= Second-order reaction
A <second-order reaction> has overall <order of reaction> two: a <rate law> may depend on $[\mathrm A]^2$, or on $[\mathrm A][\mathrm B]$ with partial <orders of reaction> one and one. Doubling all relevant <reactant> <concentrations> multiplies the <reaction rate> by four at fixed conditions. This experimentally determined dependence does not establish <molecularity> or a unique <reaction mechanism>. For an <elementary reaction> obeying the <law of mass action>, <molecularity> and overall <order of reaction> agree; a net <chemical equation> alone cannot establish that the conversion is elementary.<Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994): N and O entries by International Union of Pure and Applied Chemistry>[Glossary of Terms Used in Physical Organic Chemistry, IUPAC Recommendations 1994, entry “order of reaction”, final paragraphs: elementary versus stepwise reactions]{ref}
For the single-<concentration> model, explicitly choose the <disappearance rate> convention
$$
-\frac{d[\mathrm A]}{dt}=k_{\mathrm{obs}}[\mathrm A]^2,\qquad [\mathrm A](0)=[\mathrm A]_0>0,\qquad k_{\mathrm{obs}}>0.
$$
Here $k_{\mathrm{obs}}$ is the coefficient for loss of A, held fixed at the specified <temperature> and other conditions. Its subscript labels the chosen observed <disappearance rate>; it does not imply a <pseudo-first-order reaction>. Assume a uniformly mixed <closed system>, fixed <volume>, and no appreciable reverse or competing <chemical reactions>. The <integrated rate law> is
$$
\boxed{\frac1{[\mathrm A](t)}=\frac1{[\mathrm A]_0}+k_{\mathrm{obs}}t},\qquad
[\mathrm A](t)=\frac{[\mathrm A]_0}{1+k_{\mathrm{obs}}[\mathrm A]_0t}.
$$
The reciprocal-<concentration> plot is a straight line with slope $k_{\mathrm{obs}}$ and intercept $1/[\mathrm A]_0$. A brief derivation uses the <chain rule>: $d(1/[\mathrm A])/dt=-[\mathrm A]^{-2}d[\mathrm A]/dt=k_{\mathrm{obs}}$, then applies the <initial condition>. This is a hyperbolic decay, rather than the <exponential decay> of a <first-order reaction>.<Chemistry 2e>[§12.4, “Second-Order Reactions”, integrated equation and straight-line form; Table 12.2]{ref}
As an original prescribed example, choose $[\mathrm A]_0=0.020\,\mathrm{mol\,L^{-1}}$ and $k_{\mathrm{obs}}=0.50\,\mathrm{L\,mol^{-1}\,s^{-1}}$. At $t=100\,\mathrm s$, the reciprocal-<concentration> calculation is $50+50=100\,\mathrm{L\,mol^{-1}}$, giving $[\mathrm A]=0.010\,\mathrm{mol\,L^{-1}}$. The initial <disappearance rate> is $0.50(0.020)^2=2.0\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}$; after $100\,\mathrm s$ it is $5.0\times10^{-5}\,\mathrm{mol\,L^{-1}\,s^{-1}}$. Halving the <concentration> quarters the <disappearance rate>.
<Dimensional analysis> checks every formula: $k_{\mathrm{obs}}[\mathrm A]^2$ has units $\mathrm{mol\,L^{-1}\,s^{-1}}$, $k_{\mathrm{obs}}t$ has units $\mathrm{L\,mol^{-1}}$, and $k_{\mathrm{obs}}[\mathrm A]_0t$ is a <dimensionless quantity>. Changing <concentration> units changes the numerical <rate constant>: the same example is $[\mathrm A]_0=20\,\mathrm{mol\,m^{-3}}$, $k_{\mathrm{obs}}=5.0\times10^{-4}\,\mathrm{m^3\,mol^{-1}\,s^{-1}}$, giving $[\mathrm A](100\,\mathrm s)=10\,\mathrm{mol\,m^{-3}}$.
\Image[/second-order-concentration-and-reciprocal-time.png]
{title=<Second-order reaction> <concentration> decay and reciprocal-<concentration> lines for three prescribed starting <concentrations>}
{id=image-second-order-concentration-and-reciprocal-time}
{description=Original model curves, not measured data. All use the <disappearance rate> coefficient $k_{\mathrm{obs}}=0.50\,\mathrm{L\,mol^{-1}\,s^{-1}}$. The reciprocal lines share a slope while their intercepts differ; marked points give each run's first <half-life>.}
For an <elementary reaction> $2\mathrm A\longrightarrow\mathrm P$, instead define the stoichiometrically normalized <reaction rate> $r=k[\mathrm A]^2$. Two A entities disappear per P formed, so
$$
r=-\frac12\frac{d[\mathrm A]}{dt}=\frac{d[\mathrm P]}{dt}=k[\mathrm A]^2,\qquad
-\frac{d[\mathrm A]}{dt}=2k[\mathrm A]^2,\qquad k_{\mathrm{obs}}=2k.
$$
Consequently the reciprocal-<concentration> slope is $2k$, not $k$. In the prescribed example, the normalized <rate constant> would be $k=0.25\,\mathrm{L\,mol^{-1}\,s^{-1}}$; initially $r=1.0\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}$ while the <disappearance rate> is twice that. With $[\mathrm P](0)=0$, <stoichiometry> gives $[\mathrm P](100\,\mathrm s)=([\mathrm A]_0-[\mathrm A])/2=0.0050\,\mathrm{mol\,L^{-1}}$. Both <rate constants> have the same dimensions, so a units check alone cannot catch their factor-of-two difference. The symbol $r$ denotes the same normalization as $v$ in <reaction rate>.
The <half-life>, meaning the time to reach half a specified starting <concentration>, follows by setting $[\mathrm A]=[\mathrm A]_0/2$:
$$
\boxed{t_{1/2}=\frac1{k_{\mathrm{obs}}[\mathrm A]_0}}.
$$
For $r=k[\mathrm A]^2$ in $2\mathrm A\longrightarrow\mathrm P$, this becomes $t_{1/2}=1/(2k[\mathrm A]_0)$. Its units are seconds because $k_{\mathrm{obs}}[\mathrm A]_0$ has units $\mathrm{s^{-1}}$. Doubling the starting <concentration> halves the first <half-life>; a quoted <half-life> needs its starting <concentration>. This differs from a <first-order reaction>, whose <half-life> is independent of starting <concentration>.<Chemistry 2e>[§12.4, “Half-Life of a Reaction”, second-order derivation and Table 12.2]{ref}
Within the prescribed run, $[\mathrm A]$ reaches $0.010$, $0.0050$, and $0.0025\,\mathrm{mol\,L^{-1}}$ at $100$, $300$, and $700\,\mathrm s$. The successive halving intervals are $100$, $200$, and $400\,\mathrm s$: each begins at a smaller <concentration>. More generally, reaching fraction $f$ of the initial <concentration>, with $0<f<1$, requires $t=(f^{-1}-1)/(k_{\mathrm{obs}}[\mathrm A]_0)$, so $90\%$ consumption takes nine initial <half-lives>, rather than a fixed sequence of equal waiting times.
\Image[/second-order-halving-intervals-and-initial-concentration.png]
{title=Lengthening <second-order reaction> halving intervals and the dependence of the first <half-life> on initial <concentration>}
{id=image-second-order-halving-intervals-and-initial-concentration}
{description=Original model predictions, not measurements. Left: the prescribed $0.020\,\mathrm{mol\,L^{-1}}$ run. Right: $t_{1/2}=1/(k_{\mathrm{obs}}[\mathrm A]_0)$ at fixed $k_{\mathrm{obs}}=0.50\,\mathrm{L\,mol^{-1}\,s^{-1}}$, compared with a <first-order reaction> chosen to have $t_{1/2}=100\,\mathrm s$. That comparison uses $k_1=\ln2/(100\,\mathrm s)$, with different units from $k_{\mathrm{obs}}$.}
A straight reciprocal-<concentration> plot tests this particular single-<concentration> <rate equation>. It does not uniquely identify a <reaction mechanism>, nor does curvature rule out every <second-order reaction>: two independently changing <reactant> <concentrations> require their own <integrated rate law>. At low <concentration>, a small <measurement error> $\delta[\mathrm A]$ becomes approximately $-\delta[\mathrm A]/[\mathrm A]^2$ in the reciprocal, magnifying late-time errors. Use the line as a diagnostic; a <nonlinear regression> of the original <concentrations> can retain their measurement scale and explicit error model.
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