Consider the single‑step, bimolecular reaction. CH3Br+NaOH⟶CH3OH+NaBr When the concentrations of CH3Br and NaOH are both 0.120 M, the rate of the reaction is 0.0080 M/s. What is the rate of the reaction if the concentration of CH3Br is doubled? rate: M/s What is the rate of the reaction if the concentration of NaOH is halved? rate: M/s What is the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of 5? rate: M/s

CH₃Br+NaOH⟶CH₃OH+NaBr

It is a single step bimolecular reaction so order of reaction is 2, one for CH₃Br and one for NaOH.

rate of reaction = k x [CH₃Br] [ NaOH]

.008 = k x. 12 x. 12

k =.55555

when concentration of CH₃Br is doubled

rate of reaction =.555555 x [.24] [.12 ]

=.016 M/s

when concentration of NaOH is halved

rate of reaction =.555555 x [.12] [.06 ]

=.004 M/s

when concentration of both CH₃Br and Na OH is made 5 times

rate of reaction =.555555 x. 6 x. 6

= 0.2 M/s