A differential rate law tells how fast concentration is changing at a given composition. An integrated rate law instead relates concentration directly to elapsed time.

For a single-reactant process

$$\mathrm A\rightarrow\text{products},$$

three common idealized cases are especially useful.

Zero order

If

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

then concentration decreases linearly:

$$\boxed{[A]_t=[A]_0-kt}.$$

A plot of $[A]$ versus $t$ is linear with slope $-k$. Setting $[A]_t=[A]_0/2$ gives

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

First order

If

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

this is the shared first-order exponential-decay pattern:

$$\boxed{[A]_t=[A]_0e^{-kt}},$$

or equivalently

$$\boxed{\ln[A]_t=\ln[A]_0-kt}.$$

A plot of $\ln[A]$ versus $t$ is linear with slope $-k$, and

$$t_{1/2}=\frac{\ln2}{k}.$$

The concentration-independent half-life is therefore a distinctive signature of ideal first-order behavior.

Second order in one reactant

If

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

then

$$\boxed{\frac1{[A]_t}=\frac1{[A]_0}+kt}.$$

A plot of $1/[A]$ versus $t$ is linear with slope $k$. The half-life is

$$t_{1/2}=\frac1{k[A]_0}.$$

Using concentration-time data to identify order

For a reaction that follows one of these simple models, test which transformed concentration gives a straight line against time:

• $[A]$ versus $t$: zero order;
• $\ln[A]$ versus $t$: first order;
• $1/[A]$ versus $t$: second order.

Example

A first-order decomposition has

$$k=0.0200,\mathrm{min^{-1}},\qquad [A]_0=0.800,\mathrm M.$$

After $30.0$ min,

$$[A]_t=(0.800)e^{-(0.0200)(30.0)}\approx0.439,\mathrm M.$$

Its half-life is

$$t_{1/2}=\frac{0.693}{0.0200}\approx34.7,\mathrm{min}.$$

Integrated rate laws describe how composition evolves when a particular differential rate law remains valid. They do not determine the rate law from the overall balanced equation; that dependence must still come from experiment or from a justified elementary-step mechanism.