BITSAT2019ChemistryChemical KineticsActual
For the chemical reaction 2 O 3 ⟶ 3 O 2 , the reaction proceeds as follows O 3 ⇌ O 2 + O (Fast) O + O 3 ⟶ 2 O 2 , (Slow) the rate law expression should be given as
Options
- Ar = k O 3 2 O 2 - 1
- Br = k O 3 2
- Cr = k O 3 O 2
- Dunpredictable
Correct answer
A. r = k O 3 2 O 2 - 1
Step-by-step solution
Let k be the rate constant of given reaction 2 O 3 ⟶ k 3 O 2 From the slowest step r = k ' O O 3 ... i To eliminate O , (From fast step) k eq = O 2 O O 3 O = k eq . O 3 O 2 Now, substituting value of O in Eq i r = k ' k eq O 3 O 3 O 2 Let k ' k eq = k Thus, r = k O 3 2 O 2 - 1