Kinetics  IB syllabus > kinetics (hl) > 16.1 

16.1  The rate expression16.1.1: Define the terms rate constant and order of reaction. 
Throughout this topic it is essential to keep in mind that this is an experimentally determined science. There are no theories that can determine the rate expression by merely considering the chemicals reacting.
By observation we can see that varying certain conditions changes the rate with which the reactants are used up (or the products formed). Kinetics investigates this and in doing so gives us clues as to the mechanism of the reaction.
Rate: Measured as the concentration, mass or volume (or anything else that can be directly related to one of these) of a reactant or product changes with time. Conventionally, we define rate in terms of concentration change with time... mol dm^{3} s^{1} (moles per litre per second) is the 'usual' unit
Mechanism: Is the actual route taken by reacting particles in order to make the products. It may be that there are several stages or steps between reactants and products involving the formation of intermediates. These intermediates have a varying lifespan and are often undetectable. One of the aims of kinetics is to build up a picture of the actual mechanism involved in a reaction.
When all other factors are kept constant we can investigate how the rate of a reaction depends on the concentration of the reactants.
A + B > Products 
To allow for any possibility we can express the effect of the concentrations of A and B on the rate of the reaction by the expression:
Rate = k [A]^{x}[B]^{y} 
Where does this equation come from?
Mathematically, it allows for all possible effects of changing [A] and [B]. For example, if changing the concentration of A has no effect on the rate then the value of x=0, If the rate is directly proportional to the concentration of A then x=1. If there is some other effect then x will take another number. The equation then is ust a method of expressing the rate in terms of the concentrations. It is called the rate expression, also the rate equation or law.
Rate = k [A]^{x}[B]^{y} 
The 'k' is called the rate constant. As the name implies, it has a constant value for any specific reaction at constant temperature. he value of the rate constant for a reaction gives us a measure of how fast the reaction is. A very small value for k means a very slow reaction and vice versa.
The numbers to which the concentrations are raised in the rate expression are called the orders of reaction with respect to the individual concentrations.
Rate = k [A]^{x}[B]^{y} 
X is the order with respect to [A] and Y is the order with respect to [B]. The total order of the reaction is the sum of the individual orders, in this case X + Y.
The order of the reaction with respect to a specific reagent provides information about the number of particles of this reagent involved in the rate determining step of the mechanism. The rate determining step is the slowest step.
Note: If an equilibrium feeds into the rate determining step then this will also influence the order with respect to any particles that appear on the left hand side (reactants) of the equilibrium.
The units of the rate constant depend on the total order of the reaction.
For order = 1 (first order reaction)
The rate expression is:
Rate = k [A]^{1} 
Substituting the units in each part we get:
mol dm^{3} s^{1} = k [mol dm^{3}]^{1} 
Cancelling out from both sides gives:
s^{1} = k 
For a second order reaction:
Rate = k [A]^{2} 
Substituting the units in each part we get:
mol dm^{3} s^{1} = k [mol dm^{3}]^{2} 
Rearranging and cancelling out from both sides gives:
dm^{3} mol^{1} s^{1} = k 
16.1.2: Deduce the rate expression for a reaction from experimental data. AIM 7: Virtual experiments can be used here
The rate of a reaction MUST be determined by experimental measurements. These can involve following the disappearance of reactants or the generation of products. There are many techniques that may be used.
This procedure provides experimental results of the following kind:
Experiment  [A] / mol dm^{3}  [B] / mol dm^{3}  [C] / mol dm^{3}  Rate/ mol dm^{3} s^{1} 
1  0.1  0.1  0.1  6.2 x 10^{4} 
2  0.1  0.2  0.1  1.2 x 10^{3} 
3  0.1  0.1  0.2  6.2 x 10^{4} 
4  0.2  0.1  0.2  2.5 x 10^{3} 
When a component is kept constant between two experiments we can effectively remove it from the rate expression by combining its constant value wth the rate constant to make a kind of super constant (that we are not concerned with any way). In this way we know that any change in rate is due to the concentration that changes.
Inspecting the table above we can see that in experiments 1 and 2 the concentrations of A and C are kept constant. This means that any change in rate is due to the change in the concentration of B. When [B] doubles from 0.1 to 0.2 we can see that the rate also doubles from 6.2 x 10^{4} to 1.2 x 10^{3} (these are approximate figures to simulate the 'real' situation)
This means that the rate is directly proportional to the concentration of B.
Rate = k'[B] 
So in this case the order with respect to [B] is equal to 1
Rate = k'[B]^{1} 
Repeat the process, inspecting experimental data 1 and 3. In these two experiments the concentrations of A and C are constant and so any change in the rate is due to the change in concentration of C. However we can see that the rate does not change. This means that changing C doesn't affect the rate. The order with respect to the concentration of C must be zero.
Rate = k''[C]^{0} 
Now inspect experimental data 3 and 4. The concentrations of B and C are kept constant while [A] doubles The rate increases fourfold between these two experiments. That means that any change in the concentration of A produces a corresponding squared change in the rate. The order with respect to [A] is therefore 2.
Rate = k''[A]^{2} 
Now all of the orders can be placed in the rate expression.
Rate = k [A]^{2}[B]^{1}[C]^{0} 
This can now be used to calculate the value for the rate constant y substituting into any of the experimental data above (for one of the experiment runs)
Choosing expt. 1 for convenience (most of the numbers are easy)
6.2 x 10^{4} = k [0.1]^{2}[0.1]^{1}[0.1]^{0} 
k =  6.2 x 10^{4} 
0.1 x 0.1 x 0.1 
k= 6.2 x 10^{1} dm^{6} mol^{2} s^{1} 
16.1.3: Solve problems involving the rate expression
The majority of problems involve processing a set of data and obtaining the orders of reaction:
Experiment  [A] / mol dm^{3}  [B] / mol dm^{3}  [C] / mol dm^{3}  Rate/ mol dm^{3} s^{1} 
1  0.1  0.1  0.1  6.2 x 10^{4} 
2  0.1  0.2  0.1  1.2 x 10^{3} 
3  0.1  0.1  0.2  6.2 x 10^{4} 
4  0.2  0.1  0.2  2.5 x 10^{3} 
From the above set of data it may be seen that when the concentration of B doubles (all other concentrations remaining constant) the rate also doubles. This tells us that the order with respect to [B] is 1.
If the rate should multiplies by a factor of four while the concentration only doubles, the order would be 2.
Once all orders are obtained the rate constant can be determined by 'plugging' values into any one of the experimental data sets.
16.1.4: Sketch, identify and analyse graphical representations for zero, first and second order reactions. Students should be familiar with both concentrationtime and rateconcentration graphs
The simplest plot is the concentration (of reactants or products) against the time. This is always a curve, the gradient of which is the rate of the reaction. A typical experimental setup to obtain the results is shown below:
gas collection apparatus  graph of gas volume against time 
The gradient of this graph at any point gives the rate of the reaction at that time. It would be possible to select several points on the graph and then calculate the corresponding rates. These can then be plotted on a second graph of rate against concentration (volume of gas, in this case).
The shape of concentration / rate graphs depends on the overall order of the reaction.
There is no correlation between the rate and the concentration of a zeroth order reaction. A flat line is obtained.
This is in keeping with the rate law expression: Rate = k[A]^{0}
There is direct proportionality between the rate and the concentration. A straight line is obtained passing through the origin.
This is in keeping with the rate law expression: Rate = k[A]^{1}
The gradient (slope) of the graph gives the rate constant k, whose units are now s^{1}
When rate is plotted against concentration a curve is obtained. However, when the order is any other than 0th or 1st a curve is also given. The graph is of limited value in this case and the data must be further processed.
Inspection of the rate expression for a second order reaction
Rate = k[A]^{2}
shows that a plot of rate against concentration would give a curve. If logs are taken throughout however this gives:
log Rate = log k + 2log[A]
This has the same form as a straight line graph y = mx +c
Consequently, a plot of log rate against log[A] gives a straight line graph whose intercept is the value for log k and the gradient is equal to the order of the reaction.
This treatment is valid for any order values. The gradient (slope) of the graph line is always equal to the order.
Resources
Model
to explore the change in concentration during the decomposition of dinitrogen
pentoxide, N_{2}O_{5}, at 338K.
2N_{2}O_{5}(g) > 4NO_{2}(g)
+ O_{2}(g)
Useful links




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