99 Percentile Qs Bank for JEE MainChemistryChemical Kinetics
Consider the following gaseous decomposition of ozone: 2 O 3 g ⟶ 3 O 2 g . The observed rate law is r = - d O 3 dt = k O 3 2 O 2 If the above reaction occurs through the following mechanism: O 3 ⇌ K O 2 + O fast equilibrium O + O 3 ⟶ k 1 2 O 2 slow Then, which of the following relation is correct?
Options
- Ak = k 1
- Bk = Kk 1
- Ck = K
- DNone of these
Correct answer
B. k = Kk 1
Step-by-step solution
In Complex Reaction Rate law is defined by the slowest step. So According to slowest step Rate Law Expression can be represented as: Rate law = r = k 1 O O 3 ...........(1) Now Intermediate term ie [O] can never comes in Rate Law Expression.So to eliminate it,we apply Equilibrium constant Expression on Reversible Equation. K = O 2 O O 3 . . . . . . . . . . . . . ( 2 )         ( K = Equilibrium   constant ) O = K O 3 O 2 On Puting value of [O] in Equation 1 , we get ⇒