A number of experiments are carried out by varying concentration of a reactant w.r.t. which the order is to be determined keeping the concentrations of all other reactants constant. The initial rate of the reaction at each concentration is determined by concentration-time curves. The order of the reactant is then calculated from the rates at various concentrations. The experiments are repeated with other reactants in a similar way. Likewise, the orders w.r.t. all reactants are determined. The overall order is the sum of the orders of all the reactants.

In this method we use the integrated rate law equations. Substituting the various experimental values of `a, x, t` etc. As you can see, each order of reaction has a unique input and output variable that produces a straight line. For example: if we graph the following rate data for the decomposition of `H_2O_2` assuming that it could be zero, first, or second order, we find that the graph for a `2nd` order reaction (`1//[A]` versus `t`) gives a straight line. Therefore, the reaction has the rate law rate `= k [H_2O_2]^2`.

We calculate the rate constant. If we get a constant value of `k_1` (rate constant for first order), it is a first-order reaction; and similarly, for constant values of `k_2` and `k_3`, the reaction are of second and third order respectively. Rate data for decomposition of `H_2O_2`

This data fit to a straight line only for `2nd` order reaction in integrated rate law.

(a) A graph is plotted between time `t` and the concentration of reactant `(a-x)` or product `(x)`, slope of which gives the rate of reaction `(dx)/(dt)` for the selected time instant. (figure 1 & 2)The various values of the rate `((dx)/(dt))` are now plotted against the corresponding concentration `(a-x)` or `(a-x)^2` or `(a-x)^3` from which we draw the following graphical conclusions : See fig.1.


(b) Alternatively, the order (`n`) can also be determined from the slope of the curve plotted
between `log [(dx)/(dt)]` and `log (a-x)` : See fig.2.

`(dx)/(dt) = k(a-x)^n`

`log[(dx)/(dt)] = logk +nlog(a-x)`

(c)We know that the time required to complete a definite fraction (say one half) of the reaction depends on the initial concentration of the reactant in the following way :

For zero-order reaction, `t_(1//2) prop a`

i.e., `t_(1//2) prop 1/a^(-1)`

For first-order reaction, `t_(1//2)` is independent of `a`

i.e., `t_(1//2) prop 1/a^(0)`

For Second-order reaction, `t_(1//2) prop 1/a`

For third-order reaction, `t_(1//2) prop 1/a^2`

`:.` for `n^(th)`-order reaction, `t_(1//2) prop 1/a^(n-1)`

Thus we get the fo llowing plots for reactions of various orders : See fig.3.

(d) From integrated rate law equations of various reactions of different orders, we may have the following plots : See fig.4.

In case of gaseous reactions, concentration may be replaced by pressure.

As we know that

`t_(1//2) prop (1/a)^(n-2)` where `n` is the order We may have the following relation

`(t_1/2)/(t_(1/2))={a_2/a_1}^(n-2)`

Where, `(t_(1/2))` and `(t_(1/2))`, are the half-life periods, or time for a definite fractional change of a reaction when the respective initial concentrations of the reactants are `a_1` and `a_2`. Taking log we have,

`n=1+(log(t_(1/2)))/((log(a_2/a_1)))`