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For reaction RX + OH^(-) rarr ROH + X^(-...

For reaction `RX + OH^(-) rarr ROH + X^(-)`, rate expression is `R = 4.7 xx 10^(-5) [RX][OH^(-)] + 2.4 xx 10^(-5)[RX]`.
What `%` of reactant react by `S_(N)2` mechanism when `[OH^(-)] = 0.001` molar?

A

`1.9`

B

`66.2`

C

`95.1`

D

`16.4`

Text Solution

Verified by Experts

The correct Answer is:
A
Fraction of reactant decomposing
by `S_(N)2` mechanism
`= (4.7 xx 10^(-5)[RX][OH^(-)])/(4.7 xx 10^(-5) [RX][OH^(-)] + 2.4 xx 10^(-5)[RX])`
Put `[OH-] = 0.001 M`.
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Knowledge Check

  • For the reaction R - X +OH^(-) rarr R - OH+ X^(-) , the rate expression is given as rate = 4.7 xx10^(-5) [ R-X ] [ OH^(-) ]+ 0.24 xx 10^(-5)[R-X] What \% of R -X react by S_(N) 2 mechanism when [OH^(-) ] = 0.001 M

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