Practice Problem 10
Use the experimental data found in the table at the beginning of this lesson to determine whether the reaction between phenolphthalein (PHTH) and the OH- ion is a first-order or a second-order reaction.
Answer
The first step in solving this problem involves calculating the natural log of the phenolphthalein concentration, ln (PHTH), and the reciprocal of the concentration, 1/(PHTH), for each point at which a measurement was taken:
| (PHTH) (mol/L) | ln (PHTH) | 1/(PHTH) | Time (s) |
| 0.0050 | -5.30 | 200 | 0.0 |
| 0.0045 | -5.40 | 222 | 10.5 |
| 0.0040 | -5.52 | 250 | 22.3 |
| 0.0035 | -5.65 | 286 | 35.7 |
| 0.0030 | -5.81 | 333 | 51.1 |
| 0.0025 | -5.99 | 400 | 69.3 |
| 0.0020 | -6.21 | 500 | 91.6 |
| 0.0015 | -6.50 | 667 | 120.4 |
| 0.0010 | -6.91 | 1000 | 160.9 |
| 0.00050 | -7.60 | 2000 | 230.3 |
| 0.00025 | -8.29 | 4000 | 299.6 |
| 0.00015 | -8.80 | 6670 | 350.7 |
| 0.00010 | -9.21 | 10,000 | 391.2 |
We then construct graphs of ln (PHTH) versus time (Figure 1)and 1/(PHTH) versus time (Figure 2). Only one of these graphs, Figure 1, gives a straight line. We therefore conclude that these data fit a first-order kinetic equation, as noted in the beginning of this lesson.
Rate = k(phenolphthalein)
Figure 1. A plot of the natural log of the concentration of phenolphthalein versus time for the reaction between phenolphthalein and excess OH- ion is a straight line, which shows that this reaction is first order in phenolphthalein.
Figure 2. A plot of the reciprocal of the concentration of phenolphthalein versus time for the reaction between phenolphthalein and excess OH- ion is not a straight line, which shows that this reaction is not second order in phenolphthalein.
