Molarity of 3 hydrogen peroxide

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)3CBr]exp2)m ([OH-]exp2)n

0.040 M/s = k([1.0 M])m (0.20 M)n simplifies to: 4.0 = (2.0)n 0.010 M/s k([1.0 M])m (0.10 M)n

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

Ln 4.0 = (Ln 2.0) (m) (Ln 4.0) / (Ln 2.0) = 1.39 / 0.69 = 2.01 = n

Since “n” = 2 the reaction is second order with respect to (CH3)3CBr.

Now that the order of the reaction for each reactant has been determined the rate law can be written for this equation:

Rate = k [(CH3)3CBr] [OH-]2

(B) Sample Exercise for Determining Molarity of a Diluted Solution

When two solutions are mixed in an experiment the total volume of the solutions is increased and the concentration of each solution is diluted. Recall:

Moles solute before dilution = moles solute after dilution

so; (Molarity)conc x (Volume) conc = (Molarity)dil x (Volume)dil

The Decomposition of Hydrogen Peroxide

Advanced Chemistry with Vernier 3

What is the concentration of each compound when 6.0 mL of a 0.60 M solution of H2O2 is mixed with 2.0 mL of a 0.25 M potassium iodide solution?

(1) First consider the total volume of the solution after mixing:

6.0 mL H2O2 + 2.0 mL KI = 8.0 mL total volume of diluted solution

(2) Determine the concentration of H2O2 after mixing:

[0.75 M]conc x (6.0 mL) conc = [H2O2]dil x (8.0 mL)dil

(0.60 M)(6.0 mL) / (8.0 mL) = 0.45 M = [H2O2]dil

(3) Determine the concentration of KI after mixing:

[0.25 M]conc x (2.0 mL) conc = [H2O2]dil x (8.0 mL)dil

(0.40 M)(2.0 mL) / (8.0 mL) = 0.10 M = [KI]dil

The concentration of the H2O2 decreased from 0.60 M to 0.45 M and the concentration of KI decreased from 0.25 M to 0.10 M after the two solutions are mixed. The solutions will be diluted as the reactions begin so be sure to use the diluted concentrations of each reactant when determining the reaction orders or rate constants.

(C) Sample Exercise for Converting Pressure Rate Data to Molarity Units

The data rate recorded in this experiment is in kilopascals per second (kPa/s). Since pressure is a unit of concentration in the gas phase it can be used to determine rate when the change of pressure is measured over the change in time. However, in this experiment the reactants are in the aqueous phase and the concentration unit for the solutions is in molarity (M). kPa/s can be converted to M/s by using a derivation from the ideal gas law:

Molarity = P / RT

In which P = kPa/s, R = 8.314 L * kPa / K*mol (one of ideal gas constants), and T = temperature on Kelvin.

In one experiment the rate of O2 production was 0.22 kPa/s when the temperature of the water bath was 19.0 oC. What is this rate in M/s?

Molarity = (0.22 kPa/s) / (8.314 L*kPa/K*mol)(292.0 K) = (0.22 kPa / 2.43 x 103 L*kPa / mol)

Molarity = 9.05 x 10-5 M/s

The units kPa and Kelvin cancel leaving mol/L*s or M/s.