Chemistry · Reaction kinetics · integrated rate laws

Integrated First-Order Rate Law

Q: How do I calculate first-order half-life?

Set [A]_t=[A]_0/2 in the integrated equation. This gives t_1/2=ln(2)/k. With k=0.20 min⁻¹, half-life is 0.693/0.20=3.47 min. It does not depend on [A]_0 as long as the reaction remains first-order and k is constant. This distinguishes it from many zero- or second-order reactions. Check units: if k is in s⁻¹, t_1/2 comes out in seconds. Check whether concentration data over the studied time window are consistent with a straight line in a log-concentration-versus-time plot. This supports the model choice but alone does not prove a molecular mechanism or apply automatically beyond the tested time range.

The integrated first-order rate law calculates how reactant concentration decreases over time when the rate is proportional to its concentration.

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Formula

ln \frac{[ A ] _{t}}{[ A ] _{0}} = - k t

k in s⁻¹ or another inverse-time unit · t in the matching time unit · [A] in the same concentration unit

Variables & units

Variables & units
Symbol	Meaning	Unit
$[A]_{\mathrm{t}}$	Concentration of A at time t	mol/L
$[A]_0$	Initial concentration of A at t=0	mol/L
$k$	First-order rate constant	s⁻¹
$t$	Elapsed reaction time	s

LaTeX

\ln \frac{[A]_t}{[A]_0} = -k t

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Derivation & background

Integrating the differential rate law −d[A]/dt=k[A] gives ln([A]_t/[A]_0)=−kt or [A]_t=[A]_0e^(−kt). A plot of ln[A] against t is linear with slope −k. The half-life is t₁/2=ln 2/k and does not depend on initial concentration. This applies to a first-order reaction over the range considered, such as a unimolecular step or pseudo-first-order kinetics when another reactant is in large excess. Concentrations must remain positive; the equation alone does not identify the mechanism from measurements.

Exam blueprint

Validity range

Applies to v=k[A] at constant temperature for a dominant first-order step; pseudo-first order may apply when another reactant is in large excess.

Derivation steps

Integrating the proportional decay gives a linear relation between ln[A] and time.

1From −d[A]/dt=k[A], d[A]/[A]=−k dt.
2Integrating from [A]₀ to [A]_t and 0 to t gives ln([A]_t/[A]₀)=−kt.

Rearrangements

Concentration after time

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

Concentration decays exponentially and remains positive.

Half-life

t_{1/2}=\frac{\ln 2}{k}

For first order, it is independent of [A]₀.

Task variant

k=0.231 min⁻¹, [A]₀=0.100 mol/L. Find [A] after 10 min.

[A]=0.100e^(−2.31)=0.00994 mol/L; ln(0.00994/0.100)=−2.31=−kt.

What is t₁/₂ for k=0.231 min⁻¹?

t₁/₂=0.693/0.231=3.00 min.

Common mistakes

Mixing minutes in t with seconds⁻¹ in k.

Time and rate constant must use reciprocal units.

Applying the half-life formula to other reaction orders.

t₁/₂=ln2/k applies only to first order; for second order it depends on [A]₀.

Exam context

Determining k from concentration-time data, calculating half-life and interpreting a linear ln[A]-time plot.

Formula cluster

Kinetics and integrated rate laws

The integrated law connects concentration and reaction time for a first-order reaction.

v = k·[A]^m·[B]^n

Rate Law (Reaction Order)

k = A·e^(−EA/RT)

Arrhenius Equation

v₂ = v₁·Q₁₀^(ΔT/10 K)

RGT Rule (Q10 Rule)

Worked example

For k=0.231 min⁻¹ and [A]_0=0.100 mol/L after t=10.0 min, [A]_t=0.100·e^(−2.31)=0.00994 mol/L. Check: ln(0.00994/0.100)=−2.31=−0.231·10.0. The half-life is ln 2/0.231≈3.00 min.

Applications

Degradation and decay processes, reaction kinetics, pharmacokinetic concentration decrease and estimating k from time series

How do you calculate with Integrated First-Order Rate Law?

Here is how to work through a typical Integrated First-Order Rate Law (ln([A]ₜ/[A]₀) = −kt) task step by step:

1
Task
k=0.231 min⁻¹, [A]₀=0.100 mol/L. Find [A] after 10 min.
Solution path
[A]=0.100e^(−2.31)=0.00994 mol/L; ln(0.00994/0.100)=−2.31=−kt.
2
Task
What is t₁/₂ for k=0.231 min⁻¹?
Solution path
t₁/₂=0.693/0.231=3.00 min.

Frequently asked questions about Integrated First-Order Rate Law

What is the integrated first-order rate law?+

For decay A→products with rate constant k, ln([A]_t/[A]_0)=−kt. Equivalently, [A]_t=[A]_0e^(−kt). With k=0.20 min⁻¹ and t=5.0 min, the remaining fraction is e^(−1)=0.368; an initial 0.50 mol/L becomes about 0.184 mol/L. k has units of inverse time, and t must use the same time unit. The law assumes first-order kinetics and constant conditions. Check whether concentration data over the studied time window are consistent with a straight line in a log-concentration-versus-time plot. This supports the model choice but alone does not prove a molecular mechanism or apply automatically beyond the tested time range.

How can I recognize first-order kinetics in data?+

Plot ln([A]) against time. If data are approximately linear over the period considered, this supports the integrated first-order form; the slope is −k. Alternatively, check whether half-life remains independent of initial concentration. Measurement error, multiple pathways or a mechanism change can limit linearity. A straight plot alone does not prove the mechanism, but it is a useful kinetic test. Use concentrations from the same measurement method and account for detection limits. Check whether concentration data over the studied time window are consistent with a straight line in a log-concentration-versus-time plot. This supports the model choice but alone does not prove a molecular mechanism or apply automatically beyond the tested time range.

How do I calculate first-order half-life?+

Set [A]_t=[A]_0/2 in the integrated equation. This gives t₁/2=ln(2)/k. With k=0.20 min⁻¹, half-life is 0.693/0.20=3.47 min. It does not depend on [A]_0 as long as the reaction remains first-order and k is constant. This distinguishes it from many zero- or second-order reactions. Check units: if k is in s⁻¹, t₁/2 comes out in seconds. Check whether concentration data over the studied time window are consistent with a straight line in a log-concentration-versus-time plot. This supports the model choice but alone does not prove a molecular mechanism or apply automatically beyond the tested time range.

How do I determine k from two concentration measurements?+

Rearrange ln([A]_t/[A]_0)=−kt to get k=(1/t)ln([A]_0/[A]_t). For example, if concentration falls from 0.80 to 0.40 mol/L in 10 min, k=ln(2)/10=0.0693 min⁻¹. Both concentrations must use the same unit so their ratio is dimensionless. More measurements are usually better, because linear regression of ln([A]) against t averages measurement noise. Check whether background signal or reversibility distorts the apparent decay. Check whether concentration data over the studied time window are consistent with a straight line in a log-concentration-versus-time plot. This supports the model choice but alone does not prove a molecular mechanism or apply automatically beyond the tested time range.

When does the first-order equation not apply?+

It does not apply when rate is not proportional to concentration of A or k changes during measurement. A reaction with multiple reactants can be pseudo-first-order under conditions such as one partner being in large excess and effectively constant. Reversible reactions also often approach equilibrium rather than decay indefinitely. Check concentration-time data, temperature control and mechanism. An apparently good exponential model over a short measurement window need not hold outside that range. Check whether concentration data over the studied time window are consistent with a straight line in a log-concentration-versus-time plot. This supports the model choice but alone does not prove a molecular mechanism or apply automatically beyond the tested time range.

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Which formula describes Integrated First-Order Rate Law?

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How do you rearrange Integrated First-Order Rate Law for Concentration after time?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax. Chemistry 2e, Abschnitt 12.4: Integrated Rate Laws (Integration einer Reaktion erster Ordnung).

Related formulas

[A]_t=[A]_0-kt

Integrated Zero-Order Rate Law

t_{1/2}^{(0)}=\frac{[A]_0}{2k};\quad t_{1/2}^{(1)}=\frac{\ln2}{k};\quad t_{1/2}^{(2)}=\frac{1}{k[A]_0}

Half-Life by Reaction Order

More formulas: Reaction kinetics

\frac{1}{[A]_t}=\frac{1}{[A]_0}+kt

Integrated Second-Order Rate Law

k=\frac{k_BT}{h}e^{-\Delta G^\ddagger/(RT)}

Eyring Equation

\frac{d[I]}{dt}\approx0\Rightarrow[I]\approx\frac{k_1[A][B]}{k_{-1}+k_2[C]}

Steady-State Approximation for a Reaction Intermediate

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Integrated First-Order Rate Law

The integrated first-order rate law calculates how reactant concentration decreases over time when the rate is proportional to its concentration.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

[A]_{\mathrm{t}}

Concentration of A at time t

mol/L

[A]_0

Initial concentration of A at t=0

mol/L

k

First-order rate constant

s⁻¹

t

Elapsed reaction time

Derivation & background

Exam blueprint

Validity range

Applies to v=k[A] at constant temperature for a dominant first-order step; pseudo-first order may apply when another reactant is in large excess.

Derivation steps

Integrating the proportional decay gives a linear relation between ln[A] and time.

1From −d[A]/dt=k[A], d[A]/[A]=−k dt.
2Integrating from [A]₀ to [A]_t and 0 to t gives ln([A]_t/[A]₀)=−kt.

How do you calculate with Integrated First-Order Rate Law?

Here is how to work through a typical Integrated First-Order Rate Law (ln([A]ₜ/[A]₀) = −kt) task step by step:

Task

k=0.231 min⁻¹, [A]₀=0.100 mol/L. Find [A] after 10 min.

Solution path

[A]=0.100e^(−2.31)=0.00994 mol/L; ln(0.00994/0.100)=−2.31=−kt.

Task

What is t₁/₂ for k=0.231 min⁻¹?

Solution path

t₁/₂=0.693/0.231=3.00 min.

Frequently asked questions about Integrated First-Order Rate Law

What is the integrated first-order rate law?+

How can I recognize first-order kinetics in data?+

How do I calculate first-order half-life?+

How do I determine k from two concentration measurements?+

When does the first-order equation not apply?+

Retain Integrated First-Order Rate Law for exams

Create a curated FSRS exam set for ln([A]ₜ/[A]₀) = −kt: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential