Statement: The elementary reaction is single step reaction and does not possess mechanism.
Explanation: An elementry reaction has order of reaction and molecularity same.

Elementry Reactions

The order of a single step reaction can be

Does a zero order reaction has zero molecularity?

Molecularity of an elementary reaction

Molecularity of an elementary reaction

The rate and mechamical reaction are studied in chemical kinetics. The elementary reactions are single step reaction having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction aA + bBrarr cC+dD is given as: rate =((dx)/(dt))=-1/a(d[A])/(dt)=-1/b(d[B])/(dt)=1/c(d[C])/(dt)=1/d(d[D])/(dt) or expression can be written as : rate =K_(1)[A]^(a)[B]^(b)-K_(2)[C]^(c )[D]^(d) . At equilibrium, rate = 0 . The constants K, K_(1), K_(2) are rate constants of respective reaction. In case of reactions governed by two or more steps reaction mechanism, the rate is given by the slowest step of mechanism. At the point of intersection of the two curve shown for the reaction: Ararr nB the concentration of B is given by:

The rate and mechamical reaction are studied in chemical kinetics. The elementary reactions are single step reaction having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction aA + bBrarr cC+dD is given as: rate =((dx)/(dt))=-1/a(d[A])/(dt)=-1/b(d[B])/(dt)=1/c(d[C])/(dt)=1/d(d[D])/(dt) or expression can be written as : rate =K_(1)[A]^(a)[B]^(b)-K_(2)[C]^(c )[D]^(d) . At equilibrium, rate = 0 . The constants K, K_(1), K_(2) are rate constants of respective reaction. In case of reactions governed by two or more steps reaction mechanism, the rate is given by the slowest step of mechanism. For a gaseous reaction, the rate is expressed in terms of (dP)/(dt) in place of (dC)/(dt) or (dn)/(dt) where C is concentration, n is number of moles and 'P' is pressure of reactant. The three are related as: