Derive the expression for integrated rate law (equation) for the first order reaction.

Explanation

Consider the following first order reaction, A`to` B.
The rate of the chemical reaction is given by the rate law expression as,
Rate R = K [A] where [A] is the concentration of the reactant A and k is the velocity constant or specific rate of the reaction.
The instantaneous rate is given by,
`R = k[A] = (-d[A])/(dt)`
` (-d[A])/([A]) = k.dt`
If `[A_(0)]` is the initial concentration of the reactant and `[A]_(t)` at time t. then by integrating the above equation,
`-underset([A_(0)])overset([A]_(t))int (d[A])/([A]) = k underset(0)overset(t)int dt`
` therefore - log_(e).([A_(0)])/([A]_(t)) = kt " "therefore log_(e).([A_(0)])/([A]_(t)) = kt`
`therefore k = (1)/(t) log_(e).([A_(0)])/([A]_(t)) " "therefore k = (2.303)/(t) log_(10).([A_(0)])/([A]_(t))`
This is the integranted rate equation for the first order reaction .
This is also called integrated rate law.