The second order reaction 2C2H4 ---> C4H8 has a half life of 1.51 min. when [C2H4] = 0.250 M. How long would it take for [C2H4] to drop from 0.187 M to 0.092 M?
`2C_2H_4 -----gt C_4H_8`
This is a second order reaction. Therefore r is given by,
`r = k[C_2H_4]^2`
Therefore rate of decomposition of `C_2H_4` is given by,
`(d([C_2H_4]))/(dt) = -k[C_2H_4]^2`
`(d([C_2H_4]))/([C_2H_4]^2) = -kdt`
`((-1)/([C_2H_4]))_(N_0)^(N_t) = -kt`
The half time when `[C_2H_4] = 0.25 M` is given as 1.51 min.
`-(1/0.125 - 1/0.25) = -k xx 1.51`
`4 = k xx 1.51`
`k = 2.649` per min.
Therefore time for [C_2H_4] drop from 0.187 M to 0.092 M is,
`-(1/0.092 - 1/0.187) = -2.649xx t`
`t = 2.085` mins.
The time for [C_2H_4] drop from 0.187 M to 0.092 M is 2.085 minutes.