Define an expression for the rate constant of a 1st order reaction. Define half life period. A first order reaction takes 69.3 minutes for 50% completion. How much time will be needed for 80% completion?

Explanation

For first reactions , the rate of the reaction is proportional to the first power of the concentration of the reactant R.
For the reaction `R to P `
Rate = `-(d[R])/(dt) =k[R] or (d[R])/([R]) = - kdt `
On integrating the equation ,we get
In `[R] = - kt+l `
where , l is the constant of integration
In the beginning when t = 0 , `R = [R]_(0)` ,where `[R]_(0)` is the initial concentration of the reactant
Eq. (i) can be written as
In `[R]_(0) = - k xx 0 +l `
In `[R]_(0) = l `
On substituting In `[R]_(90)=l ` in the Eq. (i) we get
In [R] ` = - kt + In [R]_(0)" "...(ii) `
or In `([R])/([R]_(0)) =-kt " "...(iii)`
or ` In ([R])/([R]_(0)) = - k or k = 1/t" in "([R]_(0))/([R]) " "...(iv) `
or `k = (2.303)/t log. ([R]_(0))/([R]) " "...(iv)`
or `k = (2.303)/t log. ([R]_(0))/([R]) " "...(v)`
Thus ,Eq. (v) is usually written as :
`k = (2.303)/t log. ([N]_(0))/([N])`
or `k = (2.303)/t log. a/((a-x))`
where , a is the initial concentration of reactant ,R and `(a-x) `is the concentration of reactant R , left after time t as x moles of it get changed to products after time t .
i.e ., `[R]_(0) = [N]_(0) = a and [R] = [N] = (a-x)`
Half - period is defined as the time in which the concentration of a reactant is reduced to one half its initial concentration .It is represented as `t_(1/2)`
For first order reaction `t_(1//2)` is represented as `t_(1//2)`
For first order reaction `t_(1//2)` is represented as
`t_(1//2) = (0.693)/k`
Given a first order reaction is 50 % complete in `69.3` min.
` :. t_(1//2) = 69.3 ` min
We know that , n for 80 % completion of reaction ,
Let a = 100 , thus , `a-x = 100 - 80 = 20 `
`t = (2.303)/k log. a/(a-x)`
` :. t = (2.303)/(0.01) log. 100/20`
`t = (2.303)/(0.01) log 5 rArr (2.303)/(0.01) xx 0.6990`
`= 160.98 approx 161 ` min