JEE Main 2022 · 24 June · evening · Single correct · +4 / −1

For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is
(Given : ln 10 = 2.303 and log 2 = 0.3010)

  1. (a) 1.12
  2. (b) 2.43
  3. (c) 3.32
  4. (d) 33.31
Answer and solution

Answer (c) 3.32

Solution

Chemistry ma’am · 3 steps

Write both times as a log over k: the k cancels, and x is just log 10 divided by log 2.

  1. First order time. For a first order reaction, t = 1/k ln[A]₀/[A]. After 90% reaction, only 10% is left, so [A]₀/[A] = 100/10 = 10.
  2. Both times. So t_(90%) = (ln 10)/k. At half life half is left, so t_(1/2) = (ln 2)/k.

    x = t_(90%)/t_(1/2) = (ln 10)/(ln 2) = (log 10)/(log 2) = 1/0.3010

  3. So. x = 1/0.3010 = 3.32, which is option (c).

The idea to keep. In first order kinetics, comparing two times means comparing two logs, because the same k cancels. Remember t_(99.9%) = 10 t_(1/2) and t_(90%) ≈ 3.32 t_(1/2).

Answer from the NTA final answer key. Solution by PrepWiser.

Study Chemical Kinetics

Asked about 150 times in JEE Main since 2020. This question is from 2022.