Visual walkthrough — Differential rate equations for 0, 1st, 2nd order — derivations

This page rebuilds the three differential rate equations from the parent note nothing but a picture of molecules disappearing. We start from the one honest question — "how fast is stuff vanishing right now?" — and let the pictures do every step. No symbol appears before it is drawn.

Step 1 — What "rate" even means: a slope on a shrinking-crowd graph

WHAT. Picture a jar of reactant. Call the amount of it per litre the concentration, written $[\text{A}]$ — read "concentration of A". The square brackets just mean "how much A per litre", measured in $\text{M}$ (molar = ${\text{mol L}}^{-1}$). As time $t$ ticks forward, $[\text{A}]$ falls.

WHY. Before we can ask "how fast", we need a picture of what is changing. The figure plots $[\text{A}]$ going down as $t$ goes right.

PICTURE. Look at the blue curve sliding downhill. Pick two nearby instants. The concentration drops by a tiny amount $d[\text{A}]$ over a tiny time $dt$. The steepness of that little drop is the derivative $\frac{d[\text{A}]}{dt}$ — the tool that answers "how much does one quantity change per tiny change in another?". We use a derivative and not plain division because the slope keeps changing along the curve; a derivative is the slope at one instant.

$\text{Rate}=\underset{\text{crowd shrinks, so make it positive}}{\underbrace{-}}\underset{\begin{array}{c}\text{slope of the blue curve:}\\ \text{drop in }[\text{A}]\text{ per tiny second}\end{array}}{\underbrace{\frac{d[\text{A}]}{dt}}}$

The minus sign is there because $[\text{A}]$ is falling, so $\frac{d[\text{A}]}{dt}$ is a negative number; the minus flips it positive so a rate is never negative.

Step 2 — The one design choice: how does the slope depend on the crowd?

WHAT. We now guess that the rate is controlled by the current crowd size raised to some power: $-\frac{d[\text{A}]}{dt}=k[\text{A}{]}^n$

WHY. Everything downstream is just this one equation with $n=0$, $1$, or $2$. So we must read each symbol carefully.

PICTURE. Three little crowds are drawn, and beside each an arrow whose length is the rate. In the left crowd the arrow ignores the crowd size; in the middle it grows with the crowd; on the right it grows with the square.

$-\frac{d[\text{A}]}{dt}=\underset{\begin{array}{c}\text{rate constant:}\\ \text{fixed for this reaction}\\ \text{at this temperature}\end{array}}{\underbrace{k}}\underset{\begin{array}{c}\text{crowd size}\\ \text{raised to the}\\ \text{order }n\end{array}}{\underbrace{[\text{A}{]}^n}}$

• $k$ = the rate constant. It does not change as the reaction runs — see it only changes with temperature.
• $n$ = the order, an experimentally measured number. It is the whole story.

Step 3 — The universal trick: separate the two worlds, then integrate

WHAT. Every one of the three derivations uses the same two moves. Move 1: shove all the $[\text{A}]$ stuff to the left, all the $t$ stuff to the right. Move 2: integrate — add up all the tiny slices from the start to time $t$.

WHY. The equation mixes $[\text{A}]$ and $t$ inside one derivative. We cannot solve for $[\text{A}]$ while they are tangled. Separating them lets each side be summed on its own. Integration is the tool that answers "if I know the tiny change at every instant, what is the total accumulated change?" — it is the reverse of the derivative from Step 1.

PICTURE. The figure shows the area under a curve being chopped into thin strips and stacked: that stacking is what the integral sign $\int$ means.

$\underset{\begin{array}{c}\text{sum over all}\\ \text{concentrations from}\\ \text{start }[\text{A}{]}_0\text{ to now }[\text{A}{]}_t\end{array}}{\underbrace{{\int }_{[\text{A}{]}_0}^{[\text{A}{]}_t}}}f([\text{A}])d[\text{A}]=-k\underset{\begin{array}{c}\text{sum over all}\\ \text{time from }0\text{ to }t\end{array}}{\underbrace{{\int }_0^t}}dt$

Here $[\text{A}{]}_0$ is the concentration at the starting whistle ($t=0$) and $[\text{A}{]}_t$ is the concentration at the clock reading $t$. The shape of $f([\text{A}])$ on the left is the only thing that differs between the three orders. We now do each.

Step 4 — Zero order ($n=0$): a straight ramp down

WHAT. Put $n=0$. Since anything to the power $0$ is $1$, the crowd size vanishes from the equation: $-\frac{d[\text{A}]}{dt}=k[\text{A}{]}^0=k$ Separate and integrate (here $f([\text{A}])=1$): ${\int }_{[\text{A}{]}_0}^{[\text{A}{]}_t}d[\text{A}]=-k{\int }_0^{t}dt⟹[\text{A}{]}_t-[\text{A}{]}_0=-kt$

WHY. With rate glued to the constant $k$, the crowd falls by the same amount every second — a perfectly straight downhill line.

PICTURE. The straight pink line drops at a fixed steepness. Its slope is $-k$; its height at $t=0$ is $[\text{A}{]}_0$.

$\underset{\text{now}}{\underbrace{[\text{A}{]}_t}}=\underset{\begin{array}{c}\text{y-intercept:}\\ \text{starting height}\end{array}}{\underbrace{[\text{A}{]}_0}}-\underset{\begin{array}{c}\text{slope size:}\\ \text{drop per second}\end{array}}{\underbrace{k}}t$

Degenerate case (a real trap). A straight ramp cannot go below zero — but the formula would happily hand you a negative concentration for large $t$. Physically the reactant simply runs out at $t_{\text{end}}=[\text{A}{]}_0/k$, and the line stops. Look at where the pink line hits the axis in the figure: past that point the law no longer applies.

Step 5 — First order ($n=1$): a curve that never quite reaches zero

WHAT. Put $n=1$, so $f([\text{A}])=1/[\text{A}]$: $-\frac{d[\text{A}]}{dt}=k[\text{A}]⟹\frac{d[\text{A}]}{[\text{A}]}=-kdt$ Integrating $1/[\text{A}]$ gives the natural logarithm $\ln$: $\ln [\text{A}{]}_t-\ln [\text{A}{]}_0=-kt⟹\ln \frac{[\text{A}{]}_t}{[\text{A}{]}_0}=-kt$

WHY the logarithm? The integral of $1/x$ is $\ln x$ — that is the specific tool that undoes "one over the crowd". The natural log answers "to what power must $e\approx 2.718$ be raised to get this number?". Undoing the log (raising $e$ to both sides) gives the clean exponential: $[\text{A}{]}_t=[\text{A}{]}_0{e}^{-kt}$

PICTURE. Left panel: the yellow exponential curve sags fast at first, then flattens, approaching zero but never touching. Right panel: plotting $\ln [\text{A}]$ instead straightens it into a line of slope $-k$.

$\underset{\text{now}}{\underbrace{[\text{A}{]}_t}}=\underset{\text{start}}{\underbrace{[\text{A}{]}_0}}\underset{\begin{array}{c}\text{shrink factor:}\\ \text{multiply by less than 1}\\ \text{as time grows}\end{array}}{\underbrace{e^{-kt}}}$

Special feature. The half-life $\ln 2/k$ has no $[\text{A}{]}_0$ in it — every halving takes the same time, no matter how much you start with. That is why radioactive dating works.

Step 6 — Second order ($n=2$): the reciprocal climbs in a straight line

WHAT. Put $n=2$, so $f([\text{A}])=1/[\text{A}{]}^2=[\text{A}{]}^{-2}$: $-\frac{d[\text{A}]}{dt}=k[\text{A}{]}^2⟹[\text{A}{]}^{-2}d[\text{A}]=-kdt$ Integrating $[\text{A}{]}^{-2}$ gives $-[\text{A}{]}^{-1}$ (add one to the power, divide by the new power): $[-[\text{A}{]}^{-1}{]}_{[\text{A}{]}_0}^{[\text{A}{]}_t}=-kt⟹\frac{1}{[\text{A}{]}_t}-\frac{1}{[\text{A}{]}_0}=kt$

WHY. Here integration uses the ordinary power rule, not a log, because the power $-2$ is not $-1$. The natural variable that comes out straight is $1/[\text{A}]$, the reciprocal — "one divided by the crowd".

PICTURE. Left: the blue $[\text{A}]$ curve drops even more sluggishly at the tail than first order. Right: plotting $1/[\text{A}]$ turns it into a rising straight line of slope $+k$ (note: positive).

$\underset{\begin{array}{c}\text{reciprocal now}\\ \text{(the y-axis)}\end{array}}{\underbrace{\frac{1}{[\text{A}{]}_t}}}=\underset{\begin{array}{c}\text{y-intercept:}\\ \text{reciprocal at start}\end{array}}{\underbrace{\frac{1}{[\text{A}{]}_0}}}+\underset{\begin{array}{c}\text{slope}\\ \text{(positive!)}\end{array}}{\underbrace{k}}t$

Worked check (from the parent's ${\text{NO}}_2$ example). With $[\text{A}{]}_0=0.0100\text{M}$, $k=0.543{\text{M}}^{-1}{\text{s}}^{-1}$, $t=300\text{s}$: $\frac{1}{[\text{A}{]}_t}=100+0.543\times 300=262.9\Rightarrow [\text{A}{]}_t=3.80\times 10^{-3}\text{M}$

Step 7 — Why the graphs never lie: reading order off a straight line

WHAT. The three integrated laws each become a straight line when you plot the correct thing on the y-axis. The left-hand side of the integrated law is your y-axis.

WHY. A straight line is the only shape a human eye can judge instantly. So we bend each law into $y=(\text{intercept})+(\text{slope})t$.

PICTURE. All three linear plots stacked: $[\text{A}]$ (zero), $\ln [\text{A}]$ (first), $1/[\text{A}]$ (second). Only the matching transform is straight; the other two curve.

The one-picture summary

Below, all three worlds sit side by side: same starting height, same clock, three fates. The pink zero-order line marches straight to zero and stops; the yellow first-order curve halves in equal steps forever; the blue second-order curve crawls, its tail the longest of all.

Retrieval check