The decomposition of `N_(2)O_(5)(g)` is a first order reaction with a rate constant of `5xx10^(-4)s^(-1)` at `45^(@)C`. i.e., `2N_(2)O_(5)(g) to 4NO_(2)(g)+O_(2)(g)`. If initial concentration of `N_(2)O_(5)` is 0.25M, calculate its concentration after 2 min. Also calculate half-life for decomposition of `N_(2)O_(5)` (g).

Explanation

`k=5xx10^(-4)s^(-1), [R]_(0)=0.25 M, t=2 "min"=120s`
Using
`k=(2.303)/(t)"log"([R]_(0))/([R])" or log"([R]_(0))/([R])=(kt)/(2.303)`
Substituting the values, we get
`"log"(0.25)/([R])=(5xx10^(-4)xx120)/(2.303)=0.02605 or (0.25)/([R])=` Antilog 0.02605 = 1.061
or `[R]=(0.25)/(1.061)=0.23M`
`t_(1//2)=(0.693)/(k)=(0.693)/(5xx10^(-4)) or t_(1//2)=1386`