Meaning of “why do second order reactions increase”
In chemical kinetics, a “second-order reaction” is one whose rate depends on concentration with a total exponent of 2. The phrase “why do second order reactions increase” is answered by the rate law: when concentration rises, the rate rises more than proportionally because the concentration term is squared (or multiplied by a second reactant concentration).
Second-order rate law and why the rate grows rapidly
A second-order rate law commonly appears in one of these forms:
The key feature is the exponent sum \(2\). This is why second order reactions increase strongly with concentration:
If \([A]\) doubles in case (A), then \([A]^2\) becomes \((2[A])^2 = 4[A]^2\), so the rate becomes \(4\times\) larger.
If both \([A]\) and \([B]\) double in case (B), then \([A][B]\) becomes \((2[A]) \cdot (2[B]) = 4[A][B]\), so the rate becomes \(4\times\) larger.
Molecular interpretation: collision frequency and the rate-determining step
Many second-order behaviors arise when the rate-determining step is bimolecular: two reactant entities must meet with suitable orientation and enough energy to react. Increasing concentration increases the number density of reactant particles, which increases the frequency of bimolecular encounters. Because two reactant “counts” are involved, the dependence becomes quadratic (or a product), matching the observed second-order rate law.
Numerical example showing the strong increase
Assume a single-reactant second-order law \(\text{rate} = k[A]^2\) with \(k = 0.25\ \text{M}^{-1}\text{s}^{-1}\).
| Trial | \([A]\) (M) | Rate calculation | Rate (M·s\(^{-1}\)) | Rate relative to Trial 1 |
|---|---|---|---|---|
| 1 | 0.10 | \(0.25 \cdot (0.10)^2 = 0.25 \cdot 0.010\) | \(0.0025\) | \(1.0\) |
| 2 | 0.20 | \(0.25 \cdot (0.20)^2 = 0.25 \cdot 0.040\) | \(0.010\) | \(4.0\) |
| 3 | 0.30 | \(0.25 \cdot (0.30)^2 = 0.25 \cdot 0.090\) | \(0.0225\) | \(9.0\) |
The quadratic dependence is visible: tripling \([A]\) from \(0.10\) to \(0.30\) multiplies the rate by \(3^2 = 9\).
Integrated second-order rate law: how concentration changes with time
For a second-order reaction in a single reactant \(A\) (rate \(= k[A]^2\)), the integrated rate law is:
This form explains an additional “increase” effect: higher initial concentration makes the concentration drop faster early on, and it also shortens the half-life.
Second-order half-life depends on the initial concentration
Setting \([A] = \frac{[A]_0}{2}\) in the integrated law gives the half-life:
A larger \([A]_0\) makes \(t_{1/2}\) smaller. For example, doubling \([A]_0\) halves \(t_{1/2}\). This concentration-dependent half-life is a hallmark of second-order kinetics.
Visualization: rate vs concentration for a second-order rate law
Key takeaways
- A second-order rate law (\(\text{rate} = k[A]^2\) or \(\text{rate} = k[A][B]\)) makes the rate very sensitive to concentration.
- The molecular reason is increased frequency of effective bimolecular encounters as concentration rises.
- The integrated second-order law \(\frac{1}{[A]} = kt + \frac{1}{[A]_0}\) and \(t_{1/2} = \frac{1}{k[A]_0}\) show that higher initial concentration reduces the half-life.