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?

Rate will increase by a factor of 2.

Explanation:

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

Thus when concentration of is doubled ,

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

the ratio of the conjugate base to the acid required to attain the desired ph.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2