Second-order reactions
A second-order reaction is one where the rate changes strongly with reactant concentration, typically proportional to the square of a single reactant or to the product of two concentrations. The Second-Order Reactions and the Integrated Rate Law is used to compute concentration at a later time and related quantities from the integrated form.
Core definitions and essential formulas
For a single-reactant second-order process \(A \rightarrow \text{products}\), the rate law is
\[
\text{rate} = -\frac{\mathrm{d}[A]}{\mathrm{d}t} = k[A]^2.
\]
Here \([A]\) is concentration, \(t\) is time, and \(k\) is the second-order rate constant. The minus sign indicates that \([A]\) decreases while the forward rate is treated as positive. If \([A]\) is in \(\mathrm{mol\cdot L^{-1}}\) and time is in seconds, then \(k\) has units \(\mathrm{L\cdot mol^{-1}\cdot s^{-1}}\) (equivalently \(\mathrm{M^{-1}\cdot s^{-1}}\)).
Separating variables and integrating from \([A]_0\) at \(t=0\) to \([A]_t\) at time \(t\) gives the integrated rate law:
\[
\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}.
\]
This form is linear in \(t\) when plotting \(1/[A]\) versus \(t\): slope \(k\), intercept \(1/[A]_0\). The half-life follows by setting \([A]_{t_{1/2}}=\tfrac{1}{2}[A]_0\):
\[
t_{1/2}=\frac{1}{k[A]_0}.
\]
Larger \(k\) or larger \([A]_0\) leads to faster depletion and a smaller half-life. For typical calculator outputs such as \([A]_t\), fraction remaining \([A]_t/[A]_0\), fraction consumed \(1-[A]_t/[A]_0\), and \(t_{1/2}\), the integrated law makes the meaning direct: as \(t\) increases, \(1/[A]_t\) increases linearly, so \([A]_t\) decays more slowly at later times than at early times.
Common pitfalls
- Using \(-\Delta[A]/\Delta t\) as a constant rate; second-order rate changes with \([A]\).
- Mixing time units so \(k\) and \(t\) are inconsistent (seconds vs minutes).
- Entering \([A]\) in \(\mathrm{mol\cdot m^{-3}}\) while interpreting results as \(\mathrm{mol\cdot L^{-1}}\).
- Applying this form to reactions that are second order overall but involve two reactants with changing concentrations.
Micro example
If \([A]_0=0.50\ \mathrm{mol\cdot L^{-1}}\) and \(k=0.40\ \mathrm{L\cdot mol^{-1}\cdot s^{-1}}\), then
\[
t_{1/2}=\frac{1}{0.40(0.50)}=5.0\ \mathrm{s}.
\]
Use this tool when a single-reactant second-order model fits the system and the goal is predicting \([A]_t\) or extracting \(k\) from a linear \(1/[A]\) versus \(t\) relationship. Avoid using it for multi-reactant kinetics without justified simplifications; a next-step concept is deriving the correct integrated law for \(A+B\rightarrow \text{products}\) or using the method of initial rates to determine reaction orders.
