The iodination of acetone, \ m CH_3COCH_3, in aqueous solution is catalyzed by t

ID: 746183 • Letter: T

Question

The iodination of acetone, m CH_3COCH_3, in aqueous solution is catalyzed by the hydrogen ion m H^+: matrix { & m _{H^+} & cr m I_2+CH_3COCH_3 & longrightarrow & m HI+CH_3COCH_2I} The reaction can be followed visually by adding starch. The purple starch-iodine complex forms immediately, then disappears when the iodine has been consumed. The kinetics of the reaction may be determined by recording the time required for the color to disappear, which is the time required for the iodine, m I_2, to be consumed. The following data were acquired in an iodination experiment involving acetone. All reaction times are in terms of the rate of disappearance of m I_2. Trial Volume of 0.0010 M~ m I_2 ( m mL) Volume of 0.050 M~ m HCl ( m mL) Volume of 1.0 M acetone ( m mL) Volume of water ( m mL) Temperature (^{circ} m C) Reaction time ( m s) 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 Part A In the general rate law m rate={it k}[I_2]^{it X}[H^+]^{it Y}[CH_3COCH_3]^{it Z} what are the values of X, Y, and Z?

Explanation / Answer

HOPE THIS HELPS !! http://gyazo.com/a75cd7520bafb5718bac2d0…

Activation Energy from Experimental Data

The following data were acquired in an iodination experiment involving acetone. All reaction times are in terms of the rate of disappearance of I_2. there are only two temperatures 25 and 42.4 degree C ...there should be atleast three temperatures ...

anyways let me try..

reaction is CH3COCH3 + I2 + H+ ------> CH3COCH2 + HI

let rate law be,r = k[CH3COCH3]^m [H+]^n [I2]^p

since we are using starch as a method of investigating rate .....so lets concentrate on iodine only at the moment ..

you should be knowing that rate can also be expressed as one single reactant by the following expression ..

r = - delta[I2]/delta t

in these datas I2 is the limiting reagent .....as it is present in lesser quantity than H+ and acetone ....

so it will be entirely consumed ...

so r = -( [I2]final - [I2]initial )/delta t

r = - (0 -[I2]initial )/delta t

r = [I2]initial/delta t

now given concentration of I2 = 0.001 M ...
in exp-1 5ml is used so its concentration will change...

so using M1V1 = M2V2 in order to find out concentration of I2 in experiment 1

M1 = 0.001 M
V1 = 5 ml

M2 = ? M
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/s

similarly for experiment 2

M1 = 0.001
V1 = 10 ml

M2 = ? M
V2 = 10 + 10 + 10 + 20 = 50 ml

M2 = 0.001 X 10 / 50 = 2 X 10^-4 M

rate = 2 X 10^-4/249 = 8.03 X 10^-7 M/s

rate for experiment -2 is 8.03 X 10^-7 and for experiment -1 is 7.79 X 10^-7...which is almost equal ...

notice that we have doubled the concentration of I2 from experiment -1 to experiment -2 ...
M2 in exp.1 was 1 X 10^-4 and M2 in exp -2 is 2 X 10^-4 .....

but the rate is still almost equal so order with respect to I2 is 0...as there is no change in rate with change in concentration of I2

for experiment- 3

using the same process as above
M2 for iodine = 2 X 10^-4

so rate = 2 X 10^-4/128 = 1.6 X 10^-6

similarly for experiment-4

rate = 2 X 10^-4/131 = 1.53 X 10^-6

for experiment-5
rate = 2 X 10^-4/38 = 5.37 X 10^-6

now in experiment -2 and experiment-3 ...

conc. of I2 = 2X10^-4
conc. of acetone = 1X10/50 = 0.2 M

conc. of H+ in exp-2 = 10X0.05/50 = 0.01 M
conc. of H+ in exp-3 = 20X0.05/50 = 0.02 M

see conc. of I2 and acetone remains constant in exp-2 and exp-3 ....but conc. of H+ has doubled ..

rate in exp-2 = 8.03 X 10^-7
rate in exp-3 = 1.6 X 10^-6

as 1.6 X 10^-6/8.03 X 10^-7 = 1.99 or almost 2

so rate doubles when conc. of H+ doubles keeping conc. of acetone and I2 constant so order with respect to H+ is 1

similarly for experiment-2 and experiment-4
conc.of I2 = 2X10^-4
conc. of H+ = 0.01 M

conc. of acetone in exp-2 = 10X1/50 = 0.2
conc. of acetone in exp-4 = 20X1/50 = 0.4 M

rate in exp-2 = 8.03 X 10^-7
rate in exp-4 = 1.53 X 10^-6

1.53 X 10^-6 / 8.03 X 10^-7 = 1.91 which is almost equal to 2

so order with respect to acetone is also equal to 1 as rate doubles when acetone doubles keeping cocn . of H+ and I2 constant

so rate law is r = k [acetone] [H+]

now for experiment -1
[acetone]initiial = 1 X 10 / 50 = 0.2 M
[H+]initial = 0.05 X 10 / 50 = 0.01 M

calculating rate constant k

exp-1
rate = k[aceton] [H+]

7.69 X 10^-7 = k X 0.2 X 0.01

k = 3.85 X 10^-4

for exp-2
[acetone] initial = 0.2
[H+] initial = 0.01

r = k [acetone] [H+]

8.03 X 10^-7 = k X 0.2 X 0.01

k = 4.015 X 10^-4

for exp-3
[acetone]initial = 0.2
[H+] initial = 0.05 X 20/50 = 0.02 M

rate = k [acetone] [H+]
1.6 X 10^-6 = k X 0.2 X 0.02
k = 4 X 10^-4

for exp-4
[acetone initial = 20 X 1/50 = 0.4 M
[H+] initial = 0.01 M

rate = k [acetone] [H+]

1.53 X 10^-6 = k X 0.4 X 0.01

k = 3.8 X 10^-4

now averaging k value for exp-1 to exp-4 as these were done at same temperature ...

average value of k at 298 K = [ (3.85 X 10^-4) + (4.015 X 10^-4) + (4 X 10^-4) + (3.8 X 10^-4) ] / 4 = 3.92 X 10^-4

so average value of k at 298 K = 3.92 X 10^-4

now for exp-5

[acetone] initial = 0.2 M
[H+] initial = 0.01 M

so rate= k [acetone] [H+]
5.37 X 10^-6 = k X 0.2 X 0.01

k = 2.68 X 10^-3

so k at 298 K = 3.92 X 10^-4 ...and k at 315.4 K = 2.68 X 10^-3

now using arrhenius eqn ..

log k2/k1 = Ea/2.303R [T2-T1/T1T2]

where k2 = 2.68 X 10^-3
T2 = 315.4

k1 = 3.92 X 10^-4
T1 = 298 K

Ea = activation energy
R = 8.314 J/K/mole

so log 2.68 X 10^-3 / 3.92 X 10^-4 = Ea/2.303X8.314 [315.4-298/315.4X298]

log 6.84 = Ea/19.15 X [17.4/93989.2 ]

Ea = 0.84 X 19.15 X 93989.2 / 17.4 = 86891.39 J

so activation enegy = 86891.39 J or almost equal to 86 kj

please check the maths as it was a very lengthy question ...

also i know that you will have some confusion regarding these ...please feel free to ask any question on these ....you can mail me or add additional details ...i will definitely help there are only two temperatures 25 and 42.4 degree C ...there should be atleast three temperatures ...

anyways let me try..

reaction is CH3COCH3 + I2 + H+ ------> CH3COCH2 + HI

let rate law be,r = k[CH3COCH3]^m [H+]^n [I2]^p

since we are using starch as a method of investigating rate .....so lets concentrate on iodine only at the moment ..

you should be knowing that rate can also be expressed as one single reactant by the following expression ..

r = - delta[I2]/delta t

in these datas I2 is the limiting reagent .....as it is present in lesser quantity than H+ and acetone ....

so it will be entirely consumed ...

so r = -( [I2]final - [I2]initial )/delta t

r = - (0 -[I2]initial )/delta t

r = [I2]initial/delta t

now given concentration of I2 = 0.001 M ...
in exp-1 5ml is used so its concentration will change...

so using M1V1 = M2V2 in order to find out concentration of I2 in experiment 1

M1 = 0.001 M
V1 = 5 ml

M2 = ? M
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/s

similarly for experiment 2

M1 = 0.001
V1 = 10 ml

M2 = ? M
V2 = 10 + 10 + 10 + 20 = 50 ml

M2 = 0.001 X 10 / 50 = 2 X 10^-4 M

rate = 2 X 10^-4/249 = 8.03 X 10^-7 M/s

rate for experiment -2 is 8.03 X 10^-7 and for experiment -1 is 7.79 X 10^-7...which is almost equal ...

notice that we have doubled the concentration of I2 from experiment -1 to experiment -2 ...
M2 in exp.1 was 1 X 10^-4 and M2 in exp -2 is 2 X 10^-4 .....

but the rate is still almost equal so order with respect to I2 is 0...as there is no change in rate with change in concentration of I2

for experiment- 3

using the same process as above
M2 for iodine = 2 X 10^-4

so rate = 2 X 10^-4/128 = 1.6 X 10^-6

similarly for experiment-4

rate = 2 X 10^-4/131 = 1.53 X 10^-6

for experiment-5
rate = 2 X 10^-4/38 = 5.37 X 10^-6

now in experiment -2 and experiment-3 ...

conc. of I2 = 2X10^-4
conc. of acetone = 1X10/50 = 0.2 M

conc. of H+ in exp-2 = 10X0.05/50 = 0.01 M
conc. of H+ in exp-3 = 20X0.05/50 = 0.02 M

see conc. of I2 and acetone remains constant in exp-2 and exp-3 ....but conc. of H+ has doubled ..

rate in exp-2 = 8.03 X 10^-7
rate in exp-3 = 1.6 X 10^-6

as 1.6 X 10^-6/8.03 X 10^-7 = 1.99 or almost 2

so rate doubles when conc. of H+ doubles keeping conc. of acetone and I2 constant so order with respect to H+ is 1

similarly for experiment-2 and experiment-4
conc.of I2 = 2X10^-4
conc. of H+ = 0.01 M

conc. of acetone in exp-2 = 10X1/50 = 0.2
conc. of acetone in exp-4 = 20X1/50 = 0.4 M

rate in exp-2 = 8.03 X 10^-7
rate in exp-4 = 1.53 X 10^-6

1.53 X 10^-6 / 8.03 X 10^-7 = 1.91 which is almost equal to 2

so order with respect to acetone is also equal to 1 as rate doubles when acetone doubles keeping cocn . of H+ and I2 constant

so rate law is r = k [acetone] [H+]

now for experiment -1
[acetone]initiial = 1 X 10 / 50 = 0.2 M
[H+]initial = 0.05 X 10 / 50 = 0.01 M

calculating rate constant k

exp-1
rate = k[aceton] [H+]

7.69 X 10^-7 = k X 0.2 X 0.01

k = 3.85 X 10^-4

for exp-2
[acetone] initial = 0.2
[H+] initial = 0.01

r = k [acetone] [H+]

8.03 X 10^-7 = k X 0.2 X 0.01

k = 4.015 X 10^-4

for exp-3
[acetone]initial = 0.2
[H+] initial = 0.05 X 20/50 = 0.02 M

rate = k [acetone] [H+]
1.6 X 10^-6 = k X 0.2 X 0.02
k = 4 X 10^-4

for exp-4
[acetone initial = 20 X 1/50 = 0.4 M
[H+] initial = 0.01 M

rate = k [acetone] [H+]

1.53 X 10^-6 = k X 0.4 X 0.01

k = 3.8 X 10^-4

now averaging k value for exp-1 to exp-4 as these were done at same temperature ...

average value of k at 298 K = [ (3.85 X 10^-4) + (4.015 X 10^-4) + (4 X 10^-4) + (3.8 X 10^-4) ] / 4 = 3.92 X 10^-4

so average value of k at 298 K = 3.92 X 10^-4

now for exp-5

[acetone] initial = 0.2 M
[H+] initial = 0.01 M

so rate= k [acetone] [H+]
5.37 X 10^-6 = k X 0.2 X 0.01

k = 2.68 X 10^-3

so k at 298 K = 3.92 X 10^-4 ...and k at 315.4 K = 2.68 X 10^-3

now using arrhenius eqn ..

log k2/k1 = Ea/2.303R [T2-T1/T1T2]

where k2 = 2.68 X 10^-3
T2 = 315.4

k1 = 3.92 X 10^-4
T1 = 298 K

Ea = activation energy
R = 8.314 J/K/mole

so log 2.68 X 10^-3 / 3.92 X 10^-4 = Ea/2.303X8.314 [315.4-298/315.4X298]

log 6.84 = Ea/19.15 X [17.4/93989.2 ]

Ea = 0.84 X 19.15 X 93989.2 / 17.4 = 86891.39 J

so activation enegy = 86891.39 J or almost equal to 86 kj

please check the maths as it was a very lengthy question ...

also i know that you will have some confusion regarding these ...please feel free to ask any question on these ....you can mail me or add additional details ...i will definitely help there are only two temperatures 25 and 42.4 degree C ...there should be atleast three temperatures ...

anyways let me try..

reaction is CH3COCH3 + I2 + H+ ------> CH3COCH2 + HI

let rate law be,r = k[CH3COCH3]^m [H+]^n [I2]^p

since we are using starch as a method of investigating rate .....so lets concentrate on iodine only at the moment ..

you should be knowing that rate can also be expressed as one single reactant by the following expression ..

r = - delta[I2]/delta t

in these datas I2 is the limiting reagent .....as it is present in lesser quantity than H+ and acetone ....

so it will be entirely consumed ...

so r = -( [I2]final - [I2]initial )/delta t

r = - (0 -[I2]initial )/delta t

r = [I2]initial/delta t

now given concentration of I2 = 0.001 M ...
in exp-1 5ml is used so its concentration will change...

so using M1V1 = M2V2 in order to find out concentration of I2 in experiment 1

M1 = 0.001 M
V1 = 5 ml

M2 = ? M
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/s

similarly for experiment 2

M1 = 0.001
V1 = 10 ml

M2 = ? M
V2 = 10 + 10 + 10 + 20 = 50 ml

M2 = 0.001 X 10 / 50 = 2 X 10^-4 M

rate = 2 X 10^-4/249 = 8.03 X 10^-7 M/s

rate for experiment -2 is 8.03 X 10^-7 and for experiment -1 is 7.79 X 10^-7...which is almost equal ...

notice that we have doubled the concentration of I2 from experiment -1 to experiment -2 ...
M2 in exp.1 was 1 X 10^-4 and M2 in exp -2 is 2 X 10^-4 .....

but the rate is still almost equal so order with respect to I2 is 0...as there is no change in rate with change in concentration of I2

for experiment- 3

using the same process as above
M2 for iodine = 2 X 10^-4

so rate = 2 X 10^-4/128 = 1.6 X 10^-6

similarly for experiment-4

rate = 2 X 10^-4/131 = 1.53 X 10^-6

for experiment-5
rate = 2 X 10^-4/38 = 5.37 X 10^-6

now in experiment -2 and experiment-3 ...

conc. of I2 = 2X10^-4
conc. of acetone = 1X10/50 = 0.2 M

conc. of H+ in exp-2 = 10X0.05/50 = 0.01 M
conc. of H+ in exp-3 = 20X0.05/50 = 0.02 M

see conc. of I2 and acetone remains constant in exp-2 and exp-3 ....but conc. of H+ has doubled ..

rate in exp-2 = 8.03 X 10^-7
rate in exp-3 = 1.6 X 10^-6

as 1.6 X 10^-6/8.03 X 10^-7 = 1.99 or almost 2

so rate doubles when conc. of H+ doubles keeping conc. of acetone and I2 constant so order with respect to H+ is 1

similarly for experiment-2 and experiment-4
conc.of I2 = 2X10^-4
conc. of H+ = 0.01 M

conc. of acetone in exp-2 = 10X1/50 = 0.2
conc. of acetone in exp-4 = 20X1/50 = 0.4 M

rate in exp-2 = 8.03 X 10^-7
rate in exp-4 = 1.53 X 10^-6

1.53 X 10^-6 / 8.03 X 10^-7 = 1.91 which is almost equal to 2

so order with respect to acetone is also equal to 1 as rate doubles when acetone doubles keeping cocn . of H+ and I2 constant

so rate law is r = k [acetone] [H+]

now for experiment -1
[acetone]initiial = 1 X 10 / 50 = 0.2 M
[H+]initial = 0.05 X 10 / 50 = 0.01 M

calculating rate constant k

exp-1
rate = k[aceton] [H+]

7.69 X 10^-7 = k X 0.2 X 0.01

k = 3.85 X 10^-4

for exp-2
[acetone] initial = 0.2
[H+] initial = 0.01

r = k [acetone] [H+]

8.03 X 10^-7 = k X 0.2 X 0.01

k = 4.015 X 10^-4

for exp-3
[acetone]initial = 0.2
[H+] initial = 0.05 X 20/50 = 0.02 M

rate = k [acetone] [H+]
1.6 X 10^-6 = k X 0.2 X 0.02
k = 4 X 10^-4

for exp-4
[acetone initial = 20 X 1/50 = 0.4 M
[H+] initial = 0.01 M

rate = k [acetone] [H+]

1.53 X 10^-6 = k X 0.4 X 0.01

k = 3.8 X 10^-4

now averaging k value for exp-1 to exp-4 as these were done at same temperature ...

average value of k at 298 K = [ (3.85 X 10^-4) + (4.015 X 10^-4) + (4 X 10^-4) + (3.8 X 10^-4) ] / 4 = 3.92 X 10^-4

so average value of k at 298 K = 3.92 X 10^-4

now for exp-5

[acetone] initial = 0.2 M
[H+] initial = 0.01 M

so rate= k [acetone] [H+]
5.37 X 10^-6 = k X 0.2 X 0.01

k = 2.68 X 10^-3

so k at 298 K = 3.92 X 10^-4 ...and k at 315.4 K = 2.68 X 10^-3

now using arrhenius eqn ..

log k2/k1 = Ea/2.303R [T2-T1/T1T2]

where k2 = 2.68 X 10^-3
T2 = 315.4

k1 = 3.92 X 10^-4
T1 = 298 K

Ea = activation energy
R = 8.314 J/K/mole

so log 2.68 X 10^-3 / 3.92 X 10^-4 = Ea/2.303X8.314 [315.4-298/315.4X298]

log 6.84 = Ea/19.15 X [17.4/93989.2 ]

Ea = 0.84 X 19.15 X 93989.2 / 17.4 = 86891.39 J

so activation enegy = 86891.39 J or almost equal to 86 kj

please check the maths as it was a very lengthy question ...

also i know that you will have some confusion regarding these ...please feel free to ask any question on these ....you can mail me or add additional details ...i will definitely help

Navigate

The iodination of acetone, \ m CH_3COCH_3, in aqueous solution is catalyzed by t

The iodine clock experiment, which consists of the following three reactions, is

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.

The iodination of acetone, \ m CH_3COCH_3, in aqueous solution is catalyzed by t

Question

Explanation / Answer

Related Questions

Navigate