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JEE Advanced Chemistry Chemical Kinetics 2021 JEE Advanced 2021 (Paper 2)

JEE Advanced Chemistry Question (2021) — Solution

Question

For the following reaction  2 X + Y → k P  the rate of reaction is d [ P ] dt = k [ X ] . Two moles of X are mixed with one mole of Y to make 1 . 0   L of solution. At 50   s ,   0 . 5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are)  ( Use:  ln   2 = 0 . 693 )

Options

  1. A. The rate constant,  k , of the reaction is  13 . 86 × 10 - 4   s - 1 .
  2. B. Half-life of X is 50   s .
  3. C. At 50   s ,   - d [ x ] dt = 13 . 86 × 10 - 3   mol   L - 1   s - 1 .
  4. D. At  100   s , - d Y dt = 3 . 46 × 10 - 3   mol   L - 1   s - 1

Answer

D. At  100   s , - d Y dt = 3 . 46 × 10 - 3   mol   L - 1   s - 1

Step-by-step solution

\( array lllll & 2 X + & y & & P \\ t =0 & 2 & 1 & & 0 \\ t =50 & 2-2 0.5 & 1-0.5 & & \\ & 1mole & 0.5 & & array \) rate \(=- 1 2 d x d t =- d y d t = d P d t =K[X]\) \(- 1 2 d x d t =K[X]\) \(- d x d t =2 K[X]=K^1[X]\) Half life is \(t=50 sec \) \(2 K= 0.653 L 50 \) \(K= 0.6932 100 =6.332 10^ -3 \) \(t=50 sec \) \(- d x d t =2 K[X]\) \(- d x d t =2 6.332 10^ -3 1\) \(=13.864 10^ -3 ~mole / L / Sec \) \(- d y d t =K[X]=6.332 10^ -3 ( 1 2 )\) \(=3.46 10^ -3 ~mole / L / Sec ^ -1 \)

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Related: Chemistry — Chemical Kinetics · All PYQ Banks