Derive the law of chemical equilibrium from law of mass action.

Explanation

Consider a general reversible reaction, `A + B hArr X + Y`
According to law of mass action, rate of forward reaction `(r_(f)) = K_(f)[A][B]` ….(i)
Where, `K_(f)` is rate constant for forward reaction.
Similarly, rate of backward reaction `(r_(b)) = K_(b)[X][Y]` .......(ii)
At equilibrium, the rate of forward reaction = the rate of backward reaction. .....(iii)
Therefore, from equation (i) and (ii), we have
`K_(f) [A][B] = K_(b)[X][Y]` or, `(K_(f))/(K_(b)) = ([X][Y])/([A][B])`
Since `K_(f)` and `K_(b)` are constants, so their ratio `(K_(f)//K_(b))` will also give a new constant represented by K.
`therefore K = ([X][Y])/([A][B])` .......(iv)
K is called equilibrium constant and equation (iv) is known as law of chemical equilibirum.
For a geneeral reaction of type, `aA + bB hArr c C + d D`
According to law of chemical equilibrium, `K = ([C]^(c )[D]^(d))/([A]^(a)[B]^(b))`