Question

Hydrolysis of an alkyl halide $(RX)$ by dilute alkali $[\stackrel{\ominus }{O}H]$ takes place simultaneously by $S{N}^2$ and $S{N}^1$ pathway. A plot of $-\frac{1}{[RX]}\frac{d[R-x]}{dt}VS[\stackrel{\ominus }{O}H]$ is a straight line of the slope equal to $2\times 10^{3}mo{l}^{-1}L{h}^{-1}$ and intercept equal to $1\times 10^2{h}^{-1}$. Calculate the initial rate $\left(mole{L}^{-1}mi{n}^{-1}\right)$ of comsumption of $RX$ when the

Answer 1

Answer: 5

$\frac{-d[RX]}{dt}=k_2[RX][\stackrel{\ominus }{O}H]$ (by $S{N}^2$ path way)
$k_2=$ rate constant of $S{N}^2$ reaction
$\frac{-d[RX]}{dt}=k_1[RX](b^{2}S{N}^1$ path way)
$k_1=$ rate constant of $S{N}^1$ reaction
$\frac{-d[RX]}{dt}=k_2[RX][\stackrel{\ominus }{O}H]+k_1[RX]$
$-\frac{1}{[RX]}\frac{d[RX]}{dt}=k_2[\stackrel{\ominus }{O}H]+k_1$
This is the equation of a straight line for $-\frac{1}{[RX]}\frac{d[RX]}{dt}$
vs $[\stackrel{\ominus }{O}H]$ plot with slope equal to $k_2$ and intercept equal to $k_1$
from question:
$k_2=2\times 10^{3}mo{l}^{-1}Lh{r}^{-1},k_1=1\times 10^{2}h{r}^{-1}$
$[RX]=1.0M$ and $[\stackrel{\ominus }{\mathrm{OH}}]=0.1M$
Hence,
$\frac{-d[RX]}{dt}=2\times 10^3\times 1\times 0.1+1\times 10^2\times 1$
$=300mol{L}^{-1}h{r}^{-1}$
$=5mol{L}^{-1}mi{n}^{-1}$