JEE Mains Chemistry · Chemical Kinetics

First Order Reactions and Half-Life

k = (2.303/t) log([A]₀/[A]) and t½ = 0.693/k: a first-order half-life does not depend on the starting amount, so every time is a multiple of it set by a logarithm.

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Why this matters

Twenty-eight PYQs, twenty-one of them numerical, and five from 2026 — the chapter's largest page. Twelve find a time, a fraction left or a rate constant from the integrated law; ten compare the times to two levels of completion, such as 99.9% against 90%, or two reactants decaying at different speeds; six use the exponential form, read a plot, turn a ratio of rates into a ratio of concentrations, or test statements. Three ideas cover the page.

Concept 1 of 3: The integrated first-order law and the half-life

For

r = k[A]

, integrating gives

\ln\dfrac{[A]_0}{[A]} = kt

. Only the RATIO by which the concentration falls matters, so each half-life takes the same time from any starting amount:

t_{1/2} = \dfrac{0.693}{k}

. Put in what REMAINS, not what has reacted.

Definition

$k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ , or $t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$ .
$[A]$ is what remains: 70% decomposed means $\dfrac{[A]_0}{[A]} = \dfrac{100}{30}$ .
$t_{1/2} = \dfrac{0.693}{k}$ , independent of $[A]_0$ . Unit of $k$ : time $^{-1}$ .
Moles, masses or partial pressures can stand in for concentration when the volume is fixed.
After $n$ half-lives, $\left(\tfrac12\right)^n$ remains. When the ratio is a power of 2, count half-lives instead of taking logs.
Logs to keep at hand: $\log 2 = 0.301$ , $\log 3 = 0.477$ , $\log 5 = 0.699$ , $\ln 10 = 2.303$ .

Integrated first-order law

k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}\qquad t_{1/2} = \frac{0.693}{k}

Worked example

A first-order reaction has

t_{1/2} = 20

min. How long does it take for 70% of the reactant to decompose? (

\log 3 = 0.477

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q50Moderate

Example 1 · Chemical Kinetics · First Order Reactions and Half-Life

If the half-life of a first order reaction is 6.93 minutes then the time required for completion of

99\%

of the reaction will be

\_\_\_\_

minutes. (Given :

\log2 = 0.3010

)

The percent decomposed used as [A]

70% decomposed leaves 30%. Use

\log\dfrac{100}{30}

, not

\log\dfrac{100}{70}

; the second gives a time far too short.

Expiry time of a drug

A drug that stops working at 50% decomposition expires after ONE half-life. If it falls to one-eighth in 18 months, that is three half-lives, so

t_{1/2} = 6

months and the expiry is 6 months — not 18, and not 9.

Drill 11 more on the integrated first-order law and the half-life

Concept 2 of 3: Comparing the times to two levels of completion

Every first-order time is

\dfrac{2.303}{k}

times a logarithm. Divide two such times and

k

cancels: the ratio of times is the ratio of the logs. The same idea handles two reactants with different half-lives — write each as

[X]_0\left(\tfrac12\right)^{t/t_{1/2}}

and set them equal.

Definition

$\dfrac{t_1}{t_2} = \dfrac{\log([A]_0/[A]_1)}{\log([A]_0/[A]_2)}$ for one reaction.
Landmarks in half-lives: $t_{75\%} = 2t_{1/2}$ , $t_{87.5\%} = 3t_{1/2}$ , $t_{90\%} = 3.32t_{1/2}$ , $t_{99\%} = 6.64t_{1/2}$ , $t_{99.9\%} \approx 10t_{1/2}$ .
$t_{99\%} = 2t_{90\%}$ and $t_{99.9\%} = 3t_{90\%}$ , since $\log 100 = 2$ and $\log 1000 = 3$ .
For two different reactions, bring in their $k$ values: $k_1 : k_2$ is the inverse of $t_{1/2,1} : t_{1/2,2}$ .
Two decaying reactants: $[A]_0\left(\tfrac12\right)^{t/t_A} = [B]_0\left(\tfrac12\right)^{t/t_B}$ , then solve for $t$ .

Ratio of two completion times

\frac{t_1}{t_2} = \frac{\log([A]_0/[A]_1)}{\log([A]_0/[A]_2)}

Worked example

For a first-order reaction, find

\dfrac{t_{99\%}}{t_{75\%}}

. (

\log 2 = 0.301

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q32Moderate

Example 2 · Chemical Kinetics · First Order Reactions and Half-Life

An organic compound undergoes first order decomposition. The time taken for decomposition to

\left( \frac{1}{8} \right)^{\text{th~}}

and

\left( \frac{1}{10} \right)^{\text{th~}}

of its initial concentration are

t_{1/8}

and

t_{1/10}

respectively. What is the value of

\frac{t_{1/8}}{t_{1/10}}\times 10

(\log2 = 0.3)

67% complete read as two-thirds left

67% complete leaves about one-third, so

t = \dfrac{2.303}{k}\log 3 \approx 1.58\,t_{1/2}

. Using

\log 1.5

(two-thirds left) answers the question for 33% completion.

Times scaled like the percentages

99.9% completion does not take about twice as long as 90%: the logs are 3 and 1, so it takes three times as long.

Drill 9 more on comparing the times to two levels of completion

Concept 3 of 3: Exponential form, straight-line plots and rate ratios

The same law can be written as an exponential, as a straight line, or through the rate.

[A] = [A]_0e^{-kt}

gives the fraction decomposed.

\ln[A]

against

t

is a straight line of slope

-k

. And because

r = k[A]

, a ratio of two rates is a ratio of two concentrations.

Definition

$[A] = [A]_0e^{-kt}$ ; fraction decomposed $= 1 - e^{-kt}$ .
$\ln\dfrac{[A]}{[A]_0}$ (or $\ln\dfrac{p}{p^0}$ ) against $t$ : straight line through the origin, slope $-k$ . With $\log$ : slope $-\dfrac{k}{2.303}$ .
$\dfrac{r_1}{r_2} = \dfrac{[A]_1}{[A]_2}$ for first order, so rates at two times feed straight into $k = \dfrac{2.303}{t}\log\dfrac{r_1}{r_2}$ .
$r = k[A]^{1/2}[B]^{1/2}$ with $[A]_0 = [B]_0$ in a 1 : 1 reaction: $[A] = [B]$ throughout, so $r = k[A]$ , first order.
A first-order reaction never reaches 100%: $e^{-kt}$ is never zero.

Exponential form

[A] = [A]_0\,e^{-kt}\qquad \ln\frac{[A]}{[A]_0} = -kt

Worked example

The rate of a first-order reaction is 0.090 mol L

^{-1}

^{-1}

at 10 min and 0.030 mol L

^{-1}

^{-1}

at 30 min. Find its half-life. (

\log 2 = 0.301

\log 3 = 0.477

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q32Moderate

Example 3 · Chemical Kinetics · First Order Reactions and Half-Life

Consider the first order reaction

R \rightarrow P

.The fraction of molecules decomposed in the given first order reaction can be expressed as

A first-order reaction that 'completes'

[A] = [A]_0e^{-kt}

is never zero, so a first-order reaction never reaches 100% in a finite time. A statement that it completes in 1000 s is false.

Rate ratio used for any order

A ratio of rates equals a ratio of concentrations only for first order. For second order the rate ratio is the SQUARE of the concentration ratio.

Drill 5 more on exponential form, straight-line plots and rate ratios

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The integrated first-order law and the half-life
Integrated first-order law
$k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}\qquad t_{1/2} = \frac{0.693}{k}$
Comparing the times to two levels of completion
Ratio of two completion times
$\frac{t_1}{t_2} = \frac{\log([A]_0/[A]_1)}{\log([A]_0/[A]_2)}$
Exponential form, straight-line plots and rate ratios
Exponential form
$[A] = [A]_0\,e^{-kt}\qquad \ln\frac{[A]}{[A]_0} = -kt$

Watch out for (6)

The percent decomposed used as [A]
→ The integrated first-order law and the half-life
Expiry time of a drug
→ The integrated first-order law and the half-life
67% complete read as two-thirds left
→ Comparing the times to two levels of completion
Times scaled like the percentages
→ Comparing the times to two levels of completion
A first-order reaction that 'completes'
→ Exponential form, straight-line plots and rate ratios
Rate ratio used for any order
→ Exponential form, straight-line plots and rate ratios

Test yourself on Chemical Kinetics

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