Zero Order

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

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

Units of k: $\text{mol}\,\text{L}^{-1}\,\text{s}^{-1}$

Linear plot: $[A] \text{ vs } t$ → slope = $-k$, intercept = $[A]_0$

Time to complete reaction: $t_{\text{complete}} = \frac{[A]_0}{k}$

First Order

$\boxed{\ln[A] = \ln[A]_0 - kt} \quad \Leftrightarrow \quad \boxed{k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}}$

$[A] = [A]_0\,e^{-kt}$

$t_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}$

Units of k: $\text{s}^{-1}$

Linear plot: $\ln[A] \text{ vs } t$ → slope = $-k$, intercept = $\ln[A]_0$

Second Order (A → products)

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

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

Units of k: $\text{L}\,\text{mol}^{-1}\,\text{s}^{-1}$

Linear plot: $\frac{1}{[A]} \text{ vs } t$ → slope = $+k$ (positive), intercept = $\frac{1}{[A]_0}$

Half-Life Summary

$t_{1/2}^{(0)} = \frac{[A]_0}{2k} \quad \text{(directly } \propto [A]_0\text{)}$

$t_{1/2}^{(1)} = \frac{0.693}{k} \quad \text{(independent of } [A]_0\text{)}$

$t_{1/2}^{(2)} = \frac{1}{k[A]_0} \quad \text{(inversely } \propto [A]_0\text{)}$

General Units of k (nth order)

$k = \left(\frac{\text{mol}}{\text{L}}\right)^{1-n} \cdot \text{s}^{-1} = \text{mol}^{1-n}\,\text{L}^{n-1}\,\text{s}^{-1}$