High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Chemistry Textbook > Kinetics

Steady State Approximation

EnzymesPrevious Section
Pre-EquilibriumNext Section

Steady State Approximation

  • The steady state approximation assumes that after some time (an induction period) the concentration of any intermediates is constant. ie. d[I]/dt = 0
  • Why is this useful? It allows us to express the rate of a reaction in terms of only reactants and products. We can then compare these results to experiment. An example is shown below.
  • We can use this approximation if we are not told which step is slow or fast as well or if the steps have similar rates and one is not clearly the slow step or fast step!
0:00 / 0:00
Find a rate law for the following reaction using the provided mechanism which contains only reactants and rate constants. You may use the steady state approximation for any intermediate.

overall reaction: 2NO+H2→H2O+N2O‾\underline{2NO+H_2 \to H_2O+N_2O}
elementary step #1: 2NO⇄k−1k1N2O22NO \xtofrom[k_{-1}]{k_1}N_2O_2 (fast)
elementary step #2: N2O2+H2→k2H2O+N2ON_2O_2+H_2 \xrightarrow{k_2}H_2O +N_2O (slow)

Step 1: Figure out the rate of the overall reaction

Note: This is the same as saying find the formation of the product N2O (or H2O):
Hint: which step do we get this from??
Rate determining step/slow step!
d[N2O]dt=k2[N2O2][H2]\cfrac{d[N_2O]}{dt}=k_2[N_2O_2][H_2]

Step 2: Look at the elementary steps to see if there are any intermediates.


Are there any intermediates here?
Yes!
If so, what is it?
N2O2

Step 3: Since there is an intermediate, we need to first write out the net rate of formation for the intermediate:

  • to do this, it would help to first write out the rates of each elementary step!
  • Note: For fast steps, write out the rate for the forward AND backward reaction. For slow steps the backward reaction can be ignored.
  • rate elementary step #1 (fwd): v1=
    k1[NO]2
  • rate elementary step #1 (bwd): v1'=
    k1'[N2O2]
  • rate elementary step #2: v2=
    k2[N2O2][H2]

d[N2O]dt=k1[NO]2−k−1[N2O2]−k2[N2O2][H2]\cfrac{d[N_2O]}{dt}=k_1[NO]^2-k_{-1}[N_2O_2]-k_2[N_2O_2][H_2]

Step 4: Now we can make the steady state approximation by setting the net rate of formation of any intermediate equal to 0. Once we do this, we can now solve for our intermediate's concentration.

0=k1[NO]2−k−1[N2O2]−k2[N2O2][H2]0=k_1[NO]^2-k_{-1}[N_2O_2]-k_2[N_2O_2][H_2]
k1[NO]2=k−1[N2O2]+k2[N2O2][H2]k_1[NO]^2=k_{-1}[N_2O_2]+k_2[N_2O_2][H_2]
k1[NO]2=[N2O2](k−1+k2[H2])k_1[NO]^2=\left[N_2O_2\right](k_{-1}+k_2[H_2])
[N2O2]=k1[NO]2(k−1+k2[H2])[N_2O_2]=\cfrac{k_1[NO]^2}{(k_{-1}+k_2[H_2])}

Step 5: Now that we have an expression for the concentration if the intermediate [N2O2] we can substitute it into the overall reaction rate expression/product formation expression:


d[N2O]dt=k2[N2O2][H2]\frac{d\left[N_2O\right]}{dt}=k_2\left[N_2O_2\right]\left[H_2\right]
d[N2O]dt=k2k1[NO]2(k−1+k2[H2])[H2]\cfrac{d[N_2O]}{dt}=k_2\cfrac{k_1[NO]^2}{(k_{-1}+k_2[H_2])}[H_2]

Step 6: Simplify the rate equation:

What is the rate law when H2 is in vast excess?
  • If H2 is in vast excess then...
k2[H2]≫k−1∴k−1+k2[H2]≈k2[H2]k_2[H_2] \gg k_{-1} \therefore k_{-1} +k_2[H_2] \approx k_2[H_2]
d[N2O]dt=k2k1[NO]2(k2[H2])[H2]\cfrac{d[N_2O]}{dt}=k_2\cfrac{k_1[NO]^2}{(k_2 [H_2])}[H_2]
d[N2O]dt=k1[NO]2\cfrac{d[N_2O]}{dt}=k_1[NO]^2
What about if we weren't told anything was in excess?
  • We want to try and simplify the denominator... k' + k2[H2]
  • Which is bigger, k' or k2?
    k'
  • Since
    k'
    is so much bigger: k'~k' +k2[H2]
  • Our simplified rate expression becomes...

d[N2O]dt=k2 k1[NO]2k′ + k2[H2][H2]\frac{d\left[N_2O\right]}{dt}=k_2\ \frac{k_1\left[NO\right]^2}{k'\ +\ k_2\left[H_2\right]}\left[H_2\right]

d[N2O]dt=k2 k1[NO]2k′[H2]\frac{d\left[N_2O\right]}{dt}=k_2\ \frac{k_1\left[NO\right]^2}{k'}\left[H_2\right]

rewritten: d[N2O]dt=k1k2[NO]2[H2]k′\frac{d\left[N_2O\right]}{dt}=\frac{k_1k_2\left[NO\right]^2\left[H_2\right]}{k'}