[kelvinLTR]

from the suggested mechanism for decompostion reaction

    O₃ O₂ +'O'          (fast)
    O₃ + 'O'  2O2                     (slow)

the 2nd step is said to be the rate determining step. My question is, we use the steady state for 'O' in finding the

• . But steady state approximation can be used for consecutive reactions where k₁<<k₂ where k1 is the rate constant of first reaction and k2 that of the 2nd.
  
  But in this mechanism, the slow reaction is the second step!
  
  So how can we still use the steady state approximation?
  
  Thanks in advance 

[kelvinLTR]

the post is a bit incomplete. The second line reads "My question is My question is, we use the steady state for 'O' in finding the rate of the overall reaction"

[Babcock_Hall]

Were you told to use the steady-state approximation, or are you wondering whether or not it is valid in this instance?  Also, the first step is reversible, and that will change things relative to the case in which the first step is not.

[kelvinLTR]

I was told to use the Steady state approximation.

[Babcock_Hall]

One strategy would be to go ahead and use the steady-state approximation, then come back to the question of its justification later.  That is how I would approach the problem (I can take a guess as to why it is applicable here, but I am not sure).  When you apply the steady-state approximation, were you able to get an answer?  I may or may not be able to verify it.