Decomposing Hydrogen Peroxide Essay

1987 words - 8 pages

Introduction
The purpose of this experiment is to gain an understanding of factors which influence the rates of reactions. Several experiments will be conducted using various combinations of concentrations of the reactant (hydrogen peroxide) and a catalyst (potassium iodide), as well as one pairing in which a change in temperature is the only variable. Data collected will be used to determine the rate law, rate constant, and activation energy for each of the reactions Hydrogen peroxide (H_2 O_2) naturally decomposes into water (H_2 O) and oxygen gas (O_(2 )) according to the following balanced equation.
2 H_2 O_2 (aq) ↔ 2 H_2 O (l)+O_(2 ) (g)
Normally this reaction takes a long period of time, but the rate of reaction can be increased with the addition of a catalyst, which lowers the activation energy needed for a reaction to begin, but is not consumed in the reaction. In this experiment, potassium iodide (KI) will be used:
2H_2 O_2 (aq) □( □(↔┴KI )) 2 H_2 O (l)+O_(2 ) (g)
It is expected that the reaction will not occur at a significant rate until the addition of the catalyst.
The rate of a reaction can be expressed as the change in concentration of either reactants or products divided by the change in time, and will decrease as the reaction progresses(Chang 558).
rate= -1/2 (∆ [H_2 O_2])/(∆ t)= 1/2 (∆ [H_2 O])/(∆ t)= (∆ [O_2])/(∆ t)
A reaction’s rate is proportional to the concentration of the reactant, and the value of this ratio (called the rate constant, k) remains the same throughout the reaction, regardless of concentration. Only a change in temperature will change the rate constant for a given reaction. It is hypothesized that an increase in temperature will result in an increase in the value of k. The general rate law shows this relationship as follows:
rate= k〖[H_2 O_2 ] 〗^x 〖[KI] 〗^y
The concentration of each reactant is raised to a power which is determined experimentally and the sum of these powers determines the overall order of the reaction. If the concentration of a reactant is doubled and the initial rate also doubles (21), the reaction is first order with respect to that reactant, and its exponent is 1. If the concentration of a reactant is doubled and the initial rate quadruples (22), the reaction is second order with respect to that reactant, and its exponent is 2. If the concentration of a reactant is doubled and the initial rate increases eight-fold (23), the reaction is third order with respect to that reactant, and its exponent is 3. Should the change in concentration not affect the rate, the reaction is not dependent on the concentration of that reactant, and its order and exponent are both 0 to reflect this (Chang 566-567).
Because the solutions will be provided in just one concentration each, the effective doubling of reactant concentrations will actually be the result of dilutions, with the new solutions prepared using the dilution formula: M1V1 = M2V2.
In one pairing of...

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