Question

In: Chemistry

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

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

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

Calculate the average rate coefficient in te above question.

A.0.003
B.0.013
C.0.025
D.0.034

Expert Solution

-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 1^st order w.r.t ketone.

From trial 2 & 3 : [H⁺] increased by 3 and rate has also increased by 3 so 1^st 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

queen_honey_blossom answered 1 year ago

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