Topic III- Pre-class Problem
As discussed in the previous recitation, the rate law for the reaction
2 H2(g) + 2NO(g) ↔ N2(g) + 2H2O(g) is Rate=k[H2][NO]2 (at least at low
concentrations of H2). A possible mechanism for this reaction with two steps and an
oxygen atom intermediate might take the form:
H2 + 2NO → N2 + X + O (Step 1; k1)
O+Y→Z
(Step 2; k2)
A. Identify the chemical species X, Y, and Z in the mechanism.
Write down your logic.
Answer the question.
What concepts are being challenged?
B. Which step must be the rate limiting step of the mechanism to yield the observed rate
law? Answer the question.
What concepts do you have to understand to answer this question?
Rewrite the question so that the answer you gave above is incorrect.
C. J. Parker Siburt and R. MacPhail, Duke University 2008
Topic III In-Class Group Problem
An alternative mechanism for the reaction
Rate = kobs[NO]2[H2] is:
2NO + 2H2 → N2 + 2H2O with
2NO → N2O2
k1 forward; k-1 backward
N2O2 + H2 → N2O + H2O
k2
N2O + H2 → N2 + H2O
k3
A) Show that this mechanism yields the overall reaction stoichiometry.
B) Identify any intermediates.
C) If the second step is the slow, rate determining step, and the first step is in rapid
equilibrium, find the predicted the rate law and show it is consistent with the
observed rate law.
C. J. Parker Siburt and R. MacPhail, Duke University 2008
Topic III Problem Manipulation
At high concentrations of [H2] the observed rate law is found to change to the form:
Rate = k[NO]2 i.e. the rate becomes independent of [H2]. Assess whether the two step
mechanism, you should to be plausible in the pre-problem, can explain this additional
data.
Reexamine the mechanism in the Pre-recitation Problem:
2NO + H2 → O + H2O + N2 slow
k1
O + H2 → H2 O
k2
Write the forward rates for both steps.
Based on these rates, would you expect the mechanism to change as [H2] increases?
(Hint look at the ratio)
Can this mechanism give the rate law Rate = k[NO]2 observed at high [H2]?
What other aspect of this two step mechanism might make it less likely to be the
correct mechanism?
C. J. Parker Siburt and R. MacPhail, Duke University 2008
Name:
Quiz 1
TA:______________________
Rec Time: 8:45 10:20 11:55
At high concentrations of [H2] the observed rate law for the reaction 2NO + 2H2 → N2
+ 2H2O is found to be: Rate = kobs[NO]2 i.e. the rate becomes independent of [H2].
Assess whether the following three step mechanism can explain this observation.
Examine the mechanism:
2NO → N2O2
k1 forward; k-1 backward
N2O2 + H2 → N2O + H2O
k2
N2O + H2 → N2 + H2O
k3
Which step would you predict to be the slow step at high [H2] and why? Support
your answer by comparing the forward rate laws for the first two steps.
Based on your answer above, what overall rate law would you predict under these
high [H2] conditions? Does your predicted rate law agree with the observed rate law
at high [H2]?
What are the units of k1, k-1, and kobs?
This problem is a manipulation of the first group problem from recitation that dealt
with the same three step mechanism. Identify one underlying concept common to
both problems (group and quiz) and BRIEFLY explain how the quiz and the group
problem approach that concept differently.
C. J. Parker Siburt and R. MacPhail, Duke University 2008