In a zero order reaction, rate is independent of the concentration of the reactants. Mathematically, a reaction is said to be of zero order when the rate is proportional to zero power of the reactant concentration. Suppose a zero order reaction is of the form

`A ->`Product

Initial concentration `a` `0`

Concentration after timet `(a-x)` `x`

Differential rate law of this reaction will be

`-(d[A])/(dt)=(d[text(product)])/(dt)=k_0[A]^0`

or `-(d(a-x))/(dt)=+(dx)/(dt)=k_0(a-x)^0=k_0`

On integrating, we get the integrated rate law as

`k_0 = x/t` or `x =k_0 t`

`k_0 t = a-(a-x)`

Where `k_0` is the rate constant of a zero order reaction, the unit of which is concentration per time. In zero order reactions, the rate constant is equal to the rate of the reaction at all concentrations.

The time required for half of their reactants to change into product is known as half-life period. Let us see how the half-life period depends on the initial reactant concentration 'a' in reaction of various orders. In a zero-order reaction we know,

`k_0=x/t` or `t=x/k_0`

Let `t_(1//2)` be the half-life period.

i.e., when , `t = t_(1//2)`, `x= a/2`

`:.` `t_(1//2) = a/(2k_0)` or `t_(1//2) prop a` (`1//(2k_0)` is constant)

Thus `t_(1/2)` for zero order reaction is proportional to `a`.

A first order reaction is one whose rate varies as 1st power of the concentration of the reactant i.e. the rate increases as number of times as the number of times the concentration of reactant is increased . Let us consider a unimolecular first order reaction represented by the general equation.

`A-> ` Product

`a x= 0` at time `t= 0`

`a- x x = x` at time `t = t`

The initial concentration of `A` is `a` mole `L ^(- 1)` and its concentration after any time `t` is `(a - x)` mole `L ^(-1)` . This means during the time interval `t`, `x` mole `L^(-1)` of `A` has reacted. The rate of reaction at any time `t` is given by the following first - order kinetics.

`-[d(a-x)]/dt prop (a-x)` or `(dx)/(dt) prop (a-x)` or `(dx)/(dt) = k (a-x)`

(`(d(a-x))/(dt)= 0` a has a given value for a given expt.) where `k` is the rate constant of the reaction. `(dx)/(a-x) = kdt`

This is differential rate equation and can be solved by integration.

`int[(dx)/(a-x)] = k int(dt)` or `ln(a-x) = kt +C` ...............(1)

Where `C` is integration constant

The constant `C` can be evaluated by applying the initial condition of the reaction i.e. when `t = 0`, `x = 0`. Putting these in equation (1), we get `C = - lna`

Putting the value of `C` in equation (1), we get

`- ln(a - x) = kt - ln a` or `k =1/t ln [a/(a-x)] = 2.303/t log [a/(a-x)]`....... (2)

If `[A_0]` and `[A]` be the concentrations of reactant at zero time and time `t`, respectively then Eq. 2 may be put as

`k=1/t ln ([A_0]/[A])`

This is the integrated rate expression for first order reaction. Unit of the rate constant is `s^(-1)`

`k = ln a - ln (a-x)`

`kt = ln a - ln (a - x)`

`a - x= a e ^(-kt)`

`x = a[1 - e^(-kt)]`

The half-time of a reaction is defined as the time required to reduce the concentration of the reactant to half of its initial value. It is denoted by the symbol `t_(1//2)`. Thus,

When `x=a/2,t=t_(1//2)`

Putting these in equation 2 mentioned above, we get

`k=2.303/t_(1//2) log a/(a-a/2)-2.303/(t_1//2) xx 0.30103` `( log 2=0.30103)`

`t_(1//2)=0.693/k`.....(3)

Since `k` is a constant for a given reaction at a given temperature and the expression lacks any concentration term so from equation (3) it is evident that half-time of a `1st` order reaction is a constant independent of initial concentration of reactant . This means if we start with `4` mole `L^(-1)` of a reactant reacting by first-order kinetics and after `20` minute it is reduced to `2` mole `L ^(- 1)`. Its half life will be `20` minute. That is, after `20` minutes from the start of reaction the concentration of the reactant will be `2` mole `L^(-1)`, after `40` minutes from the start of reaction of concentration is `1` mole `L^(-1)`. After `60` minutes from the start of reaction the concentration of the reactant will be reduced to `0.5` mole `L ^(- 1)`. In other words, if during `20` minute `50%` of the reaction completes, then in `40` minute `75%`, in `60` minute `87.5%` of the reaction and on will complete as shown in the fig.1.

Fraction left after `n` half-lives `= (1/2)^n` Concentration left after `n` half lives `a_n =(1/2)^n a_0`

It is also to be noted that equation (3) helps to calculate `t_(1//2)` or `k` with the knowledge of `k` or `t_(1//2)`.

A general expression for `t_(1//2)` is as follows

`t_(1//2) prop 1/(a^(n-1))`

Where `n =` order of reaction.

Graphical Representation for `n^(th)` Order Reaction :

So, from this it is evident that a plot of `(dx)/(dt)` vs `(a -x)^n` will be a straight line passing through the origin and will have its slope equal to `k`, the rate constant of reaction. Thus, for a first order reaction one will get straight line passing through the origin of `(dx)/(dt)` i.e. rate of reaction be plotted against `a - x` as shown in fig.2.

Taking logarithm of the above equation `log= n log (a - x) +log k` This equation tells that a plot of `log (dx)/(dt)` vs `log (a -x)` will be straight line of the slope equal to `n`, order of reaction and intercept equal to `log k`. For first order reaction this slope will be `1` as
shown below. Equation may be rearranged as `log (a - x) = (k/2.303) t + log a`

Thus, a plot of log `(a - x)` vs. `t` will be straight line with slope equal to `-k/2.303` and intercept equal to `log a`, if the reaction is of first order.