DERIVATION OF EQUILIBRIUM CONSTANT EXPRESSION - PUNJAB-BOARD / 10TH-CLASS / CHEMISTRY / 10-CHEMISTRY-1-19-10TH-CLASS-NBF-2024 / CHEMICAL-EQUILIBRIUM / DERIVATION-OF-EQUILIBRIUM-CONSTANT-EXPRESSION

9.2 Law of Mass Action (Derivation for General Reaction)

Let us apply the law of Mass Action for a general reaction.

\mathbf{a}\mathbf{A} + \mathbf{b}\mathbf{B}\quad \rightleftharpoons \quad \mathbf{c}\mathbf{C} + \mathbf{d}\mathbf{D}

This reaction consists of two reactions; forward and reverse reactions. According to this law, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of its reactants raised to power equal to their number of moles in the balanced chemical equation of the reaction. Let us first discuss the forward reaction. A and B are the reactants whereas 'a' and 'b' are their number of moles.

The rate of forward reaction according to law of Mass Action is:

R_{f} \propto [A]^{a} [B]^{b}

R_{f} = k_{f} [A]^{a} [B]^{b}

where $\mathsf{k}_{\mathsf{r}}$ is the rate constant for the forward reaction. Similarly, the rate of the reverse reaction $\mathsf{R}_{\mathsf{r}}$ is directly proportional to the product of $[\mathsf{C}]^{\mathsf{c}}[\mathsf{D}]^{\mathsf{d}}$ , where 'c' and 'd' are the number of moles as given in the; of $[\mathsf{C}]^{\mathsf{c}}[\mathsf{D}]^{\mathsf{d}}$ , where 'c' balanced chemical equation. Thus,

R_{r} \propto [C]^{c} [D]^{d}

R_{r} = k_{r} [C]^{c} [D]^{d}

where $k_{r}$ is the rate constant for the reverse reaction . We know that at equilibrium state the rates of both the reactions are equal.

The rate of forward the reaction = The rate of the reverse reaction Such as:

R_{r} = R_{r}

and putting the values of $R_{r}$ and $R_{r}$