The following data were acquired in an iodination experiment involving acetone.

Question

The following data were acquired in an iodination experiment involving acetone. All reaction times are in terms of the rate of disappearance of .
Trial Volume of 0.0010
() Volume of 0.050
() Volume of 1.0 acetone
() Volume of water
() Temperature
() Reaction time
()
A 5.0 10.0 10.0 25.0 25.0 130
B 10.0 10.0 10.0 20.0 25.0 249
C 10.0 20.0 10.0 10.0 25.0 128
D 10.0 10.0 20.0 10.0 25.0 131
E 10.0 10.0 10.0 20.0 42.4 38

What is the rate constant at 25.0 based on the data collected for trial B?

I've asked this a few times and nobody can get me a step by step with an answer with the appropriate units?!

Explanation / Answer

CH3COCH3 + I2 + H+ ------> CH3COCH2 + HI
r = k[CH3COCH3]^m [H+]^n [I2]^p
r = - delta[I2]/delta t
in these datas I2 is the limiting reagent
so r = -( [I2]final - [I2]initial )/delta t
r = - (0 -[I2]initial )/delta t
r = [I2]initial/delta t
so using M1V1 = M2V2 in order to find out concentration of I2 in experiment 1
M1 = 0.001 M ,V1 = 5 ml
V2 = 5 + 10 + 10 + 25 = 50 ml
so 0.001 X 5 = M2 X 50
M2 = 0.001 X 5 / 50 = 1 X 10^-4
so [I2]initial = 1 X 10^-4 M
so rate = 1 X 10^-4 / 130 = 7.692 X 10^-7 M

now you can use same process for rest of question you will get ans definately.

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