Lab Report Materials Thermodynamics : Hee's Law: determine the enthalpy

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  • 8/19/2019 Lab Report Materials Thermodynamics : Hee's Law: determine the enthalpy

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    Materials Thermodynamics (EBT324) Exp. 1

    1. OBJECTIVE

    1.1 To develop a simple calorimeter.

    1.2 To determine enthalpy of reaction that difficult to be carried out straight

    forwardly.

    2. INTRODUCTION

    Hess’s Law states that the heat evolved or absorbed in a chemical process is the

    same whether the process takes place in one or in several steps. This is also known as

    the law of constant heat summation. [1 To illustrate Hess’s Law! the e"ample thermal

    e#uations are shown below$

    % & ' ( %'   ∆H ( dH1

    %' & ' ( %'2   ∆H ( dH2

    Then!

    % & 2' ( %'2   ∆H ( dH1 & dH2

    ∆H or the change in enthalpy arose as a unit of measurement meant to calculate

    the change in energy of a system or the change in the internal energy of a system! by

    simultaneously measure the amount of heat and work e"changed. [2 )nthalpy of the

    reaction or energy change of a reaction ∆H! is the amount of energy or heat absorbed

    in a reaction. *f the energy is re#uired! ∆H is positive while if the energy is released!

    ∆H is negative. [+

    % calorimeter is a device used to measure heat of reaction. % styrofoam cup is

    used as a calorimeter which used to determine the heat released or absorbed in a

    chemical reaction. ,ue to its good insulated walls! the heat e"change with the

    environment can be prevented. *n order to measure heat of reaction! we often enclose

    reactants in a calorimeter! initiate the reaction! and measure the temperature difference

     before and after the reaction. [-

    3. EQUIPMENT/APPARATUS

    +.1 lass stir stick 

    +.2 'eaker /100ml and -00ml

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    Materials Thermodynamics (EBT324) Exp. 1

    +.+ Thermometer 

    +.- raduated cylinder 

    +. 3hemicals

    +..1 Hydrochloric acid solution /H3* 24

    +..2 5atrium carbonate /5a236+

    +..+ 5atrium hydrogen carbonate /5aH36+

    4. PROCEDURE

    4.1 Determination o! ent"a#$% !or rea&tion 'i(

    1 +0ml of H3* was put into calorimeter and was stirred carefully with a

    glass stir stick until a constant temperature is needed. The temperature

    was measured and recorded as a starting temperature! T1.

    2 % mass of 2.00 gram of 5a236+ was accurately found and recorded.

    + 7olid 5a236+ was placed into the H3* in the calorimeter and stirred

    gently with the glass stir stick.- The temperatures were recorded for every +0 second until the

    temperatures are constant.

    % graph of temperatures vs times was plotted and the highest

    temperature! T2 was calculated.

    8 The enthalpy for reaction /i was calculated.

    4.2 Determination o! ent"a#$% !or rea&tion 'ii(

    1 +0ml of H3* was put into calorimeter and was stirred carefully with a

    glass stir stick until a constant temperature is needed. The temperature

    was measured and recorded as a starting temperature! T1.2 % mass of +.00 gram of 5aH36+ was accurately found and recorded.

    + 7olid 5aH36+ was placed into the H3* in the calorimeter and stirred

    gently with the glass stir stick.

    - The temperatures were recorded for every 1 second until the

    temperatures are constant.

    % graph of temperatures vs times was plotted and the highest

    temperature! T2 was calculated.

    8 The enthalpy for the reaction /ii was calculated.9 The enthalpy for the reaction /iii was calculated with using of enthalpy

    from the reaction /i and /ii.

    ). *AB QUESTIONS

    ).1 +"at i, t"e -e,,, *a0

    Hess:s law states that energy changes are state functions! which allows

    us to calculate the overall change in enthalpy by simply summing up the

    changes for each step of the way! until product is formed. %ll steps have to

     proceed at the same temperature and the e#uations for the individual steps

    must balance out. The sum of enthalpy in the two steps is e"actly same as the

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    Materials Thermodynamics (EBT324) Exp. 1

    one;step reaction. The principle underlying Hess:s law does not

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    Materials Thermodynamics (EBT324) Exp. 1

    5or rea&tion 'i( 6

    4ass of Na

    2CO

    3 ( 2.00g

    D44 of  Na2CO3  ( /2"2+ & 12 & /+"18 (108

    4ass of H3l ( 1.1B? " +0 ( +.89g

    ET ( +0.? F 2?.- (1.C3

    # ( s "

     RMM of Na2

    CO3

    massof Na2 CO3 x (massof Na2CO3+massof HCl ) x△T   

    # ( -.1B "106

    2.00 x (2.00+35.67 ) x 1.5

      ( &12.2 k@Amol

      ( ;enthalpy

      ( ;12.2 k@Amol /)"othermic

    5or rea&tion 'ii( 6

    4ass of NaHCO

    3  ( +.0g

    D44 of NaHCO

    3  ( 2+ & 1 & 12 & /+"18 ( B-

    4ass of H3l ( 1.1B? " +0 ( +.89g

    ET ( 2- F 2B.B ( ;-.BC3

    # ( s "

     RMM of Na2

    CO3

    massof Na2 CO3 x (massof Na2CO3+massof HCl ) x△T   

    # ( -.1B "84

    3.00 x (3.0+35.67 ) x−4.8

      ( ;21.92 k@Amol

      ( ;enthalpy

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    Materials Thermodynamics (EBT324) Exp. 1

      ( &21.92 k@Amol /)ndorthermic

    Gor reaction /iii!

    25a3l/a# & H 

    2O

    /l &CO

    2 /g = Na

    2CO

    3 /s & 2H3L/a#

     ΔH 1=+12.52kJ /mol

    2  NaHCO

    3 /s & 2H3l/a# = 25a3l/a# &2 H 

    2O

    /l &2CO

    2 /g

     ΔH 2=+2(21.72)kJ /mol

      ( &-+.-- k@Amol

    2   NaHCO3 /s =  Na2CO3 /s &  H 2O /l & CO2 /g

     ΔH 3=+55.96kJ /mol

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    Materials Thermodynamics (EBT324) Exp. 1

    0 20 -0 80 B0 100 120 1-0 180 1B0 200

    2B

    2B.

    2?

    2?.

    +0

    +0.

    +1

    +1.

    +2

    raph of Temperature versus Time

    Time /s

    Temperature /C3

    Gigure 1$ raph of Temperature versus time for reaction/i

    0 10 20 +0 -0 0 80 90 B0 ?0 100

    0

    10

    1

    20

    2

    +0

    +

    raph of Temperature versus Time

    Time /s

    Temperature /C3

    Gigure 2$ raph of temperature versus time for reaction /ii

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    Materials Thermodynamics (EBT324) Exp. 1

    The enthalpy change of reaction of solid natrium carbonate! Na

    2CO

    3  and

    solid natrium hydrogen carbonate! NaHCO

    3  was determined by reaction /i and

    reaction /ii respectively.

    'ased on the result calculated! it can be obtained that the enthalpy of reaction

    /i is 12.2k@Amol while the enthalpy of reaction /ii is &21.92k@Amol. The enthalpy  

    of reaction /iii can be calculated by adding up the reverse of reaction /i with reaction

    /ii. 7ince reaction /i was reversed and reaction /ii was doubled to get reaction /iii!

    the enthalpy of reaction /i become &12.2k@Amol and reaction /ii become

    -+.--k@Amol. Therefore! the enthalpy of reaction /iii is .?8k@Amol.

    *n reaction /i! the enthalpy shows a negative sign means that it was an

    e"othermic reaction. Heat will be released from the system to the surrounding. *n

    reaction /ii! positive sign was shown by the enthalpy. *t was an endothermic reaction!

    means that heat will be absorb from surrounding into the system. Deaction /iii is an

    endothermic reaction.

    Grom the e"periment have been done! e"perimental results was deviate from the

    theoretical results. This is due to some errors that occur when e"periment was carried

    out. hen solid was poured into calorimeter! time was needed to close the lip of 

    calorimeter. However! there was some heat loss to the environment. This will cause

    the measurement of temperature inaccurate. 'esides that! impurity of calorimeter used

    will also affect the accuracy of temperature measured. Therefore! to enhance the

    results of e"periment! it should be introduce a more closed calorimeter to prevent heat

    loss and before start an e"periment! make sure the apparatus used was clean properly.

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    Materials Thermodynamics (EBT324) Exp. 1

    7. CONC*USION

    %s a conclusion! we can conclude the e"periment was carried out successfully and

    the ob