Integrated Rate Law: Second Order

Picture two molecules of NO₂ colliding to make NO and O₂. The rate depends on the chance that two NO₂ molecules meet, not just one. That single fact changes the whole shape of the concentration-versus-time curve. Second-order reactions are common in gas-phase decompositions, radical recombination, and many organic reactions like ester saponification. Recognizing second-order behavior lets a chemist predict how long a reactant will last, extract a rate constant from raw data, and compare mechanisms across different reactions. This article builds the integrated rate law for a single second-order reactant. It then extends the treatment to two different reactants. You will see how to linearize the data. You will learn how the half-life changes as the reaction proceeds. Finally, you will work two full numeric examples end to end.

Key takeaways

For $A \to$ products with rate $= k[A]^2$ , the reciprocal concentration grows linearly with time.
The diagnostic plot is $1/[A]$ versus $t$ ; a straight line with slope $k$ confirms second order.
Unlike first order, the second-order half-life is not constant — it depends on the starting concentration.
For two reactants $A + B \to$ products with unequal concentrations, the integrated law involves a logarithm of a concentration ratio.
When $[A]_0 = [B]_0$ , the two-reactant case collapses back to the single-reactant second-order form.
Key idea: Second order means the rate depends on the square of one concentration or the product of two, so the reciprocal of concentration — not its logarithm — marches linearly with time.

The core insight

Because the rate of a second-order reaction falls off as concentration squared, integrating $-d[A]/dt = k[A]^2$ produces a term in $1/[A]$ rather than $\ln[A]$ ; this reciprocal dependence is the mathematical fingerprint that distinguishes second order from first order and dictates that its half-life lengthens as the reaction proceeds, in contrast to the constant half-life of first-order kinetics.

The physical picture

Rate laws describe how fast concentration changes right now (see Rate Laws and Reaction Order). Integrated rate laws answer a different, often more practical, question: given a starting concentration, what is the concentration at any later time? For a first-order reaction, one molecule decides to react on its own, so the rate constant carries units of inverse time and the decay is exponential. A second-order reaction requires two reactant particles to interact — either two molecules of the same species colliding, or one molecule of $A$ meeting one of $B$ . Because the rate depends on a product of two concentrations (or a concentration squared), the depletion of reactant is more sluggish at high concentration and speeds up relatively less as concentration falls, compared with first order. The signature consequence: plotting $1/[A]$ against time gives a straight line, not $\ln[A]$ . This has a direct physical meaning. The "encounter frequency" between reactant particles scales with the product of their number densities. Halve the concentration of $A$ and you don't just halve the rate — you quarter it (for $\text{rate} = k[A]^2$ ). This is why second-order reactions dilute out much faster at high concentration but linger stubbornly once concentration drops.

:::svg[Reciprocal concentration versus time for a second-order decay, k = 0.10 M⁻¹s⁻¹, [A]₀ = 1.0 M]

Figure

Concentration-time curves compared: first order versus second order, same initial rate

The core relations

-\frac{d[A]}{dt} = k[A]^2

where: $k$ is the second-order rate constant with units of M⁻¹ s⁻¹ (or M⁻¹ time⁻¹).

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

where: $[A]_0$ is the initial concentration and $[A]_t$ is the concentration at time $t$ .

t_{1/2} = \frac{1}{k[A]_0}

where: the second-order half-life depends inversely on the starting concentration, unlike first order.

\frac{1}{[A]_0-[B]_0}\ln\!\left(\frac{[B]_0[A]_t}{[A]_0[B]_t}\right) = kt

where: this is the two-reactant case for $A+B\to$ products with rate $=k[A][B]$ and $[A]_0 \ne [B]_0$ .

The core relations — derivations

:::derivation[Single-reactant second order: 1/[A] versus t] Goal. Integrate the rate law $-d[A]/dt = k[A]^2$ to obtain an explicit relation between $[A]$ and $t$ .

Step 1. Start from the differential rate law for the disappearance of $A$ in $A \to$ products:

-\frac{d[A]}{dt} = k[A]^2

Step 2. Separate variables, putting all concentration terms on one side and time on the other:

-\frac{d[A]}{[A]^2} = k\,dt

Step 3. Integrate both sides. The left side runs from $[A]_0$ at $t=0$ to $[A]_t$ at time $t$ ; the right side runs from $0$ to $t$ :

\int_{[A]_0}^{[A]_t} -\frac{d[A]}{[A]^2} = \int_0^t k\,dt

Step 4. The antiderivative of $-1/[A]^2$ with respect to $[A]$ is $1/[A]$ , so evaluating the limits gives:

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

Step 5. Rearranging puts the result in the standard linear form, $y = mx + b$ , with $y = 1/[A]_t$ , slope $k$ , and intercept $1/[A]_0$ :

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

Derivation AConcentration-dependent half-life for second order

Goal. Find the time at which $[A]_t$ has fallen to half its initial value, and show that this time is not constant.

Step 1. Substitute $[A]_t = \frac{1}{2}[A]_0$ into the integrated law from the previous derivation:

\frac{1}{\tfrac{1}{2}[A]_0} = \frac{1}{[A]_0} + k\,t_{1/2}

Step 2. Simplify the left side:

\frac{2}{[A]_0} = \frac{1}{[A]_0} + k\,t_{1/2}

Step 3. Subtract $1/[A]_0$ from both sides:

\frac{2}{[A]_0} - \frac{1}{[A]_0} = k\,t_{1/2} \quad\Longrightarrow\quad \frac{1}{[A]_0} = k\,t_{1/2}

Step 4. Solve for $t_{1/2}$ , arriving at the boxed result showing explicit dependence on $[A]_0$ :

\boxed{t_{1/2} = \dfrac{1}{k[A]_0}}

This means each successive half-life is twice as long as the one before it (since $[A]_0$ halves each time), the opposite trend from first-order kinetics where $t_{1/2}$ is constant.

Go deeper · Derivation B · 1 minTwo unequal reactants: A + B → products

Derivation BTwo unequal reactants: A + B → products

Goal. Integrate $-d[A]/dt = k[A][B]$ when $[A]_0 \ne [B]_0$ , using stoichiometry to eliminate one variable.

Step 1. Since one $A$ and one $B$ are consumed together, define the extent variable $x$ such that $[A]_t = [A]_0 - x$ and $[B]_t = [B]_0 - x$ . The rate law becomes:

\frac{dx}{dt} = k([A]_0-x)([B]_0-x)

Step 2. Separate variables:

\frac{dx}{([A]_0-x)([B]_0-x)} = k\,dt

Step 3. Apply partial fractions. Because $[A]_0 \ne [B]_0$ , write:

\frac{1}{([A]_0-x)([B]_0-x)} = \frac{1}{[B]_0-[A]_0}\left(\frac{1}{[A]_0-x} - \frac{1}{[B]_0-x}\right)

Step 4. Integrate both sides from $x=0$ to $x=x$ (equivalently $t=0$ to $t=t$ ). Each term integrates to a natural logarithm:

\frac{1}{[B]_0-[A]_0}\left[-\ln([A]_0-x) + \ln([B]_0-x)\right]_0^x = kt

Step 5. Evaluate the limits and combine the logarithms into a single ratio, then substitute back $[A]_t = [A]_0-x$ and $[B]_t = [B]_0-x$ :

\frac{1}{[B]_0-[A]_0}\ln\!\left(\frac{[B]_t/[A]_t}{[B]_0/[A]_0}\right) = kt

Step 6. Rearranging the sign convention (multiplying numerator and denominator inside the log by $-1$ swaps $[A]_0-[B]_0$ into the front) gives the standard textbook form:

\boxed{\dfrac{1}{[A]_0-[B]_0}\ln\!\left(\dfrac{[B]_0[A]_t}{[A]_0[B]_t}\right) = kt}

As a check, if $[A]_0 \to [B]_0$ , this expression becomes $0/0$ ; applying L'Hôpital's rule recovers exactly the single-reactant result $1/[A]_t - 1/[A]_0 = kt$ , confirming internal consistency.

Worked examples

Example 1 — Foundational: dimerization of a diene. A gas-phase reaction $2\,\text{NOBr} \to 2\,\text{NO} + \text{Br}_2$ obeys second-order kinetics in NOBr with $k = 0.80\ \text{M}^{-1}\text{s}^{-1}$ at 300 K. Starting from $[\text{NOBr}]_0 = 0.0250\ \text{M}$ , find (a) the concentration after 90 s, and (b) the half-life at this starting concentration. Step 1 — set up the integrated law.

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

Step 2 — plug in numbers.

\frac{1}{[A]_0} = \frac{1}{0.0250\ \text{M}} = 40.0\ \text{M}^{-1}

kt = (0.80\ \text{M}^{-1}\text{s}^{-1})(90\ \text{s}) = 72.0\ \text{M}^{-1}

\frac{1}{[A]_{90}} = 40.0\ \text{M}^{-1} + 72.0\ \text{M}^{-1} = 112.0\ \text{M}^{-1}

Step 3 — invert.

[A]_{90} = \frac{1}{112.0\ \text{M}^{-1}} = 8.93\times10^{-3}\ \text{M}

Sanity check: the concentration dropped from 0.0250 M to 0.00893 M — well over half gone, consistent with $kt=72$ being much larger than the initial reciprocal contribution of 40. Step 4 — half-life.

t_{1/2} = \frac{1}{k[A]_0} = \frac{1}{(0.80\ \text{M}^{-1}\text{s}^{-1})(0.0250\ \text{M})} = \frac{1}{0.0200\ \text{s}^{-1}} = 50.0\ \text{s}

So after 90 s (1.8 half-lives) the concentration should be a bit less than a quarter of the original; $8.93\times10^{-3}/0.0250 = 0.357$ , which is indeed between $1/2$ and $1/4$ — consistent, since half-lives lengthen as reaction proceeds (unlike first order where 1.8 half-lives always gives $2^{-1.8}=0.287$ regardless of starting value).

Example 2 — Applied, two unequal reactants: saponification of an ester. Ethyl acetate reacts with hydroxide, $\text{CH}_3\text{COOC}_2\text{H}_5 + \text{OH}^- \to \text{CH}_3\text{COO}^- + \text{C}_2\text{H}_5\text{OH}$ , with rate $= k[\text{ester}][\text{OH}^-]$ and $k = 6.3\ \text{M}^{-1}\text{min}^{-1}$ at 25 °C. A reaction mixture starts with $[\text{ester}]_0 = 0.0500\ \text{M}$ and $[\text{OH}^-]_0 = 0.0250\ \text{M}$ . Find the time required for the ester concentration to fall to $0.0350\ \text{M}$ . Step 1 — use stoichiometry to find $[\text{OH}^-]_t$ . Every mole of ester consumed removes one mole of hydroxide:

x = [\text{ester}]_0 - [\text{ester}]_t = 0.0500 - 0.0350 = 0.0150\ \text{M}

[\text{OH}^-]_t = 0.0250 - 0.0150 = 0.0100\ \text{M}

Step 2 — apply the two-reactant integrated law.

\frac{1}{[A]_0-[B]_0}\ln\!\left(\frac{[B]_0[A]_t}{[A]_0[B]_t}\right) = kt

with $A=$ ester, $B=\text{OH}^-$ : $[A]_0-[B]_0 = 0.0500-0.0250 = 0.0250\ \text{M}$ . Step 3 — evaluate the argument of the logarithm.

\frac{[B]_0[A]_t}{[A]_0[B]_t} = \frac{(0.0250)(0.0350)}{(0.0500)(0.0100)} = \frac{8.75\times10^{-4}}{5.00\times10^{-4}} = 1.750

\ln(1.750) = 0.5596

Step 4 — solve for $t$ .

\frac{0.5596}{0.0250\ \text{M}} = kt = (6.3\ \text{M}^{-1}\text{min}^{-1})\,t

22.38\ \text{M}^{-1} = (6.3\ \text{M}^{-1}\text{min}^{-1})\,t

t = \frac{22.38\ \text{M}^{-1}}{6.3\ \text{M}^{-1}\text{min}^{-1}} = 3.55\ \text{min}

Sanity check: hydroxide, the minority reagent, drops from 0.0250 M to 0.0100 M — a 60% depletion — while ester drops only 30%, exactly as expected since equal moles are consumed but hydroxide started with less. The answer, 3.6 minutes, is a physically reasonable timescale for a rate constant of order 1–10 M⁻¹min⁻¹ near millimolar concentrations.

Limiting cases & common traps

Situation	What happens	Watch out for
$[A]_0 = [B]_0$ in the two-reactant law	Reduces exactly to the single-reactant form; the general formula gives 0/0	Do not plug equal concentrations directly into the boxed two-reactant equation — use the single-reactant form instead
Very dilute solution	$t_{1/2}$ becomes very long since $t_{1/2} \propto 1/[A]_0$	Do not assume a "fixed" half-life the way you would for first order
Very concentrated solution	Reaction depletes rapidly at first, then half-life stretches out as $[A]_0$ drops	Early-time data can look deceptively fast; always check linearity of $1/[A]$ vs $t$ , not just an initial rate
Units of $k$	M⁻¹ time⁻¹, not time⁻¹	Mixing up first- and second-order $k$ units is the single most common transcription error
Plotting the wrong axis	Plotting $\ln[A]$ vs $t$ for second-order data	Gives a curve, not a line — confirms the reaction is NOT first order

The chemist behind it

The reciprocal-concentration method for second-order kinetics grew out of the broader 19th- and early 20th-century effort to quantify reaction rates, following the pioneering rate studies of Jacobus van 't Hoff and Svante Arrhenius. But it was the meticulous kinetic measurements of Augustus George Vernon Harcourt and William Esson in the 1860s — analyzing the reaction between hydrogen peroxide and hydrogen iodide — that first established the mathematical machinery for treating reaction orders systematically, laying groundwork later formalized into the integrated rate laws used today, including the second-order case central to this article.

Origins · 1 minAugustus George Vernon Harcourt and how this was discovered

Augustus George Vernon Harcourt

1834-1919 · Pioneered systematic kinetic rate measurements and co-developed the mathematical treatment of reaction order with William Esson

:::quiz[A reaction obeys rate = k[A]^2. If [A]₀ is doubled, what happens to the initial half-life?]

It doubles
It stays the same
It is cut in half ✓
It quadruples

:::quiz[For A + B → products with rate = k[A][B] and [A]₀ ≠ [B]₀, what happens to the general integrated rate law as [A]₀ approaches [B]₀?]

It becomes undefined and has no limiting form
It reduces to the single-reactant second-order law via L'Hôpital's rule ✓
It becomes first order instead
The rate constant changes units

:::faq[Why does 1/[A] increase linearly instead of [A] decreasing linearly?] Integrating $-d[A]/dt = k[A]^2$ produces an antiderivative of $1/[A]^2$ , which is $1/[A]$ , not $[A]$ itself. This falls directly out of calculus: the power rule for $\int [A]^{-2}\,d[A]$ gives $-[A]^{-1}$ . Physically, as $[A]$ shrinks, its reciprocal grows, and the second-order rate law happens to make that growth perfectly linear in time — a mathematical coincidence of the specific exponent 2 that does not hold for other reaction orders.

Ask the tutor: How does the second-order half-life formula change if the reaction is third order instead?

How can I tell from data whether a reaction is first or second order?

Plot both $\ln[A]$ versus $t$ and $1/[A]$ versus $t$ using the same concentration-time data. Whichever plot comes out as a straight line reveals the true order — first order gives a straight line for $\ln[A]$ , second order gives a straight line for $1/[A]$ . This graphical method is one of several standard tools; the article on Methods for Determining Reaction Order expands on isolation and initial-rates techniques too.

Why does the second-order half-life get longer as the reaction proceeds?

Because $t_{1/2} = 1/(k[A]_0)$ depends explicitly on whatever concentration you start each half-life interval from. As the reaction proceeds, $[A]$ drops, so each successive half-life uses a smaller " $[A]_0$ " and therefore takes proportionally longer. This contrasts sharply with first-order kinetics, where $t_{1/2} = \ln 2 / k$ contains no concentration term at all and is constant throughout the reaction.

Does the two-reactant formula work if the stoichiometric coefficients are not both 1?

The boxed two-reactant law assumes a 1:1 stoichiometry between A and B, so that a single extent variable $x$ tracks both concentrations symmetrically. For reactions like $2A+B\to$ products, you must rewrite the rate law with the correct stoichiometric factors before separating variables, since $d[A]/dt \ne d[B]/dt$ in that case. The general method of partial fractions still applies, but the specific boxed formula given here needs modification first.

What real reactions actually follow simple second-order kinetics?

Classic textbook examples include the gas-phase decomposition of NOBr and NO₂, radical-radical recombination such as two •CH₃ radicals combining to ethane, and saponification of esters by hydroxide ion in solution. In practice, many reactions that look second order over a limited concentration range are really more complex mechanisms whose rate-determining step happens to be bimolecular; the integrated law here describes the overall kinetics only if that simple picture holds throughout the run.