Second Order ReactionsA reaction is said to be of second order if its reaction rate is determined by the variation of two concentration terms.

The kinetics of second order reactions are given as follows:

(i) When concentration of both reactants are equal or two molecules of the same reactant are involved in the change, i.e.,

A + B → products

or 2A → products

dx/dt = k(a-x)

^{3}On solving this equation,

k = 1/t.x/a(a-x)

where a = initial concentration of the reactant or reactants and

x = concentration of the reactant changed in time t.

(ii) When the initial concentrations of the two reactants are different, i.e.,

A + B → products

Initial conc. a b

dx/dt = k(a-x)(b-x)

k = 2.303/t(a-b) log

_{10}b(a-x)/a(b-x)(a - x) and (b - x) are the concentrations of A and B after time interval, t.

Characteristics of the second order reactions

(i) The value of k(velocity constant) depends on the unit of concentration. The unit of k is expressed as (mol/litre)^{-1}time^{-1}or litre mol^{-1}time^{-1}.(ii) Half life period (t

_{1/2}) = 1/k.0.5a/(a×0.5a) = 1/kaThus, half life is inversely proportional to initial concentration.

(iii) Second order reaction conforms to the first order when one of the reactants is present in large excess.

Taking k = 2.303/t(a-b) log

_{10}b(a-x)/a(b-x); if a>>> b then(a-x) = a and (a-b) = a

Hence,, k = 2.303/ta log

_{10}ba/a(b-x)or ka = k' = 2.303/t log

_{10}b/((b-x))(since 'a' being very large, may be treated as constant after the change). Thus the reaction follows first order kinetics with respect to the reactant taken relatively in small amount.

Examples of second order reactions1. Hydrolysis of ester by an alkali (saponification).

CH

_{3}COOC_{2}H_{5}+ NaOH → CH_{3}COONa + C_{2}H_{5}OH2. The decomposition of NO

_{2}into NO and O_{2}.3. Conversion of ozone into oxygen at 100

^{o}C2NO

_{2}→ 2NO + O_{2}4. Thermal decomposition of chlorine monoxide.

2Cl

_{2}O → 2Cl_{2}+ O_{2}