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Lab-Hess's Law

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					Laboratory

I. Title: The Heats of Solution and Neutralization: A Demonstration of Hess’s Law

II. Discussion:

Hess’s Law states that the total enthalpy change for a chemical or physical change is independent of the pathway. That is the reaction
may take place in one step or several steps, H will be the same. Another way of looking at this law is to say that the overall change
in internal energy or enthalpy for a reaction is dependent only upon the potential energy of the products when compared to the
potential energy of the reactants. The path or set of reactions we decide upon to re-arrange reactants into products will have no
bearing on the PE positions of products vs. reactants. Thus the enthalpy of reaction or H is equal to the sum of the enthalpy changes
for the individual steps in the reaction or process.
The enthalpies of reaction for a series of reactions, which will take specific reactants to specific products, may be measured
experimentally as will be observed in this laboratory. The reaction being studied is the neutralization of an acid solution by a solid
crystalline hydroxide in one step. The two reactions which precede this reaction are; reaction I, the dissolving of the base (Heat of
Hydration or Heat of Solution) and reaction II, the bond formation between H + and OH- to form water (Heat of Neutralization). These
add together to give the same overall reaction. Thus the Hsolution added to the Hneutralization should equal the Hrxn for the one step
reaction.

         I        Hsolution                                                        NaOH(s)  Na+(aq) + OH-(aq)
         II       Hneutralization                     Na+(aq) + OH-(aq)     + H+(aq) + Cl-(aq)  Na+(aq) + Cl-(aq) + H2O(l)

         III      Hrxn                                           NaOH(s) + H+(aq) + Cl-(aq)  Na+(aq) + Cl-(aq) + H2O(l)



In this lab you will use a calorimeter as the reaction vessel to measure the amount of heat evolved or absorbed during three reactions.
You may assume that the heat of the reaction will be used to change the temperature of the aqueous solution and the glass of the
container. The specific heat capacity of the water is 4.184 J/g oC, while the specific heat capacity of the beaker is 0.8368 J/g oC. Also,
the density of water as well as the acid and base solutions used in this lab is 1.00 g/mL at 25 oC.

III. Purpose: To measure the enthalpy changes for three reactions and demonstrate Hess’ Law of the Additivity
of Heats of Reaction.

IV. Terms to Know: Enthalpy of Neutralization, Enthalpy of Solution, Enthalpy of Reaction, and Hess’ Law

V. Materials: Make a list of all the materials that you used in this lab

VI. Procedure:
       A. Reaction I – Heat of Solution = the heat of dissolving and dissociation of NaOH(s) in water.
             1. Mass a clean, dry 150 mL beaker to the nearest 0.01 g. Record mass.
             2. Measure 50.0 mL of tap water and transfer this to the beaker. Mass both the water and the
                 beaker. Record mass.
             3. Measure and record the temperature of this sample of water.
             4. Assemble Calorimeter as instructed.
             5. Mass 2.00 g (+/- 0.01 g) of NaOH(s) on the same analytical balance. Record mass.
             6. Add the NaOH to the calorimeter. IMMEDIATELY close the calorimeter and swirl the
                 mixture. Monitor the temperature and record the highest temperature that is achieved by the
                 solution. (You will know the NaOH has dissolved when the maximum temperature is
                 reached.)
             7. Rinse and dry the beaker thoroughly and move on to reaction II.
        B. Reaction II – Heat of Neutralization = heat of neutralizing HCl(aq) with NaOH(aq)
              1. Assemble your calorimeter.
              2. Measure exactly 25.0 mL of 2.00 M HCl(aq) and add this to the 150 mL beaker in your
                  calorimeter. Record volume on your data table.
              3. Measure and record the temperature of the HCl solution in the calorimeter.
              4. Clean your graduated cylinder.
              5. Measure exactly 25.0 mL of 2.00 M NaOH(aq). Record volume on your data table.
              6. Measure and record the temperature of the NaOH solution in the graduated cylinder. This
                  temperature should be the same as the HCl solution.
              7. Add the NaOH solution to the calorimeter and IMMEDIATELY close the lid to the
                  calorimeter. Swirl and monitor the temperature.
              8. Record the highest temperature achieved by the chemical reaction.
              9. Rinse the beaker thoroughly, dry it and move on to reaction III.

        C. Reaction III –Heat of Reaction = heat of dissolving and neutralizing NaOH(s) in HCl(aq)
               1. Assemble your calorimeter.
               2. Measure exactly 50.0 mL of 1.00 M HCl(aq) and add this to the 150 mL beaker in your
                   calorimeter. Record volume on your data table.
               3. Measure and record the temperature of the HCl solution in the calorimeter.
               4. Mass 2.00 g (+/- 0.01 g) of NaOH(s) on the same analytical balance. Record mass.
               5. Add the NaOH to the calorimeter. IMMEDIATELY close the calorimeter and swirl the
                   mixture. Monitor the temperature and record the highest temperature that is reacted. (You
                   will know the NaOH has dissolved when the maximum temperature is reached.)
               6. Record the extreme temperature reached.
               7. Rinse the beaker thoroughly, dry it and clean up the rest of your lab area.

VII. Data Tables
       1. Draw and label the calorimeter that is used in this lab.
       2. Fill in the appropriate data.

Reaction 1                                      Value (with units)
Mass of Beaker       (+/-            g)
Mass of Beaker & H2O (+/-            g)
Mass of H2O          (+/-           g)
                                  o
T initial of H2O     (+/-          C)
                                   o
T final of H2O      (+/-            C)
Mass of NaOH         (+/-            g)

Reaction 2
Volume of 2.00 M HCl(aq) (+/-             mL)
                                           o
T initial of HCl(aq)         (+/-           C)
Volume of 2.00 M NaOH(aq) (+/-            mL)
                                           o
T initial of NaOH(aq)        (+/-          C)
                                           o
T final of acid/base mixture (+/-          C)

Reaction 3
Volume of 1.00 M HCl(aq)        (+/-       mL)
                                           o
T initial of HCl(aq)            (+/-        C)
Mass of NaOH                   (+/-          g)
                                           o
T final of acid/base mixture   (+/-          C)
VIII. Calculations                             Show all of your work with units in the boxes below!!!

                                            Reaction 1               Reaction 2                Reaction 3


    T = (Tfinal – Tinital)




kJ of energy absorbed
  by just the water.


*Remember, DH2O = ____________

**Treat the volumes of the acid and
the base as the volumes of the water.




kJ of energy absorbed
   by just the glass.




  Total kJ absorbed in
    the calorimeter
     (water + glass)




   Moles (n) of NaOH




kJ/mol NaOH ( = H)



          Hess’ Law                     Reaction 1         +      Reaction 2            = Reaction 3
IX. Results
       1. Explain why Hrxn 1 plus Hrxn II should equal Hrxn III .

       2. Calculating % Error.
              a) allow the sum of Hrxn I + Hrxn II to equal the true value
              b) Hrxn III = trial value
              c) calculate the % error
              d) Discuss possible sources of error.

X. Follow-up:
       1. Calculate the Hrxn III using the following formula:

       Hrxn = (sum of Hf products) – (sum of Hf reactants) or Hrxn =  (Hf products) –  (Hf reactants)
              (Include this formula that is used to solve the problem)

       The values for Hf are provided below:

               NaCl (aq)       -407 kJ/mol
               HCl (aq)        -167 kJ/mol
               H2O (l)         -286 kJ/mol
               NaOH(s)         -427 kJ/mol
               NaOH (aq)       -470 kJ/mol

       Answer: -99 kJ

       2. Define Heat of Formation.
       3. Does Hess’ Law apply to the use of Heats of Formation? Explain your answer.

				
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posted:11/30/2011
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