experiments that this situation is described quantitatively by the following formula for the rate constant k = Ae-Ea/RT in which E α and A are called an activation energy and a pre-factor. ‡  R is the gas constant (= 8.314 × 10−3 kJ mol−1 K−1). By taking the natural logarithms (log e or ln) of both sides,
Therefore, by plotting ln k against 1/T , we will obtain a straight line as shown in the next diagrams. From the slope and the intercept of the plot we can calculate the activa- tion energy E a and the pre-factor A. Equations (20) and (21) are called the Arrhenius equation and the plot of ln k against 1/T the Arrhenius plot. The Arrhenius law is closely related to the Boltzmann's law .  In thermal equilib- rium at temperature T , the molecules can take various energies. 8 The distribution of the number of molecules plotted against energy E will look the following diagrams. As seen, the number of molecules decreases as the energy increases (in the high energy region). According to the picture of the activation energy barrier lying between the reac- tant and product states, the reaction rate will be proportional to the number of molecules having higher energy than E α , i.e., in the shaded area of the diagram.  By raising the temperature the molecules will receive energy from the surroundings, and then the dis- tribution graph will shift toward the right, i.e., toward the higher energy. This results in an increase of the number of molecules above E α and hence the increase of the reaction rate. This behaviour is described quantitatively by the Arrhenius equation (20) or (21).
Data analysis
3
Suppose that we have carried out a series of experiments to observe changes along time of the concentrations of the reactants and products as in the diagrams in Section 2.1 From these and other supplementary data, we attempt to determine the rate law, i.e., the orders with respect to the species involved and the rate constant.
Unimolecular reaction A → P
3.1
We begin with a very simple reaction in which only one specie A is involved in the reactant. → Product(s) (22) We do not consider the backward reaction. We also assume that there is no intermediate species or intermediate steps involved in the reaction.  That is, (22) is a single step reaction. This is an example of the elementary reaction. In this case, the rate law will contain only the concentration of A,
Because the reaction rate is that of the loss of A, −d[A]/dt, the above equation is a simple differential equation
which can be integrated mathematically.
Zero order reaction
3.1.1
Let us first consider the zero order case, n = 0. Because [A] 0 = 1, (24) becomes
This means that the gradient d[A]/dt is a constant − k .  Therefore, [A] decays at a constant rate and the plot of [A] along time will be a straight line.
Here and hereafter, we denote the concentration of A by [A] t when we wish to empha- size its dependence on time. The initial concentration at t = 0 is [A] 0 .

First order reaction
For n = 1, we have
The solution 1 is an exponential function
This is equivalent to
Therefore, the plot of ln[A] t against time t is a straight line. ¶ For simplicity we write A( t ) instead of [A] t . It follows from (27) that
Putting t = 0, we find C = 1n A(0). We thus get (29) which is equivalent to (28).
3.1.3
Second order reaction
For n = 2,
Therefore, the plot of 1/[A] t against time t is a straight line.
3.1.4
Summary
.
.
In this way, the order n and the rate constant k can be determined from the graphs of [A], ln[A] or 1/[A] against time. Suppose we have obtained from experiments the con- centration of A as a function of time.  We then plot [A], ln[A] or 1/[A] against time.  If one of these was found to be a straight line, the order n is 0, 1 or 2, respectively. ‖ From (30),
Putting t = 0, we find C = 1/A(0). This immediately gives (32) and after rearranging we get (31).
3.1.5
Half-life Method
The half-life is the time taken for the concentration of the reactant [A] to halve.  It may depend on the starting concentration [A]. For example, for zero order reactions where [A] t is a linear function of t , the half life decreases along time, i.e., as the starting concentration decreases.  (Confirm this by drawing a graph.)  In contrast, for the first order reactions the half life is a constant that is independent of the starting concentration. See the diagrams below.
By dividing both sides by [A] 0 and taking the natural logarithms, we find
As seen in the previous diagram, the decay profile of the concentration [A] for the second order reaction is quite different from the first order reaction. It is easy to show from Eq (32) that the half life for the second order reaction is proportional to the inverse of the starting concentration, τ = 1/k[A] 0 . This means that the half-life increases as the reaction proceeds. In summary, Oth order: half-life = [A] 0 /2k . 1st order: . half-life = ln 2/k
2nd order: .
Question: Confirm these behaviours (qualitatively) on graphs.  Derive these mathe- matical formulas of the half-life from Eqs (26) - (32).