Lab 2

Kinetics: Method of Initial Rates

Background

The Differential Rate Law. For the general reaction (1), the rate is expressed as the change of concentration with respect to time, t, of any reactant or product.

(1): aA + bB <==> dD + fF

(2): Rate = -(1/a) d[A]/dt = (1/f) d[F]/dt

Since A is consumed and F is produced by (1), their derivatives have opposite signs. Thus we introduce a minus sign in front of d[A]/dt to ensure a positive rate. Also, unless a = f, -d[A]/dt and d[F]/dt are different. In general, the rate of reaction is a function of the concentrations of reactants and products, as indicated in (3).

(3): Rate = -(1/a) d[A]/dt = (1/f) d[F]/dt = k[A]^m[B]ⁿ[D]^s[F]^v

k is the rate constant of the reaction, and [A], [B], etc., are the molar concentrations of reactants and products in (1). The exponents m,n... appearing in (3) are usually either positive or negative integers or zero.

Because (3) is a differential equation, it is called the differential rate law for (1). The goal of a kinetics study is to determine the differential rate law for the reaction (i.e., the values of the exponents m, n, s, and v) and the numerical value of the rate constant k at each of several temperatures. Suppose that appropriate kinetics experiments carried out on (1) give m = 2, n = 1, s = 0, and v = -1. The differential rate law for the reaction is then

-1/a d[A]/dt = k[A]²[B]/[F]

The exponents m, n, etc. are called orders. We say that the reaction is second order in A, first order in B, and inverse first order in F. It is important to realize that there is no necessary connection between the reaction orders m, n, s, and v, in (3) and the stoichiometric coefficients a, b, d, f, in (l).

The Method of Initial Rates. A straightforward way to determine experimentally the specific form of the rate law (3) is to measure the initial rate, -(d[A]/dt)₀, as a function of each of the initial concentrations [A]₀, [B]₀, etc, and from the data deduce the reaction orders. Suppose that we found that -(d[A]/dt)₀ quadrupled (hence time to completion is one fourth of the original time needed to complete the reaction) when we doubled the [A]₀ while keeping the other concentrations constant. We would conclude that m = 2. Similarly, if -(d[A]/dt)₀ decreased by a factor of 2 (hence the time to completion is double the original time) when [B]₀ was cut in half while keeping other concentrations the same, we would conclude that n = 1.

Experiment

The balanced equation for the reaction you will explore today is

2 I ^- + S₂O₈ ^2- ® I₂ + 2 SO₄ ^2-

This equation is not elementary, but rather the sum of its mechanistic elementary steps. This is clear because the overall rate suggests that three particles collide to give the products. This is statistically improbable. More likely, 2 species collide to make an intermediate, which in turn collides with the third species to form the products. This leaves two possibilities:

Mechanism 1

The iodides collide with each other to form I₂^2-, which then collides with S₂O₈^2-.

2 I^- ® I₂ ^2-

I₂ ^2- + S₂O₈ ^2- ® I₂ + 2 SO₄ ^2-

if the first step is slow, rate = k₁ [I^-]²

if the second step is slow, rate = k₂ [I₂^2-][S₂O₈^2-]

if the second step is slow, an intermediate is present [I₂^2-], and must be substituted out of the rate law using the first reaction as fast and reversible:

rate forward = rate reverse Þ k_1fwd[I^-]² = k_1rev[I₂^2-], solving for [I₂^2-] = (k_1fwd/k_1rev)[I^-]²

and then substituting in to the rate law

rate = (k₂k_1fwd/k_1rev) [I^-]²[S₂O₈^2-]

Mechanism 2

One iodide, I^-, collides with the S₂O₈^2- to form an intermediate, SO₄I^- which collides with the other I^- to give the final products:

I^- + S₂O₈^2- ® SO₄^2- + SO₄I^-

SO₄I^- + I^- ® I₂ + SO₄^2-

If the first step is slow, rate = k₁ [I^-][S₂O₈^2-]

If the second step is slow, rate = k₂[SO₄I^-][I^-]

Again, if the second step is slow here, there is an intermediate that needs to be substituted out using the first reaction as fast and reversible.

Rate forward = rate reverse Þ k_1fwd[I^-][S₂O₈^2-] = k_1rev[SO₄^2-][SO₄I^-], solving for [SO₄I^-] and then substituting into the rate law gives

YOU TELL ME! Answer this question as part of your prelab.

The goal of the experiment, then, is to determine the rate law using the method of initial rates and then deduce the mechanism from the determined rate law.

We will determine the reaction rate by using a starch indicator. When I₂ begins to appear it reacts with the starch to produce a purple color. The thiosulfate (S₂O₃^2-) is used to suppress the reaction slightly. The thiosulfate reacts with I₂ as shown below

I₂(aq) + 2S₂O₃^2-(aq) --> 2I^-(aq) + S₄O₆^2-(aq)

Once the thiosulfate ion is used up then the iodine reacts with the starch and you see the blue color. Quantitatively, we know that when the solution finally turns purple the reaction has produced a molar amount of I₂ equal to half the molar amount of S₂O₃^2- we added. The average rate at which I₂ is produced during this time (∆ [I₂]/ ∆t) is thus just [S₂O₃^2–]_o/∆t (the subscript '_o' indicates 'initial value').

To summarize how this all works: various amounts of the reactants are mixed together, along with a small amount of S₂O₃^2-, which suppresses the I₂ produced initially. When the S₂O₃^2- is all used up, the purple color appears. Measuring the time it takes for the color to appear gives us the average ∆ [I₂]/ ∆t, which is inverse to the rate because the coefficient for I₂ in the balanced equation is 1. Since two thiosulfate ions react with one I₂ molecule, the final expression for the initial rate of reaction is:

Rate = - ∆ [I₂]/ ∆t = - ∆ [S₂O₃^2-]_o/∆t.

Procedure

A. Concentration effects

For each run, pipet the exact amount of solution indicated on the chart. Solutions 1 and 2 should be added to a 125 mL flask (A) and all the other solutions to another flask (B). Carefully pour flask B into flask A and begin timing, while swirling the flask. If you spill, start again. Record the time change (Dt) of the reaction by noting when a pale purple color first appears constant throughout the solution. THE SOLUTION WILL KEEP TURNING PURPLE. It is important that you are consistent in taking a time measurement at the same purple color each time.

Solution		Run amount (mL)
		1^*		2		3		4		5		6		7		8		9		10		11
.20M KI		20		15		10		5		20		20		20		15		10		5		20
.0050 M Na2S2O3 in a .4% starch solution		10		10		10		10		10		10		10		10		10		10		10
.20 M KCl				5		10		15		5		10		15
.1 M K2SO4																5		10		15
.1 M CuSO4																						1 drop
.1 M K2S2O8	20		20		20		20		15		10		5		20		20		20		20

Time-

*Perform reaction 3 times to practice consistency and take an average time for this run.

B. Temperature effects (if we have time)

To study the effects of temperature you will repeat run #1 at 2 more temperatures. Prepare flask A and flask B the same way as before but then you will heat or cool both flasks to the same desired temperature. It will then be assumed that quick mixing will result in a uniform reaction temperature. Measure the reaction temperature immediately after the lasting pale purple color.

Perform the reaction at 40^oC by heating both flasks. You can do this by heating directly on a hot plate, but you must be VERY careful not to heat to hot. Alternatively you can try running hot water over the flask (this may not get the mixture hot enough.) To lower the temperature, submerge the flasks into an ice bath.

Record the time elapsed for each reaction.

Data

Record all your information in data tables. Your reaction table should also be in your notebook.

Calculations/Questions

1) Calculate the rate of each reaction using the method described above in the introduction. Remember we are judging the rate by the reciprocal time for disappearance of thiosulfate and hence appearance of purple Iodine.

1) You must calculate [A] for each chemical in all your runs (they were diluted

when you added other reagents!!...its Ok to just cross out the amount and

replace them with concentrations in the above table)

2) Determine the order of I^-, S₂O₈^2-, and SO₄^2-.

3) Write the rate law for the reaction.

4) Select the mechanism from above that is consistent with this rate law.

5) Explain why (justify) this mechanism is the appropriate choice.

6) Solve for the rate constant (using the definition of rate and the concentration values at a given condition).

7) Did the catalyst have a significant effect on the rate of reaction? Explain.

8) Use runs at two different temperatures to calculate the activation energy of the uncatalyzed reaction (if we have time).

To do this use the Arrhenius equation at two different temperatures

k₁ = Ae ^-E_a ^{/ RT}₁ at T₁ and

k₂ = Ae ^-E_a ^{/ RT}₂ at T₂ .

take the ln of both sides of each equation and then SUBTRACTING the two equations will give you an expression that contains no A factor. This is similar to what we did with the Clausius Clapeyron equation when we subtracted the two equations. Try and figure this out! You can do it if you take your time.

1) Comment on temperature effects on the rate of reaction