3.4.1 Kinetics - Determination of the rate equation
Students should:
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Solving the rate equation
If a series of experimental results are obtained for rate at different reactant concentrations the change in rate can be ascertained for those reactions when the concentration of one of the reactants is kept constant while the other reactant concentration is changed.
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experiment
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concentration of A
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concentration of B
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Rate
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1
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0.1M
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0.1M
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6 x 103
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2
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0.2M
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0.1M
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1.2 x 104
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3
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0.4M
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0.1M
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2.4 x 104
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4
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0.1M
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0.2M
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6 x 103
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5
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0.2M
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0.2M
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1.2 x 104
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In experiments 1,2 and 3 the concentration of A changes while the concentration of B is kept constant. This means that the rate equation can be written as:
Rate = k'[A]m
For these three experiments, if we inspect the rate of experiments 1,2 and 3 we see that as the concentration of A is doubled so the rate doubles. In other words the order of reaction must be 1, so that whatever happens to the concentration must also happen in equal amounts to the rate.
Similar inspection of experiments 1 and 4 ( or 2 and 5) show that while A is kept constant there is no effect on the rate when the concentration of B is changed. The order with respect to B must be 0.
The orders can now be substituted into the rate equation:
Rate = k[A]1[B]0
To obtain a value for the rate constant we simple substitute the values for any one of the experiments above, using the newly determined orders. (It 's always easier to choose the experiment with the simplest numbers - in this case experiment 1)
Values from experiment 1:
6 x 103 = k x [0.1]1 x [0.1]0
k = 6 x 103 / 0.1
k = 6 x 104
Why solve the rate equation?
The main reason for investigating the rate of chemical reactions is to gain information about how the actual reactions occur in terms of collisions etc.. Thermodynamics investigations and data give us information about why a process occurs and kinetics investigations help us to understand how processes occur.
The series of steps taken by the particles in a reaction is called the mechanism of the reaction.
By finding out the orders of a reaction with respect to their individual component concentrations, we get clues about the possible collisions that are occuring in each stage of the mechanism.
NOTE It should be pointed out here that mechanisms are often speculation rather than fact. However, if the evidence is strong enough to support the speculation or hypothesis then we can feel comfortable about using it.
The mechanism
As stated above, many reactions proceed via a series of steps, the set of which is called the mechanism of the reaction. It would be very unlikely if each step were the same speed and the most likely scenario is that one of the steps is considerably slower than the other(s).
The slowest step effectively has the greatest effect on the overall rate, and we can approximate our kinetics measurements and say that our data actually represents the slowest step (rate determining step).
The orders of the rate equation show us the number of particles of that kind that are involved in the rate determing step.
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Example: A reaction between A and B is found to have an order of 2 with respect to the concentration of A, but an order of 0 with respect to the concentration of B. This suggest that the rate determining step does not involve B at all. Wwe can postulate: 2 A intermediate + B |
Collision theory
The simplest interpretation of collision theory tells us logically that for particles to react they must collide. The mechanism of the reaction shows us the actual collisions that are taking place.
Amongst the billions of collisions that occur between particles there will be an infintesimally small number of three particle collision. This means that to all intents and purposes three particle collisions cannot affect the rate of a reaction.
The conclusion is that each step of a mechanism can only involve a MAXIMUM of two particles. In other words each step must be either a two particle collision, or one particle breaking apart.
Third order reactions
If three particle collisions are impossible, how can a reaction have an order of three or more?
The only way that this is possible is if the slow rate determining step involves an intermediate that has to be generated in an earlier step:
For example:
2 A 
intermediate
(fast)
intermediate + B
products (slow, rate determining)
In this case the intermediate must be formed by a previous step and in turn that previous step involves 2 particles of A. The rate equation for this reaction would look like the following:
Rate = k[A]2[B]1
The reaction is second order with respect to the concentration of A and first order with respect to the concentration of B. It is third order overall.
The effect of temperature on the rate constant.
The rate constant, k, is only constant at constant temperature. Increasing the temperature always causes the value of the rate constant to increase. This is because there are more particles with the available activation energy to react at higher temperature.
This is shown by the Maxwell-Boltzmann distribution:
Activation energy
The actual dependence of the rate constant on the temperature is given by the Arrhenius equation:
k = Ae(-Ea/RT)
where:
- A is a constant related to the number, orientation and frequency of collisions occurring between the particles in the reaction.
- k is the rate constant
- R is the universal gas constant
- T is the absolute temperature
Determination of the activation energy from practical results
If rates experiments are carried out at different temperatures the results can be plotted on a graph to obtain a value for the activation energy for a specific reaction.
| k = Ae(-Ea/RT) | ||
| therefore: | ln k = lnA - Ea/RT |
A plot of natural log of k against (1/T) gives a straight line of gradient - Ea/R, from which the activation energy can be found. The intercept of the graph on the Y axis gives us the value of lnA.