Simulation · Chemistry · Class 12

How does a reactant run out: steadily or by halves?

From the lesson Half-life and pseudo first order in Chemical Kinetics. Change the values and watch what happens.

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].

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