Determining the Rate Law for the Crystal Violet-Hydroxide Ion Reaction
Adam Capriola
CHM 1122 Section 155
February 14, 2007
Introduction
The kinetics of a chemical equation is determined by its rate. The rate is the speed at
which the reactants form into products. The rate is dependent on the concentrations and the
orders of the reactants. One way to find the order is by first measuring the concentration of the
products as time passes. A spectrophotometer is one tool that can measure relative concentration
if the reactants change color as the form products. Graphing absorbance versus time,
ln(absorbance) versus time, and 1/absorbance versus time will determine the order depending on
which graph produces a straight line. In this experiment, crystal violet and NaOH form a
complex that changes from transparent blue to colorless over time. The absorbance is measured
using a spectrophotometer, and the rate law is then determined using this information.
Experimental
First, a spectrophotometer was turned on and set at a wavelength of 595 nm. Next, a
cuvet was obtained, rinsed, and filled with deionized water. The outside of the cuvet was
cleaned Kimwipe to get rid of smudges. The cuvet was then inserted into the spectrophotometer
and the spectrophotometer was zeroed. Next, 10.0 mL of 0.010 M NaOH solution was dispensed
into one clean 25 mL graduated cylinder and 10.0 mL of 1.50 x 10-5 M crystal violet solution
was dispensed into another clean 25 mL graduated cylinder. The solutions were then
simultaneously poured into a clean 50 mL beaker. This mixture was mixed with a glass stirring
rod for a few moments to ensure consistency. The cuvet was then rinsed with the mixture two or
three times and was then filled with the mixture. The cuvet again cleaned with a Kimwipe and
was inserted into the spectrophotometer. The absorbance reading was measured every minute
for twenty minutes, starting when the cuvet was first put in. This process was then repeated,
replacing the 0.010 M NaOH solution with 0.020 M NaOH solution.
Results
0.010 M NaOH Solution:
Time
Absorbance
0
0.743
1
0.728
2
0.713
3
0.703
4
0.691
5
0.680
6
0.673
7
0.664
8
0.654
9
0.644
10
0.636
11
0.624
12
0.610
13
0.603
14
0.593
15
0.580
16
0.571
17
0.560
18
0.548
19
0.539
20
0.530
lnA
-0.297
-0.317
-0.338
-0.352
-0.370
-0.386
-0.396
-0.409
-0.425
-0.440
-0.453
-0.472
-0.494
-0.506
-0.523
-0.544
-0.560
-0.580
-0.601
-0.618
-0.635
1/A
1.35
1.37
1.40
1.42
1.45
1.47
1.49
1.51
1.53
1.55
1.57
1.60
1.64
1.66
1.69
1.72
1.75
1.79
1.82
1.86
1.89
Absorbance vs. Time
0.8
Absorbance
0.7
0.6
0.5
y = -0.0104x + 0.7364
R2 = 0.9985
0.4
0.3
0.2
0.1
0
0
5
10
15
20
25
15
20
25
Time
lnA vs. Time
0
-0.1
0
5
10
lnA
-0.2
-0.3
-0.4
-0.5
-0.6
y = -0.0165x - 0.2979
R2 = 0.9971
-0.7
Time
1/A
1/A vs. Time
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
y = 0.0264x + 1.3326
R2 = 0.9906
0
5
10
15
20
25
Time
0.020 M NaOH Solution:
Time
Absorbance
0
0.697
1
0.650
2
0.606
3
0.569
4
0.530
5
0.492
6
0.459
7
0.428
8
0.399
9
0.373
10
0.351
11
0.328
12
0.308
13
0.288
14
0.269
15
0.254
16
0.236
17
0.221
18
0.206
19
0.192
20
0.179
lnA
-0.361
-0.431
-0.501
-0.564
-0.635
-0.709
-0.779
-0.849
-0.919
-0.986
-1.047
-1.115
-1.178
-1.245
-1.313
-1.370
-1.444
-1.510
-1.580
-1.650
-1.720
1/A
1.43
1.54
1.65
1.76
1.89
2.03
2.18
2.34
2.51
2.68
2.85
3.05
3.25
3.47
3.72
3.94
4.24
4.52
4.85
5.21
5.59
Absorbance vs. Time
0.8
Absorbance
0.7
0.6
0.5
0.4
0.3
0.2
y = -0.0251x + 0.6333
R2 = 0.9646
0.1
0
0
5
10
15
20
25
15
20
25
Time
lnA vs. Time
0
-0.2 0
-0.4
5
10
lnA
-0.6
-0.8
-1
-1.2
-1.4
-1.6
-1.8
-2
y = -0.0675x - 0.3685
R2 = 0.9998
Time
1/A vs. Time
6
y = 0.2008x + 1.0732
R2 = 0.9695
5
1/A
4
3
2
1
0
0
5
10
15
20
25
Time
Discussion/Conclusions
For the graphs using 0.010 M NaOH, the plot of Absorbance vs. Time had the straightest
line (R2 = 0.9985), but the plot of lnA vs. Time also had a very straight line (R2 = 0.9971). The
plot of 1/A had the least straight line with R2 equaling 0.9906. For the graphs using 0.020 M
NaOH, lnA vs. Time had the straightest line (R2 = 0.9998). Absorbance vs. Time and 1/A vs.
Time were not nearly as straight, with their R2 equaling 0.9646 and 0.9695, respectively. I think
it is fairly safe to say that the rate equation is first order because lnA vs. Time overall yielded the
straightest line in the two runs. The rate equation is rate = k [NaOH].