| 1. | Safety |
| 2. |
Objectives/Overview |
| 3. |
Procedures |
|
4. |
Observations |
| 5. | |
| 6. | |
| 7. | Conclusions |
| 8. | Calculations/Set-Ups |
| 9. | Grading Scale |
| 10. | Review Prelab Questions |
|
11. |
Review Postlab Questions |
Experiment
15
Kinetics:
Rates of Reaction and Activation Energies
OBJECTIVES/OVERVIEW
Kinetics
is the study of the rates of chemical reactions.
Chemists characterize reactions by order depending upon how
the reaction rate varies with the concentration of reactants. The rate law is an equation expressing the observed behavior
of a chemical reaction in terms of the powers or exponents of the
concentrations of reactants needed to quantify the observed rates
of reaction. The order
of a chemical reaction with respect to a reactant is the exponent
to which that reactant's concentration must be raised to match the
reaction's rate behavior. Assume the following chemical reaction.
C
→ products
One of the sets of equations below will apply depending on whether the reaction is zero, first or second order for the disappearance of the reactant. The observed dependence of the rate of reaction on concentration and dependence of concentration on time follows the equations below in differential and integrated forms:
, is the
instantaneous decrease in concentration (–DC)
per change in time (Dt);
[Ct] is the concentration of reactant C at time t, [Co] is the initial
concentration of reactant C (concentration when t = 0), and k is the
value of the rate constant. The equations on the right are linear equations
of the form: y = b + mx, where the rate constant k can be found from
the slope m. The time (t) is the independent variable (x). The dependent
variable (y) and intercept b are functions of the concentration as tabulated
below: 
The correct order for a reaction can be determined by plotting graphs of the functions (y) above versus time and observing which one of the functions produces a straight line. For example, if plotting [C] vs. t gives a straight line, the reaction is zero order.
Consider the following example:
|
Time
(t), min |
Concentration ([Ct]),
M |
Time
(t), min |
Concentration ([Ct]),
M |
|
0 |
1.50 |
50 |
0.90 |
|
10 |
1.38 |
60 |
0.78 |
|
20 |
1.26 |
70 |
0.66 |
|
30 |
1.14 |
80 |
0.54 |
|
40 |
1.02 |
90 |
0.42 |
The three functions have been calculated and are tabulated below:
|
|
Zero
Order |
First
order |
Second
Order |
| Time
(t), min |
([Ct]), M |
ln[Ct] |
1/[Ct] |
|
0 |
1.50 |
0.405 |
0.667 |
|
10 |
1.38 |
0.322 |
0.725 |
|
20 |
1.26 |
0.231 |
0.794 |
|
30 |
1.14 |
0.131 |
0.877 |
|
40 |
1.02 |
0.02 |
.980 |
|
50 |
0.90 |
-0.11 |
1.1 |
|
60 |
0.78 |
-0.25 |
1.3 |
|
70 |
0.66 |
-0.42 |
1.5 |
|
80 |
0.54 |
-0.62 |
1.9 |
| 90 |
0.42 |
-0.87 |
2.4 |
The
graph shows all three functions versus time. Inspection of the three lines reveals
that the zero order plot ([C] versus t) is the one that follows a
straight line. Notice
that we need to observe the reaction through at least one, and preferably
for more than two half-lives to confirm the non-linearity of the other
plots. (Also note that the slope is negative
for zero and first order plots but positive for the second order plot.)
Having determined that the reaction
order is zero for the reaction above, we may now evaluate the rate constant
k for this reaction by determining the value for the constant slope
of the line. The slope is defined as the change in y divided by the
change in x or 
whereDy =(yf
- yi) and Dx
= (xf - xi). The graph of concentration versus time is re-drawn below
with arbitrary xi, yi, xf and yf values chosen to maximize the range
of these values.This will minimize the effect of any errors in measurement
on the results of the calculations. Note that the chosen values correspond
to line values that are NOT data points. The line has been drawn to
best represent all the data, whereas individual points are subject to
random experimental error that would bias the data.
| |
Dy =
yf
- yi
= 0.36 - 1.44 = -1.08 Dx =
xf
- xi
= 95.0 - 5.0 = 90.0 slope = k
= - slope = -0.0120 |
The rate constant k for this zero order reaction is equal to minus
the slope:
Reaction rate constants vary with temperature in a manner predicted by the Arrhenius equation which is given in two forms below:
where Ea defines the activation energy and A defines the frequency factor for the reaction. There are two unknowns in these equations (Ea and A); therefore, we must have at least two and preferably more data values to evaluate these constants. The logarithmic form of the equation will be used since it is of the same linear form dealt with previously (y mx) where the independent variable x corresponds to the reciprocal of the absolute temperature (1/T) and the dependent variable y corresponds to the natural logarithm of the rate constant [ln k] measured at that temperature. The slope of this line will then have the value ÐEa/R where R is the Ideal Gas Constant. Thus, we have the equation slope = ÐEa/R. (Although the y intercept of the plot is equal to the natural logarithm of the frequency factor (ln A), A is rarely determined graphically because the x axis does not extend to zero.) Another form of the Arrhenius equation is useful when comparing rates at different temperatures: ln(k2/k1) = Ea/R (1/T1-1/T2).
The zero order rate constant calculated previously has been combined with two additional values at different temperatures in the table below. The additional functions needed to construct the graph are also tabulated.
|
Temp., ¼C |
Temp. (T), K |
1/T, 1/K |
Rate constant (k), M/min |
ln(k) |
|
25.0 |
298.1 |
0.003354 |
0.0120 |
-4.423 |
|
35.0 |
308.1 |
0.003245 |
0.225 |
-3.794 |
|
45.0 |
318.1 |
0.003143 |
0.406 |
-3.204 |
The appropriate graph is plotted below using 1/T as the independent variable x and ln(k) as the dependent variable y. (Note that to obtain greater accuracy in determining the slope, the (1/T) and ln k values do not extend to zero on the graph.)
Again points may be selected from the line to calculate the slope of the data. Choosing xi = 0.00310 (Ð1/K), yi = -2.98 and xf = 0.00340 (Ð1/K), yf = - 4.63 gives us a Dx of 0.00030 (-1/K) and a Dy of ‑ 1.65. The resulting slope is calculated to be ‑ 5.5 x 103 K, from which we may calculate the activation energy
