This work was created by Burke Scott Williams, Claremont Colleges (swilliams@jsd.claremont.edu) for the Virtual Inorganic Pedagogical
Electronic Resource (VIPEr) and licensed under Creative Commons (Attribution, Share-Alike, Non-Commercial, 2009)
Two Step Mechanisms and the Steady State Approximation
1. Consider the conversion of cis-[Co(en)2Cl2]+ to trans-[Co(en)2Cl2]+. One can envision two plausible mechanisms.
Mechanism : (associative)
+
Cl
H2
N
Cl
Co
N
NH2
H2
H2N
k1[Cl-]
k-1
[Co(en)2Cl3]
7C
k2
(-Cl-)
H2
N
Cl
H2
N
+
Co
N
H2
cis
Cl
N
H2
trans
Mechanism(dissociative)
H2
N
+
Cl
Cl
k1
(-Cl-)
2+
[Co(en)2Cl]
Co
N
H2
NH2
H2N
cis
-
k-1[Cl ]
5C
k2[Cl-]
H2
N
Cl
H2
N
+
Co
N
H2
Cl
trans
N
H2

For each mechanism,
a. Derive the rate law
b. Show what happens to the rate law both when the first step is rate limiting and the second step is rate
limiting
c. Give kobs for each of the four cases.
This work was created by Burke Scott Williams, Claremont Colleges (swilliams@jsd.claremont.edu) for the Virtual Inorganic Pedagogical
Electronic Resource (VIPEr) and licensed under Creative Commons (Attribution, Share-Alike, Non-Commercial, 2009)
2. Now for a different sort of case. Let’s imagine you wanted to convert your [CoA 5Cl]+ to [CoA6+]. We can, again,
imagine either an associative or dissociative mechanism:
Mechanism : (associative)
[CoA5Cl]+
k1[A]
k-1
[CoA6Cl]+
k2
(-Cl-)
[CoA6]2+
Mechanism(dissociative)
k1
+
[CoA5Cl]
(-Cl-)
k-1[Cl-]
[CoA5]2+
k2[A]
[CoA6]2+

For each mechanism,
a. Derive the rate law
b. Show what happens to the rate law both when the first step is rate limiting and the second step is rate
limiting
c. Give kobs for each of the four cases.
d. Determine what happens to kobs as [A] goes toward either 0 or ∞ (leaving [Cl-] constant). Draw a graph of
kobs vs. [A].
e. Finally, do the same thing you did in d. for 1/kobs, and plot 1/kobs vs. 1/[A].
This work was created by Burke Scott Williams, Claremont Colleges (swilliams@jsd.claremont.edu) for the Virtual Inorganic Pedagogical
Electronic Resource (VIPEr) and licensed under Creative Commons (Attribution, Share-Alike, Non-Commercial, 2009)
3. Michaelis-Menten Kinetics: Now let’s consider a catalytic reaction. In this particular case, we’ll imagine an
enzyme (a protein catalyst), but the math is the same regardless of the nature of the catalyst. Imagine an enzyme
E, a substrate S, a substrate bound to enzyme ES, and a product, P.
E + S
k1
k-1
ES
k2
E + P
a. First, apply the steady-state approximation, and find the concentration of [ES] in terms of [E], [S], and the
fundamental rate constants. For simplification’s sake, in your final answer, let the Michaelis Constant, Km be
k  k2
equal to: 1
.
k1
b. Second, since the knowable quantity in this case is the total enzyme concentration, [E]0, and since [E]0 = [E]
+ [ES], replace [E] in your expression from Part a., and solve for [ES] again.
c. Depending on the form of your Part b. answer, it may make your equation simpler to multiply both top and
bottom by [S]. If so, go ahead and do that.
d. The rate of reaction in biology is called the velocity, V. Please, using your expression for [ES], give an
expression for V.
e. How does this simplify in the limit as [S] → ∞ ? Rename this value Vmax, and rewrite your expression for V.
f.
Draw a plot of V vs. [S]. This should be linear at both high and low [S]. Does this resemble your saturation
kinetics for cobalt on the last page?
g. Just like in the cobalt case on the last page, let’s make one final plot: 1/V vs. 1/[S]. This is called a
Lineweaver-Burk plot in biology. It’s used in enzymology, cell biology, pharmacology, and more. What are
the x-intercept, y-intercept, and slope of the line equal to?