Using the PSSH, develop a rate law for the rate of formation of HBr for the mechanism given by Equations (CD73), (CD74), and (CD76) through (CD78). 

SolutionEach reaction is to be considered elementary. First we write down the net rate of formation of the product resulting from all the reaction steps: 


(CDE72.1) (CDE72.2) 




The intermediate species Brand Hare free radicals, which are present in low concentrations and highly reactive. Consequently, the use of the PSSH is justified. We next write the net rate of formation of the intermediate species and set these rates equal to zero (r_{H} = 0, r_{Br} = 0). 




(CDE72.3) (CDE72.4) 




Adding Equations (CDE72.3) and (CDE72.4) gives 



(Br) as a function of (Br_{2}) 
(CDE72.5)  

Subtracting Equation (CDE72.3) from (CDE72.2) gives 

r_{HBr}= 2k_{3} (H)(Br_{2}) 
(CDE72.6)  
To eliminate (H) from this equation, rewrite Equation (CDE72.3) as 

(CDE72.7)  
and solve for (H): 

(CDE72.8)  

To eliminate (Br), we use Equation (CDE72.5) to obtain 



(H) as a function of reactants and products 


Substituting Equation (CDE72.9) into Equation (CDE72.6), we see that 

(CDE72.10)  
We can rewrite Equation (CDE72.10) in the form 

(CDE72.11)  
By comparing Equations (CDE72.6) and (CDE72.11) we again see that the mechanism produces a rate law consistent with experimental observation! 