You are studying the reaction of iodine with a ketone to produce iodoketone with

ID: 1051878 • Letter: Y

Question

You are studying the reaction of iodine with a ketone to produce iodoketone with the following equation:

I2 + ketone iodoketone + H+ + I-

Data for initial rates and concentrations are given in the table below:

-d[I2]/dt [I2] [ketone] [H+]

mol-1 L s-1 M M M

7 x 10-5 5 x 10-4 0.2 1.0 x 10-2

7 x 10-5 3 x 10-4 0.2 1.0 x 10-2

1.7 x 10-5 5 x 10-4 0.5 1.0 x 10-2

5.4 x 10-5 5 x 10-4 0.5 3.2 x 10-2

0, 1, 0

Calculate the average rate coefficient in te above question.

0, 1, 0

B.0, 1, 1 C.1, 1, 1 D.1, 2, 0

-d[I2]/dt [I2] [ketone] [H+]

mol-1 L s-1 M M M

0.7 x 10-5 5 x 10-4 0.2 1.0 x 10-2

0.7 x 10-5 3 x 10-4 0.2 1.0 x 10-2

1.7 x 10-5 5 x 10-4 0.5 1.0 x 10-2

5.4 x 10-5 5 x 10-4 0.5 3.2 x 10-2

Q.1) Answer is (B) 0, 1, 1

From trial 1 & 2 : [I2] changed but rate has not changed so zero order w.r.t I2.

From trial 2 & 3 : [ketone] increased by 2.5 and rate has also increased by 2.5 so 1st order w.r.t ketone.

From trial 2 & 3 : [H+] increased by 3 and rate has also increased by 3 so 1st order w.r.t H+.

Q.2)

rate = k x [ketone][H+] or, rate coefficient = k = rate / {[ketone][H+]}

For trial #1: k = 0.7 x 10-5 /(0.2 x 1.0 x 10-2) = 0.0035 s-1M-1

For trial #3: k = 1.7 x 10-5 /(0.5 x 1.0 x 10-2) = 0.0034 s-1M-1

For trial #4: k = 5.4 x 10-5 /(0.5 x 3.2 x 10-2) = 0.0034 s-1M-1

So the answer is (D) 0.0034

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