Initial Rates
1. A series of experiments are completed for the following reaction and the results are outlined in the table below.
H2 + I2 → HI
| Experiment | [H2] / mol dm-3 | [I2] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.05 | 0.05 | 8.00×10-5 |
| 2 | 0.1 | 0.05 | 1.60×10-4 |
| 3 | 0.2 | 0.05 | 3.20×10-4 |
| 4 | 0.2 | 0.1 | 6.40×10-4 |
H2 – 1st order. When you look at experiments 1 and 2, the concentration of H2 doubles (whereas nothing else changes) and the rate of reaction doubles.
I2 – 1st order. When you look at experiments 3 and 4, the concentration of I2 doubles (whereas nothing else changes) and the rate of reaction doubles.
Rate = k [H2] [I2]
k = 8×10-5 / (0.05 x 0.05) = 3.2×10-2 mol-1 dm3 s-1
2. A series of experiments are completed for the following reaction and the results are outlined in the table below.
CO(g) + NO2(g) → CO2(g) + NO(g)
| Experiment | [CO] / mol dm-3 | [NO2] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 2.5 | 0.8 | 3.33×10-3 |
| 2 | 5 | 0.8 | 3.33×10-3 |
| 3 | 3.75 | 0.4 | 8.32×10-4 |
CO – zero order. When you look at experiments 1 and 2, the concentration of CO doubles (whereas nothing else changes) and the rate remains constant.
NO2 – second order. When you look at either experiment 1 or 2 and experiment 3, the concentration of NO2 is divided by 2 and the rate is divided by 4. You can ignore the changes in [CO] because CO is zero order.
Rate = k [NO2]2
k = 3.33×10-3 / 0.82 = 5.2×10-3 mol-1 dm3 s-1
3. A series of experiments are completed for the following reaction and the results are outlined in the table below.
CH4(g) + H2O(g) → 3H2(g) + CO(g)
| Experiment | [CH4] / mol dm-3 | [H2O] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.5 | 0.75 | 2.25×10-1 |
| 2 | 0.1 | 0.75 | 4.50×10-2 |
| 3 | 0.2 | 1.5 | 1.80×10-1 |
CH4 – First order. When you look at experiments 1 and 2, the concentration of CH4 is divided by 5 and the rate is also divided by 5.
H2O – First order. When you look at experiments 2 and 3, the concentration of CH4 and H2O both double. The rate quadruples. Given that one doubling of the rate is due to the CH4, then the other doubling is due to the H2O. H2O is therefore first order.
Rate = k [CH4] [H2O]
k = 2.25×10-1 / (0.5 x 0.75) = 0.6 mol-1 dm3 s-1
4. A series of experiments are completed for the following reaction and the results are outlined in the table below.
I2(aq) + CH3COCH3(aq) → CH3COCH2I(aq) + H+(aq) + I–(aq)
| Experiment | [H+] / mol dm-3 | [CH3COCH3] / mol dm-3 | [I2] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|---|
| 1 | 0.5 | 4 | 0.1 | 8.00×10-5 |
| 2 | 1 | 4 | 0.1 | 1.60×10-4 |
| 3 | 1 | 4 | 0.2 | 1.60×10-4 |
| 4 | 0.5 | 2 | 0.2 | 4.0×10-5 |
H+ – First order. When you look at experiments 1 and 2, the rate doubles as [H+] doubles.
I2 – Zero order. When you look at experiments 2 and 3, the rate doesn’t change as [I2] doubles.
CH3COCH3 – First order. When you look at experiments 1 and 4, the rate halves as [CH3COCH3] halves. You can ignore the change in I2 as it is zero order.
Rate = k [H+] [CH3COCH3]
k = 8×10-5 / (0.5 x 4) = 4×10-5 mol-1 dm3 s-1
5. A series of experiments are completed for the following reaction and the results are outlined in the table below.
H2O2(aq) + 2I–(aq) + 2H+(aq) → 2H2O(l) + I2(aq)
| Experiment | [H2O2] / mol dm-3 | [I–] / mol dm-3 | [H+] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|---|
| 1 | 0.5 | 0.75 | 0.8 | 0.60 |
| 2 | 1.5 | 0.75 | 0.8 | 1.80 |
| 3 | 1.0 | 1.5 | 0.8 | 2.40 |
| 4 | 1.5 | 1.5 | 1.6 | 3.60 |
H2O2 – first order. When you look at experiments 1 and 2, the rate triples as [H2O2] triples.
I– – first order. When you look at experiments 1 and 3, the rate quadruples as [H2O2] and [I–] both double.
H+ – zero order. When you look at experiments 2 and 4, rate rate doubles when [I–] and [H+] double. The doubling is caused by the [I–] doubling.
Rate = k [H2O2] [I–]
k = 0.6 / (0.5 x 0.75) = 1.6 mol-1 dm3 s-1
6. A series of experiments are completed for the following reaction and the results are outlined in the table below.
6H+(aq) + 5Br–(aq) + BrO3–(aq) → 3Br2(aq) + 3H2O(l)
| Experiment | [H+] / mol dm-3 | [Br–] / mol dm-3 | [BrO3–] / mol dm-3 | Initial Rate / mol dm-3 s-1 |
|---|---|---|---|---|
| 1 | 0.2 | 0.4 | 0.25 | 9.60×10-7 |
| 2 | 0.2 | 0.8 | 0.5 | 3.84×10-6 |
| 3 | 0.2 | 1.6 | 0.5 | 7.68×10-6 |
| 4 | 0.4 | 0.2 | 0.25 | 1.92×10-6 |
Br– – First order. When you look at experiments 2 and 3, the rate doubles when [Br–] doubles.
BrO3– – First order. When you look at experiments 1 and 2, the rate quadruples when [Br–] and [BrO3–] double. As one doubling is due to the [Br–] change, [BrO3–] doubling causes the rate to double.
H+ – Second order. When you look at experiments 1 and 4, the rate doubles when [H+] doubles and [Br–] halves. This means that [H+] doubling must be causing the rate to quadruple.
Rate = k [H+]2 [Br–] [BrO3–]
k = 9.6×10-7 / (0.22 x 0.4 x 0.25) = 2.40×10-4 mol-3 dm9 s-1