# Rates of Chemical Reactions II: Rate and Order of H2O2 Decomposition

Let's determine the rate and order of reaction for the decomposition of hydrogen peroxide.

You need this equipment: Mohr buret, leveling bulb, 125-mL Erlenmeyer flask, buret clamp, graduated cylinder, rubber tubing, test tube, medicine droppers (2), barometer, ring stand, iron ring, wristwatch with timer, thermometer, pneumatic trough, No. 00 and No. 6 one-hole stoppers, clay triangle

You need these materials: 3% H2O2, .10 M Hg(NO3)2, .10 M KI

The area of chemistry concerned with rates of chemical reactions is called chemical kinetics. Experiments show that rates of homogeneous reactions in solution depend on the nature of the reactants, the concentrations of the reactants, the temperature, and catalysis. Before a reaction can occur, the reactants must come into direct contact via collisions of the reacting particles. However, even then, the reacting particles must collide with sufficient energy to result in a reaction. If they do not, their collisions are ineffective.

Changing the concentration of a solution alters the number of particles per unit volume. The more particles present in a given volume, the greater the probability of their colliding. Increasing the concentration of a solution increases the number of collisions per unit time and therefore the rate of reaction.

Since temperature is a measure of the average kinetic energy, an increase in temperature increases the average kinetic energy of the particles. An increase in kinetic energy increases the velocity of the particles and therefore the number of collisions between them in a given period of time. It also results in a greater proportion of the collisions having the required energy for reaction. Thus, the rate of reaction increases. As a rule of thumb, for each 10° increase in temperature (Celsius or Kelvin), the rate of reaction doubles.

Catalysts, in some cases, increase reaction rates by bringing particles into close juxtaposition in the correct geometrical arrangement for reaction to occur. In other cases, catalysts offer an alternative route to the reaction, one that requires less energetic collisions between reactant particles. If less energy is required for a successful collisions, then a larger percentage of the collisions will result in a reaction. Although a catalyst may take place in a reaction, it can always be recovered at the end of the reaction, chemically unchanged.

Rate of reaction can be measured by observing the rate of disappearance of the reactants or the rate of appearance of the products. In practice, one measures the change of concentration with time of one of the participants in the reaction. Whatever species is most convenient to measure can yield a valid rate of reaction. In general, the rate of reaction will depend on the concentration of the reactants. Thus, the hypothetical rate of a reaction A + B → C + D can be expressed as: k[A]x[B]y, where [A] and [B] are the molar concentrations of A and B, x and y are the powers to which the respective concentrations must be raised to describe the rate, and k is the specific rate constant.

One of the objectives of chemical kinetics is to determine the rate law. The rate law is determined by determining the numerical values of the exponents in the rate equation. The powers to which the concentrations in the rate law are raised are termed the order of the reaction. For example, if the rate law is: rate = k[A]2[B], then the reaction is said to be second order in A and first order in B. The overall order of the reaction is the sum of the exponents; for example, the preceding rate law is a third-order reaction.

The specific rate constant, k, has a definite value that is independent of the concentration. It is characteristic of a given reaction and depends only on temperature. Once the rate is known, the value of k can be calculated.

In this experiment, the following reaction will be observed:

2H2O2 (aq) → O2(g) + 2H2O (l)

By observing the volume of oxygen formed as a function of time, once can determine how the rate of reaction is affected by different initial concentrations of hydrogen peroxide and iodide ion (which is introduced into the reaction in the form of the strong electrolyte KI). Although iodide does not appear in the chemical equation, it does have a pronounced effect on the rate of the reaction, for it serves as a catalyst.

The ultimate goal of the experiment is to deduce a rate law for the reaction, showing the dependence of the rate on the concentrations of H2O2 and I. The rate law will be of the following form:

rate of oxygen production = k[H2O2]x[I]y

The objective is to determine the numerical values of the exponents x and y.

Procedure:

A. Order of Reaction
1. The following apparatus was assembled using the labeled parts: (view) 2. A clay triangle was placed on the iron ring to support the leveling bulb. The trough was filled with room temperature water. The temperature of the water was recorded.
3. Room temperature water was added to the assembly until the height of water in the buret was approximately 10 mL from the top when the water in the leveling bulb was at the same level.
Solution 1
4. 10.0 mL of .10 M KI and 15.0 mL of distilled water were added to a clean Erlenmeyer flask. The flask was carefully swirled in the trough for several minutes so that the solution attained the bath temperature. 5.0 mL of 3% H2O2 was added to the flask. The flask was quickly stoppered.
5. One lab partner swirled the flask consistently as vigorously as possible throughout the experiment. The other lab partner observed the volume of oxygen produced during the reaction.
6. After approximately 2 mL of gas had been evolved, the recording of volume and time was commenced. The leveling bulb was brought to the level of water at the time of each reading. Time and volume readings were taken at approximately 2 mL intervals until 14 mL of oxygen had been evolved.
Solution 2
7. The Erlenmeyer flask was rinsed. The bath temperature was checked for consistency with the first experiment. The first solution's procedure was repeated, this time using 10.0 mL of .10 M KI and 10.0 mL of distilled water, followed by 10.0 mL of 3% H2O2.
Solution 3
8. The Erlenmeyer flask was rinsed. The bath temperature was checked for consistency with the first and second experiment. The first solution's procedure was repeated, this time using 20.0 mL of .10 M KI and 5.0 mL of distilled water, followed by 5.0 mL of H2O2.
B. Effect of Temperature
1. The bath temperature was adjusted so that it was approximately 10° to 12° higher than what it was for part A. The experiment from Part A was repeated, using the procedure for Solution 1, but the amounts of substances outlined for Solution 2. After 14 mL of O2 was collected, no more volume-time readings were collected, but the reaction was allowed to continue to completion. The solution was saved for Part C.
C. Identification of the Catalyst
1. Approximately 1 mL of the reaction mixture from Part B was placed into a small test tube. Approximately 10 drops of .1 M Hg(NO3)2 were added to the test tube. The tube was shaken. Qualitative observations were recorded. The test tube was photographed.
2. Approximately .2 mL of .10 M KI was diluted in approximately 5 mL of distilled water. Two drops of this solution were added to 10 drops of .1 M Hg(NO3)2. Qualitative observations were recorded. The test tube was photographed.

Observations:

A. Order of Reaction

Bath temperature for all three solutions: 19.0°C

Solution 1 observations:

Buret reading (mL) ± .1 Volume of O2 (mL) ± .1 Time (sec)
10.5 0.0 0
12.5 2.0 95
14.5 4.0 185
16.5 6.0 265
18.5 8.0 360
20.5 10.0 455
22.5 12.0 550
24.5 14.0 670
26.5 16.0 810

Solution 2 observations:

Buret reading (mL) ± .1 Volume of O2 (mL) ± .1 Time (sec)
10.5 0.0 0
12.5 2.0 55
14.5 4.0 110
16.5 6.0 155
18.5 8.0 205
20.5 10.0 255
22.5 12.0 300
24.5 14.0 345
26.5 16.0 400

Solution 3 observations:

Buret reading (mL) ± .1 Volume of O2 (mL) ± .1 Time (sec)
11.0 0.0 0
13.0 2.0 45
15.0 4.0 95
17.0 6.0 165
19.0 8.0 215
21.0 10.0 270
23.0 12.0 330
25.0 14.0 390
27.0 16.0 450
B. Effect of Temperature

Bath temperature: 31.0°C

Solution observations:

Buret reading (mL) ± .1 Volume of O2 (mL) ± .1 Time (sec)
11.0 0.0 0
13.0 2.0 35
15.0 4.0 65
17.0 6.0 95
19.0 8.0 115
21.0 10.0 135
23.0 12.0 160
25.0 14.0 190

Buret reading after total evolution of oxygen: approximately 54 mL; however, the reading extended past the bottom of the markings on the buret. This reading is questionable.

Barometric pressure: 762 mm Hg

Vapor pressure of water at 31.0°C: 33.7 mm Hg

C. Identification of the Catalyst

a. When mercuric nitrate is added to the solution in which the reaction had gone to completion, an opaque, bright orange precipitate forms. The color, as photographed, is shown here: (view) b. When mercuric nitrate is added to the original KI solution, an opaque, dark orange/red precipitate forms. The color, as photographed, is shown here: (view) Results:

A. Order of Reaction

Slopes of best-fit lines, by solution, as determined by computer:

```    Solution 1: 47.659 sec/mL O2
Solution 2: 25.123 sec/mL O2
Solution 3: 27.574 sec/mL O2
```

However, since the desired figure is mL O2/sec, the inverse of these slopes must be determined.

Rate of formation of O2:

```    Solution 1: .02098 mL O2/sec
Solution 2: .03980 mL O2/sec
Solution 3: .03627 mL O2/sec
```

When the amount of H2O2 is doubled while KI is kept constant, the rate of reaction doubles. When the amount of KI is doubled and the amount of H2O2 is halved, the rate stays nearly constant.

According to the equation: rate = k[A]x[B]y, if A is doubled, then the rate changes by the ratio R2/R1. The following equation can be used to determine "x", the exponent of A, under conditions of doubled concentration:

```    exponent = ln (R2/R1) ÷ ln 2
```

When applied to the data presented above, the following figures are produced.

Solution 1-2. The concentration of H2O2 is doubled. Thus, the exponent of H2O2 is:

```    exponent = ln (.03980/.02098) ÷ ln 2
exponent = .92375
```

Solution 1-3. The concentration of KI is doubled. Thus, the exponent of KI is:

```    exponent = ln (.03627/.02098) ÷ ln 2
exponent = .78976
```

The results indicate that the rate law is as follows:

```    rate = k[H2O2][KI]
```

Both x and y are equal to 1.

B. Effect of temperature
```    Slope of best-fit line, as determined by computer: 13.839 sec/mL O2
Rate of formation of O2: .07226 mL O2/sec
```

This rate of formation compares to the rate of formation of solution 2 as follows:

```    .07226 mL O2/sec ÷ .03980 mL O2/sec =
1.816 times greater
```

Using the ideal-gas law, the moles of O2 collected can be calculated. The volume of O2, as will be required for this equation, is 43 mL of O2, because O2 evolution started at the 11 mL mark and ended with approximately 54 mL of gas.

```    PV=nRT
(762 mm Hg + 33.7 mm Hg) × .043 L O2 = n × 62.36 mm Hg-L/mol-K × 304 K
n = .001805 mol O2
```

To find the molar concentration of the original H2O2 solution, a multipartite equation must be applied.

First, calculate the number of moles of H2O2 based on the stoichiometric equation.

```    2H2O2 → 2H2O + O2
```

Thus, the number of moles of H2O2 is twice the number of moles of O2.

```    2 × .001805 = .003610 mol H2O2
```

This trial used 10 mL of H2O2. Thus, .003610 moles of H2O2 exist per .010 L of solution. The molar equation is as follows:

```    molarity = .003610 mol H2O2 ÷ .010 L solution molarity = .361 M H2O2
```
C. Identification of the Catalyst

KI is a strong electrolyte. The color produced by the first solution was produced because of I ions distributed in solution. A similar color was produced by the diluted KI solution because of the I ions distributed in solution. Thus, KI breaks up into K+ and I ions when placed into solution.

Discussion: The known values for x and y are 1. Two percent error calculations can be performed based on the values obtained above. (view)

```    Percent error, H2O2 exponent.
(observed − true) ÷ true × 100 =
(.92375 − 1) ÷ 1 × 100 =
-7.63 percent error.

Percent error, KI exponent.
(observed − true) ÷ true × 100 =
(.78976 − 1) ÷ 1 × 100 =
-21.02 percent error.
```

The error in this experiment is the result of several sources of error. If the flask were stirred inconsistently, then the rate of production of O2 would have been adversely affected. This would have skewed the calculation of the slope, thereby affecting the rate of reaction that was calculated, thereby affecting the exponents calculated. If any tap water had been in the flask as a result of rinsing, then other potential catalysts for the reaction may have entered into the equation, affecting the rate of reaction. Any residual substances in the Erlenmeyer flask could have affected the calculated rates. Any contaminants left behind in graduated cylinders could have also affected calculated rates. Furthermore, the graduated cylinders are highly imprecise measuring tools. The amounts poured into and out of the graduated cylinders are simply estimates of appropriate amounts. They are rarely accurate to more than .2 mL, which is a significant amount when dealing with small amounts of solution. The Mohr buret used did not contain sufficient graduation to measure the final reading in Part B, which means that the calculations stemming from that observations are highly questionable.

The theory associated with this experiment is the collision theory. The collision theory states that reactions are caused by the collisions of particles. The more collisions that occur, and the more forceful the collisions, the greater the rate of reaction. This is supported by the data produced in this experiment. When the temperature was increased, the rate of reaction increased because of more collisions occurring. When the concentrations of the reactants increased, the rates of reaction increased, because more collisions were occurring.

There are many ramifications for this experiment. First, important laboratory experience was gained, as always. In research-based science and industry, it is very important to know or to be able to determine the rates of reactions. All chemical manufacturing is the result of chemical reactions, caused by particle collisions. Knowing how fast a reaction will occur is critical to controlling the reaction and therefore the production. Being able to calculate the relationship between concentration and rate on the 10-mL scale means that the reaction will be controllable on the 100,000-L scale. Knowing the role of catalysts means that they can be used effectively to create favorable rates and production situations. The rate of reaction and its relationship to concentrations has innumerable applications in science and industry, among other fields. It is critical to maintaining control over chemical reactions.

Questions:

1a. The rate of formation of O2 is half of the rate of formation of H2O.

1b. The rate of formation of O2 is half of the rate of disappearance of H2O2.

2. The levels of water in the leveling bulb and buret should be kept the same to ensure that the same pressure is acting on all liquids. The pressure in the buret can be changed with the leveling bulb. By putting the leveling bulb at the same level as the buret, the gas inside the closed-system buret is at the same pressure as the external world.

3. The Erlenmeyer flask was continually swirled to ensure equal mixing of the reactant and the catalyst. This ensures a uniform reaction, and therefore a uniform rate of reaction for all parts of the solution.

4a. Using .20 M KI would result in slopes of curves that were twice as steep as the original slopes.

4b. The rates of the reactions would have been twice as fast as the original rates.

4c. The numerical value of k for the reaction would remain unchanged.

Conclusion: This experiment was a success. The rate and order of reaction for the decomposition of hydrogen peroxide was determined to a reasonable rate of success. Furthermore, laboratory experience and practice with using universal formulas were gained.

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