First-Order Differential Equations — Maths for Thermodynamics & Kinetics

Equations for how things change

Where the integrated rate laws come from

A differential equation relates a quantity to its own rate of change. The most important first-order ODE in chemistry says “the rate is proportional to how much is left”:

\frac{d [ A ]}{d t} = - k [A]

To solve it, we use separation of variables: move all the $[A]$ terms to one side and all the $t$ terms to the other, then integrate both sides. The left integral gives a logarithm; the right gives $- k t + C$ . Exponentiating and applying the initial condition $[A] (0) = [A]_{0}$ yields the integrated rate law:

[A] = [A]_{0} e^{- k t}

The same three-step recipe (separate → integrate → apply initial condition) solves radioactive decay, Newton's law of cooling, and many other physical problems. The exponential decay shape is universal because the underlying equation is always $d y / d t = - k y$ .

Key first-order results

Integrated rate law

[A] = [A]_{0} e^{- k t}

or equivalently

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

Half-life (1st order)

t_{1/2} = \frac{ln 2}{k} = \frac{0.693}{k}

— independent of

[A]_{0}

Newton cooling

T (t) = T_{env} + (T_{0} - T_{env}) e^{- k t}

Radioactive decay

N (t) = N_{0} e^{- λ t}, t_{1/2} = ln 2/ λ

Kinetics lives here

Every integrated rate law is the solution of a differential equation. First order gives e^(−kt); the same maths covers radioactive decay and Newton's law of cooling. Open the separation solver to see the curve and the half-life update live.

Pseudo-first-order conditions arise when one reactant is present in large excess. For the bimolecular reaction A + B → products with $[B] ≫ [A]$ , the rate law $r = k [A] [B]$ simplifies to $r = k_{obs} [A]$ where $k_{obs} = k [B]_{0} \approx const$ . This converts a two-variable problem into a standard first-order ODE, giving a pseudo-first-order half-life $t_{1/2} = ln 2/ k_{obs}$ .

Parallel first-order reactions (A → B and A → C simultaneously) each with rate constants $k_{1}$ and $k_{2}$ lead to $d [A] / d t = - (k_{1} + k_{2}) [A]$ . The solution is still exponential with an effective rate constant $k_{eff} = k_{1} + k_{2}$ , and the product ratio is constant: $[B] / [C] = k_{1} / k_{2}$ at all times.

Second-order integrated law

\frac{1}{[ A ]} = \frac{1}{[ A ] _{0}} + k t, t_{1/2} = \frac{1}{k [ A ] _{0}}

Pseudo-first-order

k_{obs} = k [B]_{0}

when

[B] ≫ [A]

. Plot

ln [A]

t

to find

k_{obs}

Parallel reactions

[A] = [A]_{0} e^{- (k_{1} + k_{2}) t}, [B] / [C] = k_{1} / k_{2}

(ratio constant)

Radioactive series

Parent

N_{1} = N_{0} e^{- λ_{1} t}

; daughter

N_{2} = \frac{λ _{1} N _{0}}{λ _{2} - λ _{1}} (e^{- λ_{1} t} - e^{- λ_{2} t})

Common pitfalls

Forgetting the initial condition sets the integration constant. After integrating, you get $ln [A] = - k t + C$ ; the constant $C$ is not zero — it is $ln [A]_{0}$ , fixed by setting $t = 0$ .
Mixing up reaction order. A straight line in $ln [A]$ vs $t$ means first order; a straight line in $1/ [A]$ vs $t$ means second order. Do not confuse them.
Sign of k. The rate constant $k$ is always positive. The negative sign enters through the rate law $d [A] / d t = - k [A]$ . Writing $d [A] / d t = + k [A]$ would give exponential growth, not decay.
Half-life units. The half-life $t_{1/2} = ln 2/ k$ has units the reciprocal of $k$ . If $k$ is in s⁻¹, then $t_{1/2}$ is in s — not in mol L⁻¹ s⁻¹.

Separation solver

Choose a scenario and drag the rate constant — watch the exponential decay or cooling curve update alongside the three separation steps.

Scenario:

\frac{d [ A ]}{d t} = - k [A] \Rightarrow [A] = [A]_{0} e^{- k t}

Step 1separate: dy/y = −k dt

Step 2integrate: ln y = −kt + C

Step 3exponentiate + apply y(0)

rate constant k0.40

half-life t½1.73 s

The same equation — rate of change proportional to current amount — describes radioactive decay, first-order kinetics and a cooling cup of coffee. That's why e^(−kt) is everywhere.

Worked example 1Deriving the first-order integrated rate law

Starting from $d [A] / d t = - k [A]$ , use separation of variables to derive $[A] = [A]_{0} e^{- k t}$ and hence show that $ln [A] = ln [A]_{0} - k t$ is a straight line when plotted against time.

Worked example 2Half-life and rate constant for N₂O₅ decomposition

The gas-phase decomposition of N₂O₅ is first order with $k = 1.8 \times 1 0^{- 5} s^{- 1}$ at 25 °C. (a) Calculate the half-life. (b) After how long will $[N_{2} O_{5}]$ fall to 10% of its initial value?

Worked examples

Radioactive decay, Newton cooling, and second-order kinetics — try each before revealing the solution.

Worked example 3Radioactive decay of carbon-14

Carbon-14 decays with a half-life of 5730 years. A charcoal sample contains 72% of the $^{14} C$ activity of a living tree. Estimate the age of the sample.

Worked example 4Newton's law of cooling — time to safe temperature

A flask of reaction mixture at 95 °C is placed in a 20 °C water bath. The cooling constant is $k = 0.030 mi n^{- 1}$ . How long until the mixture reaches 37 °C (safe to handle)?

Worked example 5Second-order kinetics: dimerisation of NO₂

The dimerisation $2 N O_{2} \to N_{2} O_{4}$ is second order in NO₂ with $k = 0.54 L mo l^{- 1} s^{- 1}$ at 300 K. If $[N O_{2}]_{0} = 0.100 M$ , find $[N O_{2}]$ after 10.0 s and the half-life.

Worked example 6Pseudo-first-order hydrolysis of sucrose

The acid-catalysed hydrolysis of sucrose (C₁₂H₂₂O₁₁) in excess water is pseudo-first-order. At 298 K with $[H^{+}] = 0.50 M$ , the pseudo-first-order rate constant is $k_{obs} = 6.2 \times 1 0^{- 3} mi n^{- 1}$ . How long until 75% of the sucrose has reacted?

Worked example 7Parallel first-order reactions — product ratio

Azomethane (AZM) decomposes via two parallel first-order pathways: AZM → ethane + N₂ ( $k_{1} = 2.5 \times 1 0^{- 4} s^{- 1}$ ) and AZM → methane + diazomethane ( $k_{2} = 1.5 \times 1 0^{- 4} s^{- 1}$ ) at 600 K. (a) Find the effective half-life of AZM. (b) What fraction of product comes from pathway 1?

ChallengeChallenge — consecutive reactions A → B → C: concentration of B(t)

Species A decays to B with rate constant $k_{1}$ , and B decays to C with rate constant $k_{2}$ . Starting from pure A ( $[A]_{0}$ , no B or C), derive $[B] (t)$ using an integrating factor.

Check yourself

Four quick questions tying separation of variables to chemical kinetics.

Question 1 of 4 · Score 0

A first-order differential equation relates a function to its:

Choose an answer.