Answer: Rate will increase by a factor of 2.
Explanation:
As the reaction proceeds by an [tex]SN_2[/tex] mechanism, both the reactants take part in the reaction and the rate law for the reaction will be:
[tex]Rate=k[NaN_3]^1[CH_3I]^1[/tex]
Thus when concentration of [tex]NaN_3[/tex] is doubled ,
[tex]Rate'=k[2NaN_3]^1[CH_3I]^1[/tex]
[tex]Rate'=2\times k[NaN_3]^1[CH_3I]^1[/tex]
[tex]Rate'=2\times Rate[/tex]
Thus the rate of the reaction would increase by a factor of 2.
