In order to write a rate equation (rate law) for a reaction we need to know the orders of reaction with respect to each reactant i.e. how changing the concentration of each reactant in turn affects the initial rate of the reaction. The actual details of the experiment are not important, we simply need the results!
e.g. How does the rate of the reaction depend on the concentration of dinitrogen pentaoxide in the following decomposition reaction?
2N2O5(g) ⇾ 4NO2(g) + O2(g)
| [N2O5] / mol dm -3 | Initial rate / mol dm-3 h-1 |
| 0.01 | 0.016 |
| 0.02 | 0.032 |
| 0.04 | 0.064 |
If we look at the results of this series of experiments, we can see that doubling the [N2O5] from 0.01 to 0.02 mol dm-3 causes the initial rate of reaction to double from 0.016 to 0.032 mol dm-3 h-1. Similarly we can see that doubling the [N2O5] from 0.02 to 0.04 mol dm-3 causes the initial rate of reaction to double from 0.032 to 0.064 mol dm-3 h-1.
The reaction is first order with respect to dinitrogen pentaoxide and so …
rate = k [N2O5]
Since we have experimental results we can determine a value for k, the rate constant, using any of the sets of results:
k = rate / [N2O5] = 0.032 / 0.02 = 1.6 h-1
To figure out the units we repeat the calculation in terms of the units:
k = rate / [N2O5] = mol dm-3 h-1 / mol dm-3 = h-1 (as the units for concentration cancel out)
We can confirm all of this graphically. If we plot a graph of initial rate vs. concentration we will get a straight line passing through the origin if the reaction is first order with respect to that particular reactant, and the gradient of that line is the value for k, the rate constant in this example as there are no other [reactants] in the rate equation.
Calculating a value for k using the gradient of the line is the more accurate method, as a line of best fit averages out any anomalous data.
e.g. How does the rate of the reaction depend on the concentration of nitrogen(IV) oxide in the reaction with carbon monoxide?
NO2(g) + CO(g) ⇾ NO(g) + CO2(g)
| [NO2] / mol dm -3 | Initial rate / mol dm-3 s-1 |
| 0.15 | 0.010 |
| 0.30 | 0.040 |
| 0.60 | 0.16 |
| 0.90 | 0.32 |
If we look at the results of this series of experiments, we can see that doubling the [NO2] from 0.15 to 0.30 mol dm-3 causes the initial rate of reaction to quadruple from 0.010 to 0.040 mol dm-3 s-1. Similarly we can see that doubling the [NO2] from 0.03 to 0.06 mol dm-3 causes the initial rate of reaction to quadruple from 0.040 to 0.16 mol dm-3 s-1.
The reaction is second order with respect to nitrogen(IV) oxide and given that the reaction is first order with respect to carbon monoxide we can write the rate equation …
rate = k [NO2]2 [CO]
Once again, we can confirm all of this graphically. If we plot a graph of initial rate vs. concentration then we should get a graph where the gradient of the line increases with increasing concentration and a straight-line graph if we plot initial rate vs. (concentration)2.
e.g. Chloromethane is hydrolysed by water in the atmosphere:
CH3Cl(g) + H2O(g) ⇾ CH3OH(g) + HCl(g)
The rate of this reaction was investigated by varying the starting concentrations of CH3Cl and H2O and measuring the formation of HCl over time:
| Experiment | [CH3Cl] / mol dm-3 | [H2O] / mol dm-3 | Initial rate of formation of HCl / mol dm-3 s-1 |
| 1 | 0.0250 | 0.250 | 2.838 |
| 2 | 0.0375 | 0.250 | 4.256 |
| 3 | 0.0250 | 0.125 | 0.709 |
We can write a rate equation for this reaction and calculate a value for the rate constant if we figure out the order of reaction with respect to chloromethane and water.
- chloromethane: if we take experiments 1 and 2 (because the concentration of water is the same so cannot affect the rate of reaction) we see that increasing the [CH3Cl] by 1.5 times (0.0375 / 0.025 = 1.5) causes the rate to increase by 1.5 times (4.256 / 2.838 = 1.5). The reaction is first order with respect to chloromethane.
- water: if we take experiments 3 and 1 (because the concentration of chloromethane is the same so cannot affect the rate of reaction) we see that doubling the [H2O] from 0.125 to 0.250 mol dm-3 causes the rate to quadruple from 0.709 to 2.838 mol dm-3 s-1. The reaction is second order with respect to water.
rate = k [CH3Cl] [H2O]2
k = rate / [CH3Cl] [H2O]2 = 0.709 / (0.0250 x 0.1252) = 1815 dm6 mol-2 s-1
If the units are a complete mystery to you then read on …
Practice questions
- Use the following results for a series of reactions between hydrogen and nitrogen(II) oxide to determine the order of reaction with respect to each reactant, the rate equation and a value for the rate constant.
2H2(g) + 2NO(g) ⇾ 2H2O(g) + N2(g)
| Experiment | Initial [H2] / mol dm-3 | Initial [NO] / mol dm-3 | Initial rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.01 | 0.025 | 2.4 x 10-5 |
| 2 | 0.005 | 0.025 | 1.2 x 10-5 |
| 3 | 0.01 | 0.0125 | 0.6 x 10-5 |
- Use the following results for a series of reactions between two gases, A and B, to determine the order of reaction with respect to each reactant, the rate equation and a value for the rate constant.
A(g) + 3B(g) ⇾ AB3(g)
| Experiment | Initial [A] / mol dm-3 | Initial [B] / mol dm-3 | Initial rate / mol dm-3 min-1 |
|---|---|---|---|
| 1 | 0.1 | 0.1 | 0.00200 |
| 2 | 0.1 | 0.2 | 0.00798 |
| 3 | 0.3 | 0.1 | 0.00601 |
- The reaction between hydrogen and iodine was investigated and it was determined that the rate equation for the reaction was
rate = k [H2] [I2]
and that at 610K the value for the rate constant was 8.58 x 10-5 dm3 mol-1 s-1. Determine the rate of reaction for each of the following experiments and comment on the effect of the concentration of hydrogen and iodine on the reaction rate.
| Experiment | Initial [H2] / mol dm-3 | Initial [I2] / mol dm-3 | Initial rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.01 | 0.05 | |
| 2 | 0.02 | 0.05 | |
| 3 | 0.02 | 0.1 |
Answers
- The reaction is first order with respect to hydrogen (experiments 3 and 1, concentration doubles and rate doubles) and second order with respect to nitrogen(II) oxide (experiments 3 and 1, concentration doubles and rate quadruples)
rate = k [H2] [NO]2
k = rate / [H2] [NO]2 = 0.6 x 10-5 / (0.01 x 0.01252) = 3.84 dm6 mol-2 s-1
- The reaction is first order with respect to A (experiments 1 and 3, concentration trebles and rate trebles) and second order with respect to B (experiments 1 and 2, concentration doubles and rate quadruples)
rate = k [A] [B]2
k = rate / [A] [B]2 = 0.0079 / (0.1 x 0.22) = 2.0 dm6 mol-2 s-1
| Experiment | Initial [H2] / mol dm-3 | Initial [I2] / mol dm-3 | Initial rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.01 | 0.05 | 4.29 x 10-8 |
| 2 | 0.02 | 0.05 | 8.58 x 10-8 |
| 3 | 0.02 | 0.1 | 1.72 x 10-7 |
Doubling the initial concentration of hydrogen (experiments 1 and 2) causes the initial rate of reaction to double and doubling the initial concentration of iodine (experiments 1 and 3) causes the initial rate of reaction to double. This confirms the orders of reaction with respect to each reactant and the rate equation given.