Verified Solution • 2025–26 Session moderate

QJEEAdv08Ch50be: JEE Advanced 2008 Chemistry - Class 12 JEEAdv08Ch50be Solution: Under the same reaction conditions, initial concentration of…

Under the same reaction conditions, initial concentration of 1.386 mol dm −3^{-3} of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio (k1k0)\left( {{{{k_1}} \over {{k_0}}}} \right) of the rate constants for first order ( k1k_1 ) and zero order ( k0k_0 ) of the reactions is:

Multiple Choice Options

  • A 0.5 mol −1^{-1} dm 3^3
  • B 1.0 mol dm −3^{-3}
  • C 1.5 mol dm −3^{-3}
  • D 2.0 mol −1^{-1} dm 3^3

Step-by-Step Verified Solution

Rate constant for first order kinetics:

k1=0.693t1/2=0.69340 s−1{k_1} = {{0.693} \over {{t_{1/2}}}} = {{0.693} \over {40\,{s^{ - 1}}}} ..... (i)

Rate constant for zero order kinetics:

k0=A02t1/2=1.3862×20{k_0} = {{{A_0}} \over {2{t_{1/2}}}} = {{1.386} \over {2 \times 20}} mol dm −3^{-3} s −1^{-1} ..... (ii)

Divide (i) by (ii)

k1k0=0.69340×401.386=0.6931.386{{{k_1}} \over {{k_0}}} = {{0.693} \over {40}} \times {{40} \over {1.386}} = {{0.693} \over {1.386}} mol −1^{-1} dm −3^{-3} = 0.5 mol −1^{-1} dm 3^{3}

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