Chemistry: 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?

The reaction of methyl iodide with sodium azide,

jandrew8007

20.01.2022

Rate will increase by a factor of 2.

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

$Rate'=2\times k[NaN_3]^1[CH_3I]^1$

$Rate'=2\times Rate$

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

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Jalyn6883

29.01.2021

It’s a and e

Explanation:

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