Using Initial Rate Data
1. A series of experiments were conducted on the following reaction. The table below contains the data from the experiments.
A + 2B → C + D
| Experiment | [A] / mol dm-3 | [B] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.500 | 0.500 | 1.50 |
| 2 | 0.500 | 1.00 | 3.00 |
| 3 | 0.250 | 0.500 | 0.375 |
| 4 | 0.250 | 2.00 |
Rate = k [A]2 [B]
Compare experiments 1 and 4. [A] divides by 2, so divides the rate by 4. [B] multiplies by 4, so multiples the rate by 4. Therefore the rate doesn’t change. The answer is 1.5 mol dm-3 s-1.
2. A series of experiments were conducted on the following reaction, which involves the catalyst, Z. The table below contains the data from the experiments.
X + Y → E + F
| Experiment | [X] / mol dm-3 | [Y] / mol dm-3 | [Z] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|---|
| 1 | 2.5 | 2 | 4 | 52 |
| 2 | 2.5 | 4 | 4 | 52 |
| 3 | 5 | 8 | 4 | 104 |
| 4 | 2.5 | 4 | 6 | 117 |
| 5 | 2 | 8 | 104 |
Rate = k [X] [Z]2
Compare to experiment 1. [Z] doubles, so rate quadruples. However, the rate only doubles, so X must halve. The answer is 1.25 mol dm-3.
3. A series of experiments were conducted on the following reaction, which involves the catalyst, Z. The table below contains the data from the experiments.
2NO + 2H2 → N2 + 2H2O
| Experiment | [NO] / mol dm-3 | [H2] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.50 | 0.5 | 7.50×10-3 |
| 2 | 0.75 | 0.5 | 1.69×10-2 |
| 3 | 1.5 | 1.0 | 1.35×10-1 |
| 4 | 0.75 | 3.38×10-2 |
Rate = k [NO]2 [H2]
Compare to experiment 2. [NO] stays the same but rate doubles. Therefore, [H2] must double. The answer is 1.0 mol dm-3.
4. A series of experiments were conducted on the following reaction, which involves the catalyst, Z. The table below contains the data from the experiments.
NO2 (g) + CO (g) → NO (g) + CO2 (g)
| Experiment | [NO2] / mol dm-3 | [CO] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 1.6 | 1.6 | 5.12×10-3 |
| 2 | 0.8 | 0.8 | 1.28×10-3 |
| 3 | 0.4 | 1.6 | 3.20×10-4 |
| 4 | 0.2 | 8.00×10-5 |
Rate = k [NO2]2
[CO] is irrelevant. Compare experiment 3 and 4. The rate was divided by 4. Therefore, [NO2] was divided by 2. The answer is 0.2 mol dm-3.
5. A series of experiments were conducted on the following reaction between iodine chloride and hydrogen. The table below contains the data from the experiments.
2ICl (l) + H2 (g) → 2HCl (g) + I2 (s)
| Experiment | [ICl] / mol dm-3 | [H2] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.8 | 0.6 | 0.211 |
| 2 | 1.2 | 0.6 | 0.317 |
| 3 | 1.8 | 1.8 | 1.43 |
| 4 | 0.96 | 0.48 |
Rate = k [ICl] [H2]
Compare experiments 2 and 4. [ICl] was multiplied by 0.8 and [H2] was also multiplied by 0.8. Therefore, multiply the rate from experiment 2 by 0.8 and then by 0.8. The answer is 0.203 mol dm-3 s-1.