Question

The reaction of methyl iodide with sodium azide, NaN3, proceeds by an SN2 mechanism. What is the effect of doubling the concentration of NaN3 on the rate of the reaction?

Answer 1

Answer: Rate will increase by a factor of 2.

Step-by-step explanation:

As the reaction proceeds by an
`SN_2` mechanism, both the reactants take part in the reaction and the rate law for the reaction will be:

`Rate=k[NaN_3]^1[CH_3I]^1`

Thus when concentration of
`NaN_3` is doubled ,

`Rate'=k[2NaN_3]^1[CH_3I]^1`

`Rate'=2* k[NaN_3]^1[CH_3I]^1`

`Rate'=2* Rate`

Thus the rate of the reaction would increase by a factor of 2.