ChemistryISC Class 12 · Chemical Kinetics

Integrated Rate Equations and Half-Life for ISC Class 12

Updated 15 October 2026 · Fact-checked

Integrated rate equations link concentration to time. Zero order: [A] = [A]₀ − kt, half-life [A]₀/2k. First order: k = (2.303/t) log([A]₀/[A]), half-life 0.693/k, independent of starting concentration. A straight line of [A] or log[A] against time shows which order applies.

Zero order falls in a straight line with slope −k; first order curves down and halves in the same t½ each time.

Illustration drawn for QuestionBuddy.

Understand Integrated Rate Equations and Half-Life

A rate law gives the rate at one instant. To answer 'how much is left after 30 minutes?' we need the integrated rate equation, obtained by integrating the rate law. The CISCE syllabus asks for it only for zero and first order reactions.

Zero order. Rate $= - \frac{d [ A ]}{d t} = k [A]^{0} = k$ . Integrating from $[A]_{0}$ at $t = 0$ to $[A]$ at time $t$ : $[A] = [A]_{0} - k t$ , or $k = \frac{[ A ] _{0} - [ A ]}{t}$ . A plot of [A] against t is a straight line of slope $- k$ and intercept $[A]_{0}$ . The concentration falls by the same amount in each equal interval. Half-life: put $[A] = \frac{[ A ] _{0}}{2}$ to get $t_{1/2} = \frac{[ A ] _{0}}{2 k}$ , proportional to the initial concentration. The reaction is complete at $t = \frac{[ A ] _{0}}{k}$ . Examples: decomposition of NH₃ on a hot platinum surface at high pressure (the surface is saturated, so the rate does not depend on pressure) and the thermal decomposition of HI on a gold surface. Units of k: mol L⁻¹ s⁻¹.

First order: derivation. Rate $= - \frac{d [ A ]}{d t} = k [A]$ . Separating variables: $\frac{d [ A ]}{[ A ]} = - k d t$ . Integrating: $ln [A] = - k t + I$ . At $t = 0$ , $[A] = [A]_{0}$ , so $I = ln [A]_{0}$ . Therefore $ln [A] = ln [A]_{0} - k t$ , which gives

$k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}$ and $[A] = [A]_{0} e^{- k t}$ .

In terms of initial amount $a$ and amount reacted $x$ : $k = \frac{2.303}{t} lo g \frac{a}{a - x}$ . A plot of $lo g [A]$ (or $ln [A]$ ) against t is a straight line of slope $- \frac{k}{2.303}$ (or $- k$ ). Equivalently, a plot of $lo g \frac{[ A ] _{0}}{[ A ]}$ against t is a straight line through the origin with slope $\frac{k}{2.303}$ .

Half-life of a first order reaction. Put $[A] = \frac{[ A ] _{0}}{2}$ : $t_{1/2} = \frac{2.303}{k} lo g 2 = \frac{0.693}{k}$ . It does not depend on the initial concentration: every half-life halves what is left. After $n$ half-lives the fraction remaining is $(\frac{1}{2})^{n}$ ; 75% completion takes 2 half-lives, 87.5% takes 3, and 99.9% takes about 10. Characteristics of first order reactions: k has units of time⁻¹ (independent of concentration units), $t_{1/2}$ is constant, and the log plot is linear. Examples: hydrogenation of ethene, decomposition of N₂O₅ and N₂O, and all natural and artificial radioactive decay.

Gas-phase first order reactions. For A(g) → B(g) + C(g) at constant volume and temperature, partial pressures replace concentrations. If $p_{i}$ is the initial pressure of A and $p_{t}$ the total pressure at time t, then $p_{A} = 2 p_{i} - p_{t}$ and $k = \frac{2.303}{t} lo g \frac{p _{i}}{2 p _{i} - p _{t}}$ .

Which order? Use the graph or the half-life. If [A] against t is straight, zero order; if log[A] against t is straight, first order. If successive half-lives get shorter as the reaction proceeds, zero order; if they stay constant, first order.

Integrated Rate Equations and Half-Life: formulas and key facts

Zero order: $[A] = [A]_{0} - k t, k = \frac{[ A ] _{0} - [ A ]}{t}$; Plot of [A] against t: straight line, slope $- k$ .
Zero order half-life: $t_{1/2} = \frac{[ A ] _{0}}{2 k}$; Proportional to $[A]_{0}$ ; completion time $= [A]_{0} / k$ .
First order: $k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}, [A] = [A]_{0} e^{- k t}$; Plot of log[A] against t: straight line, slope $- k /2.303$ .
First order half-life: $t_{1/2} = \frac{0.693}{k}$; Independent of $[A]_{0}$ ; fraction left after $n$ half-lives $= (1/2)^{n}$ .
First order gas-phase reaction: $k = \frac{2.303}{t} lo g \frac{p _{i}}{2 p _{i} - p _{t}}$; For A(g) → B(g) + C(g) at constant volume.

How to solve Integrated Rate Equations and Half-Life questions

Half the ISC kinetics numericals use these equations. Decide the order first, then choose the matching equation; most first order questions can also be done by counting half-lives.

1Decide the order: from units of k (s⁻¹ means first order), from the wording ('first order'), or from data (constant half-life or straight log plot).
2Write the integrated equation for that order before substituting.
3For first order, use any consistent measure of amount: concentration, mass, moles, pressure, or percentage left ( $a$ and $a - x$ ).
4If the time is a whole number of half-lives, use $(1/2)^{n}$ as a check on your logarithm answer.
5For gas-phase data, convert total pressure to the pressure of reactant left before using the first order equation.
6Give k with units (time⁻¹ for first order, mol L⁻¹ time⁻¹ for zero order) and round only at the end.

Integrated Rate Equations and Half-Life in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Pesticide residue decays by first order kinetics

Suppose a farmer sprays a vegetable crop and the pesticide residue on the leaves then breaks down by first order kinetics with a half-life of 5 days. Imagine the residue just after spraying is 6.4 ppm and the safe limit for harvest is 0.4 ppm.

Calculate the rate constant and the least number of days the farmer must wait before harvesting.

Try it first, then show the solutionSolution

$k = \frac{0.693}{t _{1/2}} = \frac{0.693}{5} = 0.139$ day⁻¹.
The residue must fall by a factor of $\frac{6.4}{0.4} = 16 = 2^{4}$ , i.e. four half-lives.
Time $= 4 \times 5 = 20$ days.
Check with the equation: $t = \frac{2.303}{0.139} lo g 16 = 16.6 \times 1.204 = 20$ days.

Answer: $k \approx 0.139$ day⁻¹; the farmer must wait at least 20 days.

Memory hook: For first order decay, count half-lives: each one halves what is left, whatever the starting amount.

Tricks to remember Integrated Rate Equations and Half-Life

Short memory aids. Say each one aloud once, then use it on a question the same day.

Zero subtracts, first divides
Zero order removes the same AMOUNT every minute ( $[A]_{0} - k t$ ); first order removes the same FRACTION every minute (each half-life halves).
Use it when: Choosing the equation and reading graphs.
0.693 is ln 2
$t_{1/2} = \frac{l n 2}{k} = \frac{0.693}{k}$ : half-life is 'log of two over k'.
Use it when: Writing the half-life formula from memory.
2, 3, 10
First order: 75% done in 2 half-lives, 87.5% in 3, 99.9% in about 10.
Use it when: Quick answers and checks.

Integrated Rate Equations and Half-Life: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Straight-line plot for first order · 1 mark · MCQBoard level

For a first order reaction $A \to$ products, which plot is a straight line?
(a) [A] against t
(b) log[A] against t
(c) $\frac{1}{[ A ]}$ against t
(d) $t_{1/2}$ against [A]₀ with positive slope

Try it first, then show the solutionSolution

$lo g [A] = lo g [A]_{0} - \frac{k}{2.303} t$ has the form $y = c + m x$ .
So log[A] against t is a straight line of slope $- k /2.303$ .

Answer: (b).

Example 2: Half-life from k, then time to one-eighth · 2 marks · Short answerBoard level

A first order reaction has $k = 1.155 \times 1 0^{- 3}$ s⁻¹. Calculate its half-life and the time taken for the reactant to fall to one-eighth of its initial concentration.

Try it first, then show the solutionSolution

$t_{1/2} = \frac{0.693}{1.155 \times 1 0 ^{- 3}} = 600$ s (1 mark).
$\frac{1}{8} = (\frac{1}{2})^{3}$ : three half-lives, $3 \times 600 = 1800$ s (1 mark).

Answer: $t_{1/2} = 600$ s (10 min); 1800 s (30 min) to reach one-eighth.

Example 3: First order from two concentrations · 3 marks · Short answerHarder board level

In a first order reaction the concentration of the reactant falls from 0.60 mol L⁻¹ to 0.15 mol L⁻¹ in 50 minutes. Calculate (a) the rate constant, (b) the half-life and (c) the time needed for the concentration to fall to 0.050 mol L⁻¹.

Try it first, then show the solutionSolution

(a) $k = \frac{2.303}{50} lo g \frac{0.60}{0.15} = \frac{2.303 \times 0.602}{50} = 0.0277$ min⁻¹ (1 mark).
(b) $t_{1/2} = \frac{0.693}{0.0277} = 25$ min; check: 0.60 → 0.30 → 0.15 is two half-lives in 50 min (1 mark).
(c) $t = \frac{2.303}{0.0277} lo g \frac{0.60}{0.050} = 83.1 \times 1.079 = 89.6$ min (1 mark).

Answer: (a) $k \approx 0.0277$ min⁻¹. (b) 25 min. (c) About 90 min.

Example 4: A zero order reaction · 3 marks · Short answerHarder board level

A reaction is zero order in A. The concentration of A falls from 0.50 mol L⁻¹ to 0.35 mol L⁻¹ in 20 minutes. Calculate (a) the rate constant with units, (b) the half-life and (c) the time for the reaction to be complete.

Try it first, then show the solutionSolution

(a) $k = \frac{[ A ] _{0} - [ A ]}{t} = \frac{0.50 - 0.35}{20} = 7.5 \times 1 0^{- 3}$ mol L⁻¹ min⁻¹ (1 mark).
(b) $t_{1/2} = \frac{[ A ] _{0}}{2 k} = \frac{0.50}{2 \times 7.5 \times 1 0 ^{- 3}} = 33.3$ min (1 mark).
(c) $t = \frac{[ A ] _{0}}{k} = \frac{0.50}{7.5 \times 1 0 ^{- 3}} = 66.7$ min (1 mark).

Answer: (a) $7.5 \times 1 0^{- 3}$ mol L⁻¹ min⁻¹. (b) 33.3 min. (c) 66.7 min.

Example 5: First order gas-phase decomposition from pressure · 5 marks · Long answerHOTS / competency

A gas A decomposes by first order kinetics at constant volume: A(g) → B(g) + C(g). Pure A is introduced at 0.40 atm, and after 120 s the total pressure is 0.52 atm.
(a) Show that the pressure of A at time t is $2 p_{i} - p_{t}$ . [1]
(b) Calculate the rate constant and the half-life. [3]
(c) Predict the total pressure after two half-lives. [1]

Try it first, then show the solutionSolution

(a) If A falls by $x$ atm, B and C each rise by $x$ : $p_{t} = (p_{i} - x) + x + x = p_{i} + x$ , so $x = p_{t} - p_{i}$ and $p_{A} = p_{i} - x = 2 p_{i} - p_{t}$ (1 mark).
(b) $p_{A} = 2 (0.40) - 0.52 = 0.28$ atm (1 mark).
$k = \frac{2.303}{120} lo g \frac{0.40}{0.28} = \frac{2.303 \times 0.155}{120} = 2.97 \times 1 0^{- 3}$ s⁻¹ (1 mark).
$t_{1/2} = \frac{0.693}{2.97 \times 1 0 ^{- 3}} = 233$ s (1 mark).
(c) After two half-lives $p_{A} = 0.10$ atm, so $x = 0.30$ and $p_{t} = 0.40 + 0.30 = 0.70$ atm (1 mark).

Answer: (b) $k \approx 2.97 \times 1 0^{- 3}$ s⁻¹; $t_{1/2} \approx 233$ s. (c) 0.70 atm.

Where students lose marks in Integrated Rate Equations and Half-Life

The slips examiners see most often, and what to write instead.

Mistake: Stopping at k when the amount left is asked
ISC examiners noted that candidates found k but could not compute the percentage of a radioactive substance remaining (2024).
Do this: After finding k, either use $lo g \frac{[ A ] _{0}}{[ A ]} = \frac{k t}{2.303}$ or count half-lives, $(1/2)^{n}$ , and convert to a percentage.
Usually costs: 1 mark.
Mistake: Using the zero order equation for a first order reaction
Students subtract concentrations for a reaction stated to be first order.
Wrong working
$k = \frac{0.60 - 0.15}{50}$ for a first order reaction
Right working
$k = \frac{2.303}{50} lo g \frac{0.60}{0.15}$
Do this: Read the order first; first order always needs the logarithm of a ratio.
Usually costs: The whole numerical.
Mistake: Inverting the ratio in the log
Writing $lo g \frac{[ A ]}{[ A ] _{0}}$ gives a negative k.
Do this: Initial over final: $lo g \frac{[ A ] _{0}}{[ A ]}$ is always positive.
Usually costs: 1 mark.
Mistake: Using total pressure as the pressure of the reactant
In gas-phase problems students put $p_{t}$ straight into the log.
Wrong working
$k = \frac{2.303}{t} lo g \frac{p _{i}}{p _{t}}$
Right working
$k = \frac{2.303}{t} lo g \frac{p _{i}}{2 p _{i} - p _{t}}$ for A → B + C
Do this: Write the pressure table (initial, change, at time t) before substituting.
Usually costs: 2 marks.

Exam tips

ISC asks for the derivation of the first order equation; practise it in five lines: rate law, separate variables, integrate, find the constant, convert to log base 10.
When a time is a whole number of half-lives, show the half-life chain (100 → 50 → 25 → 12.5%) as well; it earns marks quickly.
State units of k: min⁻¹ or s⁻¹ for first order.
For graph questions, name the axes and the slope: log[A] against t, slope $- k /2.303$ .
Remember radioactive decay is first order; ISC often sets half-life questions on it.

Practice questions on Integrated Rate Equations and Half-Life

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
For a first order reaction, which quantity does not depend on the initial concentration of the reactant?
(a) initial rate
(b) half-life
(c) amount reacted in the first minute
(d) concentration after one half-life
Hint
$t_{1/2} = 0.693/ k$ .
Answer
(b) half-life.
Q2 · 2 marks · Short answerBoard level
Starting from the integrated rate equation of a zero order reaction, derive an expression for its half-life. How does the half-life change if the initial concentration is doubled?
Hint
Put $[A] = [A]_{0} /2$ in $[A] = [A]_{0} - k t$ .
Answer
$\frac{[ A ] _{0}}{2} = [A]_{0} - k t_{1/2}$ , so $t_{1/2} = \frac{[ A ] _{0}}{2 k}$ . Doubling $[A]_{0}$ doubles the half-life.
Q3 · 2 marks · Short answerHarder board level
A radioactive isotope has a half-life of 8.0 days. What percentage of a sample remains after 40 days?
Hint
Radioactive decay is first order; count half-lives.
Answer
$n = \frac{40}{8} = 5$ ; fraction left $= (\frac{1}{2})^{5} = \frac{1}{32}$ , i.e. 3.125%.
Q4 · 3 marks · Short answerHarder board level
A first order reaction is 30% complete in 25 minutes. Calculate its half-life and the time for 90% completion.
Hint
If 30% has reacted, 70% is left.
Answer
$k = \frac{2.303}{25} lo g \frac{100}{70} = 0.01427$ min⁻¹. $t_{1/2} = \frac{0.693}{0.01427} = 48.6$ min. $t_{90%} = \frac{2.303}{0.01427} lo g 10 = 161$ min.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The time taken for 50% completion of a first order reaction is the same whatever the initial concentration.
Reason (R): For a first order reaction, $t_{1/2} = \frac{0.693}{k}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Does the formula contain $[A]_{0}$ ?
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Show that for a first order reaction the time for 75% completion is twice the half-life, and the time for 87.5% completion is three times the half-life.
Hint
Use $t = \frac{2.303}{k} lo g \frac{a}{a - x}$ .
Answer
$t_{75} = \frac{2.303}{k} lo g 4 = \frac{2.303 \times 2 \times 0.301}{k} = 2 \times \frac{0.693}{k} = 2 t_{1/2}$ . $t_{87.5} = \frac{2.303}{k} lo g 8 = 3 \times \frac{0.693}{k} = 3 t_{1/2}$ .
Q7 · 3 marks · Short answerHOTS / competency
Two reactions start with [A]₀ = 1.0 mol L⁻¹ and both have a first half-life of 20 minutes. Reaction P is zero order and reaction Q is first order. (a) Find [A] in each after 30 minutes. (b) Find [A] in each after 40 minutes. (c) Comment on the result.
Hint
For P, $k = \frac{[ A ] _{0}}{2 t _{1/2}}$ ; for Q, $[A] = [A]_{0} (1/2)^{t / t_{1/2}}$ .
Answer
P: $k = 0.025$ mol L⁻¹ min⁻¹, so [A] = 0.25 M at 30 min and 0 at 40 min. Q: $[A] = (0.5)^{1.5} = 0.354$ M at 30 min and 0.25 M at 40 min. The zero order reaction keeps the same speed and finishes; the first order reaction slows as [A] falls and never quite reaches zero.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a doctor gives a patient 500 mg of a drug that is removed from the body by first order kinetics with a half-life of 4.0 hours. The drug stops working when less than 125 mg remains in the body.
(i) After how many hours should the next dose be given? [1]
(ii) How much drug remains after 10 hours? [1]
(iii) If the dose were doubled to 1000 mg, would the time to fall to half change? Explain. [1]
Hint
500 → 250 → 125 is two half-lives.
Answer
(i) Two half-lives: 8 hours. (ii) $500 \times (1/2)^{10/4} = 500 \times 0.177 = 88$ mg. (iii) No. For first order kinetics the half-life (4 h) is independent of the starting amount.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Integrated rate equation of a zero order reaction?Show answer
$[A] = [A]_{0} - k t$ .
2.Half-life of a first order reaction?Show answer
$t_{1/2} = \frac{0.693}{k}$ , independent of $[A]_{0}$ .
3.Slope of log[A] against t for a first order reaction?Show answer
$- \frac{k}{2.303}$ .
4.Give one example of a zero order reaction.Show answer
Decomposition of NH₃ on a hot platinum surface at high pressure.
5.How many half-lives for 99.9% completion of a first order reaction?Show answer
About 10.
6.Pressure of reactant at time t for A(g) → B(g) + C(g)?Show answer
$2 p_{i} - p_{t}$ .

Integrated Rate Equations and Half-Life: frequently asked questions

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

Because t_{1/2} = 0.693/k contains only the rate constant. A first order reaction always removes the same fraction in the same time, whatever the starting amount.

How do you identify a first order reaction from data?

Its half-life stays constant, log[A] against time is a straight line, and the values of k calculated from \frac{2.303}{t}\log\frac{[A]_0}{[A]} at different times agree.

What is the unit of k for a zero order reaction?

mol L⁻¹ s⁻¹ (concentration per unit time).

Is radioactive decay a first order process?

Yes. Its rate depends only on the number of nuclei present, so it has a constant half-life.

Can a zero order reaction be completed?

Yes, in a finite time [A]_0/k, unlike a first order reaction, which only approaches completion.