JEE Main 2020 · 4 September · morning · Numerical answer · +4 / −0

If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes) _______.
(Take : log 2 = 0.30; log 2.5 = 0.40)

Answer and solution

Answer 60

Solution

Chemistry ma’am · 3 steps

Write the first-order time formula for both cases and divide; k and 2.303 both cancel.

  1. First-order time. For a first order reaction, t = 2.303/k log[A]₀/[A]. Use the fraction left, not the fraction completed: 75% done leaves 25%, and 60% done leaves 40%.

    t_(60%)/t_(75%) = (log 100/40)/(log 100/25) = (log 2.5)/(log 4)

  2. Put in logs. log 2.5 = 0.40 and log 4 = 2 log 2 = 0.60. So the ratio is 0.40/0.60 = 2/3.
  3. So. t_(60%) = 2/3 × 90 = 60 minutes.

The common slip: 72. Scaling the time in direct proportion, 60/75 × 90, gives 72, not 60. That straight-line rule fits a zero order reaction, not a first order one.

The idea to keep. For first order, the time for any fraction depends only on log[A]₀/[A]. Comparing two times means taking a ratio of logs, so you never need the value of k.

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 2020.