How does a reactant run out: steadily or by halves?
"How does a reactant run out: steadily or by halves?" is a free, interactive Class 12 Chemistry simulation for NEET and JEE, built on the lesson "Half-life and pseudo first order" in Chemical Kinetics. Blue dots are reactant, orange are product; each dot is 0.025 mol L⁻¹. Drag time forward. Switch between zero order (a constant rate) and first order (rate proportional to [R]) and change k and the starting concentration. Zero order: a straight line, [R] = [R]₀ − kt, and t½ = [R]₀/2k. First order: an exponential that loses the same fraction in each interval, [R] = [R]₀e^(−kt), and t½ = 0.693/k whatever the concentration. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Half-life and pseudo first order in Chemical Kinetics. Change the values and watch what happens.
In JEE Main
42 shifts analysed
48 questions from Chemical Kinetics across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §3.3.3
- Half-life t½ is the time for the concentration of a reactant to fall to half its starting value.
- Zero order: t½ = [R]₀/2k. It is proportional to the starting concentration, so each successive half-life is half as long as the one before.
- First order: t½ = 0.693/k, independent of concentration. Every half-life of a first-order reaction takes the same time.
- Example: for k = 5.5 × 10⁻¹⁴ s⁻¹, t½ = 0.693/(5.5 × 10⁻¹⁴) = 1.26 × 10¹³ s.
- For first order, 99.9% completion takes (2.303/k) log 1000 = 6.909/k, which is about 10 half-lives; 99% completion takes 4.606/k, about 6.6 half-lives.
- After n half-lives of a first-order reaction, the fraction remaining is (½)ⁿ.
- A pseudo first order reaction is truly of higher order, but one reactant is in such large excess that its concentration barely changes. Its term is absorbed into k.
- Acid hydrolysis of ethyl acetate in a large excess of water behaves as first order: with 0.01 mol ester and 10 mol water, water only drops to 9.99 mol when the ester is gone. Rate = k′[CH₃COOC₂H₅], where k′ = k[H₂O].
- Inversion of cane sugar in acid (C₁₂H₂₂O₁₁ + H₂O → glucose + fructose) is another pseudo first order reaction: rate = k[C₁₂H₂₂O₁₁].
Questions
3 answers
- What does the "How does a reactant run out: steadily or by halves?" simulation show?
- Blue dots are reactant, orange are product; each dot is 0.025 mol L⁻¹. Drag time forward. Switch between zero order (a constant rate) and first order (rate proportional to [R]) and change k and the starting concentration.
- Does JEE Main ask questions from Chemical Kinetics?
- Yes. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Half-life and pseudo first order" in Chemical Kinetics (Class 12 Chemistry).
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