Findingf concentration and time : The rate constant for the decomposition of gasous `N_(2)O_(5) at 55^(@)C is 1.7xx10^(-3)s^(-1)` . If the initial concentration of `N_(2)O_(5)` after five half-lives ? How long will it take for the `N_(2)O_(5)` concentration to fall to `12.5%` of its initial values ?
Strategy : Because the unit of rate constant is `time^(-1)` , the decomposition of `N_(2)O_(5)` is a first order reaqction. To find `[N_(2)O_(5)]` after n half lives, multiply its initial concentration by `(1//2)^(n)` since `[N_(2)O_(5)]` drops by a factor of 2 during each successive haslf-life.

Explanation

After five half-lives `(t=5t_(1//2)), [N_(2)O_(5)]` will bw `(1//2)^(5) = (1//32)` of its initial value. Therefore
`[N_(2)O_(5)]= (0.020M)/(32) = 0.00062M`
Since `12.5%` of the initial concentration corresponds to `1//8 or (1//2)^(3)` of the initial concentration, the time required is three half-lives:
`t=3t_(1//2)=3((0.693)/(k))`
`=3((0.693)/(1.7xx10^(-3)s^(-1)))`
`= 3(4.1xx10^(2)s)`
`= 1.2xx10^(3)s(20 min)`