chem 17 expt 3

CHEMICAL KINETICS

Z.C. G. HACHERO

DEPARTMENT OF MINING, METALLURGICAL, AND MATERIALS ENGINEERING, COLLEGE OF ENGINEERING

UNIVERSITY OF THE PHILIPPINES, DILIMAN, QUEZON CITY, PHILIPPINESDATE PERFORMED: FEBRUARY 5, 2015INSTRUCTOR’S NAME: LAWRENCE JOHN PAULO TRINIDAD

METHODOLOGY

There were three parts in the experiment, namely; (1) Effect of Concentration on the Rate of Reaction, (2) Effect of Temperature on the Rate of Reaction and (3) Catalysis.

For the first part, a 50-mL beaker was marked with an “X” on the bottom. 6 different runs were done, each with different volumes and molarity of Na2S2O3, HCl, and water.

For runs 1 to 3, required volumes of 0.15 M Na2S2O3 and water was mixed in the beaker. The required volume of 3.0 M HCl, was then added and the time from the moment the acid until the X mark disappears is recorded. For runs 4 to 6, the same procedures were applied, with the difference of adding only water and 3.0 M HCl to the beaker, and 0.15 M Na2S2O3 was added instead of HCl.

Table 1. Required volumes of solution per run

RUN Volume of 0.15 M Na2S2O3 (mL)

Volume of water

Volume of 3M HCl, mL

1 10.0 3.0 2.02 5.0 8.0 2.03 2.5 10.5 2.04 5.0 1.0 1.55 5.0 1.5 1.06 5.0 2.0 0.5

For the second part, run 3 from part 1 was subjected to three temperatures: once in an ice bath, one at ambient temperature and another at elevated temperature. The required amount of 0.15 M Na2S2O2 and 50-

mL water, was added to the beaker with the “X” mark. The 50-mL beaker and test tube was then placed in a 500-mL beaker at the desired temperature. HCl was added to the solution, while the beaker was in the water bath. The time from the moment the acid is added until the “X” mark was no longer visible, was recorded.

The third part had two further divisions. The first was the oxidation of Tatrate by Hydrogen Peroxide. The first step was the preparation of hot water bath, maintained with a temperature of 65°C. To two 6-inch test tubes, 5mL 0.3 M sodium tartrate and 2 mL 6% H2O2 was added, and to test tube 2, an additional 8 drops of 0.3 M CoCl2 was added. Both test tubes were placed in the hot water bath, and observed for color changes in the solutions.

The second part was the reaction of oxalate with permanganate. To a 4-inch test tube, 1 mL saturated Na2C2O4, 5 drops of water, and 1 mL 3.0 M H2So4 was mixed. The mixture was then split into equal portions in separate test tubes. Test tube 1 was added with a drop of 0.01 M KMnO4, and the time it took for the permanganate to decolorize was recorded. A second drop of 0.01 M KMnO4 was added to test tube 1, and again record the time it took permanganate to decolorize. To Test tube 2, 5 drops of 1% MnSO4 was added and the solution was mixed thoroughly. A drop of 0.01 M KMnO4 was added, and the time it took for permanganate to decolorize was recorded.

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RESULTS AND DISCUSSION

Effect of Concentration on the Rate of Reaction

The balanced chemical equation for the reaction under study is:

Na2S2O3(aq) + 2HCl(aq) 2NaCl(aq) + S(s) + SO2(g) + H2O(l) (1)

with the net ionic equation:S2O3

2-(aq) + 2H+

(aq) S(s) + SO2(g) + H2O(l) (2)

As the reactants were mixed in the beaker, which contained specific volume of and solution, precipitates were formed. As the precipitates continued to develop, the time it took for it to visually cover the mark “X” under the beaker was measured. The results were then tabulated as shown below.

Table 2. Experimental rate data from the reaction of Na2S2O3

and HCl

RUN [S2O32-]init [H+]init TIME (s) 1/t (s-1)

1 0.1 M 0.40 M 29.71 0.0342 0.05 M 0.40 M 68.90 0.0153 0.025 M 0.40 M 388.41 0.00034 0.1 M 0.60 M 33.70 0.0305 0.1 M 0.40 M 55.69 0.0186 0.1 M 0.20 M 55.83 0.018

For Runs 1 to 3, from the data it could be incurred that the concentration is directly proportional to the rate of reaction. When the concentration of S2O3

2 is increased, the rate of reaction also increased which also indicated less time needed to form enough precipitate to cover the X mark. For Runs 4 to 6, changing the concentration of HCl did not make a significant change in reaction rate.

The order of reaction with respect to [S2O32-]

approaches the value of 1. On the other hand, the order of reaction of [H+] approaches 0, meaning the initial rate is independent of the concentration of H+. The overall kinetic order

of reaction is 1, the sum of the order of reactions of [S2O3

2-] and [H+].

The possible sources of errors encountered in the experiment: for each run it is necessary to clean the beaker used. However, doing this increases the chance of acquiring a higher volume if the glassware used was not properly dried. This may result to smaller concentration which leads to a smaller rate of reaction.

Effect of Temperature on the Rate of Reaction

As the temperature of the set-up increases, the faster a reaction will end. This can be explained through the Arrhenius equation:

k = A e-Ea/RT (3)which implies that rate of reaction is directly proportional with temperature.

Since the reaction is in first order, we assume 1/T to be the rate constant because initial concentrations are constant while the temperatures are varied.

-3.47 -5.3 -7.030

0.001

0.002

0.003

0.004

ln (1/time)

1/te

mpe

ratu

re (

1/K

)

Figure 1. Effect of temperature on reaction rate.

The graph shows a linear relation between the inverse of temperature vs the natural logarithmic of the inverse time. As it takes the equation of a line, y = mx + b (where y = ln k, m = (Ea)/R, and b = ln A), we can determine that activation energy Ea is . The activation energy is the minimum required energy to start a reaction.

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Catalysis

The third part of the experiment is further divided into two parts. For the first part, our balanced equation is:

C4H4O6 2– + 5H2O2 → 4CO2 + 6H2O + 2OH–

without the catalyst. (4)

CoCl2 is the catalyst. The action of the cobalt catalyst can be followed by observing the color changes of the solution. The solution starts out pink and the mixture then quickly turns green. The production of effervescence is due to oxidation of the tartrate ions occurs almost immediately after the green color has been observed. As the tartrate ions are consumed and the amount of gas production subsides, the color of the solution returns to the original pink color. When all the tartrate has been consumed, the color of the solution reverts back to pink, indicating that only Co(II) ions are present— and that the cobalt chloride catalyst is not used up in the reaction. [3]

For the second part, our balanced equation is:

5C2O42- + 2MnO4

- + 16H+ 10CO2 + 8H2O + 2Mn+ (5)and its balanced equation in an acidic médium is:

2MnO4- + 5C2O4

2- + 6H3O+ 2Mn2+ + 10CO2

+ 8H2O (6) with Mn as the catalyst.

A strong acid is needed in part 2 so that MnO4

could be converted to Mn. It involves autocatalysis which means that the reaction produces the compound that is used as a catalyst for the second reaction. The reaction has ‘ended’ if we see the decolorization of the solution. [4]

The presence of a catalyst, either by manual addition or by the production of the reactants themselves lowers the activation energy of the reaction. By the Arrhenius equation, we

know that lower activation energy means a faster reaction rate.

REFERENCES

[1] Petrucci, R., Herring, F., Madura, J., & Bissonnette, C. General Chemistry: Principles and Modern Applications, 10th ed.; Pearson: Canada, 2011.

[2] Chang, R. General Chemistry: Essential Concepts, 6th ed.; McGraw-Hill: New York, 2011.

[3]The Pink Catalyst; http://www.flinnsci.com/media/395440/cf0255.01.pdf (accessed February 9, 2015)

[4] Autocatalysis: Reaction of Permanganate with Oxalic Acid; http://www.chemeddl.org/alfresco/service/org/chemeddl/video.html?options=false&ID=vid:691&guest=true (accessed February 9, 2015)

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http://www.chemeddl.org/alfresco/service/org/chemeddl/video.html?options=false&ID=vid:691&guest=true



http://www.flinnsci.com/media/395440/cf0255.01.pdf



APPENDIX

Run 1: M2(S2 O3 2-¿¿=M1( S2 O32-¿V 1(S2 O3 2-¿¿

¿V2(S2 O3 2-¿

¿

M2(S2 O3 2-¿¿=(0.15 M)(0.01L)(0.015 L)

M2(S2 O3 2-¿¿= 0.10 M

M2(H+ ¿¿=M1 ( H+¿V 1 (H+ ¿¿¿

V 2(H+¿

¿

M2(H+ ¿¿=(3.0M ¿(0.002L) ¿(0.015 L)

M2(H+ ¿¿= 0.040 MRun 2: M2(S2 O3 2-¿¿=M1( S2 O32-¿V 1(S2 O3 2-¿¿

¿V2(S2 O3 2-¿

¿

M2(S2 O3 2-¿¿=(0.15 M)(0.005L)(0.015 L)

M2(S2 O3 2-¿¿= 0.050 M

M2(H+ ¿¿=M1 ( H+¿V 1 (H+ ¿¿¿

V 2(H+¿

¿

M2(H+ ¿¿=(3.0M ¿(0.002L) ¿(0.015 L)


¿V2(S2 O3 2-¿

¿

M2(S2 O3 2-¿¿=(0.15 M)(0.0025L)(0.015 L)

M2(S2 O3 2-¿¿= 0.025 M

M2(H+ ¿¿=M1 ( H+¿V 1 (H+ ¿¿¿

V 2(H+¿

¿

M2(H+ ¿¿=(3.0M ¿(0.002L) ¿(0.015 L)


¿V2(S2 O3 2-¿

¿

M2(S2 O3 2-¿¿=(0.15 M)(0.005L)(0.075 L)

M2(S2 O3 2-¿¿= 0.1 M

M2(H+ ¿¿=M1 ( H+¿V 1 (H+ ¿¿¿

V 2(H+¿

¿

M2(H+ ¿¿=(3.0M ¿(0.0015L) ¿(0.0075 L)


¿V2(S2 O3 2-¿

¿

M2(S2 O3 2-¿¿=(0.15 M)(0.005L)(0.0075L)

M2(S2 O3 2-¿¿= 0.1 M

M2(H+ ¿¿=M1 ( H+¿V 1 (H+ ¿¿¿

V 2(H+¿

¿

M2(H+ ¿¿=(3.0M ¿(0.001L) ¿(0.0075 L)


¿V2(S2 O3 2-¿

¿

M2(S2 O3 2-¿¿=(0.15 M)(0.005L)(0.015 L)

M2(S2 O3 2-¿¿= 0.050 M

M2(H+ ¿¿=M1 ( H+¿V 1 (H+ ¿¿¿

V 2(H+¿

¿

M2(H+ ¿¿=(3.0M ¿(0.00050 L) ¿(0.075 L)

M2(H+ ¿¿= 0.20 M

Rate 1 = k[S2O32- ] 1

m [ H + ] 1n

Rate 2 = k[S2O32-]2

m [H+]2n

Rate 4 = k[S2O32- ] 4

m [ H + ] 4n

Rate 5 = k[S2O32-]5

m [H+]5n

rate = k[S2O32- ] 1 [ H + ] 0

rate = k[S2O32-]

lnk = - EaR

( 1T

) + lnA

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chem 17 expt 3

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