JEE Main · 2025 · Shift-IhardCK-142

| [A]0 / mol L-1 | t1/2 / min | |---|---| | 0.100 | 200 | | 0.025 | 100 | For reaction R P, t1/2 is related to [A]0 as…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

| [A]0[A]_0 / mol L−1^{-1} | t1/2t_{1/2} / min | |---|---| | 0.100 | 200 | | 0.025 | 100 |

For reaction R→P\text{R} \to \text{P}, t1/2t_{1/2} is related to [A]0[A]_0 as above. Given: log⁡2=0.30\log 2 = 0.30.

Which of the following is true?

(A) The order of the reaction is 12\frac{1}{2}. (B) If [A]0[A]_0 is 1 M, then t1/2t_{1/2} is 20010200\sqrt{10} min. (C) The order changes to 1 if concentration changes from 0.100 M to 0.500 M. (D) t1/2t_{1/2} is 800 min for [A]0=1.6[A]_0 = 1.6 M.

Options
  1. a

    A and C only

  2. b

    A and B only

  3. c

    A, B and D only

    ✓
  4. d

    C and D only

Correct Answerc

A, B and D only

Detailed Solution

Order determination: t1/2∝[A]01−nt_{1/2} \propto [A]_0^{1-n}

200100=(0.1000.025)1−n⇒2=41−n⇒1−n=12⇒n=12\frac{200}{100} = \left(\frac{0.100}{0.025}\right)^{1-n} \Rightarrow 2 = 4^{1-n} \Rightarrow 1-n = \frac{1}{2} \Rightarrow n = \frac{1}{2} (A) TRUE ✓

Check (B): t1/2=200×(1/0.1)1/2=20010t_{1/2} = 200 \times (1/0.1)^{1/2} = 200\sqrt{10} min (B) TRUE ✓

Check (C): Order is a property of mechanism, NOT concentration-dependent. (C) FALSE ✗

Check (D): t1/2=200×(1.6/0.1)1/2=200×4=800t_{1/2} = 200 \times (1.6/0.1)^{1/2} = 200 \times 4 = 800 min (D) TRUE ✓

Answer: A, B and D only → Option (c)

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| [A]0 / mol L-1 | t1/2 / min | |---|---| | 0.100 | 200 | | 0.025 | 100 | For reaction R P, t1/2 is… (JEE Main 2025) | Canvas Classes