4.3 Second-order integrated rate law

Second-Order Integrated Rate Law

Second-order reactions have rates that depend on the square of one reactant's concentration (or on the product of two reactant concentrations). The integrated rate law for these reactions connects concentration to time in a single equation, letting you calculate how much reactant remains at any point, determine the rate constant from data, and confirm whether a reaction is actually second-order.

Derivation of the Second-Order Rate Law

The starting point is the differential rate law for a second-order reaction with a single reactant:

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

Here's how to get from this to the integrated form:

1. Separate variables by dividing both sides by $[A]^2$ and multiplying by $dt$: $\frac{d[A]}{[A]^2} = -k \, dt$

2. Integrate both sides. The left side is integrated from $[A]_0$ to $[A]$, and the right side from $0$ to $t$: $\int_{[A]_0}^{[A]} \frac{d[A]}{[A]^2} = -k \int_0^t dt$

3. Evaluate the integrals. The left side gives $-1/[A]$, so: $-\frac{1}{[A]} + \frac{1}{[A]_0} = -kt$

4. Rearrange to get the standard form of the second-order integrated rate law:

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

Notice this has the form $y = mx + b$. That linear relationship is the key to identifying and analyzing second-order reactions.

Applying the Rate Law

To find the concentration of a reactant at any time, plug your known values into $\frac{1}{[A]} = kt + \frac{1}{[A]_0}$ and solve for $[A]$.

Worked example: Suppose $k = 0.5 \text{ M}^{-1}\text{s}^{-1}$, $[A]_0 = 2.0 \text{ M}$, and $t = 10 \text{ s}$.

1. Calculate $\frac{1}{[A]_0}$: $\frac{1}{2.0} = 0.50 \text{ M}^{-1}$

2. Calculate $kt$: $(0.5)(10) = 5.0 \text{ M}^{-1}$

3. Add them: $\frac{1}{[A]} = 5.0 + 0.50 = 5.5 \text{ M}^{-1}$

4. Take the reciprocal: $[A] = \frac{1}{5.5} = 0.18 \text{ M}$

The concentration dropped from 2.0 M to 0.18 M in just 10 seconds. One thing to notice about second-order reactions: the rate slows down dramatically as concentration decreases, so the reactant never fully reaches zero the way a simple linear decay might suggest.

Identifying Second-Order Reactions from Plots

Because the integrated rate law has the form $y = mx + b$, plotting $\frac{1}{[A]}$ vs. time will produce a straight line if the reaction is second-order with respect to A.

• The slope of that line equals the rate constant $k$
• The y-intercept equals $\frac{1}{[A]_0}$

If the plot of $\frac{1}{[A]}$ vs. time curves instead of forming a straight line, the reaction is not second-order with respect to A. You'd then test other integrated rate laws (zero-order or first-order) to find the correct fit.

Determining the Rate Constant

Once you've confirmed a reaction is second-order by getting a linear $\frac{1}{[A]}$ vs. time plot, the slope gives you $k$ directly.

• The units of $k$ for a second-order reaction are always $\text{M}^{-1}\text{s}^{-1}$ (or more generally, $\text{concentration}^{-1} \cdot \text{time}^{-1}$). This is different from first-order ($\text{s}^{-1}$) and zero-order ($\text{M} \cdot \text{s}^{-1}$), so checking units is a quick way to verify you're using the right rate law.
• For example, if your best-fit line through the $\frac{1}{[A]}$ vs. time data has a slope of $0.25 \text{ M}^{-1}\text{s}^{-1}$, then $k = 0.25 \text{ M}^{-1}\text{s}^{-1}$.

A steeper slope means a larger $k$ and a faster reaction. Comparing $k$ values across different temperatures or catalytic conditions is one of the main ways chemists study what controls reaction speed.