Question

The follwoing reaction is a single-step, bimolecular reaction: CH₃Br+ NaOH -> CH₃OH+NaBr

When the concentrations of CH₃Br and NaOH are both 0.145M, the rate of the reation is 0.0020 M/s/.

A) what is the rate of the reaction if the concentration of CH₃Br is doubled?

B) what is the rate of the reaction if the concdntration of NaOH is halved?

C) what is the rate of the reaction if the concentraions of CH₃Br and NaOH are both increased by a factor of three?

Answer 1

since, this is single step reaction
this step wiill be rate determinig one
so,
rate = k*[CH3Br]*[NaOH]

A)
since, rate is directly propional to [CH3Br]
so, when [CH3Br] is doubled rate also become double
so,
rate = 2*(intial rate)
= 2*0.0020
= 0.0040 M/s
B)
since, rate is directly propional to [NaOH]
so, when [NaOH] is halved rate also become half
so,
rate = (1/2)*(intial rate)
= (1/2)*0.0020
= 0.0010 M/s
C)
since, rate is directly propional to [CH3Br],[NaOH]
so, when [CH3Br],[NaOH] each tripled rate also become 9 times
so,
rate = 9*(intial rate)
= 9*0.0020
= 0.018 M/s