The Rate Law in Chemical Kinetics PDF

Preview

Summary

The Rate Law in Chemical KineticsI. IntroductionThe rate of a reaction is the change in concentration of a reactant or a product per unit of time. The goal of this experiment was to determine the rate law, which only takes into account the reactants, and the constant k of the following reaction:2 I(...

Description

The Rate Law in Chemical Kinetics Experiments #8

Lab section: 20 202-NYB-05

Date the experiment was performed: Thursday, April 12 th 2018 Date of lab report submission: Thursday, April 19 th, 2018

I. Introduction The rate of a reaction is the change in concentration of a reactant or a product per unit of time. The goal of this experiment was to determine the rate law, which only takes into account the reactants, and the constant k of the following reaction: (aq) (aq) 2−¿ 2 SO ¿ 2−¿ O ¿ I 2(aq)+¿ −¿+S 2 ¿ ¿ 2 I (aq) 2−¿ n −¿ ] Rate = k [ I ¿ ]m[ S 2 O¿8 It is not possible to determine the rate law of this reaction without doing an experiment. In 2−¿ this lab, the method of initial rates was used. By maintaining the concentration of S 2 O 8¿ −¿ constant for the first five runs, the order of the reaction (m) with respect to I ¿ was found. −¿ ¿ remained constant and the concentration For the last four runs, the concentration of I 2−¿ ¿ of changed. S 2 O8 If starch is added only to the previous reaction, the color change would happen too fast. Therefore, a second reaction, which occurred at the same time as the other one, was used: (aq) (aq) 2−¿ S 4 O¿ −¿+¿ 2−¿ ¿ O ¿ I 2(aq) → 2 I (aq) 2 S2 ¿ In this reaction,

2−¿ S 2 O¿3

used up the I 2

; thus, delaying the apparition of the deep blue

2−¿ was consumed, the mixture turned blue. Since its initial S 2 O¿3 concentration could be calculated from the values already known, it was possible to determine color. When all of the

the initial rate of the reaction with the following formula: 2−¿ S2 O¿3 ¿ ∆¿ −1 R0 = ∙¿ 2

¿ ∆ t=time taken for the mixture ¿ change ¿

II. Procedure Refer to the Laboratory Manual.

III. Data and Results Table 1: Reagent Concentrations Reagent

Concentration (mol/L)

KI

0.180

Na 2 S2 O 3

0.0036

(NH 4 )2 S 2 O 8

0.180

Table 2: Time taken for each run Run

Time (s)

1

26.54

2

30.28

3

36.66

4

44.45

5

60.93

6

28.62

7

33.44

8

40.05

9

67.89

Table 3: Initial Concentrations −¿ ¿ I ¿ (mol/L) ¿ ¿

Run

1

0.0720

−¿ ¿ I Ln ¿ ¿ ¿

2−¿ S 2 O¿8 (mol/L) ¿ ¿ ¿

2−¿ S 2 O¿8 Ln ¿ ¿ ¿

-2.631

0.0720

-2.631

2

0.0600

-2.813

0.0720

-2.631

3

0.0480

-3.037

0.0720

-2.631

4

0.0360

-3.324

0.0720

-2.631

5

0.0240

-3.730

0.0720

-2.631

6

0.0720

-2.631

0.0600

-2.813

7

0.0720

-2.631

0.0480

-3.037

8

0.0720

-2.631

0.0360

-3.324

9

0.0720

-2.631

0.0240

-3.730

Table 4: The Initial Rates and the k Run

R0

Ln R0

k ( L/mol ∙ s )

1

0.000014

-11.21

0.0026

2

0.000012

-11.34

0.0028

3

0.0000098

-11.53

0.0028

4

0.0000081

-11.72

0.0031

5

0.0000059

-12.04

0.0034

6

0.000013

-11.28

0.0029

7

0.000011

-11.44

0.0031

8

0.0000090

-11.62

0.0035

9

0.0000053

-12.15

Table 5: Temperature and Average K Temperature (K)

295.5

Average K (mol/L ∙ s)

0.0030

IV. Sample Calculations Run 1 1) Number of moles of M=

2−¿ S 2 O 3¿

n V

n 0.0150 n=0.000054 mol

0.0036=

2) Concentration of M=

n V

2−¿ ¿ S 2 O3

0.0031

0.000054 0.0750 M =0.00072 mol / L M=

3) Concentration of

2−¿ reacted S 2 O¿8

2−¿ S2 O8¿ ¿ S ¿ 2−¿ ¿ 2O¿3 ¿ ¿ ¿ 2−¿ S 2 O¿8 ¿ ¿ 2−¿ ¿ S 2 O8 ¿ ¿ 4) Initial reaction rate 2−¿ ¿ S2 O 3 ¿ ∆¿ 1 R0 = ∙ ¿ 2 1 0.00072 R0 = ∙ 2 26.54 R0=0.000014 mol/ L ∙ s

5) Initial concentration of

−¿¿ I

M 1 ∙ V 1=M 2 ∙ V 2 0.180 ∙30.0 =M 2 ∙ 75.0 M 2=0.0720 mol / L 6) Initial concentration of

2−¿ ¿ S 2 O8

M 1 ∙ V 1=M 2 ∙ V 2 0.180 ∙30.0 =M 2 ∙ 75.0 M 2=0.0720 mol / L 7) k value 2−¿ 1 −¿ ¿ ] Rate = k [ I ¿ ]1[ S 2 O8 0.000014 =k ∙ 0.0720 ∙ 0.0720 k =0.0026 L/mol ∙ s

8) Average k average k=

0.0026 + 0.0028+ 0.0028 + 0.0031 + 0.0034 + 0.0029 + 0.0031 + 0.0035 + 0.0031 9

average k=0.0030 L/mol ∙ s

V. Conclusion The objective of this experiment was to determine the rate law and the constant of the following reaction: ( aq) ( aq) 2 SO2−¿ ¿ O 2−¿ I ¿ 2(aq)+ ¿ −¿+ S 2 ¿ ¿ 2 I (aq)

2−¿ −¿ It was achieved since the rate law and its constant were: Rate = k [ I ¿ ]1[ S O ¿ ]1. The 2 8 orders were found by taking the slope of the graphs. For this lab, there were some sources of error. For example, when the solution of flask 2 was poured into flask 1, it is likely that not all of the 30 mL got in the first flask, since spilling occurred. Therefore, next time, a beaker should be used instead of flask 1. Since the beaker has a larger opening, it is less probable for spilling to occur. In addition, there was always a delay between the moment the solution turned dark blue and the reaction time to record the time. This delay added more seconds to the times recorded....