Introduction
This laboratory experiment placed an emphasis on the determination of the order of reaction and ultimately the rate constant and activation energy of the bromate-bromide reaction. The rate equation of this reaction is represented by:
-d[BrO3–] / dt = k [BrO3–]x [Br–]y [H+]z
Hence, by using varying and constant amounts of bromate, bromide, and acid in different trial runs, it is possible to determine the x, y, and z exponents of this equation, which are actually the order of the reactions.
In order to better understand the bromate-bromide reaction, it is important to take note of the stoichiometric equation, which is noted as:
BrO3– + 5Br– + 6H+ —> 3Br2 + 3H2O
This reaction couples with the reaction of Br2 and methyl orange, which produces bleach. By combining the two reactions, the basis is built for determining the rate of the reaction. As bromate and bromide react, they produce Br2, which then reacts with methyl orange to dissipate the color of the solution and show completion of the reaction. This dissipation of color can be timed and used for calculation.
However, this reaction happens very quickly, and under normal circumstances would not be able to be timed. By adding phenol to the solution, the reaction of Br2 with methyl orange can be slowed down with the following side reaction:
C6H5OH + Br2 —> BrC6H4OH + H+ + Br–
This reaction of Br2 with phenol slows down the reaction with methyl orange enough that it is able to be timed. Otherwise, the reaction would happen very quickly and it would be untimable. Nitrate is added to the solution to also improve the quality of the reaction, by increasing the ionic strength of the ion. In theory, the rate equation of the reaction should equal:
-d[BrO3–] / dt = k (a x 2a x (3a)2) = 18ka4
Where a is equal to the original bromate concentration, 2a is the initial bromide concentration, and 3a is the initial hydrogen ion concentration. The plot of -∆[BrO3–] / t versus time at different temperatures should yield linear plots, with the slopes equaling 18ka4. Knowing the value of a, the rate constant (k), can be evaluated at each temperature.
Finally, once the rate constant is found, the activation energy can be determined by using the following equation:
log k = (-E / 2.303R) x (1 / T) + constant
By plotting the log k versus 1 / T° K for each temperature, the slope yielded also will provide the activation energy in terms of calorie per mole.
Procedure
To begin, aqueous solutions of the following reagents were prepared: 0.333 M potassium bromate, 0.667 M potassium bromide, 0.500 M perchloric acid, 0.030 M phenol, 0.500 M sodium nitrate, and 40 mg/L of methyl orange solution. Solutions comprised of these were then prepared according to the given specifications. However, potassium bromate, potassium bromide, perchloric acid, and water were combined in one flask while phenol, sodium nitrate, and methyl orange solution were combined in another flask. The flasks were then placed in the bath and allowed to come to a constant temperature. The flask containing the bromate and bromide was transferred into the flask with methyl orange. As soon as the two solutions made contact, timing began using a stopwatch, and the solutions were mixed using a glass stirring rod to ensure conformity. As soon as the orange color had completely dissipated, timing stopped. This process was repeated for every trial according to the specifications of the following charts:
Table #1 (At 25° C)
| BrO3– (mL) | Br– (mL) | NO3– (mL) | H2O (mL) | HClO4 (mL) | C6H5OH (mL) | MO (mL) | |
| Run 1 | 5 | 5 | 10 | 5 | 10 | 10 | 5 |
| Run 2 | 5 | 10 | 3.5 | 6.5 | 10 | 10 | 5 |
| Run 3 | 10 | 5 | 7 | 3 | 10 | 10 | 5 |
| Run 4 | 5 | 5 | 0 | 5 | 20 | 10 | 5 |
Table #2 (At 20° C)
| BrO3– (mL) | Br– (mL) | NO3– (mL) | H2O (mL) | HClO4 (mL) | C6H5OH (mL) | MO (mL) | |
| Run 1 | 5 | 5 | 10 | 12 | 10 | 3 | 5 |
| Run 2 | 5 | 5 | 10 | 9 | 10 | 6 | 5 |
| Run 3 | 5 | 5 | 10 | 6 | 10 | 9 | 5 |
| Run 4 | 5 | 5 | 10 | 3 | 10 | 12 | 5 |
| Run 5 | 5 | 5 | 10 | 0 | 10 | 15 | 5 |
Table #3 (At 25° C)
| BrO3– (mL) | Br– (mL) | NO3– (mL) | H2O (mL) | HClO4 (mL) | C6H5OH (mL) | MO (mL) | |
| Run 1 | 5 | 5 | 10 | 12 | 10 | 3 | 5 |
| Run 2 | 5 | 5 | 10 | 9 | 10 | 6 | 5 |
| Run 3 | 5 | 5 | 10 | 6 | 10 | 9 | 5 |
| Run 4 | 5 | 5 | 10 | 3 | 10 | 12 | 5 |
| Run 5 | 5 | 5 | 10 | 0 | 10 | 15 | 5 |
Table #4 (At 30° C)
| BrO3– (mL) | Br– (mL) | NO3– (mL) | H2O (mL) | HClO4 (mL) | C6H5OH (mL) | MO (mL) | |
| Run 1 | 5 | 5 | 10 | 12 | 10 | 3 | 5 |
| Run 2 | 5 | 5 | 10 | 9 | 10 | 6 | 5 |
| Run 3 | 5 | 5 | 10 | 6 | 10 | 9 | 5 |
| Run 4 | 5 | 5 | 10 | 3 | 10 | 12 | 5 |
| Run 5 | 5 | 5 | 10 | 0 | 10 | 15 | 5 |
Table #5 (At 35° C)
| BrO3– (mL) | Br– (mL) | NO3– (mL) | H2O (mL) | HClO4 (mL) | C6H5OH (mL) | MO (mL) | |
| Run 1 | 5 | 5 | 10 | 12 | 10 | 3 | 5 |
| Run 2 | 5 | 5 | 10 | 9 | 10 | 6 | 5 |
| Run 3 | 5 | 5 | 10 | 6 | 10 | 9 | 5 |
| Run 4 | 5 | 5 | 10 | 3 | 10 | 12 | 5 |
| Run 5 | 5 | 5 | 10 | 0 | 10 | 15 | 5 |
Results and Calculations
Determining the Order of Reaction:
-d[BrO3–] / dt = k [BrO3–]x [Br–]y [H+]z
Run 1 & Run 3
(0.002M) / (54.25secs) = k [0.333] x [0.0667] y [0.1] z
(0.002M) / (27.15secs) k [0.667] x [0.0667] y [0.1] z
0.5 = 0.499 x
x = 1
Run 1 & Run 2
(0.002M) / (54.25 secs) = k [0.0333] x [0.0667] y [0.1] z
(0.002M) / (30.34 secs) k [0.0333] x [0.1333] y [0.1] z
0.559 = 0.5 y
ln (0.559) = y ln (0.5)
y = 0.839
Run 1 & 4
(0.002M) / (54.25 secs) = k [0.0333] x [0.06667] y [0.1] z
(0.002M) / (24.47 secs) k [0.0333] x [0.06667] y [0.2] z
0.451 = 0.5z
ln (0.451) = z ln (0.5)
z = 1.15
Concentrations versus Recorded Times
|
25° C (Run 1) |
25° C (Run 2) |
30° C |
35° C |
||||
|
[BrO3–] |
t (s) |
[BrO3–] |
t (s) |
[BrO3–] |
t (s) |
[BrO3–] |
t (s) |
|
0.6 |
21.78 |
0.6 |
21.47 |
0.6 |
12.57 |
0.6 |
6.63 |
|
1.2 |
37.18 |
1.2 |
50.11 |
1.2 |
23.21 |
1.2 |
15.29 |
|
1.8 |
54.57 |
1.8 |
82.18 |
1.8 |
34.3 |
1.8 |
22.22 |
|
2.4 |
71.8 |
2.4 |
109.72 |
2.4 |
43.67 |
2.4 |
31.81 |
|
3 |
97 |
3 |
143 |
3 |
55.88 |
3 |
44.75 |
Determining the rate constant for each temperature
-d[BrO3–] / dt = k (a x 2a x (3a)2) = 18ka4
20° C: k = 0.0198 = 894.57
(18)(0.0333)4
25° C: k = 0.0321 = 1450.29
(18)(0.0333)4
30° C: k = 0.056 = 2530.10
(18)(0.0333)4
35° C: k = 0.0638 = 2882.51
(18)(0.0333)4
Activation Energy:
|
T (° K) |
1/T |
log (k) |
|
293 |
3.41 x 10-3 |
2.95 |
|
298 |
3.36 x 10-3 |
3.16 |
|
303 |
3.30 x 10-3 |
3.40 |
|
308 |
3.25 x 10-3 |
3.46 |
Slope = – E / (2.303 * R ) , R= 8.3145 J/mol K
-3287.7 = – E / (2.303 * 8.3145)
E = 62953.8 cal/mol
Conclusions
The first objective of this experiment was to determine the orders of the reaction. The values obtained vary in their likeness to the theoretical values. It was found that -d[BrO3–] / dt = k [BrO3–]1 [Br–]0.839 [H+]1.15 when it really should have been = k [BrO3–]1 [Br–]1 [H+]2. The calculated value for BrO3– was dead on, while the value for Br– was slightly off and the value for H+ was significantly different. This error could be attributed to a multitude of sources. For example, the concentrations of solutions used may have been different than believed. If the solutions were old, they may have lost strength, contributing towards this. This would in turn affect the timing with the stopwatch, which would throw off the calculation of the orders of reaction. To go along with this, if the solutions were not made accurately, meaning the exact aliquots were not used, this would also affect the time and ultimately rate.
The next objective was to determine the rate constant for each temperature and then the activation energy. The rate constants calculated seem fairly large, meaning the slopes that resulted from plotting [BrO3–] versus time were of small value. If the slopes were larger numbers, then the rate constants would have been smaller. This in turn made the activation energy seem excessive. With smaller rate constant values, the plot would have resulted with a smaller number for the slope, and thus a smaller rate constant. This problem can again be justified due to the fact that the believed concentrations and temperatures of solutions may not have been actually what we were working with.
