First Order Reactions

A reaction is said to be first order if its rate is determined by the change of one concentration term only.

Consider the reaction

                 A → products

Let α  be the concentration of A at the start and after time t, the concentration becomes (a-x), i.e, x has been changed into products. The rate of reaction after time 't' is given by the expression

                 dx/dt = k(a-x)

                 or dx/((a-x)) = k dt


Upon integration of above equation,

                   ∫ dx/(a-x) = k∫dt

 or             -loge (a - x) = kt + c

where c is integration constant.

When t = 0 ,        x = 0,

.·.   c = -loge a

Putting the value of 'c',

       -loge (a - x) = kt - loge a

or     loge a - loge (a - x) = kt

or     loge a/(a-x) = kt

or     k =2.303/t  log10 a/(a-x)


This is known as the kinetic equation for a reaction of the first order. The following two important conclusions are drawn from this equation:

(i)   A change in concentration unit will not change the numerical value of k. let the new unit  

So  k =  2.303/t log10 na/n(a-x)

or   k =  2.303/t log10 a/(a-x)

Thus for first order reaction, any quantity which is proportional to concentration can be used in place of concentration for evaluation of 'k'.

(i)    The time taken for the completion of same fraction of change is independent of initial concentration. For example, for half change,

x = 0.5a    and   t = t1/2

So      k = 2.303/t1/2 log10 a/0.5a = 2.303/t1/2 log10 2

             = 0.693/t1/2

or         t1/2 = 0.693/k


Thus, t1/2 is independent of initial concentration 'a'.

This time 't' in which the initial concentration becomes half is termed as half life period. Half life period of a first order reaction is independent of the initial concentration of the reactant.

Since the velocity constant is independent of concentration and depends inversely on the time, the unit of k will be time-1, i.e., sec-1 or min-1 or hour-1. The equation of the first order can also be written in the following form when initial concentration is not known.

           k  =  2.303/((t2-t1 )) log10 ((a-x1))/((a-x2))

(a-x1) is the concentration after time t1 and (a-x2) the concentration after time t2when t2>t1.

When the log of the concentration of the reactant at various intervals of time is plotted against the time intervals, a straight line is obtained (Fig.8.12). The slope of this line gives the value 2.303/k, from which k can be evaluated.




Examples of first order reactions

1.  Decomposition of H2O2 in aqueous solution

     H2O →  H2O + 1/2  O2


2.  Hydrolysis of methyl acetate in presence of mineral acids.


      CH3COOCH3 + H2O   →   CH3COOH + CH3OH


3.   Inversion of cane sugar in presence of mineral acids.


      C12H22O11 + H2O   →   C6H12O6 + C6H12O6


4.   Decomposition of ammonium nitrite in aqueous solution.

      NH4NO2  →  N2 + 2H2O


5.   Hydrolysis of diazo derivatives.

      C5H5N = NCl + H2O  →  C6H5OH + N2 + HCl

Note:    In case of gases, pressure can be used in place of concentration.

First Order Growth Kinetics

It is used for population growth and bacteria multiplication, e.g.,

     Time                       Population

       0                              a

       dt                           (a+x)

Growth rate is directly proportional to present population.

       dx/dt ∞ (a+x)

       = k(a + x)

       dx/(a+x)  = k dt                                        .... (i)


It is a differential equation of first order and first degree in variable separable form. It may be solved on integration.

       ∫dx/(a+x)  = k∫dt +c

       loge (a+x) = kt + c                                    ....(ii)

Here, c = integration constant

At   t = 0 ,         x = 0

       .·.  loge a = k × 0 + c

       c = loge a                                                  ...(iii)

Substituting the value of 'c' in (ii), we get

     loge (a+x) = kt + loge a

     kt = -loge a/(a+x)

     k = - 2.303/t  log10 a/(a+x)

This is the kinetics for first order growth kinetics


Note: (1) If volume of reagents are given in volumetric analysis then we use the following equation to determine rate constant.

           k = 2.303/t log10 (V - C0)/(V - V1)

where V0 = volume used at zero time,

V1 = volume used at time 't',

V = volume used at infinite time

Case I   When V0 is not given, we use

                        k =  2.303/t log10 v/(v - v1


Case II  When  is not given, then

                        k =  2.303/t  log10 (V0/V1)


(2)   if information is given in terms of angle of rotation of optically active compounds, measured by polarimeter with respect to time, then

                        k = 2.303/t  log10 {r-r0/r - r1} 


where r0 = angle of rotation at zero time,

r1 = angle of rotation at time 't',

r∞ = angle of rotation at infinite time.


Case I   If r0 is not given, then

              k = 2.303/t  log10  {r/r - r1}


Case II  If  is not given, then

               k = 2.303/t  log10 {r0/r1}


(1)   if pressure is given in gaseous reaction, then we use the following kinetic equation:

            k = 2.303/t  log10 {P0/(P0-x)}

where P0 = pressure of reactant at initial stage,

(P0 - x) = pressure of such a reactant at 't' time

Values of 'P0' and 'x' can be calculated using the following examples:

                                                  A(g) → B(g) + C(g) + D(g)

At t = 0                                        P0         0          0          0

Pressure after time 't'                    (P0-x)      x          x          x

Pressure after a long                        0            P0       P0        P0

time or infinite time              

Case I    If total pressure of reaction mixture is given in place of pressure of reactant, then

                    Pt = (P0 - x + x + x + x)

              where P1 = pressure of vessel at time 't'.

Case II  If pressure of vessel after a long time or infinite time is given, then

              P∞ =  P0 + P0 + P0


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