Question

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

Answer 1

Answer:

Explanation:

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