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JEE Advanced Chemistry Chemical Kinetics 2022 JEE Advanced 2022 (Paper 1)

JEE Advanced Chemistry Question (2022) — Solution

Question

Match the rate expressions in LIST-I for the decomposition of \( X \) with the corresponding profiles provided in LIST-II. \(X_s\) and \( k \) are constants having appropriate units.

Options

  1. A. I → P ; II → Q ; III → S ; IV → T
  2. B. I → R ; II → S ; III → S ; IV → T
  3. C. I → P ; II → Q ; III → Q ; IV → R
  4. D. I → R ; II → S ; III → Q ; IV → R

Answer

A. I → P ; II → Q ; III → S ; IV → T

Step-by-step solution

I  rate  = k X X S + X Case-1:  X > > X s ; X + X s ≈ X rate  = k X X = k     (Zero order w.r.t. X ) I → P , S   Case-2:  X ≪ X s ; X + X s ≈ X s ∴  rate  = k X X S = k ' X    ( 1 st  order w.r.t. X ) ∴     I → Q , T Case-3:  X ≈ X S rate  = k X X s + X In this case curve- R  given in List-II will match. ∴   I → P , Q , R , S , T   (The graph of half-life should start from origin) II  rate  = k X X S + X ∵   X ≪ X S ∴   X s + X ≈ X s ∴   rate  = k X X S = k ' X   ( 1 st  order w.r.t. X ) ∴   II → Q , T III   rate  = k X X s + X ∵   X > > X s ∴   X s + X ≈ X ∴   rate  = k X X = k    (Zero order w.r.t. X ) ∴     III → P , S IV  rate  = k X 2 X s + X ∵   X > > X S ∴   X S + X ≈ X ∴   rate  = k X 2 X = k X    ( 1 st  order w.r.t. X ) ∴   IV → Q , T

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