Differential rate equations for 0, 1st, 2nd order — derivations: Foundations

Before you can read a single derivation in the parent note, you need every squiggle it uses to feel obvious. This page builds each one from nothing — no prior chemistry, no prior calculus assumed.

1. Concentration and the square-bracket notation $[A]$

Picture a fixed 1-litre jar. Substance A is the coloured dots floating in it. $[A]$ is simply how crowded the dots are.

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

Figure 1 — Two identical 1-litre jars. The left jar holds few dots (low $[A]$ ); the right jar is packed (high $[A]$ ). Concentration is just dots-per-litre.

The subscripts you will meet:

$[A]_{0}$ — concentration at the start (the little $0$ means "time zero").
$[A]_{t}$ — concentration at some later time $t$ .

2. Time $t$ , and change $Δ$

$t$ is just elapsed time since the reaction began (seconds, or years for slow ones).

The Greek capital delta $Δ$ means "change in" — the after value minus the before value. So for any quantity, $Δ (quantity) = (later value) - (earlier value)$ .

Since reactant gets used up, $[A]_{t}$ is smaller than $[A]_{0}$ , so $Δ [A]$ is negative. Hold on to that minus sign — it is the reason a minus appears everywhere later.

3. Rate = steepness of the concentration curve

Put time $t$ on the horizontal ( $x$ ) axis and concentration $[A]$ on the vertical ( $y$ ) axis. As time runs to the right, the curve slides downward because reactant is being consumed. The rate is how steeply that curve drops.

Figure 2 — Concentration $[A]$ (y-axis) falling as time $t$ (x-axis) increases. The red dashed chord between two points gives the average rate over the interval; the yellow tangent gives the instantaneous rate at a single moment. The curve is steep (fast) early and gentle (slow) later.

That "slope at one instant" is exactly what the derivative gives us next.

4. The derivative $\frac{d [ A ]}{d t}$

Read it out loud as "dee A dee tee" — it is $Δ [A] /Δ t$ after you shrink the run $Δ t$ down toward zero, so the two points collapse into one and the straight line becomes the tangent.

Because concentration falls, this slope is negative — so the topic writes the rate with a leading minus to keep the rate itself positive:

$Rate = - \frac{d [ A ]}{d t} (a positive number)$

5. The rate constant $k$ and the order $n$

Put together, the whole topic hangs on one sentence:

The three cases the parent derives are just three choices of $n$ :

Figure 3 — Rate plotted against concentration $[A]$ for the three orders. $n = 0$ (blue) is a flat line — rate ignores crowding. $n = 1$ (yellow) is a straight line through the origin — rate doubles when crowding doubles. $n = 2$ (red) curves upward — rate quadruples when crowding doubles.

$n = 0$ : crowding raised to power $0$ is $1$ , so speed $= k$ , flat — crowding is ignored.
$n = 1$ : speed $= k [A]$ , straight through the origin — double crowding, double speed.
$n = 2$ : speed $= k [A]^{2}$ , upward-curving — double crowding, quadruple speed.

See Rate Laws and Reaction Order for how $n$ is measured, and Collision Theory for why $n = 2$ means "two molecules must meet."

6. Powers — the general idea, then the three we actually need

An exponent is just "how many times you multiply the base by itself" — $[A]^{3} = [A] \cdot [A] \cdot [A]$ . That picture extends smoothly to any number in the exponent:

Reaction orders can be fractional (some real reactions are order $\frac{1}{2}$ or $\frac{3}{2}$ ), so the general rule matters. But the three derivations in the parent only ever use the whole-number and $- 1$ cases below — that is why these three suffice for every worked example there:

Anything to the power $0$ is $1$ — this is why zero-order rate collapses to just $k$ .
Power $- 1$ means reciprocal (one-over) — this is why the second-order answer comes out as $\frac{1}{[ A ]}$ .

7. Integration: adding up all the little changes

The derivative took a curve and gave us a slope. Integration does the reverse — it takes the slope rule and rebuilds the curve. The long "S" symbol $\int$ means "sum up continuously."

Three specific integrals appear in the parent, one per order. You do not need to derive them here — just recognise them:

The middle integral introduces a brand-new function, $ln$ — meet it next.

8. The natural logarithm $ln$ and its partner $e^{x}$

You meet $ln$ in first-order because integrating $\frac{1}{[ A ]}$ produces exactly $ln [A]$ — there is no other elementary function whose slope is $\frac{1}{x}$ .

Figure 4 — The curves $y = e^{x}$ (yellow) and $y = ln x$ (blue) are mirror images across the dashed line $y = x$ . Because they are reflections, one function undoes the other: feed a number through $ln$ then through $e^{x}$ and you get the original number back.

$e^{x}$ and $ln x$ undo each other, like $+$ and $-$ :

This is the move used in the parent's carbon-14 ( $^{14} C$ ) example: when the equation reads $ln (ratio) = - 1.21$ , you apply $e^{(\dots)}$ to both sides to peel off the $ln$ and get the ratio itself, $e^{- 1.21} = 0.298$ .

9. Reading a straight-line graph: slope and intercept

Every integrated law is rearranged so a plot comes out straight. A straight line obeys:

slope = how fast $y$ climbs per step of $x$ (steepness).
intercept = where the line meets the $y$ -axis ( $x = 0$ ).

Prerequisite map

The diagram below shows how each foundation on this page feeds into the next, ending at the rate equations themselves. Read it top to bottom: concentration and time give you rate as a slope; the slope idea sharpens into the derivative; the derivative plus power rules build the rate law; integration (and its offshoots $ln$ / $e$ and straight-line reading) then turns that rate law into the usable formulas. If the diagram does not render on your device, the same chain is spelled out in words in this very paragraph.

Downstream, these feed Half-Life Calculations, the Arrhenius Equation, and the Steady-State Approximation.

Equipment checklist

Test yourself — reveal only after answering.

What does

[A]

physically measure?

How crowded substance A is — moles per litre in the mixture.

How is

Δ t

defined?

The elapsed time between two clock readings,

Δ t = t_{2} - t_{1}

Why does a rate law carry a leading minus sign?

Reactant concentration falls, so

\frac{d [ A ]}{d t}

is negative; the minus flips it so the rate is a positive speed.

What does the derivative

\frac{d [ A ]}{d t}

represent on a graph?

The slope of the tangent line — the instantaneous rate of change of concentration at one instant.

What is

[A]^{0}

and why does it matter?

It equals

1

, which is why a zero-order rate collapses to just

k

What does a fractional exponent like

[A]^{1/2}

mean?

A root —

[A]^{1/2} = [A]

— which appears for half-order reactions.

What does the symbol

\int

do?

It sums up all the tiny changes

d [A]

— it rebuilds concentration from the slope rule (the reverse of differentiation).

Why does the constant of integration

+ C

not appear in the final rate laws?

We integrate between definite limits, so the same

+ C

appears at both limits and cancels on subtraction.

What question does

ln x

answer?

e

raised to what power gives

x

?" — it is the inverse of

e^{x}

How do you remove a

ln

from an equation?

Apply

e^{(\dots)}

to both sides, since

e^{l n x} = x

y = m x + c

, what are

m

and

c

m

is the slope (steepness),

c

is the intercept where the line meets the

y

-axis at

x = 0

Why straighten each order's plot?

A straight line is instantly checkable and its slope gives

k

directly.