ChemistryCBSE Class 12 · Chemical Kinetics

Integrated Rate Equations and Half-Life: Zero and First Order

Updated 15 October 2026 · Fact-checked

Integrated rate equations link concentration directly to time. Zero order: $[R] = [R]_0 - kt$ and $t_{1/2} = \frac{[R]_0}{2k}$. First order: $k = \frac{2.303}{t}\log\frac{[R]_0}{[R]}$ and $t_{1/2} = \frac{0.693}{k}$, which does not depend on the starting concentration. Graphs of $[R]$ or $\ln[R]$ against time identify the order.

Notice that in zero order each half-life is shorter than the last, but in first order every half-life takes the same time.

Illustration drawn for QuestionBuddy.

Understand Integrated Rate Equations and Half-Life

A rate law such as rate $= k [R]$ tells you the rate at one instant, but experiments measure concentration at different times. Integrating the rate law gives an equation between $[R]$ and $t$ that can be tested directly.

Zero order. Here rate $= - \frac{d [ R ]}{d t} = k$ , independent of concentration. Integrating from $[R]_{0}$ at $t = 0$ : $[R] = [R]_{0} - k t$ , so $k = \frac{[ R ] _{0} - [ R ]}{t}$ . A plot of $[R]$ against $t$ is a straight line with slope $- k$ and intercept $[R]_{0}$ . The reaction finishes at $t = [R]_{0} / k$ . Zero order is uncommon and appears under special conditions: decomposition of gaseous ammonia on a hot platinum surface at high pressure, or thermal decomposition of HI on gold, where the metal surface is saturated and adding more gas cannot increase the rate.

First order. Here $- \frac{d [ R ]}{d t} = k [R]$ . Separating variables and integrating gives $ln [R] = ln [R]_{0} - k t$ , or $[R] = [R]_{0} e^{- k t}$ . In base-10 form, the one used in board numericals, $k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]}$ . A plot of $ln [R]$ against $t$ is a straight line with slope $- k$ ; equivalently $lo g \frac{[ R ] _{0}}{[ R ]}$ against $t$ is a straight line through the origin with slope $\frac{k}{2.303}$ . Examples: hydrogenation of ethene, decomposition of $N_{2} O_{5}$ and $N_{2} O$ , and all natural radioactive decay.

Gas-phase first order reactions. For $A (g) \to B (g) + C (g)$ at constant volume, start with pressure $p_{i}$ of A. If A's pressure falls by $x$ , the total is $p_{t} = p_{i} + x$ , so $p_{A} = 2 p_{i} - p_{t}$ and $k = \frac{2.303}{t} lo g \frac{p _{i}}{2 p _{i} - p _{t}}$ . Always write this pressure balance from the equation you are given.

Half-life. The half-life $t_{1/2}$ is the time for the concentration of a reactant to fall to half its initial value. For zero order, put $[R] = [R]_{0} /2$ : $t_{1/2} = \frac{[ R ] _{0}}{2 k}$ , so it is proportional to the initial concentration. For first order: $t_{1/2} = \frac{2.303}{k} lo g 2 = \frac{0.693}{k}$ , independent of $[R]_{0}$ . That is why after $n$ half-lives a first-order reactant is reduced to $(\frac{1}{2})^{n}$ of its starting amount, whatever that amount was.

Two useful results follow from the first-order equation: the time for 99% completion is twice that for 90%, and the time for 99.9% completion is about ten half-lives. In practice these reactions never reach exactly 100%; they keep halving.

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

Zero order integrated rate equation: $[R] = [R]_{0} - k t, k = \frac{[ R ] _{0} - [ R ]}{t}$; Plot of $[R]$ vs $t$ : straight line, slope $- k$ .
Zero order half-life: $t_{1/2} = \frac{[ R ] _{0}}{2 k}$; Directly proportional to initial concentration.
First order integrated rate equation: $k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]}, [R] = [R]_{0} e^{- k t}$; Plot of $ln [R]$ vs $t$ : straight line, slope $- k$ .
First order half-life: $t_{1/2} = \frac{0.693}{k}$; Independent of initial concentration; units of $k$ : s⁻¹ (or min⁻¹).
Gas-phase first order, A → B + C: $k = \frac{2.303}{t} lo g \frac{p _{i}}{2 p _{i} - p _{t}}$; Derive the pressure balance for each new equation.
Amount after n half-lives: $[R] = [R]_{0} (\frac{1}{2})^{n}, n = \frac{t}{t _{1/2}}$; First order only.

How to solve Integrated Rate Equations and Half-Life questions

Nearly every numerical in this topic is first order. Decide what is given (fraction left, fraction reacted, pressures or half-life) and pick the matching form of the equation.

1Confirm the order from the question (or from units of $k$ : s⁻¹ means first order; mol L⁻¹ s⁻¹ means zero order).
2Convert 'x% completed' to the ratio $\frac{[ R ] _{0}}{[ R ]} = \frac{100}{100 - x}$ .
3Substitute in $k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]}$ ; keep time units consistent with $k$ .
4For half-life questions use $t_{1/2} = 0.693/ k$ ; for 'time to reach 1/8, 1/16 …' count half-lives.
5For gas-phase data, write the pressure table (initial, change, at time t) and express the reactant's pressure as $2 p_{i} - p_{t}$ or as the equation requires.
6For tabulated data, test first order by checking whether successive half-lives are equal or whether $k$ calculated from different time pairs is constant.
7For zero order, use the straight-line equation; check that $[R]$ never becomes negative (the reaction stops at $t = [R]_{0} / k$ ).

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.

First-order half-life in everyday decay

Suppose a packaged fruit juice loses its vitamin C by a first-order reaction with a rate constant of 0.010 per day at room temperature. A shopkeeper wants to know how long it takes for half the vitamin C to be lost, and whether a larger bottle would last longer.

Calculate the half-life of vitamin C in the juice. Does the size of the bottle or the starting amount change it?

Try it first, then show the solutionSolution

First order: $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.010 day ^{- 1}}$ .
$t_{1/2} = 69.3$ days.
For a first-order reaction $t_{1/2}$ does not depend on the starting concentration or amount, so a bigger bottle loses half its vitamin C in the same 69.3 days.

Answer: About 69 days, the same for any bottle size.

Memory hook: First order forgets how much it started with: half goes in the same time, every time.

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 is a straight drop, first is a steady halving
Zero order: equal amounts lost in equal times. First order: equal fractions lost in equal times.
Use it when: Recognising order from data or graphs.
0.693 is ln 2
$t_{1/2} = ln 2/ k$ ; 0.693 = 2.303 × 0.301 (log 2).
Use it when: Never forget or misplace the constant.
90 → 99 → 99.9 needs 1, 2, 3 'log tens'
Each extra 9 after the decimal needs one more $\frac{2.303}{k}$ of time.
Use it when: Proof questions: $t_{99%} = 2 t_{90%}$ , $t_{99.9%} = 3 t_{90%} \approx 10 t_{1/2}$ .

Integrated Rate Equations and Half-Life: solved examples

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

Example 1: Half-life and concentration · 1 mark · MCQBoard level

The half-life of a first-order reaction:
(a) increases with initial concentration
(b) decreases with initial concentration
(c) is independent of initial concentration
(d) is zero

Try it first, then show the solutionSolution

$t_{1/2} = \frac{0.693}{k}$ contains no concentration term.

Answer: (c).

Example 2: Half-life from k · 2 marks · Very short answerBoard level

The rate constant of a first-order reaction is $1.155 \times 1 0^{- 3}$ s⁻¹. Calculate its half-life in seconds and minutes.

Try it first, then show the solutionSolution

$t_{1/2} = \frac{0.693}{k}$ (½ mark).
$= \frac{0.693}{1.155 \times 1 0 ^{- 3}} = 600$ s (1 mark).
$= 10$ minutes (½ mark).

Answer: 600 s (10 min).

Example 3: k and half-life from percentage decomposed · 3 marks · Short answerHarder board level

A first-order reaction is 30% complete in 40 minutes. Calculate (a) the rate constant, (b) the half-life and (c) the time taken for 75% completion. (log 10/7 = 0.1549)

Try it first, then show the solutionSolution

(a) After 30% reaction, $\frac{[ R ] _{0}}{[ R ]} = \frac{100}{70}$ . $k = \frac{2.303}{40} lo g \frac{10}{7} = \frac{2.303 \times 0.1549}{40} = 8.92 \times 1 0^{- 3}$ min⁻¹ (1 mark).
(b) $t_{1/2} = \frac{0.693}{8.92 \times 1 0 ^{- 3}} = 77.7$ min (1 mark).
(c) 75% completion means one quarter left, i.e. two half-lives: $2 \times 77.7 = 155$ min (1 mark).

Answer: (a) $8.92 \times 1 0^{- 3}$ min⁻¹. (b) 77.7 min. (c) About 155 min.

Example 4: First order from total pressure · 3 marks · Short answerHarder board level

The gas-phase reaction $A (g) \to B (g) + C (g)$ is first order. At constant volume and temperature, the initial pressure of pure A is 0.40 atm and the total pressure after 100 s is 0.50 atm. Calculate the rate constant.

Try it first, then show the solutionSolution

Pressure table: A: $0.40 - x$ , B: $x$ , C: $x$ ; total $= 0.40 + x = 0.50$ , so $x = 0.10$ atm (1 mark).
$p_{A} = 0.40 - 0.10 = 0.30$ atm (equivalently $2 p_{i} - p_{t}$ ) (½ mark).
$k = \frac{2.303}{100} lo g \frac{0.40}{0.30} = 0.02303 \times 0.1249$ (1 mark).
$k = 2.88 \times 1 0^{- 3}$ s⁻¹ (½ mark).

Answer: $k \approx 2.88 \times 1 0^{- 3}$ s⁻¹.

Example 5: Zero order versus first order · 5 marks · Long answerHOTS / competency

Reactant R starts at 0.50 mol L⁻¹. (a) If the reaction is zero order with $k = 0.020$ mol L⁻¹ min⁻¹, find the time for complete reaction and the half-life. (b) If instead the reaction were first order with the same initial rate, find its rate constant and half-life, and explain why it never strictly reaches completion. (c) Sketch, in words, the $[R]$ – $t$ graphs for both.

Try it first, then show the solutionSolution

(a) $[R] = [R]_{0} - k t = 0$ at $t = \frac{0.50}{0.020} = 25$ min (1 mark). $t_{1/2} = \frac{[ R ] _{0}}{2 k} = 12.5$ min (1 mark).
(b) Same initial rate: $k [R]_{0} = 0.020$ , so $k = 0.040$ min⁻¹ (½ mark); $t_{1/2} = \frac{0.693}{0.040} = 17.3$ min (1 mark).
The first-order rate falls as $[R]$ falls ( $[R] = [R]_{0} e^{- k t}$ ), so $[R]$ approaches zero only as $t \to \infty$ (½ mark).
(c) Zero order: straight line from 0.50 down to zero at 25 min. First order: a curve that starts with the same slope but flattens, halving every 17.3 min (1 mark).

Answer: (a) 25 min; $t_{1/2} = 12.5$ min. (b) $k = 0.040$ min⁻¹, $t_{1/2} = 17.3$ min; the rate keeps falling with concentration. (c) Straight line vs exponential decay curve.

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

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

Mistake: Inverting the ratio inside the log
$lo g \frac{[ R ]}{[ R ] _{0}}$ is written, giving a negative $k$ .
Wrong working
$k = \frac{2.303}{40} lo g \frac{70}{100}$
Right working
$k = \frac{2.303}{40} lo g \frac{100}{70}$
Do this: Initial over remaining: the ratio is always greater than 1.
Usually costs: 1-2 marks.
Mistake: Using percentage reacted instead of percentage left
Wrong working
30% complete: $\frac{[ R ] _{0}}{[ R ]} = \frac{100}{30}$
Right working
30% complete leaves 70%: $\frac{[ R ] _{0}}{[ R ]} = \frac{100}{70}$
Do this: Write ' $[R]$ = what is left' before substituting.
Usually costs: Most of the numerical.
Mistake: Using the first-order half-life formula for zero order
Do this: Zero order: $t_{1/2} = \frac{[ R ] _{0}}{2 k}$ ; first order: $t_{1/2} = \frac{0.693}{k}$ . Check the units of $k$ .
Usually costs: 1 mark.
Mistake: Wrong pressure balance in gas-phase questions
Total pressure is used directly as the reactant's pressure.
Wrong working
$k = \frac{2.303}{100} lo g \frac{0.40}{0.50}$
Right working
$p_{A} = 2 p_{i} - p_{t} = 0.30$ atm; $k = \frac{2.303}{100} lo g \frac{0.40}{0.30}$
Do this: Always build the initial/change/final pressure table from the equation.
Usually costs: 2 marks.

Exam tips

Write the formula with $2.303$ and $lo g$ (base 10) as CBSE marking schemes do; log values are usually given.
Keep time units the same as asked (s, min or h) and label $k$ accordingly.
For 'show that' questions, write both expressions separately and then divide.
When data tables are given, calculate $k$ for at least two intervals to show it is constant.
Sketch graphs with labelled axes: $[R]$ vs $t$ (zero order, straight), $ln [R]$ vs $t$ (first order, straight, slope $- k$ ).

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, a plot of $ln [R]$ against time is:
(a) a straight line with positive slope
(b) a straight line with slope $- k$
(c) a curve
(d) a line parallel to the time axis
Hint
$ln [R] = ln [R]_{0} - k t$ .
Answer
(b).
Q2 · 2 marks · Very short answerBoard level
Write the integrated rate equation for a zero-order reaction and show that its half-life is $\frac{[ R ] _{0}}{2 k}$ .
Hint
Put $[R] = [R]_{0} /2$ .
Answer
$[R] = [R]_{0} - k t$ . At $t_{1/2}$ , $\frac{[ R ] _{0}}{2} = [R]_{0} - k t_{1/2}$ , so $t_{1/2} = \frac{[ R ] _{0}}{2 k}$ .
Q3 · 1 mark · Very short answerHarder board level
A first-order reaction has a half-life of 20 minutes. What fraction of the reactant remains after 1 hour?
Hint
Count half-lives.
Answer
60 min = 3 half-lives; fraction left $= (1/2)^{3} = 1/8$ (12.5%).
Q4 · 2 marks · Very short answerHarder board level
The rate constant of a first-order reaction is 0.0462 min⁻¹. How long will it take for the concentration to fall from 0.80 mol L⁻¹ to 0.20 mol L⁻¹? (log 4 = 0.6021)
Hint
Ratio $[R]_{0} / [R] = 4$ .
Answer
$t = \frac{2.303}{0.0462} lo g 4 = 49.85 \times 0.6021 = 30.0$ min. (Check: $t_{1/2} = 15$ min, and 0.80 → 0.20 is two half-lives.)
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The half-life of a first-order reaction does not depend on the initial concentration of the reactant.
Reason (R): For a first-order reaction, $t_{1/2} = 0.693/ k$ .
(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
Look for concentration in the formula.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
For the decomposition of A, the following data were recorded at constant temperature:
t = 0 s: [A] = 0.60 M; t = 30 s: [A] = 0.30 M; t = 60 s: [A] = 0.15 M.
Show that the reaction is first order and calculate the rate constant.
Hint
Compare successive half-lives.
Answer
[A] halves every 30 s whatever its starting value (0.60 → 0.30 → 0.15), so the half-life is constant: first order. $k = \frac{0.693}{30} = 0.0231$ s⁻¹. Check: $k = \frac{2.303}{60} lo g \frac{0.60}{0.15} = 0.0231$ s⁻¹.
Q7 · 3 marks · Short answerHOTS / competency
Show that for a first-order reaction the time required for 99% completion is twice the time required for 90% completion.
Hint
$\frac{[ R ] _{0}}{[ R ]}$ is 10 for 90% and 100 for 99%.
Answer
$t_{90%} = \frac{2.303}{k} lo g 10 = \frac{2.303}{k}$ . $t_{99%} = \frac{2.303}{k} lo g 100 = \frac{2 \times 2.303}{k}$ . So $t_{99%} = 2 t_{90%}$ .
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a patient is given a 400 mg dose of a drug that is removed from the body by a first-order process with a half-life of 4.0 hours. The drug stops being effective below 25 mg.
(a) Calculate the rate constant of elimination. (1 mark)
(b) How much drug remains after 12 hours? (1 mark)
(c) After how many hours does the drug stop being effective? (1 mark)
(d) If the dose were doubled to 800 mg, would the half-life change? Give a reason. (1 mark)
Hint
400 → 25 mg is a sixteen-fold drop.
Answer
(a) $k = \frac{0.693}{4.0} = 0.173$ h⁻¹. (b) 12 h = 3 half-lives: $400 \times \frac{1}{8} = 50$ mg. (c) $400/25 = 16 = 2^{4}$ : four half-lives, 16 hours. (d) No; for a first-order process $t_{1/2}$ is independent of the starting amount (it would just take one more half-life to fall below 25 mg).

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.First-order half-life formula?Show answer
$t_{1/2} = 0.693/ k$ .
2.Zero-order half-life formula?Show answer
$t_{1/2} = [R]_{0} /2 k$ .
3.Which plot is linear for a first-order reaction?Show answer
$ln [R]$ (or $lo g [R]$ ) against $t$ .
4.Fraction left after 4 half-lives?Show answer
1/16.
5.One example of a zero-order reaction?Show answer
Decomposition of $N H_{3}$ on hot platinum at high pressure.
6. $t_{99.9%}$ is about how many half-lives?Show answer
About 10.

Integrated Rate Equations and Half-Life: frequently asked questions

What is the integrated rate equation for a first-order reaction?

k = \frac{2.303}{t}\log\frac{[R]_0}{[R]}, equivalently [R] = [R]_0e^{-kt}.

Why is the half-life of a first-order reaction constant?

Because t_{1/2} = 0.693/k depends only on the rate constant, not on how much reactant you start with.

How can I tell zero order from first order using a graph?

If [R] against t is a straight line, it is zero order. If \ln[R] against t is a straight line, it is first order.

Is radioactive decay a first-order process?

Yes. All natural and artificial radioactive decay follows first-order kinetics.

Does a first-order reaction ever finish?

In principle it approaches zero only as time tends to infinity; in practice it is taken as complete after many half-lives.