PARTIAL LAB REPORT: RATE LAWS DATA

F21UTA-742
UTA-742
Chemical Kinetics: Determining the Rate Law for a Chemical
Reaction
©University of Texas at Arlington, 2021.
Objective: The rate law for the reaction between bleach (sodium hypochlorite) and blue food
dye (FD&C Blue No. 1) will be determined using a modern spectrophotometer and standard
techniques.
INTRODUCTION
Many food colorants are synthetic dyes derived from coal tar that have been approved by the
FDA for use in foods, drugs, and cosmetics, based on numerous studies in animals. They are
commonly added to improve the appearance or appeal of a given food product, e.g., Fruit Loops!
FD&C Blue No. 1, also known as Brilliant Blue FCF (or Acid Blue 9), is a commonly used watersoluble dye whose line structure is shown below. It can be combined with red dyes to produce
various shades of green and is often found in ice cream, cereals, dairy products, sweets, and drinks.
It is also used in soaps, shampoos, and other cosmetics applications.
A common feature of these dyes is the presence of
extended conjugated bonds (alternating single and
SO3double bonds highlighted by the bold bonds in the
figure on the left,) which is responsible for their color.
O 3S
SO3- One way to get rid of the color exhibited by these
dyes is to break-up the conjugation. In fact, normal
N
N
household bleach removes stains by reacting with
C2H5
C2H5
double bonds in colored molecules to render them
Brilliant Blue (No. 1)
colorless. Household bleach is a 5-8% weight to
volume solution of sodium hypochlorite (NaOCl) in water. The solution is made basic (pH >10) by
addition of sodium hydroxide (lye) which prevents the hypochlorite ion (OCl-) from decomposing.
The reaction of bleach with brilliant blue is shown in Equation 1 and results in the loss of the blue
color as the resulting product has had the conjugation disrupted at the site of chlorination and is now
colorless.
SO3-
SO3Cl
–
O 3S
SO3N
N
C2H5
C2H5
Brilliant Blue (No. 1)
+
–
SO3-
O3S
–
OCl
N
N
C2H5
(colorless)
C2H5
(colorless)
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(1)
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The hypochlorite ion is a strong oxidant that will attack many molecules with conjugated bonds
and thus the mechanism by which bleach cleans clothes is often simply by decolorizing them. The
oxidation process also helps to solubilize the ‘dirt’ molecules and thus bleach helps to remove these
from clothes.
The reaction shown in equation 1 tells us the reactants and the products and their stoichiometry
but it does not tell us the mechanism of this chemical reaction (intimate chemical steps) nor the rate
(speed) at which this reaction occurs. Chemical kinetics involves the study of the rate and
mechanism of a chemical reaction. A general rule of thumb says that reaction rates increase with
increasing concentration of the reactants and with increasing temperature. While these rules are
generally true, they are not always so.
We can determine how concentrations affect the rate of a reaction by experiment and relate them
quantitatively using the rate law for that particular reaction. Consider the hypothetical reaction
aA + bB → cC + dD. The rate law for this reaction is written as:
rate =
where
d[M]
= k [A]x[B]y
dt
(2)
M 
(i.e., rate of formation or consumption of a specified species)
t
k = rate constant, is independent of concentrations, but dependent on temperature
x, y = order of reaction (x + y = overall order of reaction)
rate =
Note, that the coefficients a,b,c and d do not appear in the rate law. A common mistake by many
general chemistry students is to assume that the values of x and y should be the same as a and b.
While these values can be the same they are not always the same and x and y must be determined
experimentally.
In this reaction, Brilliant Blue, hereafter simply referred to as B, reacts with hypochlorite ion
(OCl-) to yield the colorless product shown on the right in equation 1. Substituting the two reactants
into equation 2 gives us the rate law (equation 3) in which x, y and k remain to be determined. These
quantities are arrived at experimentally and cannot simply be inferred from the balanced chemical
reaction.
rate =
-d[B]
dt
= k [B]x[OCl-]y
(3)
So the big problem is how do we monitor the rate of a chemical reaction if we cannot see the
molecules? In this case, we can take advantage of the change in color (from blue to colorless) that
happens when B is converted to product. Recall from one of your earlier experiments that the
absorbance (A) of a particular substance is related to its concentration by Beers Law, shown in
equation 4:
A = bc
(4)
where A = absorbance (units are in optical density OD),  = molar extinction coefficient (units are
L/mol.cm, or M-1cm-1), b is the cuvette pathlength in cm (and is 1 cm for this experiment) and c is
the concentration of the absorbing species (units are mol/L, or M).
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The important thing here is that the absorbance or intensity of the color is proportional to the
concentration of the colored species and therefore by monitoring the change in absorbance versus
time we can measure the change in concentration of the colored species versus time (the rate -d[B]/dt
shown in equation 3). Since we are following the rate at which something disappears (which will be
a negative number) we put a negative sign in front, so the final rate we report will be a positive
number, e.g. if you’re walking backward you could be said to be going at – 3 mph but if all we care
about is how fast your moving we convert this -(-3 mph) = +3 mph so as to report a positive number.
We will use a UV-Visible spectrophotometer to record the absorbance spectrum of B, to
determine its molar extinction coefficient (), and finally, to determine the rate of reaction 1 under
varying conditions. From this data, you will determine the order of the reaction in [B] given by x,
the order of the reaction in [OCl-] given by y, and the rate constant k and thus report the actual rate
law in equation 3 with real numbers.
An example data set is shown below for the reaction of a different dye, crystal violet (CV+), with
hydroxide ion (OH-) which is shown in equation 5 and the corresponding rate law in equation 6.
+
CV
(violet)
+
–
(5)
CV-OH
(colorless)
OH
-d[CV+]
rate = dt
= k [CV+]x[OH-]y
1.2
Absorbance (a. u.)
In step 1, the students measured the
absorption spectrum of the CV+ dye in the
visible region by preparing a solution of
known concentration (in this case the [CV+]
= 6.48 x 10-6 M ) and measuring the
absorbance versus wavelength spectrum
using a 1 cm pathlength cuvette. Since the
visible region is approximately 400 to 700
nm this is the region set to be recorded by the
spectrophotometer. The resulting spectrum
is shown in Figure 1.
(6)
1.0
0.8
0.6
0.4
0.2
0.0
400
450
500
550
600
650
700
Wavelength (nm)
Figure 1. Absorption Spectrum of CV+ in water.
(Arrow shows the analytical wavelength of 588 nm)
As can be seen the CV+ spectrum has
an intense peak at 588 nm. If we measure the
absorbance at this peak (the so-called
analytical wavelength) then we can
determine the  at this wavelength by substituting the absorbance (A) at 588 nm (A = 1.23),
c = [CV+] = 6.48 x 10-6 M, and b = 1 cm into equation 4.
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1.23 OD =  (1 cm)(6.48 x 10-6 M)
(7)
 = 1.90 x105 M-1cm-1
(8)
From this preliminary experiment, the students learned two things. First, the wavelength at which to
measure the disappearance of the colored species (the analytical wavelength,) should be 588 nm.
Second, the molar extinction coefficient () for CV+ at this wavelength is 1.90 x 105 M-1cm-1. The
students can now relate changes in absorbance to changes in concentration quantitatively!
In step 2, the students set out to determine the rate law for the reaction shown in equation 5.
Working solutions of CV+ and OH- were prepared with known concentration.
For example:
CV+ solution = 12.9 x10-6 M
OH- solution = 3.49 x10-2 M
Two experimental runs were planned using the volumes shown in Table 1.
Table 1
Run
Volume
CV+
solution
(mL)
Volume
OHsolution
(mL)
Volume
Water
(mL)
Total
Volume
(mL)
Initial [CV+]
(M)
Initial [OH-]
(M)
1
2
10.0
10.0
10.0
5.0
0
5.0
20.0
20.0
6.45 x10-6
6.45 x10-6
1.74×10-2
8.72×10-3
The concentrations of CV+ and OH- in the initial reaction mixture were determined using the dilution
formula
McVc = MdVd
(9)
For example:
In runs 1 and 2, the initial [CV+] is calculated as follows:
(12.9 x10-6 mol/L)(0.010 L) = (Md)(0.020L)
Md = 6.45 x10-6 M
In run 1, the students prepared the spectrophotometer for a time course measurement so that the
spectrophotometer would measure the absorbance at 588 nm every 5 sec and display the results as
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shown in Figure 2. Run 1 was started by mixing the 10 mL of CV+ with the 10 mL of OH- in a small
beaker, quickly transferring some of this solution into a clean, dry cuvette and immediately putting
this cuvette into the spectrophotometer and starting the automated data collection (on the PC).
Run 2 was conducted similarly except the [OH-] was halved, as shown in Table 1.
Figure 2. Decrease in absorbance at 588 as a function of kinetic Run 1.
From the data obtained in these two runs it is possible to determine the order of the reaction with
respect to both CV+ and OH- as well as the rate constant, k.
CALCULATIONS
After each run, the student recorded the following data on their data sheet from the computer. Even
though the computer recorded data every 5 s, they were told to record the absorbance data every 30 s
as this is enough data to get a good graph but not so much as it will take an excessive amount of time
to do all the calculations. The raw data (read from the spectrometer) from Run 1 was collected in
Table 2 and is shown in italics. This data was processed on a spreadsheet (MS Excel) and the
remaining columns calculated. The [CV+] was determined using Beers law: each value of Abs by 
(1.9×105 M-1cm-1) and b (1 cm). ln[CV+] and 1/[CV+] were calculated from the [CV+] data. The
reasons for these calculations are explained in the next section.
In addition to the data every 30 s as described above, the student should record all the data points
(every 5 seconds) over the first 30 seconds of the run. We will use this data to approximate the initial
rate of the reaction.
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Table 2: Data from Run 1
Time
(s)
Abs
(OD)
[CV+]
(x10-6 M)
ln[CV+]
1/[CV+]
(x105)
0
5
10
15
20
25
30
60
90
120
150
180
210
240
270
300
330
360
390
420
1.17
1.14
1.10
1.06
1.02
0.988
0.946
0.802
0.667
0.553
0.461
0.385
0.321
0.270
0.226
0.191
0.160
0.136
0.116
0.101
6.15
6.00
5.79
5.58
5.37
5.20
4.97
4.22
3.51
2.91
2.42
2.02
1.68
1.42
1.18
1.00
0.842
0.715
0.610
0.531
-12.0
-12.0
-12.0
-12.1
-12.1
-12.2
-12.2
-12.4
-12.6
-12.7
-12.9
-13.1
-13.3
-13.5
-13.6
-13.8
-14.0
-14.1
-14.3
-14.5
1.62
1.67
1.73
1.79
1.86
1.92
2.00
2.36
2.84
3.43
4.12
4.93
5.91
7.03
8.40
9.94
11.8
13.9
16.3
18.8
I. Determining the order, x, with respect to CV+
There are several ways to analyze the data to
determine the order of the reaction with respect to a
particular reactant. In the CV+ experiment, the
[CV+] is much smaller than [OH-] such that the
change in [OH-] over both runs is almost
negligible. Under such conditions, the [OH-] is
essentially constant and therefore it is OK to
approximate the rate law as:
+
rate =
-d[CV ]
+ x
= k [CV ]
dt
(10)
where the constant [OH-] is factored into the rate
constant k.
This simpler expression (10) can be
integrated to give us the integrated rate law under
situations where x is fixed at 0,1, or 2. Other
values of x are possible but these will be sufficient
for this case, as well as for your experiment.
When x = 0 the integrated rate law is:
[CV+] = kt + [CV+]o
(11)
When x = 1, the integrated rate law is:
ln[CV+] = -kt + ln[CV+]o
(12)
And when x = 2, the integrated rate law is:
1
1
+ = kt +
[CV ]
[CV+]o
(13)
The derivation of these three integrated rate laws is given in the kinetics chapter of your general
chemistry textbook. All three expressions have the generic form of a line: y = mx + b. With the
above data, a student can then make graphs of all three possibilities: [CV+] vs. time, ln[CV+] vs.
time, and 1/[CV+] vs. time (Graphs 1-3, respectively) of which only one will be linear. As can be
seen with this data, only graph 2 results in a straight line, therefore we can deduce the order x of the
reaction with respect to CV+ is 1 (x = 1) because this is the only one that obeys one of the rate laws
expressed in equations 11, 12, and 13 (by being linear!).
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-6
Graph 2
-12.5
-13.0
-13.5
-14.0
0 100 200 300 400
time (s)
1.6×10
1/[CV+]
5
4
3
2
1
-12.0
Graph 1
ln[CV+]
[CV+]
6×10
6
Graph 3
1.2
0.8
0.4
0
100 200 300 400
time (s)
0
100 200 300 400
time (s)
[CV+]
II. Determining the order, y, with respect to OHThe order of the reaction with respect to
-6
6×10
[OH-] was determined by comparing the
G ra ph 4
initial rate of the reaction in Run 1 and Run
5
2. The initial rate is determined by
4
drawing a line tangent to the curve at time
+
= 0 in the [CV ] vs time plot (Graph 4) and
3
determining the slope of this line. In our
experiment we will approximate the
2
initial rate by determining the slope of
1
the best-fit-line on the data from the first
30 seconds (taken every 5 s) of the run.
0
100
200
300
400
The slope of this line was determined to be
-8
time
(s)
-4.50×10 M/s. From equation 6, we see
that the initial rate is –d[CV+]/dt for Run 1, the initial rate is 4.5×10-8 M/s. The data for Run 2 is not
given here, but when the students plotted [CV+] vs time for Run 2 and determined the best-fit-line
over the first 30 s of the run, they found a slope of -1.18×10-8 M/s and therefore the initial rate for
Run 2 is 1.18×10-8 M/s.
With these initial rates, we can now determine the order with respect to [OH-] by using the
following analysis. The ratio of Rate 1 to Rate 2 is equal to the ratio of their rate laws as shown in
equation 14, where the subscripts 1 and 2 indicate Run 1 and Run 2.
x
y
k[CV]1 [OH]1
rate 1
=
x
y
rate 2
k[CV]2 [OH]2
(14)
Since the value of [CV+]1 = [CV+]2 and k is constant, these cancel out as shown in equation 15, to
give an expression only dependent on the [OH-].
k[CV]1x[OH]1y
[OH]1y
rate 1
=
x
y =
k[CV]2 [OH]2
rate 2
[OH]y2
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(15)
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Plugging in the data for rate 1, rate 2, [OH-]1 and [OH-]2, we obtain equation 16.
[OH-]
rates
(1.74×10-2)y
4.50 x 10-8
= (8.72×10-3)y
1.18 x 10-8
1.74×10-2 y
8.72×10-3
(16)
3.81 = 2y
y must be an integer value
therefore round 3.81 to 4
4 = 2y
y=2
Thus, we have now solved for both x and y and the rate law is:
Rate = k [CV+]1[OH-]2
(17)
III. Determining the rate constant k
The rate constant k is now easily determined from the data in Table 1 and the initial rates.
For Run 1:
4.50×10-8 M/s = k (6.45 x10-6 M)1(1.74 x 10-2 M)2
k1 = 23.0 M-2s-1
For Run 2:
1.18×10-8 M/s = k (6.45 x10-6 M)1(8.72 x 10-3 M)2
k2 = 24.1 M-2s-1
Average these two values:
kavg = 23.6 M-2s-1
Now the full rate law can be given as:
Rate = 23.6 M-2s-1 [CV+]1[OH-]2
48
(18)
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In this experiment, you will determine the full rate law expression for the reaction of brilliant blue
(B) with common bleach using the same type of data collection and analyses as was done for the
crystal violet (CV+) reaction.
Hazardous waste disposal: Use the large plastic beaker from your drawer to collect all waste
solutions from today’s experiment. This includes all kinetic runs as well as any extra, unreacted
bleach or dye. At the end of the experiment all this should be disposed of in the Halogenated Waste
container.
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