Thermochemical Report
enthalpy change of netralization
           Created by:
      Aris Wakhyudin (01)
             XI.A.1
         SMA Negeri 2 Cirebon
  Jln. Dr. Cipto Mangunkusumo No.1
           Preface
  Gratitude that I convey to the presence of God, Most Gracious, because thanks to the mercy
of this report I could complete as expected.
  During the finishing writing this report, ranging from preparation to finish, I get a lot of
support from other parties. So on this occasion for their assistance and moral encouragement
and material I wish to express my respect and deep gratitude to:
           1. Mrs. Tuti .S S,Pd as a chemistry teacher. Who have provided guidance.
           2. All members of my chemistry group who have helped me in the process of
              labolatorium work.
           3. And all parties involved in this practice.
  In preparing this report, I realize there are many disadvantages. Therefore I apologize for all
the shortcomings I hope this report will be particularly beneficial for the writer himself and
for others in general. Amin.
                                                             Cirebon, 12 October 2010
                                                                  Aris Wakhyudin
   1. Purpose
    To determine enthalpy change (Hn) with experimental methods, namely experiment an
acid-base neutralization reaction (HCl and NaOH).
   2.Basic theory
   3. Tools and materials
   Mixer stick               Measuring Flask           Funnel        Beaker glass
Calorimeter             solution of HCl and NaOH                 aquades
                        Pipette                   thermometer
   4.Work procedures
Creating 1liter HCl, 1M
      Prepare tools and materials
      Take a solution of concentrated HCl with the percentage of 36,46% as much as 84,84
       ml and input into the measuring flask
      Dissolve it with concentrated HCl solution until the volume reaches aquades 1 liter
      Shake the solution until evenly mixed
Acid-base neutralization reaction to determine H with experimental method (with
calorimeter)
      Prepare tools and materials.
      Measure the initial temperature of each solution, the solution of NaOH and HCl
       solution, record the result as the initial temperature of the experiment after the second
       temperature averaged solution.
      Prepare a 0.1 M NaOH solution as much as 50ml in beaker glass and 50ml of 1M HCl
       solution in other beaker glass.
      prepare a simple calorimeter and put 50ml of NaOH solution in a simple calorimetri
      put thermometer in a simple calorimetri
      put in 50ml of HCl solution into a simple calorimetri
      Mix a solution of HCl with NaOH solution that already exists in the calorimeter with
       mixer stick, stir as she noticed changes in temperature, record the temperature change
       that is read as the final temperature experiment
  5.Observation
From the experiments had been conducted, can provide the following observations:
T0 NaOH = 29, 2 oC
T0 HCl = 30,4 oC
T0 reaction = ½ (29,2 oC + 30,4 oC) = 29,8 oC
T1 reaction = 30,6 oC
T rection     = ( T1 – T2)
               = (30,6 oC – 29,8 oC)
               = 0.8 o
  6.Calculation
Calculating the volume of HCl that would be diluted to 1liter 1M solution of concentrated
HCl solution 36,46% yield
Volume of HCl solution is 1000ml
   = mass of solution : Vol of solution
mass =  x Vol
     = 1,18 x 1000
     = 1180 gram (concentrated solution)
Percentage of HCl in solution is 36,46%, so..
Mass of HCl = 36,46% x 1180
              = 430, 228
M1 of HCl solution       = gr x        1000
                          Mr vol concentrated solution
                         = 430,228 x 1000
                             36,5    1000
                         = 11,787
mol1                     = mol2
M1xVol1                  = M2 x Vol2
11,787 x Vol1            = 1 x 1000
Vol1                     = 1000/11,787
                         = 84,84 ml
Calculating Hn of the neutralization reaction of HCl and NaOH
HCl 50ml, 1M = 0,05 mole
NaOH 50ml, 0,1M = 0,005 mole
 HCl + NaOH            NaCl + H2O
 0,05     0,005
 0,005    0,005
The mole that used for calculating is 0,005
Mass of HCl      =  x volume
                 = 1 x 100
                 = 100
qsolution = m x c x t
          = 100 x 4,2 x 0,8
          = 336 J
qreaction = -qsolution = -336 J (in 0,045 mol)
so, qreaction in 1 mol is...
1mol qreaction = 1/0,005 mol x (-336)
                   = -67200 J
                   = -67,2 KJ
Hn = -67,2 KJ
   7.Conclution
From the experiments and calculations have been done, we find that Hn of the neutralization
reaction of NaOH and HCl is -67,2 KJ
                                  Reference
http://id.wikipedia.org/wiki/Hukum_Hess
http://ruzdeeweb.com/search/praktikum+kalor+kalorimeter+dan+perubahan+entalpi+suatu+re
aksi
http://muniri.com/?s=entalpi+reaksi+dengan+kalorimeter
http://www.chem-is-try.org/materi_kimia/kimia-smk/kelas_x/entalpi-dan-perubahan-entalpi/
http://kimia-asyik.blogspot.com/2009/12/penentuan-entalpi-reaksi.html
http://en.wikipedia.org/wiki/thermochemistry