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JEE Advanced Chemistry Chemical Kinetics 2019 JEE Advanced 2019 (Paper 1)

JEE Advanced Chemistry Question (2019) — Solution

Question

Consider the kinetic data given in the following table for the reaction A + B + C → Product.   Experiment No. A m o l   d m - 3 B m o l   d m - 3 C m o l   d m - 3 Rate of reaction m o l   d m - 3 s - 1 1 0.2 0.1 0.1 6.0 × 10 - 5 2 0.2 0.2 0.1 6.0 × 10 - 5 3 0.2 0.1 0.2 1.2 × 10 - 4 4 0.3 0.1 0.1 9.0 × 10 - 5 The rate of the reaction for A = 0.15   m o l   d m - 3 ,   B = 0.25   m o l   d m - 3 and C = 0.15   m o l   d m - 3 is found to be Y × 10 - 5   m o l   d m - 3 s - 1 . 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

r = k [ A ] a [ B ] b [ C ] c 6 × 10 - 5 = k ( 0.2 ) a ( 0.1 ) b ( 0.1 ) c ..... (1) 6 × 10 - 5 = k ( 0.2 ) a ( 0.2 ) b ( 0.1 ) c ...... (2) 1.2 × 10 - 4 = k ( 0.2 ) a ( 0.1 ) b ( 0.2 ) c ...... (3) 9 × 10 - 5 = k ( 0.3 ) a ( 0.1 ) b ( 0.1 ) c ...... (4) From equation (1) and (ii) : b = 0 From equation (1) and (iii) : c = 1 From equation (1) and (iv) : a = 1 Put x , y and z in equation (i) we get; 6 × 10 - 5 = k ( 0.2 ) 1 ( 0.1 ) 0 ( 0.1 ) 1 r = 3 × 10 - 3 [ A ] [ C ] For [ A ] = 0.15 mol dm - 3 [ B ] = 0.25 mol dm - 3 [ C ] = 0.15 mol dm - 3 r = 3 × 10 - 3 × 0.15 × 0.15 r = 6.75 × 10 - 5 ∴ r = 6.75

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