JEE Main 2022 · 26 July · evening · Single correct · +4 / −1

At 30° C, the half life for the decomposition of AB₂ is 200 s and is independent of the initial concentration of AB₂. The time required for 80 % of the AB₂ to decompose is
Given: log 2=0.30 log 3=0.48

  1. (a) 200 s
  2. (b) 323 s
  3. (c) 467 s
  4. (d) 532 s
Answer and solution

Answer (c) 467 s

Solution

Chemistry ma’am · 4 steps

A half-life that does not depend on concentration means first order, so the time is just a ratio of logs.

  1. Identify the order. Only a first-order reaction has a half-life that does not depend on the starting concentration. So k = 0.693/t_(1/2), which means k = (2.303 log 2)/t_(1/2).
  2. What is left. If 80% decomposes, 20% remains. So [A]₀/[A] = 100/20 = 5.

    t = 2.303/k log[A]₀/[A] = t_(1/2) × (log 5)/(log 2)

  3. Put in numbers. log 5 = log 10 - log 2 = 1 - 0.30 = 0.70. So t = 200 × 0.70/0.30 = 466.7 s.
  4. So. About 467 s is needed for 80% of AB₂ to decompose, which is option (c).

The idea to keep. For first order, t = t_(1/2) × log([A]₀/[A])/(log 2). Always use the fraction that remains, not the fraction that has reacted.

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.