I

OBJECTIVES

KINIETICS EXPERIMENT

To study the rate of reactions.To determine reaction orders.To determine the rate law for a chemical system.

INTRODUCTION

One of the methods used in studying chemical reactions is to measure the change in reactantconcentration over a given period of time. This change in concentation with respect to a time unit(seconds, minutes, hours, etc.) is called the rate of a chemical reaction. This rate of change dependson the concenffation of the reacting species; the temperature of the reaction system; and an energyfactor called the activation energy. The study of reaction rates and the factors which affect rates ofreactions is called Kinetics.

In kinetics a complete description of a reaction system starts with the writing of a general rateexpression. The following example serves to illustrate how this is done.

Given a reaction:

A + B ------->

General rate expression is:

Rare = k[A]m [B]"

where "k" is a constant, called the rate constant, and "m" and "n" are called reaction orders. Thebrackets ( t I ) are interpreted as molarity units or the number of moles per liter of reactant A and B.The general rate expression tells us that the Rate of any reaction depends on the values of; (1) therate constant; (2) the concentration of reactants; (3) the reaction orders.

One of the major objectives of Kinetics is the experimental determination of the values of thereaction orders to obtain a Rate Law. For example, if experiments indicate the the value of "m" is Iand the value of "n" is 0, the rate expression above can be re-written as:

Rate = k[A]t

(note: [B]o drops out as any number raised to the zero power is "1" and will not affect the value ofRate in the equation)

This rate expression now tells us that the rate of the reaction is directly proportional to theconcenffation of [A] and it is called a First Order reaction. Once it has been determined that it is afirst order reaction, the first order rate equation can be used to predict concentrations of reactant "A"at any time value.

If, on the other hand, experiments indicate that the value of "m" is 2 and that of "n" is 0, thefollowing rate expression can be written:

Rate = k[e1z

and the reaction is said to be a Second Order reaction and the rate law above tells us that the rate ofthe reaction will depend on the square of the concentration of "A".

If both "m" and "n" are determined to be equal to 1, the rate expression for the system becomes:

Rate -- klAlt [B]t

and the reaction system is said to be first order with respect to "A" and "B" and Second OrderOverall (overall order is the sum of "m" and "n").

The values of "m" and "n" must be determined experimentally as they cannot be predictedby the stoichiometry (the coefficients of the balanced equation) of thd reaction. The valu'es of thereaction orders need not be whole numbers although they are often rounded offto the nearest wholenumbers to make the calculations simpler.

THE REACTION TO BF STUDIED

The system you will be working with is a rather complex reaction involving three separate steps asillusfated below:

STEP 1

2IO3- +

iodateions

2 SO3-2

sulfiteions

iodide sulfiteions ions

An excess of iodate ions is allowed to react with sulfite ions to produce iodide ions in solution.When all of the sulfite ions (the limiting reagent) is consumed, the iodate reacts further with theiodide ions in the presence of acid (H+) producing iodine molecules as described in step 2:

STEP 2

IOa- + 5I-+6H+ ------------->

the iodine molecules p-roduced can be detected by the_addition of starch as they form a complexwhich exhibits a dark blue color as described in step 3.

STEP 3

starch + Iz ------------->

The rate of this reaction can be determined by measuring the time it takes for the blue color to_app€ar. ,Although the reaction is a complex one, it can be studied rather easily because of the "rate-limiting" ltep concept in Kinetics. According to this concept, the slowest stei in any multi-stepreaction determines the overall rate of the reaction. The slow step in this reaCtion has been found tope Step 1 and Steps 2 and 3 can be assumed to occur instantaneously as soon as all of the sulfiteions are consumed.

Hence, the rate expression for this reaction is that expression which describes the sloweststep.

Rate = k [iodate]m [sulfite]n

MEASUREMENT OF TIIE RATE OF REACTION

In this experiment the rate of the reaction will be based on how long it takes for the bluecolor of the starch-iodine complex to appear. We will assume that when the blue color appears, allof the sulfite ions will have been consumed. The rate of the reaction will be expressed as thechange in sulfite concenfration with time or:

Rate = Alsulfitel =time

Mseconds

PROCEDURE

You will be provided with two stock solutions, A and B. Solution A contains 0.020 M iodate ionsand solution B contains 0.010 M sulfite ions along with some acid and starch. In preparing yourreaction systems always add water first followed by solution B and then add solution A.

Use pipets to measure out the water and solutions A and B. Use a 50 Erlenmeyer flask asyour reaction vessel. After each trial simply rinse out the flask with tap water. The volumes ofsolution you are to use are those found below in the sample data.

As temperature changes affect reaction rates, you will want to minimize contact with thesolutions; after adding all solutions, grasp the top of the flask with your fingertips, swirl it gentlyand quickly, and place the flask on the lab bench. Swirl the mixture intermittently until a colorchange occurs. Time each reaction to the nearest second. Run each reaction system several timesto obtain an average time value. Record the temperature of the room for each reaction.

SAMPLE DATA

Determination of the Rate of Reaction

To determine the reaction order with respect to iodate ions and sulfite ions, you will have to measurethe rate of several reaction systems. In this example, the original solutions are 0.020 M iodate and0.010 M sulfite.

system I 0.50 mL iodate 0.50 mL sulfite 4.00 mL watersystem 2 1.00 mL iodate 0.50 mL sulfite 3.50 mL watersystem 3 0.50 mL iodate 1.00 mL sulfite 3.50 mL water

The first two systems are designed to determine the effect of increasing iodate concentrations whilemaintaining sulfite concentration const?nt. Then, a comparison of the rates of system 3 and system1 determines the effect of increasing sulfite concenffation while maintaining constant iodateconcenfration.

The following represents sample data obtained at20.0 C.

trial I ttal2 fial 3 trial4 averagesystem 1 103 s 100 s 105 s 104 s 103 ssystem2 55s 58s 54s 53s 55ssystem3 42s 46s 44s 43s 44s

The rates for systems 1 and2 are determined by dividing the average time for color appearance intothe starting colcentration of the sulfite ion. The sulfite concenfration is calculated by ionsideringthe dilution of the solution and using Mt Vt =MZYz.

M2 = MtVtY2

= (0.50 mL) (0.010 M)5.00 mL

= 0.0010 M

Rateofslstem 1 = 0.0010M/ 103 s = 9.7 xl0-5IWsRateofslstem2 = 0.0010M1 55s = 1.8x 10-3M/s

Since the amount of sulfite ions was doubled in system 3, the sulfite ion concentration must bedoubled in determining the rate of system 3:

Rateofsystem3 = 0.0020lNl/44s = 4.5x10-5Nrys

IDeterrnination of Reaction Orders

\* The reaction order with respect of iodate ions is determined in the following way. Note that thedilution calculations for the solutions have been done and are not shown.

Rate.system2 = kliodatel*Isulfite.lnRate, system 1 kliodate]*[su1fite1n

1.8x10-s = k[0.0020M1-10.001-0Mln9.7 x t0_a k [0.0010M]m [0.0010M]n

simplify by canceling out "k". We can also cancel the (0.0010M)n term because regardless of thevalue of "n", the result of the division will be 1.

Hence, l.g = (2.0),r,

solving by use of logs,

log 1.9 = m(log2.0)

solve for m,m = log 1.9 = 0.93

log2.0

This would be the reaction order with respect to the iodate ions. It can be rounded to 1, indicatingthat it is first order with respect to iodate

The reaction order with respect to the sulfite ion can be determined by comparing the rates ofsystem 3 with that of system 1.

rate.system3 = kfiodatelo'e3lsulfitelnrate, system I kliodatslo.l:[su1fite]n

4.5x10rM/s = kl0.00lMl0.e3l0.0020Mln9.7 x 10-6 tWs kt0.001M10.e3[0.0010M]n

4.6 = (1) o.rl 12; n ( Note that (t) 0.e3 = I )Using logs to solve,

log4.6= nlog(2)

n=log4.6=2.2log2

This is second order with respect to sulfite.

The reaction is therefore lst order (to the nearest whole number) in iodate ion concenffation and2nd order with respect to sulfite ions. The overall order is therefore 3.

Rate = k[iodate]l [sulfite]2

Determination of Rate Constants

Once the reaction orders have been determined, the rate constants can be calculated bysolving for "k" for each reaction system. For system 1:

k1 = Rate. system 1

liodatslo.ra lsulfitslz.z

k1 = 9.7 x 10-6 M/s(0.00 1 0M)0.e3 (0.001 0w1z.z

k1 = 2.4x104 M-2s-l

Note: To find what (0.0010)0'r: it "O,ral

to, take tltg log of the expression which gives,0.93 ln(0.0010) = -6.4 to two significant figures. Then-find the adtilog (inverse log; to'fina wnatthe number is. The antilog or inverse log of -6.4 = r.6 x l0-3 to two sig figs.

Using this approach the " k" values for the system 2 and 3 are calculated, and both came out to be2.3 x l}a M-2s-1, to 2 significant figures. The "k" value is then reported as the averageof all three"k's". The units M-2s-l are characteristic of a third order reaction.

il

1r

DATA SHEE NAME

trial 1

tr:ral2

seconds sulfite M

seconds iodate

trial 3 _seconds Average time _sectrtal4 _seconds Rate, System l_M/s

System 2 1.00 mL of 0.010 M iodate; 0.50 mL of 0.010M sulfitQ and 3.50 mL wareri!t1tii

System 1 0.50 ni[, of 0.010 M iodate, 0.50 mL of 0.010M sulfitg and 4.00 mL of water!I

M

trial I

tnal2

trial 3

_seconds sulfite

_seconds iodate-M

M

\ Ytal4 _seconds Rate, System 2 _M/s

System 3 0.50 mL of 0.010 M iodatel 1.00 mL of 0.010M sulfite;and 3.50 mL wateriltrial 1 _seconds sulfite M

_seconds Average time _sec

seconds iodate

-Mseconds Average time _secseconds Rate, System 3 _M/s

tnal2

frial3

trial 4

Calculation of orders of reaction:

Determination of "m"

Determination of "n"

Calculation of rate constants:

Rate Constant 1

Rate Constant 2

Rate Constant 3

Average Rate Constant

Rate Law

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Prelab Assignment

. Try the following as a check to see if you understand what was presented above.

1. If 200. mL of a 12M HCI solution were diluted to a final volume of 600. mL, what is its finalconcenfation?(ans. 4.0 M)

2. If 500. mL of water were added to 100. mL of a18 M solution, what is the final concenffation ofthis diluted solution?(ans. 3.0 M)

3. How much water must be added to 600. mL of an 18 M solution to decrease its concenkation to3.0 M?\- (ans. 3000 mL or 3.0liters)

4.If 250. mL of a 0.25 M solution is dilut*, doncentration of 0.10 M what must have been thefinal volume of this solution?(ans. 625 mL)

5. In the above problem (#4) how much water must have been added to starting solution? (ans. 375mL)