Integrated Rate Law for a Zero Order Reaction

Setting up the integral. A zero order reaction has a rate that does NOT depend on reactant concentration at all -- the book notes such reactions are genuinely rare, but the derivation is instructive. For $A\rightarrow\text{product}$, $\text{Rate}=k[A]^0=k$ (since anything to the power 0 is 1), so

Integrating between $t=0$ ($[A]_0$) and $t=t$ ($[A]$):

Rearranged for k: $k=\dfrac{[A]_0-[A]}{t}$ -- notice this involves the RAW concentration difference, with no logarithm anywhere, unlike the first order case.

The graphical form. Writing $[A]_0-[A]=kt$ as $[A]=-kt+[A]_0$ again matches $y=mx+c$, with $y=[A]$ (NOT $\ln[A]$ this time), $x=t$, slope $=-k$, intercept $=[A]_0$. So a plot of RAW concentration $[A]$ against time is a straight line with negative slope for a zero order reaction (Fig 7.4) -- this is the quickest visual way to tell a zero order reaction apart from a first order one: zero order gives a straight line when you plot $[A]$ itself, first order only gives a straight line when you plot $\ln[A]$. …

One formula that covers every order except exactly one. For a single reactant $A\rightarrow\text{product}$ obeying $-\dfrac{d[A]}{dt}=k[A]^n$ with any order $n\neq1$ (the case $n=1$ needs the separate logarithmic derivation of Section 7.5.1, because integrating $[A]^{-1}$ produces a natural log rather than a power), integrating between $[A]_0$ at $t=0$ and $[A]$ at time t gives the single general result

$\frac{1}{[A]^{n-1}}-\frac{1}{[A]_0^{n-1}}=(n-1)kt$ …