Question

Integrated rate law for second order unimolecular irreversible

Answer 1

The rate law for second order unimolecular irreversible reaction is

`(1)/([A]) = k.t + (1)/([A]_(0) )`

Step-by-step explanation:

A second order unimolecular irreversible reaction is

2A → B

Thus the rate of the reaction is

`v = -(1)/(2).(d[A])/(dt) = k.[A]^(2)`

rearranging the ecuation

`-(1)/(2).(k)/(dt) = ([A]^(2))/(d[A])`

Integrating between times 0 to t and between the concentrations of
`[A]_(0)` to [A].

`\int\limits^0_t -(1)/(2).(k)/(dt) =\int\limits^A_(0) _A([A]^(2))/(d[A])`

Solving the integral

`(1)/([A]) = k.t + (1)/([A]_(0) )`