In: Chemistry
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 |
-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