# Rate of Reaction

With the progress of reaction the concentration of reactants decreases while that (those) of product(s) increases. If the reaction occurs fastly, the concentration (mole L−1) of reactant will decrease rapidly while that of product will increase rapidly.
Thus, the rate of reaction is defined as the rate at which the concentration of reactant decreases or alternatively, the rate at which concentration of product increases. That is, the change in concentration of any of the substance (reactant / product) per unit time during the reaction is called rate of reaction.

If Δ C be the change in concentration of any reactant or any product during the time interval Δt then the rate may given as

$\large Rate = \pm \frac{\Delta C}{\Delta t}$

(−ve sign applies in the case of reactant whose concentration goes on decreasing with time and +ve sign applies in the case of product whose concentration goes on increasing with time).

However, the rate of reaction is not uniform. With the passage of time the concentration(s) of reactant(s) goes on decreasing and hence according to Law of Mass Action, the rate of reaction goes on decreasing. Infact, the rate of reaction decreases moment to moment. This is shown in the graph of rate vs time curve of a reaction.

Rate varies from moment to moment so rate of reaction has to be specified at a given instant of time called instantaneous rate or rate at any time t. This is as defined below.

$\large r_{inst} = \pm \frac{dc}{dt}$

Where dc is the infinitesimal change in concentration during infinitesimal time interval dt after time t i.e. between t and t + dt. The time interval dt being infinitesimal small, the rate of reaction may be assumed to be constant during the interval

The rate of reaction expressed as is actually the average rate with time interval considered.

For the reaction: 2 N2O5 → 4 NO2 + O2

The rate of reaction at any time t may be expressed by one of the following.

$\large – \frac{d[N_2 O_5]}{dt} \; , \; +\frac{d[N O_2]}{dt} \; , \; +\frac{d[O_2]}{dt}$

Where square bracket terms denote molar concentration of the species enclosed.

The above three rates are not equal to one another as is evident from the stoichiometry of the reaction. For every mole of N2O5 decomposed, 2 moles NO2 and 1/2 mole of O2 will be formed. Obviously, the rate of formation of NO2 will be four times that of O2 and it is twice the rate of consumption of N2O5. Thus, their three rates are interrelated as,

$\large +\frac{d[N O_2]}{dt} = 2(- \frac{d[N_2 O_5]}{dt} ) = 4(+\frac{d[O_2]}{dt})$

Dividing through out by 4 ,

$\large \frac{1}{4}(+\frac{d[N O_2]}{dt}) = \frac{1}{2}(- \frac{d[N_2 O_5]}{dt} ) = (+\frac{d[O_2]}{dt})$

Thus, rate of reaction expressed in terms of the various species involved in a reaction will be equal to one another if each of there is divided by the stoichiometric constituent of the species concerned. Rate divided by stoichiometric coefficient may be called as rate per mole.

The Kinetic experiment has shown that the rate of reaction mentioned above increases same number of times as the number of times the concentration of N2O5 is increased. That is, rate is doubled by doubling the concentration of N2O5. This may be mathematically expressed as

Rate ∝ [N2 O5 ]

or Rate = k [N2 O5 ]

Where k is the proportionality constant called rate constant, velocity coefficient or specific reaction rate of the reaction, and is a constant for a given reaction at a given temperature. Equation 1 showing the concentration – dependence of rate is called rate law of the reaction. Expressing rate in terms of the rate of change of concentration of various species, equation may be put as

$\large – \frac{d[N_2 O_5]}{dt} = k'[N_2 O_5]$

$\large + \frac{d[N O_2]}{dt} = k”[N O_2]$

$\large + \frac{d[ O_2]}{dt} = k”'[O_2]$

when k’, k” and k”’ are the rate constants of the reaction. These three rate constants are inter-related.

$\large \frac{k’}{2} = \frac{k”}{4} = k”’$

Thus, for a reaction represented by the general equation

a A + b B  → c C + d D

$\large \frac{1}{a}(-\frac{dC_A}{dt}) = \frac{1}{b}(-\frac{dC_B}{dt}) = \frac{1}{c}(+\frac{dC_c}{dt}) = \frac{1}{d}(+\frac{dC_D}{dt})$

We have,

$\large \frac{k^I}{a} = \frac{k^{II}}{b} = \frac{k^{III}}{c} = \frac{k^{IV}}{d}$

where  ,  kI, kII, kIII and kIV  are the rate constants of the reaction when its rate is expressed in terms A, B, C and D, respectively.