When the concentrations of CH 3 Br and NaOH are both 0.100 M, the rate of the reaction is 0.0030 M/s. What is the rate of the reaction if the concentration of CH 3 Br is doubled?

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The balanced equations will be:

[tex]CH_3Br+NaOH\rightarrow CH_3OH+NaBr[/tex]

In this reaction, [tex]CH_3Br[/tex] and [tex]NaOH[/tex] are the reactants.

The rate law expression for the reaction is:

[tex]\text{Rate}=k[CH_3Br][NaOH][/tex]

As we are given that:

[tex][CH_3Br][/tex] = concentration of [tex]CH_3Br[/tex] = 0.100 M

[tex][NaOH][/tex] = concentration of [tex]NaOH[/tex] = 0.100 M

Rate = 0.0030 M/s

Now put all the given values in the above expression, we get:

[tex]0.0030M/s=k\times (0.100M)\times (0.100M)[/tex]

[tex]k=0.3M^{-1}s^{-1}[/tex]

Now we have to calculate the rate of the reaction if the concentration of [tex]CH_3Br[/tex] is doubled.

[tex]\text{Rate}=k[CH_3Br][NaOH][/tex]

[tex]\text{Rate}=(0.3M^{-1}s^{-1})\times (2\times 0.100M)\times (0.100M)[/tex]

[tex]\text{Rate}=0.006M/s[/tex]

Thus, the rate of the reaction if the concentration of [tex]CH_3Br[/tex] is doubled is, 0.006 M/s