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Chemistry · Chemical Kinetics

Half-Life of Reactions

The half-life of a reaction is the time in which the concentration of a reactant falls to one half of its starting value. NCERT introduces it in §3.3.3 as a single number that captures reaction speed, and it carries an idea NEET tests every year: for a first-order reaction the half-life is fixed at $t_{1/2} = 0.693/k$, while for a zero-order reaction it grows with starting concentration as $t_{1/2} = [\text{R}]_0/2k$. Master how the order rewrites this number and a whole family of problems collapses to one line.

What Half-Life Means

Reaction speed can be reported as a rate, a rate constant, or — most intuitively — as a time. The half-life, symbol $t_{1/2}$, is that time-based measure. Following NCERT §3.3.3, it is defined as the time in which the concentration of a reactant is reduced to one half of its initial concentration. If a reaction starts with $[\text{R}]_0 = 0.8\ \text{mol L}^{-1}$, then $t_{1/2}$ is the moment the concentration first reaches $0.4\ \text{mol L}^{-1}$.

The power of half-life is that it folds the whole integrated rate law into one figure. But that figure is not universal — it depends on the order of the reaction. The same definition produces a half-life that is sometimes proportional to starting concentration, sometimes completely indifferent to it. Working out exactly how is the heart of this topic.

We derive $t_{1/2}$ for the two orders NEET expects you to handle algebraically — zero and first — by substituting the half-way condition $[\text{R}] = [\text{R}]_0/2$ into each integrated rate law. The contrast between the two answers is itself an examinable fact.

OrderIntegrated rate lawHalf-lifeDepends on [R]₀?
Zerokt = [R]₀ − [R]$t_{1/2} = \dfrac{[\text{R}]_0}{2k}$Yes — directly proportional
Firstkt = ln([R]₀/[R])$t_{1/2} = \dfrac{0.693}{k}$No — independent

Zero-Order Half-Life

For a zero-order reaction the rate is independent of concentration, and NCERT writes the integrated rate constant as

$$ k = \frac{[\text{R}]_0 - [\text{R}]}{t} $$

At $t = t_{1/2}$ the concentration has fallen to half its initial value, $[\text{R}] = \tfrac{1}{2}[\text{R}]_0$. Substituting,

$$ k = \frac{[\text{R}]_0 - \tfrac{1}{2}[\text{R}]_0}{t_{1/2}} = \frac{[\text{R}]_0}{2\,t_{1/2}} \quad\Longrightarrow\quad t_{1/2} = \frac{[\text{R}]_0}{2k} $$

So the half-life of a zero-order reaction is directly proportional to the initial concentration and inversely proportional to the rate constant. Start with twice as much reactant and the first half takes twice as long, because the reaction chews through reactant at a fixed absolute rate ($\text{mol L}^{-1}\,\text{s}^{-1}$) regardless of how much is present. This is exactly the behaviour NEET 2018 tested directly.

First-Order Half-Life

For a first-order reaction NCERT gives the integrated rate law in logarithmic form:

$$ k = \frac{2.303}{t}\,\log\frac{[\text{R}]_0}{[\text{R}]} $$

At $t = t_{1/2}$, set $[\text{R}] = [\text{R}]_0/2$, so the ratio $[\text{R}]_0/[\text{R}] = 2$:

$$ t_{1/2} = \frac{2.303}{k}\,\log 2 = \frac{2.303 \times 0.301}{k} = \frac{0.693}{k} $$

Notice what vanished: the initial concentration $[\text{R}]_0$ cancelled completely. The first-order half-life depends only on the rate constant, so at a fixed temperature it is a true constant of the reaction. This single result — $t_{1/2} = 0.693/k$ — is among the most frequently used formulae in the entire chapter, and it can be inverted to find $k$ whenever a half-life is quoted.

Worked Example · NCERT 3.7

A first-order reaction has a rate constant $k = 5.5 \times 10^{-14}\ \text{s}^{-1}$. Find its half-life.

$t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{5.5 \times 10^{-14}\ \text{s}^{-1}} = 1.26 \times 10^{13}\ \text{s}.$ No initial concentration is needed, and none is given — a hallmark of a first-order half-life problem.

NEET Trap

"Half-life is always 0.693/k"

The formula $t_{1/2} = 0.693/k$ is true only for first-order kinetics. Students memorise it and apply it to every reaction. For a zero-order reaction the correct expression is $t_{1/2} = [\text{R}]_0/2k$, and ignoring the $[\text{R}]_0$ term is a classic loss of a mark.

First order → $0.693/k$ (no concentration). Zero order → $[\text{R}]_0/2k$ (concentration matters).

Half-Life as an Order Fingerprint

Because the two orders treat $[\text{R}]_0$ so differently, half-life becomes a diagnostic tool — a fingerprint for the order of an unknown reaction. The general scaling, valid for any order $n$, is

$$ t_{1/2} \;\propto\; [\text{R}]_0^{\,1-n} $$

For $n = 1$ the exponent is zero, so $t_{1/2}$ is independent of starting concentration; for $n = 0$ the exponent is one, so $t_{1/2}$ is directly proportional to it. In the laboratory you exploit this by running the same reaction at two different initial concentrations and watching how the half-life responds.

If you double [R]₀ and observe…ConclusionReason
$t_{1/2}$ stays the sameFirst order$t_{1/2} = 0.693/k$, no $[\text{R}]_0$ term
$t_{1/2}$ doublesZero order$t_{1/2} = [\text{R}]_0/2k \propto [\text{R}]_0$
$t_{1/2}$ halvesSecond order$t_{1/2} = 1/(k[\text{R}]_0) \propto 1/[\text{R}]_0$

This is also why a graph of $t_{1/2}$ against initial concentration is so revealing: a horizontal line means first order, a straight line through the origin with positive slope means zero order. NEET 2022 turned exactly this graph-reading skill into a question. The figure below contrasts the two profiles side by side.

Figure 1 · t½ vs initial concentration by order Initial concentration [R]₀ → t½ → Zero order t½ = [R]₀ / 2k First order t½ = 0.693 / k (constant)

A flat half-life line is the unmistakable signature of first-order kinetics; a straight rise from the origin marks zero order.

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Build the foundation

Every half-life formula here drops out of the integrated rate equations — revise those derivations to see why $[\text{R}]_0$ survives in one case and cancels in the other.

Successive Half-Lives

What happens after the first half-life expires? For a first-order reaction the answer is elegant: because $t_{1/2}$ is independent of concentration, every successive half-life takes the same time. After one $t_{1/2}$, half the reactant remains; after two, one-quarter; after three, one-eighth — each step consuming an equal time interval.

Half-lives elapsedTimeFraction of [R]₀ remaining% completed
1$t_{1/2}$1/250%
2$2t_{1/2}$1/475%
3$3t_{1/2}$1/887.5%
4$4t_{1/2}$1/1693.75%
$n$$n\,t_{1/2}$$(1/2)^n$$100(1-2^{-n})\%$

This geometric decay is what the first-order curve below depicts. The concentration falls along an exponential $[\text{R}] = [\text{R}]_0 e^{-kt}$, and each marked drop — full to half, half to quarter, quarter to eighth — spans an identical horizontal interval of one $t_{1/2}$.

Figure 2 · First-order decay with successive half-lives Time → [R] → [R]₀ ½ ¼ ⅛ t½ t½ t½ [R] = [R]₀ e⁻ᵏᵗ

Equal time slices, halving concentration each step — the constant first-order half-life made visible.

A consequence NCERT proves in Example 3.8 is that complete consumption is mathematically gradual: the time for 99.9% completion equals $6.909/k$, which is exactly $10\,t_{1/2}$. So a reaction with a one-minute half-life still needs ten minutes to be 99.9% over — a result NEET 2025 lifted straight onto the paper.

Radioactive Decay & ¹⁴C Dating

Radioactive decay is nature's textbook first-order process: the number of nuclei disintegrating per unit time is proportional to the number present, so the decay obeys $N = N_0 e^{-kt}$ and carries a constant half-life $t_{1/2} = 0.693/k$. This shared mathematics is why a kinetics chapter can borrow the radioactivity example at all.

NCERT exercise 3.14 makes the link concrete. The half-life of $\ce{^{14}_{6}C}$ is 5730 years. A living tree maintains a steady level of $\ce{^{14}C}$; once it dies, no fresh carbon is absorbed and the isotope decays with this fixed half-life. Measuring the surviving fraction therefore dates the sample.

Worked Example · NCERT 3.14 (¹⁴C dating)

An artifact of wood contains only 80% of the $\ce{^{14}C}$ found in a living tree. Given $t_{1/2}(\ce{^{14}C}) = 5730$ years, estimate the artifact's age.

First find $k$ from the half-life: $k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{5730} = 1.21 \times 10^{-4}\ \text{yr}^{-1}.$

Then apply the first-order law with $[\text{R}]_0/[\text{R}] = 100/80 = 1.25$: $$ t = \frac{2.303}{k}\log\frac{100}{80} = \frac{2.303}{1.21\times 10^{-4}}\log 1.25 \approx 1845\ \text{years}. $$ The sample is roughly 1.8 thousand years old.

Worked Numericals

The fastest marks in this topic come from recognising which formula the data point to. If a problem gives only a rate constant and asks for half-life, it is first order. If it asks how $t_{1/2}$ changes with starting amount, decide the order first, then scale.

Worked Example · k from a measured half-life

$\ce{SO2Cl2}$ takes 60 minutes to decompose to half its initial amount. If the decomposition is first order, find the rate constant. (NCERT Intext 3.6)

For first order, $k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{60\ \text{min}} = 1.155 \times 10^{-2}\ \text{min}^{-1}.$ In SI this is $1.925 \times 10^{-4}\ \text{s}^{-1}$. The half-life and rate constant are interconvertible in one step.

Worked Example · half-life from two rate measurements

A first-order reaction has rate $0.04\ \text{mol L}^{-1}\text{s}^{-1}$ at 10 s and $0.03\ \text{mol L}^{-1}\text{s}^{-1}$ at 20 s. Find $t_{1/2}$. (NEET 2016)

Since rate $\propto [\text{R}]$ for first order, the rate ratio equals the concentration ratio. Apply the integrated law across the 10 s interval: $$ k = \frac{2.303}{10}\log\frac{0.04}{0.03} = \frac{2.303}{10}(0.1249) = 0.0288\ \text{s}^{-1}. $$ Then $t_{1/2} = \dfrac{0.693}{0.0288} \approx 24.1\ \text{s}.$

NEET Trap

Confusing "reduced to half" with "reduced by half"

A half-life is reached when the reactant is reduced to half its value, i.e. $[\text{R}] = [\text{R}]_0/2$. Phrases like "reduced to one-fourth" mean two half-lives, and "1/16th value" means four half-lives. NCERT exercise 3.16 asks for the time to reach 1/16th value precisely to test this — the answer is $4t_{1/2}$, not one.

Fraction remaining $(1/2)^n$ → that's $n$ half-lives. 1/16 → $n = 4$.

Quick Recap

Half-Life in One Screen

  • Half-life $t_{1/2}$ = time for $[\text{R}]$ to fall to $[\text{R}]_0/2$ (NCERT §3.3.3).
  • Zero order: $t_{1/2} = [\text{R}]_0/2k$ — directly proportional to initial concentration.
  • First order: $t_{1/2} = 0.693/k$ — independent of initial concentration; interconvertible with $k$.
  • General scaling $t_{1/2} \propto [\text{R}]_0^{1-n}$ lets half-life reveal the order.
  • First-order half-lives are successive and equal: $(1/2)^n$ remains after $n$ of them; 99.9% completion = $10\,t_{1/2}$.
  • Radioactive decay (e.g. $\ce{^{14}C}$, $t_{1/2} = 5730$ yr) is first order — the basis of carbon dating.

NEET PYQ Snapshot — Half-Life of Reactions

Real NEET questions on half-life behaviour and computation across orders.

NEET 2025 · Q.65

If the half-life ($t_{1/2}$) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:

  1. 10 minutes
  2. 2 minutes
  3. 4 minutes
  4. 5 minutes
Answer: (1) 10 minutes

For a first-order reaction the time for 99.9% completion is $6.909/k = 10 \times (0.693/k) = 10\,t_{1/2}$. With $t_{1/2} = 1$ min, that is 10 minutes.

NEET 2022 · Q.51

The y and x axes for zero and first order reactions, respectively, in the kinetics graph shown are:

  1. zero order (y = concentration, x = time); first order (y = rate constant, x = concentration)
  2. zero order (y = rate, x = concentration); first order (y = t½, x = concentration)
  3. zero order (y = rate, x = concentration); first order (y = rate, x = t½)
  4. zero order (y = concentration, x = time); first order (y = t½, x = concentration)
Answer: (2)

Zero order: rate $= k[\text{A}]^0 = k$, a constant — rate vs concentration is a horizontal line. First order: $t_{1/2} = 0.693/k$ is constant, so a plot of $t_{1/2}$ vs concentration is also flat.

NEET 2018 · Q.54

The correct difference between first and second order reactions is that:

  1. the rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does
  2. the half-life of a first-order reaction does not depend on [A]₀; the half-life of a second-order reaction does
  3. a first-order reaction can be catalysed; a second-order reaction cannot
  4. the rate of a first-order reaction depends on reactant concentrations; the rate of a second-order reaction does not
Answer: (2)

First order: $t_{1/2} = 0.693/k$ is independent of $[\text{A}]_0$. Second order: $t_{1/2} = 1/(k[\text{A}]_0)$ depends on initial concentration.

NEET 2018 · Q.79

When the initial concentration of the reactant is doubled, the half-life period of a zero order reaction:

  1. is halved
  2. is doubled
  3. is tripled
  4. remains unchanged
Answer: (2) is doubled

For zero order $t_{1/2} = [\text{A}]_0/2k \propto [\text{A}]_0$. Doubling the initial concentration doubles the half-life.

NEET 2016 · Q.45

The rate of a first order reaction is $0.04\ \text{mol L}^{-1}\text{s}^{-1}$ at 10 s and $0.03\ \text{mol L}^{-1}\text{s}^{-1}$ at 20 s after initiation. The half-life period is:

  1. 34.1 s
  2. 44.1 s
  3. 54.1 s
  4. 24.1 s
Answer: (4) 24.1 s

$k = \frac{2.303}{10}\log(0.04/0.03) = 0.0288\ \text{s}^{-1}$; then $t_{1/2} = 0.693/0.0288 \approx 24.1$ s.

FAQs — Half-Life of Reactions

The doubts NEET aspirants raise most often about half-life.

What is the half-life of a reaction?

The half-life of a reaction, written t½, is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is a convenient single number that summarises how fast a reaction proceeds, and it behaves differently depending on the order of the reaction.

Why is the half-life of a first-order reaction independent of the initial concentration?

For a first-order reaction the integrated rate law gives t½ = 0.693/k. The initial concentration [R]0 cancels out when you substitute [R] = [R]0/2 into the logarithmic rate equation, leaving only the rate constant k. Because k is fixed at a given temperature, every successive half is consumed in exactly the same time, regardless of how much reactant you started with.

How does the half-life of a zero-order reaction depend on the initial concentration?

For a zero-order reaction t½ = [R]0/2k, so the half-life is directly proportional to the initial concentration. If you double the initial concentration, the half-life doubles. This is the opposite behaviour to a first-order reaction and is a frequent NEET test of conceptual understanding.

How can half-life be used to find the order of a reaction?

Run the reaction at two different initial concentrations and compare the half-lives. If t½ stays the same, the reaction is first order. If t½ doubles when [R]0 doubles, it is zero order. In general t½ ∝ [R]0^(1−n), so the way t½ scales with starting concentration directly reveals the order n.

How is radioactive decay related to the half-life of chemical reactions?

Radioactive decay follows first-order kinetics, so its half-life is constant, t½ = 0.693/k, exactly like a first-order chemical reaction. This is the basis of carbon-14 dating: 14C has a half-life of 5730 years, and by measuring how much 14C remains in a sample relative to a living tree, the age of the sample can be estimated using the first-order integrated rate law.

How long does a first-order reaction take to reach 99.9% completion in terms of its half-life?

For a first-order reaction the time for 99.9% completion equals 10 half-lives. Substituting [R] = 0.001[R]0 into kt = 2.303 log([R]0/[R]) gives t = 6.909/k, and since t½ = 0.693/k, the ratio t/t½ ≈ 10. So if t½ is 1 minute, 99.9% completion takes about 10 minutes.

Related Topics
Chemical Kinetics Return to the full chapter hub and all subtopics. Integrated Rate Equations The zero- and first-order laws every half-life formula is derived from. Rate Law and Order How order is defined — the property that rewrites t½. Pseudo First-Order Reactions When higher-order reactions borrow first-order half-life behaviour.