln) + (Ccalorimeter x Tsoln) volumesoln x molarit

11、y = 100g x 4.18J/gC x (3.56C 23.0C) + (14 J/C x 12.5C) 0.500L x 2.0 mole/L = -5.4 x 103 . 0.0500L x 2.0 mol/L = - 54 kJ/mol,a,http:/www.razor-,b,NH4Cl + NaOH qrxn = -(masssoln) x (specific heat soln) x (Tsoln) + (Ccalorimeter x Tsoln) volumesoln x molarity = 100g x 4.18J/gC x (24.1C 22.9C) + (14 J/C

12、 x 1.2C) 0.500L x 2.0 mole/L = -5.2 x 102 J . 0.0500L x 2.0 mol/L = - 5.2 kJ/mol,c,NH3 + HCl qrxn = -(masssoln) x (specific heat soln) x (Tsoln) + (Ccalorimeter x Tsoln) volumesoln x molarity = 100g x 4.18J/gC x (33.1C 23.0C) + (13.6 J/C x 10.1C) 0.500L x 2.0 mole/L = -4.4x 103 J . 0.0500L x 2.0 mol

13、/L = - 44 kJ/mol,3,Write the net ionic equation, including the Hs, for the 1st two reactions studied & rearrange the equation(s) in order to created the 3rd reaction and the H value,H+(aq) + OH-(aq) H2O (l) NH3(aq) + H2O (l) NH4+ (aq) + OH-(aq) NH3(aq) + H+(aq) NH4+ (aq) H = -54 kJ mol-1 H = +5.2 kJ mol-1 H = -49 kJ mol-1,4,Use calculator to calculate the %error between the measured H & the calculated H % error = observedtheoretical x 100% theoretical = -44-(-49) x 100% -491 = -10,http:/shiar.nl/calc/tipics/ti83+01.gif,Conclusion,After doing this lab experiment, I was a