Chemical Kinetics: Conceptual Doubts Cleared (4)

easy • 2 min read
Tags Chemical Kinetics

Question

A first-order reaction has a rate constant k=2.31×10−3 s−1k = 2.31 \times 10^{-3}\text{ s}^{-1} at 298 K. Find the half-life. After how long will 75% of the reactant be consumed?

Solution — Step by Step

For first-order kinetics:

t1/2=ln⁡2k=0.693kt_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}

t1/2=0.6932.31×10−3=300 st_{1/2} = \frac{0.693}{2.31 \times 10^{-3}} = 300\text{ s}

75% consumed = 25% remaining. Two half-lives reduce concentration to 25% (first half-life: 50%, second: 25%).

t75%=2×t1/2=600 st_{75\%} = 2 \times t_{1/2} = 600\text{ s}

ln⁡[A]0[A]t=kt  ⟹  ln⁡10.25=(2.31×10−3)t\ln\frac{[A]_0}{[A]_t} = kt \implies \ln\frac{1}{0.25} = (2.31 \times 10^{-3})t

ln⁡4=1.386=2.31×10−3⋅t  ⟹  t=600 s\ln 4 = 1.386 = 2.31 \times 10^{-3} \cdot t \implies t = 600\text{ s}

Final answer: t1/2=300t_{1/2} = 300 s; t75%=600t_{75\%} = 600 s

Why This Works

For first-order reactions, the half-life is independent of initial concentration — a fingerprint of first-order kinetics. This is why t75%=2t1/2t_{75\%} = 2 t_{1/2} exactly: after one half-life, half remains; after another half-life, half of that (25%) remains.

For zero-order or second-order reactions, half-lives change as the reaction proceeds, so the doubling trick doesn’t work.

Alternative Method

Use the integrated rate law directly:

t=2.303klog⁡[A]0[A]tt = \frac{2.303}{k} \log\frac{[A]_0}{[A]_t}

For 75% consumption: [A]0/[A]t=4[A]_0/[A]_t = 4, log⁡4=0.602\log 4 = 0.602, t=(2.303/2.31×10−3)×0.602≈600t = (2.303/2.31 \times 10^{-3}) \times 0.602 \approx 600 s.

Common Mistake

Students sometimes write “75% complete = 3/4 half-lives” or similar. The half-life concept refers to halving, not arbitrary fractions. Use the integrated rate law for non-power-of-2 fractions, or recognise 25% remaining = (1/2)2(1/2)^2.

For 99% completion: log⁡(100)=2\log(100) = 2, so t=2.303×2/kt = 2.303 \times 2/k. Roughly 6.6×t1/26.6 \times t_{1/2}.

Quick first-order conversions to memorise:

  • 50% complete → 1 half-life
  • 75% complete → 2 half-lives
  • 87.5% complete → 3 half-lives
  • 99% complete → ~6.6 half-lives

Want to master this topic?

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