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JEE Advanced Chemistry Chemical Kinetics 2019 JEE Advanced 2019 (Paper 2)

JEE Advanced Chemistry Question (2019) — Solution

Question

The decomposition reaction 2 N 2 O 5 g →   ∆   2 N 2 O 4 g + O 2 g is started in a closed cylinder under the isothermal isochoric condition at an initial pressure of 1   atm . After Y × 10 3   s , the pressure inside the cylinder is found to be 1.45   a t m . If the rate constant of the reaction is 5 × 10 - 4 s - 1 , assuming ideal gas behaviour, the value of Y is

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

P 0 = 1 atm , P t = 1 .45 atm , P ∝ = 1 .5 atm From the unit of K = 5 × 10 - 4 s - 1 , The reaction is first-order t = 2.303 2 × 5 × 10 - 1 log ⁡ 1.5 - 1 1.5 - 1.45 t = 2 .303 × 10 3 = y × 10 3 ∴ y = 2.303 After round off: y = 2 .3

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