# 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?

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Expert Answers

thilina-g | Certified Educator

`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`

Separating variables,

`(d([C_2H_4]))/([C_2H_4]^2) = -kdt`

Integrating,

`((-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.**

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