What happens to the rate of an SN2 reaction when [RX] is halved, and [:Nu^-] is...

Question

Question

Answer 1

Answer #1

Answer 2

Answer #2

The rate of reaction for SN2 (Bimolecular nucleophilic substitution) reaction depends upon substrate and nucleophile as

$Rate \propto [RX][Nucleophile]$

Thats why it is second order reaction

Now if halved the concentration of RX and doubled the concentration of Nu- then overall rate of reaction will remain the same

$Rate \propto [\frac{RX}{2}][2Nucleophile]$

$Rate \propto [RX][Nucleophile]$ Hence no effect

Answer 3

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