10 min readConcept sheet2-mark derivation of t½; 3-mark numerical on half-lives; 1-mark MCQ or assertion-reason on pseudo first order; also a NEET favourite

Half-life of a reaction and pseudo first order reactions

Half-life is the time for a reactant to fall to half. For first order, t½ = 0.693/k, whatever the start. Pseudo first order reactions have one reactant in large excess.

Half-life (t½)Time in which the concentration of a reactant falls to half its initial value

Zero ordert½ = [R]₀/2k: directly proportional to the initial concentration

First ordert½ = 0.693/k: independent of the initial concentration

After n half-lives[R] = [R]₀/2ⁿ (first order); 99.9% completion takes about 10 half-lives

Derivation: half-life for zero and first order reactions

At t = t½, [R] = [R]₀/2.

1
Zero order: k = ([R]₀ − [R])/t. Put t = t½ and [R] = [R]₀/2: k = ([R]₀ − [R]₀/2)/t½ = [R]₀/(2t½).
2★
So, for zero order, t½ = [R]₀/2k.
3
First order: k = (2.303/t) log([R]₀/[R]). Put t = t½ and [R] = [R]₀/2: k = (2.303/t½) log 2.
4
t½ = (2.303 × 0.3010)/k = 0.693/k.
5★
So, for first order, t½ = 0.693/k: it does not depend on [R]₀.

★ key step

Worked example: half-life and completion time

Problem

The half-life of a first order reaction is 30 min. (a) Find k. (b) How long will it take for 87.5% of the reactant to react?

Working

(a) Formula: k = 0.693/t½ = 0.693 ÷ 30 min = 0.0231 min⁻¹.
(b) 87.5% reacted means 12.5% is left: [R] = 0.125[R]₀ = [R]₀/8.
[R]₀ → [R]₀/2 → [R]₀/4 → [R]₀/8 is three half-lives.
Time = 3 × t½ = 3 × 30 min = 90 min.

Answer

(a) k = 2.31 × 10⁻² min⁻¹; (b) 90 min

Worked example: 99.9% completion takes about 10 half-lives

Problem

Show that, for a first order reaction, the time for 99.9% completion is about ten times the half-life. (log 2 = 0.3010, log 1000 = 3)

Working

99.9% reacted means [R] = 0.001[R]₀, so [R]₀/[R] = 1000.
t(99.9%) = (2.303/k) log 1000 = (2.303 × 3)/k = 6.909/k.
t½ = 0.693/k.
Ratio = 6.909 ÷ 0.693 = 9.97, which is about 10.

Answer

t(99.9%) ≈ 10 × t½

Pseudo first order reactions

Some reactions involve two reactants, so their true order is two. But if one reactant is in large excess, its concentration hardly changes, and the reaction behaves as first order. Such a reaction is called pseudo first order. Acid hydrolysis of ethyl acetate: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (H⁺ catalyst). Rate = k′[CH₃COOC₂H₅][H₂O]. Water is the solvent and is in huge excess, so [H₂O] is almost constant. Then rate = k[CH₃COOC₂H₅], where k = k′[H₂O]. Inversion of cane sugar is another example: C₁₂H₂₂O₁₁ (sucrose) + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose), in acid; rate = k[C₁₂H₂₂O₁₁].

Half-life: zero order vs first order

Zero ordert½ = [R]₀/2k. Doubling [R]₀ doubles t½. Each later half-life is shorter than the one before.

First ordert½ = 0.693/k. t½ is the same whatever the starting amount. Every half-life is equal, as in radioactive decay.

In real life

Doctors use the half-life of a drug, such as paracetamol, to decide how many hours apart doses should be. The same idea tells how long radioactive iodine used in thyroid treatment stays in the body.

Exam angle

2 marks: “Show that the half-life of a first order reaction is independent of the initial concentration.” k = (2.303/t½) log([R]₀/([R]₀/2)) = (2.303/t½) log 2, so t½ = 0.693/k. [R]₀ cancels, so t½ does not depend on it.

2 marks: “What is a pseudo first order reaction? Give one example with its equation.” A higher order reaction that behaves as first order because one reactant is in large excess. Example: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (H⁺); rate = k[CH₃COOC₂H₅].

3 marks: “A first order reaction has k = 6.93 × 10⁻³ min⁻¹. Find its half-life and the time for 75% completion.” t½ = 0.693 ÷ 6.93 × 10⁻³ = 100 min. 75% completion leaves one quarter: two half-lives = 200 min.

Remember it

First order half-life: 0.693 over k, the same every time. Ten half-lives finish 99.9%.