For the decomposition of `N_(2) O_(5)` given as,
`N_(2) O_(5) (g) to N_(2) O_(4) (g) + 1//2O_(2) (g)`
If the rate law `=K [N_(2) O_(5) ] and K=1.68 xx 10^(-2)" s"^(-1)`, starting with 2.5 moles, of `N_(2) O_(5)(g)` in a `5.0"L"` container at 298K, how many moles of `N_(2) O_(5)` would remain after 1.0 minutes?

Explanation

Initial concentration of `N_(2) O_(5) = [ N]_0 = 2.5 = 0.5" mol/L"`
Let `[N] =` final concentration
According to first order rate equation,
`k = (2.303)/( t) log"" ([N]_0)/( [N])`
On putting values
`1.68 xx 10^(-2)" s"^(-1) = (2.303)/( 60 "sec " ) log"" (0.5)/( [N])`
On solving above equation `[N] = 0.183` moles