Group Work #17

Kinetics: 2

Group Work.17 (next to last one!)

If you can do this problem, you understand every concept we've covered in kinetics!

Below are data from 4 kinetic experiments, #1-#4.

Listed are concentration vs. time data for the various species A , B, C and D.

Exp. # Rate_overall time elapsed [A], M Rate_[A] [B], M Rate_[B] [C], M Rate_[C] [D], M Rate_[D]

#1 1.7x10^-3 0 sec 5.00 -3.3x10^-3 2.00 -1.7x10^-3 1.00 0 0.00 1.7x10^-3

60 sec 4.80 1.90 1.00 0.10

#2 3.3x10^-3 0 sec 10.0 -6.7x10^-3 2.00 -3.3x10^-3 1.00 0 0.00 3.3x10^-3

60 sec 9.60 1.80 1.00 0.20

#3 1.7x10^-3 0 sec 5.00 -3.3x10^-3 4.00 -1.7x10^-3 1.00 0 0.00 1.7x10^-3

60 sec 4.80 3.90 1.00 0.10

#4 3.3x10^-3 0 sec 5.00 -6.7x10^-3 2.00 -3.3x10^-3 2.00 0 0.00 3.3x10^-3

60 sec 4.60 1.80 2.00 0.20

a) Write a balanced equation for the reaction. (hint: study the relative concentration changes over a specific time interval.) The reaction that needs to be balanced is:

aA + bB + cC => dD

The additions to the table were calculated using the relationship:

Rate_[species] = D[species]/Dt For example, for exp #1 reactant A:

Rate_[A] = (4.80 M -5.00 M)/(60 s - 0 s) = -3.3x10^-3 M/s

Now use the values in the tables to find the coefficients of the balanced equation:

Rate_overall = -(1/a)D[A]/Dt = -(1/b)D[B]/Dt = -(1/c)D[C]/Dt = (1/d)D[D]/Dt

For example, for exp #1 reactant A:

Rate_overall = -(1/a)D[A]/Dt

1.7x10^-3 M/s = -(1/a)( (4.80 M -5.00 M)/(60 s - 0 s)) = -(1/a)(-3.3x10^-3 M/s)

a = 2 Repeat for each species to get:

2A + B D

Note that the concentration of C does not change so the coefficient is undefined and it appears over the arrow.

b) What are the reaction orders with respect to A, B, and C, and what is the overall reaction order?

Rate_overall = k[A]^x[B]^y[C]^z

Consider each species separately choosing data sets in which one species varies and all others are constant for the same time period. Compare the change in overall rate to determine x, y and z.:

For species A use exp#1(t=0) and exp#2(t=0). The rate doubled from 1.7x10^-3 M/s to 3.3x10^-3 M/s when the concentration doubled, so x = 1.

For species B use exp#1(t=0) and exp#3(t=0). The rate did not change when the concentration doubled, so y = 0.

For species C use exp#3(t=0) and exp#3(t=0). It doesn’t matter that the concentration of B is not constant because the rate does not depend on it. The rate doubled from 1.7x10^-3 M/s to 3.3x10^-3 M/s when the concentration doubled, so z = 1.

The overall reaction order is: x + y + z = 1 + 0 + 1 = 2

c) Write the rate law.

Rate_overall = k[A][C]

d) Is a catalyst involved in this reaction? Briefly justify your answer.

Catalysts are not consumed in a reaction but increase the rate of reaction by lowering the activation energy, E_a. Reactant C is a catalyst because the concentration does not change in the course of the reaction (t=0 s to t=60 s) but the rate of reaction increases with an increase in the concentration of C.

e) Calculate the rate constant for the formation of D.

Choose a data set. For example using data set #3(t = 0 s):

Rate_overall = k[A][C] from part c

1.7x10^-3 M/s = k * 5.00 M * 1.00 M

k = 3.3x10^-4 s^-1

f) Suggest a mechanism consistent with the data.

A + C => A---C Slow: rate determining step (B does not appear)

A + A---C => C---A---C fast (note that A---C and C---A---C are intermediates)

C---A---C + B => D fast

2A + B D

A + C => E Slow: rate determining step (B does not appear)

A + E => F fast (note that E and F are intermediates)

F + B => D fast

2A + B D

Or something of your own making that is made up of elementary steps that add together to yield the correct net reaction and that has a rate determining step that agrees with the rate law (Rate_overall = k[A][C]).