[Judy]

The reaction mechanism of 2N2O5 → 4NO2 + O2 is given as attached. 

using steady state approximation prove the rate law d[O2]/dt =  kexpt [N2O5]^n, calculate kexpt. ( in terms of K1, K-1, k2, k3) and n values.

The answer is :
kexpt =[(k1k2)/(k -1 + 2k2)] ]N2O5)
(as attached)
---------------

I think it should be k1k2/(k1+2k2), but the given answer is

[mjc123]

You are right (apart from what I assume is a typo). What they have given is the expression for d[O₂]/dt.
k_exp = k₁k₂/(k_-1 + 2k₂)
n = 1

[Judy]

You are right (apart from what I assume is a typo). What they have given is the expression for d[O₂]/dt.
k_exp = k₁k₂/(k_-1 + 2k₂)
n = 1

Sorry for my typo in the original statement.
I think the answer should be
kexp = k1k2/(k-1 ) not    kexp = k1k2/(k-1 + 2k2)
I wonder where  the "2k2" in the denominator comes from .

My calculation is:
R = k2[NO2][NO3]
K= k1/ k-1 = [NO2][NO3]/ [N2O5]
R= k1k2 /  k-1

[mjc123]

Your derivation neglects the fact that NO₃ is being depleted by reactions 2 and 3. If the equilibrium reaction is much faster than the others (k_-1 >> k₂), your answer is a good approximation. If the other reactions are comparable, the continual depletion of NO₃ means that equilibrium in reaction 1 is never reached, and you can't say
K= k₁/k_-1 = [NO₂][NO₃]/[N₂O₅].
You need to apply the steady state approximation, first to [NO] (which allows you to relate k₃ to k₂), then to [NO₃].

[Judy]

[/quote]
Thank you so much. I got it.