Experiment [A] [B] plot that gave straight line #1 1.0 M 1 x 10-4 M ln[B] vs tim
ID: 560081 • Letter: E
Question
Experiment [A] [B] plot that gave straight line
#1 1.0 M 1 x 10-4 M ln[B] vs time
#2 1 x 10-4 M 1.0 M 1/[A] vs time
I. A + B < = > E (fast)
E + B --> C + D (slow)
II. A + B < = > E (fast)
E + A --> C + D (slow)
III. A + A --> E (slow)
E + B --> C + D (fast)
1. A series of experiments were performed to determine the mechanism of the reaction 2A + B C + D. Given the following data, which of the following proposed mechanisms is/are most likely correct? Experiment #1 #2 [A] 1.0 M 1 x 104M 1.0 M Bl 1 x 104MIn[B] vs time 1/[A] vs time I A+BE (fast) E + B C + D (slow) 11. A+BE(fast) E + A C + D (slow) A + A E (slow) E + B C + D (fast) 111.Explanation / Answer
ntegrated rate law for 1st order reaction
ln[B] = ln[B]o - k*t
So,
order of B is 1 from experiment 1
integrated rate law for 2nd order reaction
1/[A] = 1/[A]o + k*t
order of A is 2 from experiment 2
So,
rate law must be:
rate = k [A]^2 [B]
Whichever mechanism is consistent with this rate law is correct mechanism
for mechanism I:
rate depends on the slowest step
Here 2nd step is slowest
So, rate law is:
rate = k2 [E][B]
E is an intermediate. We need to remove it.
we will use 1st step:
K = [E]/[A][B]
[E] = K*[A][B]
So,
rate = k2 [E][B]
rate = k2 K*[A][B][B]
rate = k2 K*[A][B]^2
this is not consistent
for mechanism II:
rate depends on the slowest step
Here 2nd step is slowest
So, rate law is:
rate = k2 [E][A]
E is an intermediate. We need to remove it.
we will use 1st step:
K = [E]/[A][B]
[E] = K*[A][B]
So,
rate = k2 [E][A]
rate = k2 K*[A][A][B]
rate = k2 K*[A]^2[B]
this is consistent
for mechanism III:
rate depends on the slowest step
Here 1st step is slowest
So, rate law is:
rate = k2 [A][A]
rate = k2 [A]^2
this is NOT consistent
Answer: mechanism II
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