Determination of a Rate Law
INTRODUCTION: The rate of a chemical reaction may depend
upon reactant concentrations, temperature, and the presence of catalysts. At a
fixed temperature, the rate of a reaction may be mathematically described by a rate
law. Consider the following hypothetical reaction:
A + 2 B + 3 C 
Products
(1)
The differential rate
law for this reaction represents the rate of change in concentration of
a reactant and has the following form:
rate = -d[A]
= k[A]x[B]y[C]z
(mol L-1 s-1)
(2)
dt
where k is the rate constant
and x, y, and z are the orders with respect to each reactant. The
overall order of the reaction is the sum of x+y+z.
Since A is disappearing, the derivative has a negative sign in order to make
the rate a positive quantity.
The following reaction is to be
investigated in this experiment:
H2O2(aq)
+ 2 I-(aq) + 2 H+(aq) 
2 H2O(l) + I2(aq)
(slow)
(3)
Under the conditions of the
experiment, the iodide and hydrogen ions are present in large excess.
Consequently, their concentrations remain effectively constant during the
reaction and do not appear on the right-hand side of the rate law (they are
included in the rate constant). The rate law will thus depend only upon the
concentration of hydrogen peroxide:
-d[H2O2]
= k [H2O2]n
(4)
dt
We will determine the order n
and the value of the rate constant k. Although some reactions
have quite complicated rate laws with fractional orders, we will test only for
the relatively simple cases of n = 0, 1, or 2. In order to do this, it is
necessary to integrate the rate law.
-d[H2O2]
= k dt
(5)
[H2O2]n
The exact form of the integrated
rate law depends upon the value of n. When n = 0 (zero order), the integrated
rate law is as follows:
[H2O2] = [H2O2]0
- kt
(6)
When n = 1 (first order), the
integrated rate law is as follows:
ln[H2O2]
= ln[H2O2]0
- kt
(7)
When n = 2 (second order), the
integrated rate law is as follows:
1
=
1
+ kt
(8)
[H2O2]
[H2O2]0
Note that each integrated rate
law has the form of the equation of a straight line, y = b + mx. Only one of the three equations 6-8 will actually fit
the concentration data for the reaction (i.e., only one of the three possible
orders can be correct). If one can measure the concentration of hydrogen
peroxide remaining at various times during the reaction, the data may be
plotted in the form of each of the three integrated rate laws; only one graph
will actually be a straight line, thus indicating the correct order. The value
of the rate constant k will then be related to the slope of the graph (note
that k is always a positive number, and has units which depend upon the order).
Since it is not possible to
directly analyze for the hydrogen peroxide concentration while the reaction is
taking place, we will instead analyze for the amount of iodine formed and use
the stoichiometric relationships to determine how
much hydrogen peroxide remains. After initiating the reaction, aliquots of
standard sodium thiosulfate solution are added to the
reaction mixture at intervals. The thiosulfate ion
reacts with the iodine formed by the principal reaction as follows:
I2(aq)
+ 2 S2O32-(aq) 
2 I-(aq) + S4O62-(aq)
(fast)
(9)
Since this reaction is much
faster than the main reaction, any iodine formed by reaction (3) is immediately
consumed by the thiosulfate ion. In addition, starch
is added to the mixture to indicate the presence of molecular iodine:
I2(aq)
+ starch
blue complex
(fast)
(10)
The method of analysis is
as follows. The principal reaction (equation 3) is initiated in the presence of
a starch solution and a timer is started. The mixture immediately turns blue as
iodine is formed (see the rising line beginning at zero time in Figure 1; the
rising lines indicate the formation of I2, and the vertical lines
indicate the consumption of I2 by thiosulfate).
An aliquot of standard sodium thiosulfate solution is
quickly added from a burette (first vertical part of the trace on the graph).
This amount of thiosulfate is deliberately in excess
and will immediately consume all of the iodine formed by the principal
reaction; the mixture thus becomes colorless. There is now an excess of thiosulfate ion present in the flask. As more and more
iodine is produced (second rising portion of the graph), the fixed amount of thiosulfate is eventually consumed. At this point, the blue
color reappears and the time is noted; the appearance of the blue color
signifies that the exact stoichiometry of equation 9 has
been attained (2 mmol thiosulfate:1
mmol I2).
Figure 1. Relative Amounts of I2 and S2O32-
During the Experiment
The total amount (in millimoles) of iodine produced at time t can
then be calculated from the total amount of thiosulfate
used up to that point (the total amount added prior to the color change,
since the beginning of the reaction). The total number of moles of
iodine produced is then used to calculate the total number of moles of
hydrogen peroxide consumed up to that time (see stoichiometry
of equation 3); this number is then subtracted from the initial number of moles
of hydrogen peroxide, since the number of moles of hydrogen peroxide remaining
is needed. Another excess of thiosulfate is added
(second vertical portion of the graph) and the time is again noted when the
blue color reappears (second rising portion of the graph); this process is
repeated several times to obtain sets of concentration-time data for fitting to
the three assumed rate laws. The initial amount of hydrogen peroxide is
determined by allowing the reaction to proceed to completion and determining
the total amount of iodine produced, then relating this to the total amount of
hydrogen peroxide which was initially present.
Example: The third appearance of
the blue color (fourth data point, counting zero time as the first) was at 141
seconds, after 1.73 mL total (since the
beginning of the experiment) of 0.1283 M sodium thiosulfate
had been added. There were initially 0.543 mmol
of hydrogen peroxide present in the flask. The total amount of thiosulfate added is (1.73 mL)(0.1283 mmol/mL) = 0.222 mmol. The amount of hydrogen peroxide consumed
is one-half of this amount, or 0.111 mmol. The amount of hydrogen peroxide remaining
is (0.543 - 0.111) = 0.432 mmol. The total
volume in the flask is now 206.73 mL, so the molarity of hydrogen peroxide remaining is 0.432 mmol / 206.73 mL = 0.00209 M.
EQUIPMENT: 50-mL burette, 20-mL pipet, 100-mL and 10-mL graduated cylinders, 500-mL
Erlenmeyer flask, timer, burette funnel, pipet bulb
EXPERIMENTAL: Fill a clean burette with about 20 mL of the standard sodium thiosulfate
solution, making sure that the tip is filled. Record the molarity
of the solution and the initial burette reading (to the nearest 0.01 mL).
Place 150.0 mL
of deionized water, 10.0 mL
of 1.0 M sulfuric acid solution, and 5.0 mL of
starch solution in a clean, dry Erlenmeyer flask. Pipet
in exactly 20.00 mL of hydrogen peroxide solution and
mix well.
Add 20.0 mL
of 1.00 M potassium iodide solution from a graduated cylinder as quickly
as possible, starting the timer at the same time as the addition. The
timer must run continuously throughout the experiment.
Immediately add about 1.2 mL of sodium thiosulfate solution
from the burette and record the burette reading. The blue color will disappear.
When the solution turns blue again, record the timer reading while the timer
continues to run.
Add another increment of sodium thiosulfate and again record the volume and the time at
which the color reappears. Continue to repeat this process until ten data
points have been recorded. The size of the added increments of sodium thiosulfate should be steadily decreased in order to have
the time intervals remain roughly constant.
Stop the timer and let the
mixture stand for one hour. This will be sufficient time for the reaction to
reach completion. Carefully titrate the solution with the remaining
sodium thiosulfate in your burette until the blue
color just disappears (it should not return). Record the final
burette reading, then clean up all of your equipment.
CALCULATIONS: Perform calculations in millimoles.
Note that molarity is the same value either in mol/L
or number of millimoles of solute per milliliter
of solution, so number of millimoles = (Molarity)(Volume in mL).
1. First calculate the initial
(total) number of millimoles of hydrogen peroxide
present from the total volume of thiosulfate
used in the experiment. Take note of the stoichiometries
of equations (3) and (9).
2. Set up a data table with
columns for total elapsed time (include a point for zero time), the total
volume of thiosulfate used up to that point, the
total volume of the solution in the flask, the number of mmol
of thiosulfate used, the number of mmol of hydrogen peroxide consumed, the number of mmol of hydrogen peroxide, the molarity
of hydrogen peroxide remaining (using the total solution volume), the natural
logarithm of the molarity, and the reciprocal of the molarity; be sure to include units. Show one complete set
of calculations in detail with explanations. The calculations should be done
with a spreadsheet, in which case the formulas used in each column should be
explained in the report (remember that the method of calculation is always
explained in a lab report). If you print the table in landscape format
(sideways on the paper) it should all fit on one page.
3. Prepare graphs of [H2O2]
vs. time, the natural logarithm of peroxide concentration vs. time, and the
reciprocal of peroxide concentration vs. time. Do not plot the data point
corresponding to completion of the reaction. Carry out a linear regression on
the graph which most closely shows a linear relationship. Plot the data
points and best-fit line on this graph; do a marker-plus-line plot
for the other two (i.e., connect the points). Include the point for zero time
on the graphs. Be sure to put a descriptive title (not in Y vs. X form)
on each graph. Explain in your report the correct order and how you determined
it, give the value of the rate constant with correct units and uncertainty, and
give the complete integrated rate law.
4. Calculate the time required
for the reaction to reach 99% completion, i.e., when the number of mmol of peroxide has fallen to 1% of its initial value.
Comment upon the significance of this result with respect to the waiting period
in the experiment.
Remember that the goal of the
experiment is to determine the rate law for the reaction, with an order and a
rate constant with proper units. You should give both the differential
and integrated forms of the rate law in your discussion
REPORT: In addition to reporting your data, calculations,
and the necessary graphs, you will write a conclusion for today’s
experiment. In a conclusion, you briefly
outline the results and conclusions of the experiment.