Chemical Kinetics: PYQ Walkthrough (6)

hard • 3 min read
Tags Chemical Kinetics

Question

(JEE Advanced 2022 style) For a first-order reaction, 75%75\% of the reactant decomposes in 30 minutes30 \, \text{minutes}. Find the rate constant kk and the half-life.

Solution — Step by Step

ln⁡[A]0[A]t=kt\ln\frac{[A]_0}{[A]_t} = kt

or equivalently [A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}.

If 75% decomposes, 25% remains. So [A]t/[A]0=0.25[A]_t/[A]_0 = 0.25 at t=30t = 30 min.

ln⁡(1/0.25)=ln⁡4=2ln⁡2=k×30\ln(1/0.25) = \ln 4 = 2\ln 2 = k \times 30

k=(2ln⁡2)/30=ln⁡2/15 min−1k = (2 \ln 2)/30 = \ln 2 / 15 \, \text{min}^{-1}

Numerically: k=0.693/15≈0.0462 min−1k = 0.693/15 \approx 0.0462 \, \text{min}^{-1}.

For first-order: t1/2=ln⁡2/kt_{1/2} = \ln 2 / k.

t1/2=ln⁡2/(ln⁡2/15)=15 mint_{1/2} = \ln 2/(\ln 2/15) = 15 \, \text{min}.

In 30 min (= 2 half-lives), fraction remaining =(1/2)2=0.25= (1/2)^2 = 0.25. So 75% decomposed. Matches.

Rate constant: k≈0.0462 min−1k \approx 0.0462 \, \text{min}^{-1}. Half-life: 15 min15 \, \text{min}.

Why This Works

For first-order kinetics, the half-life is independent of initial concentration — every half-life cuts the reactant by 50%. So “75% decomposed” means “two half-lives have passed.”

This shortcut bypasses the integrated rate law entirely:

  • 50% gone → 1 half-life
  • 75% gone → 2 half-lives
  • 87.5% gone → 3 half-lives
  • 93.75% gone → 4 half-lives

30-second shortcut: For first-order, fraction remaining =(1/2)n= (1/2)^n after nn half-lives. Match the given fraction to a power of 1/21/2, then t1/2=t/nt_{1/2} = t/n.

Alternative Method — Direct Substitution

Use the formula t=(2.303/k)log⁡([A]0/[A]t)t = (2.303/k) \log([A]_0/[A]_t).

30=(2.303/k)log⁡(1/0.25)=(2.303/k)(0.602)30 = (2.303/k) \log(1/0.25) = (2.303/k)(0.602)

k=2.303×0.602/30≈0.0462 min−1k = 2.303 \times 0.602 / 30 \approx 0.0462 \, \text{min}^{-1}. Same answer.

The natural log version is faster for clean fractions; the log-base-10 version matches NCERT formulas.

Common Mistake

Students often interpret “75% decomposed” as [A]t/[A]0=0.75[A]_t/[A]_0 = 0.75 — fatal error. Read carefully: “75% decomposed” means 75% is gone, leaving 25%.

Another classic: applying half-life formula to non-first-order reactions. For zero-order, t1/2=[A]0/(2k)t_{1/2} = [A]_0/(2k) (depends on concentration). For second-order, t1/2=1/(k[A]0)t_{1/2} = 1/(k[A]_0). Only first-order has constant half-life.

JEE Main 2023 had three different kinetics problems across shifts — all first-order. NEET 2024 included a radioactive decay problem (first-order in disguise). Master first-order kinetics and we cover ~80% of kinetics questions.

The “powers of 2” shortcut is the single biggest time-saver in chemical kinetics. Memorise the relationship between time and half-lives.

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