Chemical Equilibrium
3.0 Law of mass action
3.0 Law of mass action
It is proposed by Goldberg and Wage. According to this law, rate of reaction at a given temperature is directly proportional to the active masses of reactants raised to the power equal to stoichiometric coefficient of the balanced chemical equation at particular instant of reaction.
Now let us understand the terms rate of reaction and active mass.
As we all know that in a chemical reaction, substances (reactants) reacts to form another substances (products) and vice-versa. There is always a specific rate at which the reaction is proceeding.
Rate at which a substance reacts is directly proportional to the Active Mass of the substance.
- Active mass is the molar concentration i.e., number of moles per volume (in litres) of the reactants or products. The active mass of any substance $X$ is represented by square brackets $[X]$.
- Rate of a chemical reaction at a particular temperature is proportional to the product of active masses of reactants raised to the powers of their stoichiometric coefficients.
$$aA + bB \rightleftharpoons products$$
$${\text{Rate = }}k{\left[ A \right]^a}{\left[ B \right]^b}$$
where ${k}$ is the rate constant and $[A]$ and $[B]$ denotes active masses.
For a ${n^{th}}$ order reaction, rate law can be written as
$$\begin{equation} \begin{aligned} {\text{Rate = }}k{[A]^n} \\ \therefore k = \frac{{{\text{Rate}}}}{{{{[A]}^n}}} = \frac{{molarity/\sec }}{{{{\left[ {molarity} \right]}^n}}} = \frac{{mol.litr{e^{ - 1}}.{{\sec }^{ - 1}}}}{{{{\left[ {mol.litr{e^{ - 1}}} \right]}^n}}} \\\end{aligned} \end{equation} $$
Unit of rate constant is given as
$$\left( k \right) = {\left[ {moles/litre} \right]^{1 - n}}se{c^{ - 1}}$$
where $n$ is order of reaction.
