Differentiation & Rates of Change — Maths for Thermodynamics & Kinetics

A derivative is a rate of change

The one idea that powers all of kinetics and thermodynamics

Differentiation answers a single question: how fast is something changing right now? On a graph, that “how fast” is the slope of the tangent line — the steepness of the curve at one exact point. We build it from the slope of a chord (two points) and then slide the second point in until the gap vanishes:

\frac{d y}{d x} = Δ x \to 0 lim \frac{Δ y}{Δ x} = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

For a chemist this is everything: the rate of a reaction is how fast a concentration changes, $rate = - \frac{d [ A ]}{d t}$ ; a reaction reaches equilibrium where the free energy stops changing, $\frac{d G}{d ξ} = 0$ ; the steepest point of a titration curve is where $d (pH) / d V$ is largest.

Three ways to write it

\frac{d y}{d x}

f^{'} (x)

, and (for time)

\overset{x}{˙}

all mean the same thing — the derivative.

It carries units

A derivative inherits the units of

y

over

x

. So

d [A] / d t

is in mol L⁻¹ s⁻¹ — a rate.

Zero at the turning point

Positive slope = rising, negative = falling, and zero at every peak, trough or equilibrium.

The standard derivatives you will reuse constantly

Power

\frac{d}{d x} x^{n} = n x^{n - 1}

Exponential

\frac{d}{d x} e^{k x} = k e^{k x}

Natural log

\frac{d}{d x} ln x = \frac{1}{x}

Sine & cosine

\frac{d}{d x} sin x = cos x, \frac{d}{d x} cos x = - sin x

Why it matters

Without differentiation there is no rate law, no Arrhenius equation, and no way to find the minimum-energy geometry of a molecule. It is the mathematical engine under physical chemistry — head to the Slope explorer next to build the intuition by hand, then work the sample problems in each section.

Common pitfalls

Rate is a slope, not a value. The reaction rate at time $t$ is the gradient of the tangent, not the height $[A]$ nor the area under the curve.
Mind the sign. For a reactant $d [A] / d t < 0$ , so the (positive) rate is $- d [A] / d t$ . Dropping the minus sign reports a negative rate.
Power rule, not the original power. $\frac{d}{d x} x^{4} = 4 x^{3}$ — reduce the exponent by one. Leaving it as $4 x^{4}$ is the most common slip.
The constant $k$ from the exponent must come down. $\frac{d}{d t} e^{- k t} = - k e^{- k t}$ , not $e^{- k t}$ — forgetting the chain-rule factor loses the $k$ .

Slope explorer

Drag a point along the curve — the tangent's steepness is the derivative. Switch to secant mode to watch the difference quotient converge.

Drag the dot along the curve

time t2.20

[A]0.372

slope = f′(x)-0.167

rate −d[A]/dt0.167

A radioactive isotope or first-order reactant. The slope is the (negative) reaction rate.

Worked example 1Reading an instantaneous rate off a graph

A kinetics run follows the decomposition $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ . The tangent drawn to the concentration–time curve at $t = 200 s$ passes through the points $(100 s, 0.0900 M)$ and $(300 s, 0.0500 M)$ .

What is the instantaneous rate of disappearance of N₂O₅ at 200 s?

Rule builder

See each differentiation rule in action. Adjust the coefficients and watch both the function (solid) and its derivative (dashed) respond.

Pick a function type

f(x) f′(x) (derivative)

Rule

f (x) = a x^{n}

→

\frac{df}{d x} = n a x^{n - 1}

Your function

f (x) = 1 x^{3}

Derivative

f^{'} (x) = 3 x^{2}

a (coefficient)1

n (power)3

Bring the power down to the front as a multiplier, then subtract one from the power. This one rule differentiates every polynomial.

In chemistry

Concentration–time data often fits a power law; its derivative gives the rate at any instant.

Worked example 2Power rule, term by term

Differentiate $f (x) = 5 x^{4} - 2 x^{2} + 7$ with respect to $x$ .

Worked example 3Exponential — the heart of first-order kinetics

A first-order reactant follows $[A] = 0.50 e^{- 0.12 t} M$ with $t$ in seconds.

Find $d [A] / d t$ , and evaluate the reaction rate at $t = 0$ and $t = 10 s$ .

Worked example 4Chain rule on the Arrhenius equation

Show that differentiating $k = A e^{- E_{a} / RT}$ with respect to temperature gives $\frac{d k}{d T} = \frac{E _{a}}{R T ^{2}} k$ .

Kinetics lab

The payoff: a first-order reaction. The slope of the concentration–time curve is literally the reaction rate, −d[A]/dt = k[A].

First-order decay — drag the marker to read the instantaneous rate

[A] = [A]_{0} e^{- k t} - \frac{d [ A ]}{d t} = k [A]

[A]₀ (initial conc.)1.00 M

k (rate constant)0.45 s⁻¹

at t2.20 s

[A]0.372 M

slope d[A]/dt-0.167

rate = k[A]0.167 M/s

half-life t½1.54 s

Notice the slope of the tangent equals −k[A]: the steeper the curve, the faster the reaction. Because the half-life t½ = ln 2 / k doesn't depend on concentration, the gaps t½, 2t½, 3t½ are evenly spaced — the fingerprint of first-order kinetics.

Worked example 5Half-life from the rate constant

A first-order decay obeys $[A] = [A]_{0} e^{- k t}$ with $k = 2.8 \times 1 0^{- 2} s^{- 1}$ . Find the half-life — the time at which $[A] = \frac{1}{2} [A]_{0}$ .

Worked example 6Why a doubled concentration doubles the rate

For the same reaction, rate $= k [A]$ . If a fresh experiment starts at twice the initial concentration, what happens to the initial rate, and to the half-life?

Worked examples

Pull the ideas together — try each before revealing the full solution.

Worked example 7Locating a transition state and an intermediate

A reaction-coordinate model gives the energy $E (x) = x^{3} - 6 x^{2} + 9 x$ (in kJ mol⁻¹). Find the stationary points and classify each as a transition state (maximum) or a stable intermediate (minimum).

Worked example 8Why ln[A] versus t is a straight line

Show that for a first-order reaction $[A] = [A]_{0} e^{- k t}$ , a plot of $ln [A]$ against $t$ is a straight line, and state its slope.

ChallengeThe temperature sensitivity of a rate constant

The Arrhenius equation is $k = A e^{- E_{a} / RT}$ . Show that $\frac{d ln k}{d T} = \frac{E _{a}}{R T ^{2}}$ , then estimate the percentage increase in $k$ for a 10 K rise from 300 K when $E_{a} = 50 kJ mo l^{- 1}$ .

Check yourself

Five quick questions tying the maths back to the chemistry.

Question 1 of 5 · Score 0

Differentiate the following with respect to x:

f (x) = 4 x^{3}

Choose an answer.