Respuesta :
Answer:
Rate of reaction will be half of it's initial value
Explanation:
For the given [tex]S_{N}2[/tex] reaction, the rate law is -
[tex]Rate=k[1-iodo-2-methylbutane][CN^{-}][/tex]
Where k is rate constant, [1-iodo-2-methylbutane] is concentration of 1-iodo-2-methylbutane and [tex][CN^{-}][/tex] is concentration of [tex]CN^{-}[/tex]
Here nucleophile is the [tex]CN^{-}[/tex] ion
Initiallly, [tex](Rate)_{initial}=k\times [1-iodo-2-methylbutane]_{initial}\times [CN^{-}]_{initial}[/tex]
When concentration of [tex]CN^{-}[/tex] is halved then-
[tex]Rate=k\times [1-iodo-2-methylbutane]_{initial}\times \frac{[CN^{-}]_{initial}}{2}=\frac{(Rate)_{initial}}{2}[/tex]
So rate of reaction will be half of it's initial value
