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#5 Determining the Molar Enthalpy of a Solution

# #5 Determining the Molar Enthalpy of a Solution - Aaron...

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Aaron King 660556233 Section 2 Wednesday, September 10, 2007 #5 Determining the Molar Enthalpy of a Solution Table 1 Salt/Base Mass Salt Final Temperature (°C) Initial Temperature (°C) Change (°C) Exo/Endo NaOH - 29.05 21.25 7.80 Endo KNO 3 5.0004g 14.20 21.50 -7.30 Exo NH 4 Cl 5.0002g 14.51 21.20 -6.69 Exo NH 4 NO 3 5.0003g 15.04 21.67 -6.63 Exo 1. = - ∆T Tf Ti = . - . ∆T 29 05℃ 21 25℃ = . ∆T 7 80℃ =- - Cpcalorimeter qrxn mass solutionCpwater∆T∆T =-- . - . . . ( ) . Cpcalorimeter 58 3kjmol 100 0g4 18Jg℃7 80℃ 1kJ1000J 7 80℃ = . / Cpcalorimeter 7 06kj mol℃ 2. KNO 3 = - ∆T Tf Ti = . - . ∆T 14 20℃ 21 50℃ =- . ∆T 7 30℃ NH 4 Cl = - ∆T Tf Ti = . - . ∆T 14 51℃ 21 20℃ =- . ∆T 6 69℃ NH 4 NO 3

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King 2 = - ∆T Tf Ti = . - . ∆T 15 04℃ 21 67℃ =- . ∆T 6 63℃ ( - ) graphs see pages 4 7 . 3 =- qKNO3 mCpH2O∆T =- . . - . ( ) qKNO3 100 0g4 18Jg℃ 7 30℃ 1kJ1000J = . qKNO3 3 05kJ =-[( )( )( ) qNH4Cl m CpH2O ∆T =- . . - . qNH4Cl 100 0g4 18Jg℃ 6 69℃1kJ1000J = . =- qNH4Cl 2 89kJqNH4NO3 mCpH2O∆T =- . . - . ( ) qNH4NO3 100 0g4 18Jg℃ 6 63℃ 1kJ1000J = . qNH4NO3 2 77kJ ° = . . = . × . = . ∆H KNO3 3 05kJ5 0004g 0 610kJg 101 11g1mol 61 7kJmol ° = . . = . × . = . ∆H NH4Cl 2 89kJ5 0002g 0 578kJg 51 50g1mol 29 8kJmol ° = . . = . × . = . ∆H NH4NO3 2 77kJ5 0003g 0 554kJg 76 06g1mol 42 1kJmol Salt ( / ) Enthalpy kJ g ( / ) Enthalpy kJ mol KNO 3 .
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