Hydrolyiss of an alkyl halide `(RX)` by dilute alkali `[OH]^(ɵ)` takes place ismultaneously by `SN^(2)` and `SN^(1)` pathways. A plot of `-(1)/([RX]) (d[R-X])/(dt)` vs `[OH]^(ɵ)` is a straight line of the slope equal to `2 xx 10^(3) mol^(-1) L h^(-1)` and intercept equal to `1 xx 10^(2) h^(-1)`. Calculate the initial rate `("mole" L^(-1) min^(-1))` of consumption of `RX` when the reaction is carried out taking `mol L^(-1)` of `RX` and `0.1 mol L^(-1)` of `[OH]^(ɵ)` ions.

Answer

5

Explanation

`(-d[RX])/(dt)=k_(2)[RX][overset(ɵ)(O)H]`, (by `S_(N)` path way)
`k_(2)=` rate constant of `S_(N)2` reaction
`(-d[RX])/(dt)=k_(1)[RX]`, (by `S_(N)1` path way)
`k_(1)=` rate constant of `S_(N)1` reaction
`(-d[RX])/(dt)=k_(2)[RX][overset(ɵ)(O)H]+k_(1)[RX]`
`-1/([RX])(d[RX])/(dt)= k_(2)[overset(ɵ)(O)H]+k_(1)`
This is the equation of a straight line for `-1/([RX])(d[RX])/(dt)`
vs `[overset(ɵ)(O)H]` plot with slope equal to `K_(2)` and intercept equal to `k_(1)`
form question:
`k_(2)=2xx10^(3) mol^(-1) L hr^(-1), K_(1)=1 xx10^(2) hr^(-1)`
`[RX]=1.0 M and [overset(ɵ)(O)H] =0.1 M`
Hence,
`(-d[RX])/(dt)=2xx10^(3)xx1xx0.1+1xx10^(2)xx1`
`= 300 mol L^(-1) hr^(-1)`
`=5 mol L^(-1) min^(-1)`