The Rate Law for Chemical Reaction Among Hydrogen Peroxide, Iodide, and Acid

The Rate Law for Chemical Reaction Among Hydrogen Peroxide, Iodide, and Acid

To determine the rate law for a chemical reaction among hydrogen peroxide,

iodide and acid, specifically by observing how changing each of the

concentrations

Experiment 3 Chemical Kinetics

Objectives

1. To determine the rate law for a chemical reaction among

hydrogen peroxide, iodide and acid, specifically by observing how

changing each of the concentrations of H2O2, and H+ affects the rate

of reaction.

2. To observe the effects of temperature and catalyst on the rate

of reaction.

Introduction

Generally, two important questions may be asked about a chemical

reaction:

(1)How far do the reactants interact to yield products, and (2) how

fast is the reaction? “How far?” is a question of chemical equilibrium

which is the realm of chemical thermodynamics. “How fast?” is the

realm of chemical kinetics, the subject of this experiment.

In this experiment we will study the rate of oxidation of iodide ion

by hydrogen peroxide which proceeds according to the following

reaction:

H2O2 (aq) + 2 I-(aq) + 2H+(aq) I2(aq) + 2H2O(l)

By varying the concentrations of each of the three reactants (H2O2, I-

and H+), we will be able to determine the order of the reaction with

respect to each reactant and the rate law of the reaction, which is of

the form:

Rate = k [H2O2]x[I-]y[H+]z

By knowing the reaction times (†t) and the concentrations of H2O2 of

two separate reaction mixtures (mixtures A & B), the reaction order of

H2O2, x, can be calculated.

x = log(†t2/ †t1) / log ( [H2O2]1/[H2O2]2 )

The same method is used to obtain the reaction order with respect to I-

(mixtures A & C) and H+ (mixtures A & D).

Procedures

Part I) Standardization of H2O2 Solution

1. A stand, a burette clamp and a white tile were collected to

construct a titration set-up.

2. A burette was rinsed with deionized water and then with 0.05 M

Na2S2O3 solution.

3. The stopcock of the burette was closed and the sodium

thiosulphate solution was pour into it until the liquid level was near

the zero mark. The stopcock of the burette was opened to allow the

titrant to fill up the tip and then the liquid level was adjusted near

zero.

4. The initial burette reading was recorded in Table 1.

5. 1.00 cm3 of the ~0.8 M H2O2 solution was pipetted into a clean

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...te of a reaction by

providing an alternative pathway for the reaction, usually with a

pathway of lower activation energy than that of the uncatalyzed

reaction.

There are some improvements in this experiment.

First, hydrogen peroxide is unstable, it decomposes to water and

oxygen by time. Therefore do the titration as quick as possible.

2H2O2(aq) 2H2O(I) + O2(g)

Second, the concentration of iodine increase is due to the iodide can

be oxidized by oxygen which promoted by acids. Therefore do the

titration as quick as possible.

4I-(aq) + O2(g) + 4H+(aq) 2I2(aq) + 2H2O(aq)

Third, as for the human error, the problem can be minimized by

performing the titration by the same person. So, the reading can be

taken by the same person and the color change can be observed by the

same person.

Conclusion

In the experiment, the reaction was found to be zero order respect to

(H+), it is first order respect to iodide, (I-) , it is first order

respect to hydrogen peroxide, (H2O2). Hence the rate law is Rate = k[H2O2][I-].

The rate of reaction is increase when the temperature is increase and

the rate is increase when a positive catalyst is added to the

reaction.