Clock reaction kinetics lab Data trial I - BrO 3 - H + Time in seconds 1 .002M .

Question

Clock reaction kinetics lab

Data

trial

I-

BrO3-

H+

Time in seconds

1

.002M

.008M

.02M

160

2

.004M

.008M

.02M

72

3

.002M

.016M

.02M

73

4

.002M

.008M

.04M

40

5

.0015M

.006M

.00225M

138

1. Calculate the rate of each reaction by taking the concentration of thiosulfate (.0002M) and dividing by the time in seconds.

A Compare trials 1 and 2 to determine the rate law exponent for [I-].

B Compare trials 1 and 3 to determine the rate law exponent for [BrO3-].

C Compare trials 1 and 4 to determine the rate law exponent for [H+].

2. Write the rate law with all exponents.

3. Find the value and units of the rate law constant, k.

A. Substitute the values for the concentrations and rate for trial 1 and solve for k.

B. Substitute the values for the concentrations and rate for trial 2 and solve for k.

C. Substitute the values for the concentrations and rate for trial 3 and solve for k.

D. Substitute the values for the concentrations and rate for trial 4 and solve for k.

4. Calculate the average value of k.

5. As a test of the rate law, use the average value of k and the concentrations of trial 5 to calculate a predicted rate for that trial. Compare the result to your measured rate by calculating the percent difference between them. %difference = ([actual rate - predicted rate] / average of actual rate and predicted rate) x 100

trial

I-

BrO3-

H+

Time in seconds

1

.002M

.008M

.02M

160

2

.004M

.008M

.02M

72

3

.002M

.016M

.02M

73

4

.002M

.008M

.04M

40

5

.0015M

.006M

.00225M

138

Explanation / Answer

1)

trail 1 : rate = 0.0002 / 160 = 1.25 x 10-6

trail 2 : rate = 0.0002 / 72 = 2.7778 x 10-6

trail 3 : rate = 0.0002 / 73 = 2.7397 x 10-6

trail 4 : rate = 0.0002 / 40 = 5 x 10-6

trail 5 : rate = 0.0002 / 138 = 1.4493 x 10-6

now

A)

let the rate law be

rate = k [I-]^a [Br03-]^b [H+]^c

now

for trail 1 and trail 2 , we can see that

[Br03-] = constant , [H+] = constant

so

we get

rate of trail 2 / rate of trail 1 = ( [I-]2 / [I-]1 )^a

2.7778 x 10-6 / 1.25 x 10-6 = ( 0.004 / 0.002)^a

(2.222) = (2)^a

a = 1

B)

for trail 1 and trail 3

rate 3 / rate 1 = ( [Br03-]3 / [Br03-]1)^b

2.7397 x 10-6 / 1.25 x 10-6 = ( 0.016/0.008)^b

2.19 = (2)^b

b = 1

C)

for trail 1 and trail 4

rate 4 / rate 1 = ( [H+]4 / [H+]1 )^c

5 x 10-6 / 1.25 x 10-6 = ( 0.04 / 0.02)^c

4 = (2)^c

c = 2

2)

using the values of a, b and c

we get

rate = k [I-] [Br03-] [H+]^2

3)

A. for trail 1

1.25 x 10-6 = k [0.002] [0.008] [0.02]^2

k = 195.3125

B) for trail 2

2.7778 x 10-6 = k [0.004] [0.008] [0.02]^2

k = 217.014

C) for trail 3

2.7397 x 10-6 = k [0.002] [0.016] [0.02]^2

k = 214.0411

D) for trail 4

5 x 10-6 = k [0.002] [0.008] [0.04]^2

k = 195.3125

4)

so using the above obtained values

average value of k = ( 195.3125 + 217.014 + 214.0411 + 195.3125) x 10-6 / 4

average value of k = 205.42

5)

now

for trail 5

rate = 205.42 x [0.0015] [0.006] [0.00225]^2

rate = 9.35944875 x 10-9

so

the rate for trail 5 is 9.35944875 x 10-9

average of actual rate and predicted rate = (1.4493 x 10-6) + (9.359944875 x 10-9) / 2

average of actual rate and predicted rate = 7.293297244 x 10-7

now

% difference = ( 1.4493 x 10-6 - 9.35944875 x 10-9 ) x 100 / 7.293297244 x 10-7

% difference = 197.4334

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