1.

Derive half life t_((1)/(2))of first order reaction.

Answer»

Solution :For a first ORDER reaction,rate constant (k) is given by following equation .
`k=(2.303)/(t)` log `([R]_(0))/([R])`…..(b)
In this equation `t_((t)/(2))` =HALF life time.
An initial t=0 time,the concentration of reactant =`[R]_(0)`
After `t_((1)/(2))` time,the concentration of Reactant =[R]
`(1)/(2)` (initial case)=`(1)/(2)[R]_(0)`
PUT `t=t((1)/(2))` and [R]=`([R]_(0))/(2)` in equation (b),
`k=(2.303)/(t_((1)/(2)))` log `([R]_(0))/(([R]_(0))/(2))`
`thereforek=(2.303)/(t_((1)/(2)))` log 2.0
`THEREFORE k=(2.303)/(t_((1)/(2)))xx0.3010`
`therefore k=(0.693)/(t_(1)/(2))` and
`t_((1)/(2))=(0.693)/(k)`=half life of first order reaction
Derivation:It can be seen that for a first order reaction.
half life periof is constant
It is independent of initial concentration of the reacting species.
The half life `t_((1)/(2))` of a first order equation is really calculated from the rate constant & vice versa.
For zero order reaction `t_((1)/(2))prop[R]_(0)` and for first order reaction `t_((1)/(2))` is independent of `[R]_(0)`


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