Why molecularity is applicable only for elementary reactions and order is applicable for elementary as well as complex reactions ?
A complex reaction occurs through several elementary reactions. Number of molecules involved in each elementary reaction may be different i.e. the molecularity of each step may be different. Therefore, the molecularity of overall complex reaction is meaningless, on the other hand order of a complex reaction is experimentally determined by the slowest step in its mechanism and is therefore, applicable even in the case of complex reactions.
The formation of gas at the surface of tungsten due to adsorption is the reaction of order
Half-life period of a first order reaction is $1386$ seconds. The specific rate constant of the reaction is
Consider following two reaction,
$A \to {\text{Product ;}}\,\, - \frac{{d[A]}}{{dt}} = {k_1}{[A]^o}$
$B \to {\text{Product ;}}\,\, - \frac{{d[B]}}{{dt}} = {k_2}{[B]}$
Units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as
The experimental data for reaction
$2A + B_2 \longrightarrow 2AB$
Exp. | $[A]$ | $[B_2]$ | Rate $(mol\,L^{-1}\,S^{-1})$ |
$1$ | $0.50$ | $0.50$ | $1.6 \times {10^{ - 4}}$ |
$2$ | $0.50$ | $1.00$ | $3.2 \times {10^{ - 4}}$ |
$3$ | $1.00$ | $1.00$ | $3.2 \times {10^{ - 4}}$ |
The rate law
For the reaction $2A + B \to C$,the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is
$[A] (mol\,L^{-1})$ | $[B] (mol\,L^{-1})$ | Initial Rate $(mol\, L^{-1}\,s^{-1} )$ |
$0.05$ | $0.05$ | $0.045$ |
$0.10$ | $0.05$ | $0.090$ |
$0.20$ | $0.10$ | $0.72$ |