experimentally observed to be second order in Xand first order in Y:X + Y→ XYPart BIs this two-step mechanism valid?2X- X2Fastk2X2 + YXY + Xk3SlowMatch the words in the left column to the appropriate blanks in the sentences on the right.• View Available Hint(s)ReconsistentThe two-step mechanism isbecause it iswith the experimeninvalidderived rate law.validinconsistentSubmitde FeedbackNext >

As rate of reaction depends on slow step step, so rate= k [X2][Y] which is inconsistent with…

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Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

<Homework 02
Item 5
5 of 10
I Review I Constants I Periodic Table
MISSED THIS? Read Section 15.6
(Pages 656 - 661) : Watch KCV 15.6, IWE
15.9.Consider this overall reaction, which is
experimentally observed to be second order in X
and first order in Y:
X + Y→ XY
reaction occur in a single step in which X and Y collide?
e words in the left column to the appropriate blanks in the sentences on the right.
Reset Help
rate = k[X]? [Y]
The reaction
occur in a single step in which X and Y collide because this would
does not
imply a rate law of
which
the observed rate law of
aligns with
does
rate = k[X][Y]
rate = k[X][Y]²
contradicts
Request Answer

experimentally observed to be second order in X
and first order in Y:
X + Y→ XY
Part B
Is this two-step mechanism valid?
2X
- X2
Fast
k2
X2 + Y
XY + X
k3
Slow
Match the words in the left column to the appropriate blanks in the sentences on the right.
• View Available Hint(s)
Re
consistent
The two-step mechanism is
because it is
with the experimen
invalid
derived rate law.
valid
inconsistent
Submit
de Feedback
Next >

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