Displacement of the hydroxyl group of alcohol can take place with variety of reagents. One such reaction is

$\mathrm{HBr}+{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}⟶{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+{\mathrm{H}}_2\mathrm{O}$

The above reaction has a two-step mechanism.

(a) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}+{\mathrm{H}}^{+}\stackrel{k_{eq}}{\rightleftharpoons }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{OH}}_2^{+}$

(b) ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{OH}}_2^{+}+{\mathrm{Br}}^{-}\stackrel{{\kappa }_2}{⟶}{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+{\mathrm{H}}_2\mathrm{O}$

The slow and fast can be identified by the fact that rate of displacement of hydroxyl by halide ions depends on the type of halide ion.

Calculate the rate of the formation of Ethyl Bromide when concentration of alcohol is 0.5 M and HBr is 10^-1 M. Assume complete dissociation of HBr. Express your answer in terms of 10^-6 M/sec.

$\text{ [Given: }{\mathrm{K}}_{\mathrm{eq}}=4{\mathrm{M}}^{-1}\text{ and }{\mathrm{K}}_2=5\times {10}^{-2}{\mathrm{M}}^{-1}{\sec }^{-1}\text{ ] }$