If it seems that I am ignoring you in the hallways and
corridors, I apologize. It will take me time to recognize
your faces and learn your names. I will try though !
Bizarro by Dan Piraro
Read 16.7, 17.1, 17.2
Lab room assignments are posted online on the lab
webpage. 1 2
At equilibrium, the rates for the forward reaction and
Case II – A fast step is followed by a slow step.
the reverse reaction are equal.
Decomposition of ozone, O , 3o oxygen, O , 2s
believed to occur via So from step 1 k1[O 3 = k–1O ]2O]
k 1 Remember, both the forward and the reverse are
Step 1 O 3 (g) O 2 (g)+sa(g)
k k [O ]
2 [O] = 1 3
Step 2 O 3 (g) O (g) 2 w2 (g) k–1O ]2
RDS is step 2
Insert the result into the RDS rate
Rate RDS= k 2O 3[O]
Can’t leave [O] in the rate expression
k k [O ]2
No intermediate appears in rate law Rate = k 2O 3[O] = 1 2 3
k–1 [O 2
Exceptions: [H ] or [OH ] in solutions could remain if
we're unable to find substitute expressions. With the absence of an experimental rate, assume the
mechanism and the derived rate are possible.
Figure 16.22 Energy vs. rxn progress for an exothermic reaction
Energy vs. rxn progress for an exothermic reaction having a 2–step mechanism in which the first step is
having a 2–step mechanism in which the first step is fast and the second is slow (RDS)
slow (RDS) and the second is fast
In both diagrams, the intermediates are not at the crest
of the curves (that’s where the transition states for
each step are).
Both diagrams are for exothermic rxns. You could
also have an endothermic reaction with products
having higher energy than the reactants. 5 6
Fast equilibrium steps Case III – Steady state approximation
At equilibrium rate forward = rate reverse A situation can occur in which either:
Suppose A + B 2C No information is known or given about the
2 relative speeds of the rates.
Thenk 1[A][B] = k [–1
k [C]2 [products] stoic coeff When steps have relatively comparable rates
The ratio 1 = = stoic coeff
k–1 [A][B] [reactants] k1
2NO (g) N 2 2 (g)
The right hand side of the equation is simply the k –1
definition of the equilibrium constant K. N 2O 2 (g) O 2 (g) 2 NO 2 (g)
But 1 equil. step is not much faster than the 2 step.
So k–1= Equilibrium constant = K Choose the step that produces the products (usually
the last one)
Rate of rxn = k [N O ][O ]
2 2 2 2
However you can't leave an intermediate [N O ]2in2
Steady state approximation The formation and disappearance rates are equal
After some time the rxn will reach a steady state in k 1NO] = k [N–1 ]2+ 2 [N O2][O2]2 2
which the rate of change of the concentration of the 2
k 1NO] = [N O 2 (2 + k–1O ]2 2
intermediate is constant.
Solve for [N 2O 2 [N 2 ]2=
k–1 k [2 ] 2
In other words, the rate of formation of the