DETERMINATION OF REACTION RATE AND REACTION RATE CONSTANT

PRACTICAL REPORT

DETERMINATION OF REACTION RATE AND REACTION RATE CONSTANT

A. PURPOSES

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

CH3COOC2H5 + OH-                                   CH3COO- + C2H5OH

is a second-order reaction. In addition, it is 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 are 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 times, 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 concentration 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 be just calculated in experiment and 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 ethyl acetate saponification 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 concentraton OH-ion, 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 are applied 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

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) versus t is a straight line with a slope equal direction k1 premises.

In the determination of this 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 presence in the reaction mixture during the reaction can be determined, can be known with the rest titrate acid by standard alkaline solution. (Wahyuni, 2011)

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

E. PROCEDURE

Erlenmeyer 250 ml

filled

150 ml ethyl asetic 0,02 M

Erlenmeyer 250 ml

filled

150 ml NaOH 0,02 M

Mixed them by shaken well

Both of their temperature are equaled

(thermostate)

Run the stopwatch

3 mins after being mixed, pippeted 10 ml and put into a 20 ml HCl 0.02 M

Stir and immediately titrate with standard solution 0,02 M NaOH with MO indicator

Result

Done taking back at 9, 20, 28, 40, 65 minutes

Remainder mixture solution

The remaining reaction mixture is heated for several minutes.

Do titration as in step 3

Record the data

F. RESULT AND DISCUSSION

Datas of Observation

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

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

CH3COOC2H5 + 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

Results from titration are:

H+ + OH- H2O

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

  1. nHCl = 0.02 N x 0.0165 L

nHCl = 0.00033 mol

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

= 0.0011 M

  1. nHCl = 0.02 N x 0.0145 L

nHCl = 0.00029 mol

x (t= 9.00 minutes ) = 0.00029 mol : 0.3 L

= 0.00096 M

  1. nHCl = 0.02 N x 0.013 L

nHCl = 0.00026 mol

x (t= 20.00 minutes) = 0.00026 mol : 0.3 L

= 0.00086 M

  1. nHCl = 0.02 N x 0.0125 L

nHCl = 0.00025 mol

x (t= 28.00 minutes) = 0.00025 mol : 0.3 L

= 0.00083 M

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

12.35955

540

14.5

20

10

0.00096

0,01

0.00904

10.61947

1200

13

20

10

0.00086

0.01

0.00914

9.40919

1500

12.5

20

10

0.00083

0,01

0.00917

9,05125

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

DISCUSSION

Reaction order is the power of the component concentrations in the rate law. Reaction kinetics of ethylacetatesaponification are studiedby measuringthe concentration ofhydroxideionsfor reaction progress.Hydroxideion concentrationwas measuredin two ways, they are conductometricand titration. Conductometric is based on the measured conductivity of a solution. At fixed temperature conductivity value of a solution depends on ion concentration in solution. Titration is based on a particular analyte and titrant number of mutually neutralizes each other. Titration end point is marked by a change in color to suit the indicators used in the titration.

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).

  1. Top of Form

In this experiment we will determine the value of the order reaction between ethyl acetate and NaOH by titration method. The reaction is:

CH3 COOC2 H5 + 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 evaporates and prevents it contact from 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.

We mix the solution and 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 (consists 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 N

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 got the volume of NaOH that is needed to titration. After we got 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. We can see the relation between x / (a(a-x)) and time increasing but at the other time is decreasing. It means there is a mistake in our practicum. Actually the reaction rate constant is increasing at each time. Because the validity is less, it make our graph is not in linier shape. The equation y=mx+c is on the graph. From this result we can say that the saponification is order 2 reaction. The constant k1 = 0.008

G. CONCLUSIONS AND SUGGESTIONS

Conclusions

  1. The saponification reaction of:

CH3COOC2H5 + 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.

H. 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, October 8, 2012

knowing,

Practicum Lecture Practitioner

Ir. Sri Wahyuni, M.Si Dissa Feby Octafianellis

NIP. NIM. 4301410031

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.

Saponificationreactionabove is asecond order reactionwhen seen fromthe reactiontakes placewherethere are 2products thatare produced anddo nothavecoefficientsthatcan be saidthat this is asecond-order reaction

  1. Conductivity types: Ohm-1cm-1 (ῼ cm-1)
    Molar conductivity: s m
    2 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 theArrheniusequationsaysthat:

k=Ae -Ea/RT

Wherekisthe rateconstant ofthe reactionso thatthe activationenergyis determinedfrom thegradientgraphresultsbrooklnkagainst1 /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.

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