## Rate Laws and Reaction order [ENDORSED]

$\frac{d[R]}{dt}=-k; [R]=-kt + [R]_{0}; t_{\frac{1}{2}}=\frac{[R]_{0}}{2k}$

Alexandria_Leaf_2F
Posts: 30
Joined: Fri Jul 15, 2016 3:00 am

### Rate Laws and Reaction order

On page 621 of the textbook, there is an example of

2O3(g) ----> 3O2(g)

and then it says that the rate law for this reaction is

Rate= k([O3]^2/[O2]

can someone explain how that is the rate law?

Thank you!

shreya_mantri_3I
Posts: 17
Joined: Fri Jun 17, 2016 11:28 am

### Re: Rate Laws and Reaction order  [ENDORSED]

the stoichiometry of an overall reaction does not necessarily reflect the molecularity of the mechanism.
if the mechanism of the overall reaction shown above for ozone decomposition actually corresponded to a bimolecular reaction of two ozone molecules, then the expected rate law would be second-order in ozone, and we would see that − 1/2 d[O3]/ dt = k[O3]^ 2. Solving for the ozone rate, this becomes d[O3]/ dt = − 2k [O3]^2 . However, the rate law suggested by this equation is not correct. The actual rate law for ozone decomposition, deduced from experimental studies is d[O3 ] /dt = − k(exp) [O3]^2 [O2]^ (−1). The experimental rate law shows that the reaction is indeed second order in ozone. However, the reaction is also dependent upon the product concentration (O2) with an order of –1, suggesting that there is more going on in the mechanism than just a simple bimolecular reaction.
There must be an intermediate reaction which acts as the rate determining step that proves that the rate law is k(exp) [O3]^2/[O2]^-1
This is what I understood. I hope it makes sense :D

Alexandria_Leaf_2F
Posts: 30
Joined: Fri Jul 15, 2016 3:00 am

### Re: Rate Laws and Reaction order

Yes it does, Thank You!