3 h2o2 molarity

Experiment 2 Decomposition

of Hydrogen Peroxide The decomposition of hydrogen peroxide in aqueous solution proceeds very slowly. A bottle of 3% hydrogen peroxide sitting on a grocery store shelf is stable for a long period of time. The decomposition takes place according to the reaction below.

2 H2O2(aq) → 2 H2O + O2(g)

A number of catalysts can be used to speed up this reaction, including potassium iodide, manganese (IV) oxide, and the enzyme catalase. If you conduct the catalyzed decomposition of hydrogen peroxide in a closed vessel, you will be able to determine the reaction rate as a function of the pressure increase in the vessel that is caused by the production of oxygen gas. If you vary the initial molar concentration of the H2O2 solution, the rate law for the reaction can also be determined. Finally, by conducting the reaction at different temperatures, the activation energy, Ea, can be calculated.

OBJECTIVES In this experiment, you will

· Conduct the catalyzed decomposition of hydrogen peroxide under various conditions. · Calculate the average rate constant for the reaction at room temperature. · Determine the rate law expression for the reaction. · Calculate the activation energy for the reaction.

The rate law for this reaction can be determined using the observed rates of reactions from a series of different experiments. The concentration of one reactant is held constant between two different experiments, acting as the control, while the concentration of the second reactant is different between the two experiments. The rate of reaction is measured in each experiment so the impact of changing the concentration of the second reactant can be determined. The order of the reaction with respect to each reactant is determined in this fashion and once the order of each reactant is know the rate law can then be written.

(A) Sample Exercise for Determining Reaction Order

Consider the following reaction: (CH3)3CBr(aq) + OH-(aq) → (CH3)3COH(aq) + Br-(aq) A series of experiments is carried out with the following results:

Exp 1 Exp 2 Exp 3 Exp 4 Exp 5 [(CH3)3CBr] 0.50 1.0 1.5 1.0 1.0 [OH-] 0.050 0.050 0.050 0.10 0.20 Rate (M/s) 0.0050 0.010 0.015 0.010 0.040

Find the order of the reaction with respect to both (CH3)3CBr and OH-.

Advanced Chemistry with Vernier 1

2 Advanced Chemistry with Vernier

To find the order of the reaction with respect to (CH3)3CBr, choose two experiments, 1 and 3 for example, where [OH-] is constant. A similar approach can be used to find the order of the reaction with respect to OH-, comparing experiments 2 and 5, where [(CH3)3CBr]

(1) Order with Respect to (CH3)3CBr:

Rate exp 3 = k([(CH3)3CBr]exp3)m ([OH-]exp3)n Rate exp 1 k([(CH3)3CBr]exp1)m ([OH-]exp1)n

0.015 M/s = k([1.5 M])m (0.050 M)n simplifies to: 3.0 = (3.0)m 0.0050 M/s k([0.50 M])m (0.050 M)n

Using the natural log applied to both sides and solving for “m”

Ln 3.0 = (Ln 3.0) (m) (Ln 3.0) / (Ln 3.0) = 1.10 / 1.10 = 1 = m

Since “m” = 1 the reaction is first order with respect to (CH3)3CBr.

(2) Order with Respect to [OH-]:

Rate exp 5 = k([(CH3)3CBr]exp5)m ([OH-]exp5)n Rate exp 4 k([(CH3