DETERMINATION OF REACTION RATE AND REACTION RATE CONSTANT

A. PURPOSE

The purpose of this experiment is to show that the reaction of saponification of ethyl acetate by hydroxide ions.

CH3COOC2H5 + OH                                   CH3COO + C2H5OH

Is a second-order reaction. In addition, it also determined the reaction rate constants. This determination is done by titration methode.

 

B. BASIC THEORY

To determine the rate of a given chemical reaction, it should be determined how fast changes occur in the concentration of reactants or products. In general, if a reaction occurs A → B, the initial substance A and substance B did not exist. After some time, the concentration of B will increase while the concentration of A will decrease. Rate law can be determined by conducting a series of systematic experiments on the reaction A + B → C, to determine the reaction order with respect to A the concentration of A is fixed while B concentrations varied then determined the rate of the reaction on the concentration variation. As for determining the order of the reaction B, the concentration of B is fixed while the concentration of A varied then measured the rate of the reaction on the concentration variation (Partana, 2003: 49)

Order of a reaction describes the mathematical form in which the results of the changes can be demonstrated. Reaction order can only be calculated in experiment and can only be predicted if the reaction mechanism is known throughout the reaction order can be determined as the sum of the exponents for each of the reactants, whereas only exponent for each reactant known as the order of the reaction to that component. Order of reaction is the sum of the rank of the concentration factor in legal rate differential form. In general, the order of the reaction to a particular substance is not the same as the stoichiometric coefficients in the reaction equation (Hiskia, 2003).

Although the reaction of saponification of ethyl acetate by reaction of the hydroxide ion is not simple, but it turns out that this reaction is a second-order reaction with a rate law of the reaction can be given as,

 

 

Or as,

 

With,   a = initial concentration of ester, in moles liter-1

             b = initial konsentarasi OH-ions, in moles liter-1

             x = number of moles liter-1 ester or base that has reacted

             k1 = reaction rate constant

 

Both of the above equation applies to the reaction conditions that are not too close to the equilibrium state. The following equation can be integrated with respect to various initial conditions:

 

aA + bB → product, where a ≠ b dan [A]o ≠ [B]o

 

 

differential rate equation is

 

 

Which can be rearranged into :

 

 

Or,

 

If a = b and [A]o = [B]o

 

 

That can be intrgeated to become :

 

Or,

 

This last equation reveals that the brook x / a (a-x) against t is a straight line with a slope equal direction k1 premises.

In the determination of his net reaction is followed by the method of determining the concentration of OH ion at a given time is to take a certain amount of solution, and then into a solution containing excess acid. Neutralization of the bases in the reaction mixture by acid to stop the reaction. The amount of base present in the reaction mixture during the reaction dihentukan, can be known with the rest titrate acid by standard alkaline solution (Wahyuni,2012).

 

C. . EQUIPMENTS AND SUBSTANCES USED

              Equipments :

  1. Measuring flask 250 ml                            2 pc
  2. Pippete volume 20 ml & 10 ml                1 pc
  3. Erlenmeyer 250 ml                                   6 pieces
  4. Burette 50 ml                                           1 pc
  5. Spray bottle                                              1 pc
  6. Stopwatch                                                1 pc

Substances :

  1. Ethyl acetate
  2. NaOH solution 0,02 M, 150 mL
  3. HCl solution 0,02 M, 150 mL
  4. Methyl orange indicator

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

D.  PROCEDURE

a)     

       
     
   
 

 

 

 

 

                                                           

 

 

 

 

 

 

b)    Insert 20 mL 0.02 M HCl solution into 6 units Erlenmeyer

 

 

c)    Titration

 
   

 

 

 

 

 

 

 

 

 

 

 

 

d)    Do the same thing (repeat step 3 until 5 times) for the reaction mixture after 9, 20, 28, 40, and 65 minutes.

e)    Remainder mixture solution

 
   

 

 

 

 

 

 

 

 

E.  RESULT AND DISCUSSION

 

No Time (min) V HCl 0,02 M (ml) V NaOH 0,02 M (ml) V mix (ml)
1 3.01 20 16,5 10
2 9.00 20 14.5 10
3 20.00 20 13 10
4 28.00 20 12.5 10

 

Table of t , x, a, a-x and x/a(a-x)

Time (s)

VNaOH

V HCl

V Mix

x

a

 

a-x

x/a(a-x)

181

16,5

20

10

0,0011

0,01

 

0,0089

12,359

540

14.5

20

10

0,00096

0,01

 

0,00904

10.619

1200

13

20

10

0,00086

0,01

 

0,00914

9.409

2403

12.5

20

10

0,00083

0,01

 

0,00917

9.051

 

 

 

 

 

 

 

 

 

Figure. Graphic of x/a(a-x) against t ( minutes)

 

 

 

 

 

DISCUSSION

           Reaction order is the power of the component concentrations in the rate law. Order of a reaction describes the mathematical form in which the results of the changes can be demonstrated. Reaction order can only be calculated in experiment and can only be predicted if the reaction mechanism is known throughout the reaction order can be determined as the sum of the exponents for each of the reactants, whereas only exponent for each reactant known as the order of the reaction to that component. In general, the order of the reaction to a particular substance is not the same as the stoichiometric coefficients in the reaction equation (Hiskia, 2001).

Reaction kinetics of  ethyl acetate saponification is studied by measuring the concentration of hydroxide ions for reaction  progress. Hydroxide ion concentration was measured in two ways, they are conductometric and titration. Conductometric is based on the measured conductivity of a solution. At the fixed temperature conductivity value of a solution depends on ion concentration in solution. Titration based on a particular analyte and titrant number of mutually neutralize each other. Titration end point is marked by a change in color to suit the indicators used in the titration.

In this experiment we  just use Titration method.

A.Top of Form

            We will determine the value of the order reaction between ethyl acetate and NaOH. The reaction is:

CH3 COOC2 H+ OH → CH3COO + C2H5OH

 

This is the saponification reaction and the second order reaction. We will find the value of constant and the graphic.

We use the formula of a =b with the volume of NaOH 150 ml and the volume of ethyl acetate 150 ml. The erlenmeyer must be covered in order to keep the solution evaporate and prevent it contact with air.

Before we mix both of the solution, the temperature must be same (thermostate) to keep the solution when it is mixed and begin with the same initial reaction rate.

When we mix the solution, it must be shaken well to make the solution homogenous. Then, we wait until it reaches the variation of time. After that, we pour it into the erlenmeyer ( consist of 20 ml HCl 0.02 M) .

 

CH3CHOO–  + H+ → CH3COOH

 

HCl function’s is to stop the saponification reaction, then we titrate the excess of HCl with NaOH 0.02 M.

 

H+  (excess) + OH → H2O

 

     Before performing titrations, we add methyl-orange indicator (mo). Actually, in the titration of a weak acid with a strong base, mo indicator is useless. Since the titration of a weak acid and a strong base produces a pH greater than 7, while the indicator pH 3.1 to 4.4 mo had a route. Should mo red indicator will turn yellow when the titration reaches the equivalence point, but in the experiments we did indicator mo until colorless.

 

We get the volume of NaOH that is needed, to titration. After we get it we can calculate the value of x. x is number of ester or base which has reacted at time (t) ( look at the analyzed data). Based on the analyzed data, we get the value of a/a(a-x) of each the variation of time.

 

 = k1 t

 

Then we plot a/a(a-x) versus t. From the graph we can get the value of the constant from slope. The constant k1 = 0.008

 

F. CONCLUSIONS AND SUGGESTIONS

Conclusions

  1. The saponification reaction of:

CH3COOC2H+ OH → CH3COO + C2H5OH

 is a second order reaction with the value of constant (k) is 0.008

  1. The longer mixing time the more the volume of NaOH required for titration.
  2. The order of reaction can be determined by experiment.

 

Suggestions

  1. We must stir the mixture to be homogeneus solution.
  2. When we titrate the mixture and the excess HCl with NaOH , we must titrate it carefully.
  3. We should stop the titration precisely.

 

G. REFERENCES

Wahyuni, Sri, dkk.2011.Practicum Guide of Physical Chemistry. Departmen of Chemistry FMIPA UNNES: Semarang

Soeprodjo. Kimia Fisika. FMIPA IKIP Semarang: Semarang

http://id.scribd.com/doc/87426970/Kimia-Fisika-III-Kinetika-Reaksi-Saponifikasi-Etil-Asetat

http://ml.scribd.com/doc/51198868/KIMIA-FISIKA-II

 

 

                                                                                           Semarang, 8 Oktober 2012

Dosen Pengampu,                                                                          Praktikan,

 

 

Ir. Sri Wahyuni, M.Si.                                                                 Falasifah Aulia

 

 

 

ANSWER QUESTION

1.  In chemical kinetics, the order of reaction with respect to certain reactant is defined as the index, or exponent, to which its concentration term in the rate equation is raised.

For example, given a chemical reaction 2A + B → C with a rate equation

r = k[A]2[B]1

2. The stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients can be fractions, whole numbers are frequently used and often preferred. This stoichiometric coefficients are useful since they establish the mole ratio between reactants and products.

3. Saponification reaction is a second order reaction. It can be seen from the provisions of the reaction  M-1minutes-1. Reaction rate constants can not be determined theoretically but we have to go through trials.

Saponification reaction above is a second order reaction when seen from the reaction takes place where there are 2 products that are produced and do not have coefficients that can be said that this is a second-order reaction

  1. Conductivity types: Ohm-1cm-1 (ῼ cm-1)
    Molar conductivity:  s m2 mol-1,  s cm2 mol-1

5. If the titration of HCl are taken then the temperature will drop and the mixture of substances affect the outcome of the reaction rate constants.
So the temperature of the substance must be kept constant during the titration in order
If titration is delayed, the temperature must be raised to

6. To determine the order reaction :

 1. Looking units of the reaction rate constant.
 2. Comparing the half.
 3. Comparing the two equations known reaction rate data.

7. Based on the Arrhenius equation says that:

     k = Ae –Ea/RT

     Where k is the rate constant of the reaction so that the activation energy is determined from the gradient graph results brook ln k against 1 / T. Activation energy is the minimum energy needed to carry out the reaction of a reagent.
Activation energy prices will be reduced by the addition of a catalyst.

 

 

APPENDIX

 

      ANALYSIS DATA

a=b    ( 150 ml NaOH : 150 ml CH3COOC2H5)

find the value of a and b

a=

a=

a= 0.01 M

b=

b=

b=0.01 M

find the value of x:

CH3COOC2H5 : 0.02 M x 0.15 L = 0.003 mol

OH : 0.02 M  x 0.15 L = 0.003 mol

CH3COOC2H+   OH              →    CH3COO  +   C2H5OH

I:   0.003  mol       0.003 mol

R: 0.003  mol        0.003 mol            0.003 mol        0.003 mol

       E:     –                         –                    0.003 mol        0.003 mol

Take 10 ml mixture to erlenmeyer 0.02 L HCl 0.02 M

 

     CH3COO  +        H+              →    CH3COOH

I:  0.003 mol           0.0004 mol     

R: 0.00007 mol       0.00007 mol        0.00007 mol

E; 0.00003 mol       0.00033 mol        0.00007 mol

Result from titration is:

H+ + OH  →   H2O

      [HCl] excess x VHCl =  [NaOH] x VNaOH

nHCl    = 0.02 N x 0.0165 L

nHCl    =  0.00033  mol

( back to the above calculation )

x  (t= 3.01 minute )       = 0.00033 mol : 0.3 L

                                      = 0.0011  M

with the same formula, so we will get:

x (t= 9.00 minute)       = 0.00029 : 0,3 L

                                     = 0,00096 M

x (t=20.00 minute)      = 0.00086  M

x (t= 28.00 minute)     = 0.00083  M

 

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