Rate Law for IBr Decomposition: Explained Step-by-Step

[jupiter_8917]

Write the rate law for the following proposed mechanism for the decomposition of IBr to I2 and Br2.

IBr(g)---------->I(g) + Br(g) (fast)
IBr(g) + Br(g)-------->I(g) + Br2(g) (slow)
I(g) + I(g)--------->I2(g) (fast)

The answer is R=k[IBr]^2. I don't know where this answer comes from. Can anyone please explain it to me step by step? Thank you for your help!

 

[Fightfish]

We know that for an elementary reaction step, the rate is proportional to the product of the molar concentrations of each reactant.
For a reaction with a multi-step mechanism, the rate of the reaction is dependent on the rate of limiting step ie the slowest step. In this case here it is the 2nd step. So, we have a preliminary rate equation:

$$rate = k[IBr][Br]$$​

However, it does not make any sense to have an intermediate in the rate equation, so we replace [Br] with the reactant(s) that form it. Since the reaction IBr(g)----->I(g) + Br(g) occurs rapidly, [Br] is clearly a fraction of [IBr]. And thus we replace [Br] with h[IBr] where h is a constant to obtain

$$rate = k'[IBr]^2$$​

where k' denotes a modified rate constant.