Question 25 of 30 Sapling Learning Ma The following reaction is a single-step, b

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Question 25 of 30 Sapling Learning Ma The following reaction is a single-step, bimolecular reaction: CH,Br+ NaOH CH,OH + NaBr When the concentrations of CH,Br and NaOH are both 0.170 M, the rate of the reaction is 0.0010 M/s (a) What is the rate of the reaction if the concentration of CH Br is doubled? Number M/s (b) What is the rate of the reaction if the concentration of NaOH is halved? Number M/ s (c) What is the rate of the reaction if the concentrations of CHyBr and NaOH are both increased by a factor of four? Number M/ s Previous Gve Up & View Solution Check Answer Hint Next Ext. about us careers privacy policy

Explanation / Answer

a)

Given Rate law is:

rate = k[CH3Br][NaOH]

rate new / rate old = ([CH3Br]new/[CH3Br]old)*([NaOH]new/[NaOH]old)

here:

[CH3Br]new/[CH3Br]old = 2

[NaOH]new/[NaOH]old = 1

putting values

rate new / rate old = (2.0)*(1.0)

rate new / rate old = 2

rate new / 0.001 = 2

rate new = 0.002 M/s

Answer: 0.0020 M/s

b)

Given Rate law is:

rate = k[CH3Br][NaOH]

rate new / rate old = ([CH3Br]new/[CH3Br]old)*([NaOH]new/[NaOH]old)

here:

[CH3Br]new/[CH3Br]old = 1

[NaOH]new/[NaOH]old = 0.5

putting values

rate new / rate old = (1.0)*(0.5)

rate new / rate old = 0.5

rate new / 0.001 = 0.5

rate new = 5.0*10^-4 M/s

Answer: 5.0*10^-4 M/s

c)

Given Rate law is:

rate = k[CH3Br][NaOH]

rate new / rate old = ([CH3Br]new/[CH3Br]old)*([NaOH]new/[NaOH]old)

here:

[CH3Br]new/[CH3Br]old = 4

[NaOH]new/[NaOH]old = 4

putting values

rate new / rate old = (4.0)*(4.0)

rate new / rate old = 16

rate new / 0.001 = 16

rate new = 0.016 M/s

Answer: 0.016 M/s

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