Question 8 of 24 I Inco Map CH-Br+ NaOH CH-OH +NaBr When the concentra.docx

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Question 8 of 24 I Inco Map CH,Br+ NaOH CH,OH +NaBr When the concentrations of CH3Br and NaOH are both 0.170 M, the rate of the reaction is 0.0090 M/s. (a) What is the rate of the reaction if the concentration of CH3Br is doubled? Number .10 M/s (b) What is the rate of the reaction if the concentration of NaOH is halved? Number .0045 MIs (c) What is the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of six? Number .0093 M/s available! on the View this feedback b bottom divider bar. Click on the divider bar again to hide the additional feedback. Close Give Up & View Solution Try Again Next Exit Solution 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 / 0.009 = 2 rate new = 0.018 M/s Answer: 0.018 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 / 0.009 = 0.5 rate new = 0.0045 M/s Answer: 0.0045 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 = 6 [NaOH]new/[NaOH]old = 6 putting values rate new / rate old = (6.0)*(6.0) rate new / rate old = 36 rate new / 0.009 = 36 rate new = 0.32 M/s Answer: 0.32 M/s .

Question 8 of 24 I Inco Map CH,Br+ NaOH CH,OH +NaBr When the concentrations of CH3Br
and NaOH are both 0.170 M, the rate of the reaction is 0.0090 M/s. (a) What is the rate of the
reaction if the concentration of CH3Br is doubled? Number .10 M/s (b) What is the rate of the
reaction if the concentration of NaOH is halved? Number .0045 MIs (c) What is the rate of the
reaction if the concentrations of CH3Br and NaOH are both increased by a factor of six? Number
.0093 M/s available! on the View this feedback b bottom divider bar. Click on the divider bar
again to hide the additional feedback. Close Give Up & View Solution Try Again Next Exit
Solution
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 / 0.009 = 2
rate new = 0.018 M/s

Answer: 0.018 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 / 0.009 = 0.5
rate new = 0.0045 M/s
Answer: 0.0045 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 = 6
[NaOH]new/[NaOH]old = 6
putting values
rate new / rate old = (6.0)*(6.0)

rate new / rate old = 36
rate new / 0.009 = 36
rate new = 0.32 M/s
Answer: 0.32 M/s

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2. Answer: 0.018 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 / 0.009 = 0.5 rate new = 0.0045 M/s Answer: 0.0045 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 = 6 [NaOH]new/[NaOH]old = 6 putting values rate new / rate old = (6.0)*(6.0)

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