S2o8 2 2i rate law

Rates of Reactions

Objectives 1. Use a variety of calibrated glassware to make dilute solutions with known concentrations

of reactants using precision and accuracy. 2. Use color changes to monitor the rate of concentration decrease of a reactant. 3. Practice using the dilution formula to determine the concentrations of solutions diluted

from a more concentrated “stock” solution. 4. Recognize that for a reaction with a reactant coefficient of one, the initial rate of the

reaction can be determined from the negative slope of a plot of reactant concentration near the start of a reaction as a function of time.

5. Determine the order for each reactant in a rate law by using a series of trials with different initial reactant concentrations and reaction rates.

6. Determine the rate constant for each trial from initial concentration. 7. Provide the overall rate law for the reaction studied.

Introduction Chemical reactions can occur at very different rates. Compare the slow oxidation of iron

into iron oxide (rust) with the nearly instantaneous reaction of burning hydrogen gas, once the

initial spark is given. Some variables that affect the rate of a reaction include temperature,

concentration, and whether a catalyst is present. An increase in temperature increases the energy

of the collisions. The addition of a catalyst lowers the activation energy, the minimum energy

required for a reaction to take place, by orienting molecules for favorable interactions. An increase

in concentration increases the probability of a productive collision. Any of these changes may speed

up a reaction. In this experiment, the effect of concentration will be investigated.

The rate of a given reaction is gauged by either the consumption of starting materials

(reactants) or the formation of products. This measurement of concentration is related to time such

that when the concentration, X, of a reactant is monitored (Eq. 1).

rate of reaction = −- 1 N 0 change in concentration of reactant

change in time = -

1 N 0 ∆[X] ∆𝑡

Equation 1

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where N is the coefficient of X in the balanced equation. Note the negative sign in the

equation since X has been defined as a reactant and therefore is being consumed (has decreasing

concentration) during the reaction.

Often the initial rate of a reaction is investigated because it is easy to determine initial

concentrations of reactants and an equation can be written that relates reaction rate to the initial

concentrations of all the reactants. Rate may, or may not, depend on the concentration of every

species in the system. Reaction orders characterize how the rate is affected by each reactant

concentration. Consider a reaction involving reactants, X, Y, and Z, where the lower-case letters

in Equation 2 represent coefficients in the balanced equation. The initial rate of a reaction is

investigated because it is easy to determine initial concentrations of reactants and an equation can

be written that relates reaction rate to the initial concentrations of all the reactants. Rate may, or

may not, depend on the concentration of every species in the system. Reaction orders characterize

how the rate is affected by each reactant concentration. Consider a reaction involving reactants, X,

Y, and Z, where the lower-case letters represent coefficients in the balanced equation (Eq. 2).

xX + yY + zZ → qQ + rR

Equation 2

If the rate does not depend on the concentration of Z at all, it is zero order with respect to

Z, in other words, in the rate expression, its exponent is zero (Eq. 3). If the rate doubles every time

the concentration of Y is doubled, the reaction is first order with respect to Y. If doubling the

concentration of X leads to a quadrupling of the rate, it is second order with respect to X. The rate

expression for a reaction can only be determined experimentally and the rate equation for this

example would be Equation 3 where k is the rate constant for the reaction at a given temperature.

Rate = k [X]2 [Y]1

[Z]0

Equation 3

The overall order of a reaction is the summation of the orders of all the components. For

the above example, the overall order is 2 + 1 + 0 = 3.

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The order of each component must be determined experimentally. This is done by changing

the concentration of only one component at a time and then noting the change in rate. Once the

order of every component in known, then the rate constant, k, can be determined using

concentration information from one of the trials. Once k is determined along with the order of the

reaction, the rate expression can be used to calculate or predict the rate with different

concentrations.

The reaction that will be investigated is an oxidation-reduction with peroxydisulfate ion

(S2O82-), and the iodide ion (I-). The iodide is oxidized to iodine (I2) and the peroxydisulfate is

reduced to the sulfate ion (SO42-) (Eq. 4).

S!O"!#(aq) + 2I#(aq) → I!(aq) + 2 SO$!#(aq)

Equation 4

Starch is the indicator for the reaction and will turn blue in the presence of I2. Since the

starch will turn blue instantly in the presence of I2, we would not be able to determine the rate with

just these two reactants in the solution with the indicator. We would see a blue color change

instantly as soon as the first molecules of iodine would form, and we would not be able to gather

any time dependent data. Therefore, we must add something to slow the appearance of the blue

color. The thiosulfate (S2O32-) will be added for this purpose. As the iodide reacts with the S2O82-

(Eq. 4), it is converted to iodine, I2, which will immediately react with the thiosulfate (S2O32-) (Eq.

5) which is also present in solution. The thiosulfate will reduce the I2 back to I-, which will not turn the

solution blue.

2S!O%!#(aq) + I!(aq) → 2I#(aq) + S$O&!#(aq)

Equation 5

After some time, all of the S2O32- will be consumed and more I2 will be produced. The

excess I2. The excess will complex with the starch indicator, turning the solution blue. The data

you collect will be the time delay between the addition of the thiosulfate and the change in color.

You will indirectly measure the disappearance of S2O32- (Eq. 4) and, thus, be able to determine the

rate of the reaction.

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Experimental Design You will start with a relatively small amount of thiosulfate (2.0 x 10-4 mole) in solution

compared with the initial amounts of peroxydisulfate and iodide. When the 2.0 x 10-4 moles are

consumed in a reaction with iodine (at which time the solution will turn blue), 1.0 x 10-4 moles of

peroxydisulfate will have reacted with iodide because of the stoichiometry of the reactions. You

will record the time from the start of the reaction and then add another 2.0 x 10-4 moles of

thiosulfate (1.00 mL). The solution will become colorless again as the thiosulfate reacts with the

iodine. When the blue color reappears, you will record the time. You will then repeat these steps

three more times for a total of five data points. You will go through this procedure a total of three

times with different initial concentrations iodide and peroxydisulfate but with the same amount of

thiosulfate each time.

A relatively small number of moles of peroxydisulfate is consumed during each time

interval, so the reaction should be in the linear initial portion of the rate curve for the duration of

the reaction investigated. You will plot the concentration of peroxydisulfate remaining as a

function of time (Figure 1).

Figure 1. Concentration of S2O82- over time.

0.039

0.04

0.041

0.042

0.043

0.044

0.045

0.046

0.047

0.048

0.049

0 200 400 600 800 1000 1200

M ol

ar ity

o f S

2O 82

-

Time (s)

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Since peroxydisulfate (S2O82-) is the species monitored during the experiment and its

coefficient is 1 as seen in the balanced equation (Equation 4), the value of N from Equation 1 is

equal to 1 and therefore the initial rate of the reaction is the negative of the slope of the line you

have plotted (Eq. 6).

rate of reaction = −- 1 1 0 Δ[S!O"!#]

∆𝑡

Equation 6

The values of the rate of reaction from each trial will be used to determine the order and

rate constant for the reaction of peroxydisulfate with iodine. Notice the slope of the graph will be

negative because the concentration of the peroxydisulfate is decreasing. However, the rate of the

reaction is positive (Eq. 7).

Rate of reaction = –slope

Equation 7

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Procedure 1. Prepare two flasks for each experiment (Table 1) using the stock solutions (Table 2).

Remember to include these steps in your notebook. DO NOT CONFUSE the S2O32- with the S2O82-!

Table 1. Reaction Concentrations and Volumes for Experiments 1-3

Experiment 1 Experiment 2 Experiment 3

Flask #1

0.1 M EDTA 3 drops 3 drops 3 drops

1% Starch Indicator 1.0 mL 1.0 mL 1.0 mL

0.20M Na2S2O3 1.00 mL 1.00 mL 1.00 mL

KNO3 12.0 mL 24.0 mL 12.0 mL

0.20M KI 24.0 mL 12.0 mL 12.0 mL

Flask #2

0.20M (NH4)2S2O8 12.0 mL 12.0 mL 24.0 mL

Total Volume 50.0 mL 50.0 mL 50.0 mL

Table 2. Concentrations of Stock Solutions

Solution Concentration Solution Concentration

KNO3 0.20 M KI 0.20 M Na2S2O3 0.20 M Starch indicator ~1%

NH4S2O8 0.20 M EDTA ~0.1 M

The peroxydisulfate, iodide and thiosulfate ions are not found alone but as ammonium, potassium and sodium salts, respectively.

The KNO3 is added to maintain the ionic strength of the solutions.

The EDTA is added to complex out any spurious metals that might act as a catalyst for the reaction and change the overall rate.

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2. Only one experiment at a time can be performed. 3. The components in Flask #1 and Flask #2 for Experiment #1 will be mixed together and

stirred at a slow rate with a magnetic stir bar. Note the speed that the stir bar is spinning and try to complete all three experiments with the same spin rate.

4. Since temperature affects the rate, it will be important to try and maintain consistent temperatures.

5. Begin timing right when the two flasks are mixed together. Note the time when the flask contents turn blue. The color change may not be blue. The color may appear to be purple, black, blue, or even orange. Just go with it.

6. If you have not seen a color change in 10 min, then something is wrong! 7. IMMEDIATELY add 1.00 mL of the thiosulfate solution (S2O32-) with a volumetric pipet

and continue timing for the next data point. (Do NOT restart the timer; you are recording cumulative time.)

8. Repeat, adding 1.00 mL of the S2O32- solution each time the solution turns blue until five data points are collected. Depending on your technique the first data point may need to be discarded.

9. Repeat the above process for Experiments #2 and #3 being sure to change the concentration of the reactants as shown in Table 1.

Calculations During the experiment you are measuring the moles of peroxydisulfate being consumed.

You will plot the concentration (mol/L) of peroxydisulfate remaining in the flask. Since you can

determine the initial number of moles of S2O82- from your initial volumes and concentrations of

reactants (Table 1) and you know the number of moles of S2O82- being consumed as the reaction

progresses, you can determine the moles of S2O82- remaining for each color change of solution

during the reaction. You must convert this to concentration, which means you will be dividing by

the volume in the reaction flask.

The total volume of each of the experiments is determined when the reaction is initiated;

when Flask 2 is poured into Flask 1. Since you are increasing the volume in the flask by 1.00 mL

each time the solution turns blue, you must take this additional volume into account when

computing the concentration of subsequent samples (Sample Calculation 1).

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Sample Calculation 1. Calculation of Concentration in Trial 1

Initially, (Table 1) you have

0.20 mol (NH%)&S&O' L

x 0.0120 L = 0.0024 mol (NH%)&S&O'

After the solution turns blue the first time, the amount of remaining (NH4)2S2O8 is: 0.0024 mol (NH%)&S&O' − 1.0𝑥10(% (NH%)&S&O' = 0.0024 mol (NH%)&S&O'

This is dissolved in 50 mL so

0.0023 mol (NH%)&S&O' 0.050 L

= 0.046 M (NH%)&S&O'

For the concentration information on S2O82-, the first row on the table should be completed

as shown in the following example (Figure 2).

Initial moles of S2O82-

Aliquot # Cumulative time

recorded from stopwatch

(min) (sec)

Cumulative time

converted to seconds (s)

Mole of S2O82-

Consumed

Mole of S2O82-

Remaining

Concentration of S2O82-

Remaining (mol/L)

1 1.0 x10-4 mol 0.0023 mol 0.046 M

Figure 2. Sample Table from Report.

After you have completed the first three tables on the Report Sheet, you will use the dilution

formula and the information in Table 1 to determine the initial concentrations of peroxydisulfate

and iodide for each experiment to complete the fourth table on the lab report.

Graphs You will create a graph for each of the first three tables. In each graph you will plot the

remaining concentration of S2O82- (Molarity) as a function of cumulative time in seconds. Create

a scatter plot for each graph and add a trendline with an equation shown on the graph. The negative

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slope of the trendline is the rate of the reaction. Note that the rate of a reaction is always a positive

value.

Once you have the rates for all three trials from the slopes of the graphs and the initial

concentrations of the two reactants, peroxydisulfate and iodide, you can determine the order of

each reactant and the overall order of the reaction, the rate constant, and the rate law. The rate

constant should be very similar among the three experiments. The average rate constant value will

be used to write the rate law. The rate law will have the generalized form (Eq. 8).

rate = k [S2O"! #]' [I#](

Equation 8

In Eq. 8, a and b are the reaction orders; k is the rate constant. You will solve for a, b, and

k using the experimental data. Use the format option to view at least two significant figures in the

equation for the line. Remember to use the appropriate units for the rate constant, k.

Once you have the rate law, you can determine the rate of the reaction given any

concentrations of the reactants.

You will turn in 3 graphs along with your data sheet and calculations.

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Rates of Reaction Report Sheet

Name: Partner:

Instructor: Section Code: Date:

15 points total for Tables 1-3.

Experiment 1 Initial moles of S2O82-

Aliquot #

Cumulative time recorded from stopwatch

(min) (sec)

Cumulative time

converted to seconds (s)

Mole of S2O82-

Consumed

Mole of S2O82-

Remaining

Concentration of S2O82-

Remaining (mol/L)

1

2

3

4

5

Experiment 2 Initial moles of S2O82-

Aliquot #

Cumulative time recorded from stopwatch

(min) (sec)

Cumulative time

converted to seconds (s)

Mole of S2O82-

Consumed

Mole of S2O82-

Remaining

Concentration of S2O82-

Remaining (mol/L)

1

2

3

4

5

Experiment 3 Initial moles of S2O82-

Aliquot #

Cumulative time recorded from stopwatch

(min) (sec)

Cumulative time

converted to seconds (s)

Mole of S2O82-

Consumed

Mole of S2O82-

Remaining

Concentration of S2O82-

Remaining (mol/L)

1

2

3

4

5

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Values for the first two columns of the following table come from using the dilution formula and the concentrations used (Table 1). Values for the final column come from the graphs.

Data Analysis

Experiment

Initial [S2O82-] (mol/L)

2 pts each

Initial [I-] (mol/L)

2 pts each

Initial rate (mol/Ls)

5 pts each

1 __________ __________ __________

2 __________ __________ __________

3 __________ __________ __________

Calculations for the order of S2O82- (6 pts)

Calculations for the order of I- (6 pts)

Sample calculations for rate constant, k, (Exp 1) using the orders determined for reactants. (6 pts)

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Calculate the rate constant, k, for each experiment. Be sure to show units. (3 pts each)

Rate Constant for Experiment 1 ___________

Rate Constant for Experiment 2 ___________

Rate Constant for Experiment 3 ___________

Average Rate Constant ___________

Rate law for this reaction. Show values for k, a, and b in the expression (Eq. 8). (3 pts each)

Using the rate equation determined, what would be the predicted rate of reaction if the initial concentration of S2O82- were 0.085M and the initial concentration of I- were 0.030M?

Show calculations. (4 pts)