Question

The follwoing reaction is a single-step, bimolecular reaction: CH3Br+ NaOH -> CH3OH+NaBr When the concentrations of...

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?

Homework Answers

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

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