Stationary Points & Optimization — Maths for Thermodynamics & Kinetics

Finding peaks and valleys

Set the derivative to zero and let the second derivative decide

A curve is perfectly flat at its maxima and minima — the tangent line is horizontal and the slope is zero. That is why we find them by solving $f^{'} (x) = 0$ . To tell which type of stationary point we have, we check the second derivative:

f^{'} (x) = 0 (stationary), f^{''} (x) > 0 \Rightarrow minimum, f^{''} (x) < 0 \Rightarrow maximum

If $f^{''} = 0$ the second-derivative test is inconclusive — you need to check the sign of $f^{'}$ on either side (the first-derivative test). A point where $f^{''} = 0$ and $f^{''}$ changes sign is a point of inflection — neither a peak nor a valley.

This procedure is optimization — finding the best (lowest-energy, fastest, most probable) value of a quantity. Nearly every time a chemist writes “minimise the energy” or “most probable speed”, they are setting a derivative to zero.

Find stationary point

Solve

f^{'} (x) = 0

Classify: second derivative

f^{''} > 0

→ minimum (bowl up);

f^{''} < 0

→ maximum (hill).

Transition state

Maximum on an energy profile:

d E / d x = 0

d^{2} E / d x^{2} < 0

Equilibrium geometry

Minimum of

V (r)

d V / d r = 0

d^{2} V / d r^{2} > 0

Energy landscapes

Equilibrium bond lengths sit at the minimum of a potential-energy curve; transition states are maxima along a reaction path; the most-probable molecular speed is a maximum of the Maxwell–Boltzmann distribution; the minimum of the Lennard-Jones potential gives the van der Waals radius. All are

d V / d x = 0

problems. Open the finder.

The Lennard-Jones potential $V (r) = 4 ε [(σ / r)^{12} - (σ / r)^{6}]$ provides an important worked example of the complete optimisation workflow. Setting $d V / d r = 0$ gives the equilibrium separation $r_{0} = 2^{1/6} σ \approx 1.122 σ$ — the van der Waals contact distance. The force on the two atoms is $F = - d V / d r$ , and at $r_{0}$ this force is exactly zero. The second derivative $d^{2} V / d r^{2} ∣_{r_{0}} > 0$ confirms a true minimum (restoring force), and its value is proportional to the harmonic vibrational frequency of the dimer — linking optimisation directly to spectroscopy.

A subtler point: inflection points on kinetic curves carry chemical information. An autocatalytic reaction $A \to B$ with rate $r = k [A] [B]$ produces a sigmoidal concentration–time curve for $[B]$ . The inflection point — where $d^{2} [B] / d t^{2} = 0$ — marks the moment the rate of reaction is increasing most rapidly and corresponds to the maximum rate of rate change. After the inflection, substrate depletion decelerates the acceleration. Identifying this point requires the second-derivative test (set $f^{''} = 0$ , check sign change), not just looking for $f^{'} = 0$ .

LJ equilibrium separation

r_{0} = 2^{1/6} σ

, depth

V (r_{0}) = - ε

Most-probable speed (MB)

v_{mp} = 2 RT / M

— maximum of the speed distribution function.

Inflection ≠ stationary

f^{''} = 0

with sign change → inflection (slope extremum, not function extremum).

Always check end-points

On a closed interval

[a, b]

, the global optimum may be at an endpoint, not at an interior stationary point.

Common pitfalls

Not classifying the stationary point. Finding $f^{'} (x) = 0$ only locates the point — you must then use the second-derivative test (or the first-derivative sign test) to decide whether it is a maximum, minimum, or inflection point.
Missing endpoint extrema. On a closed interval, the overall extremum might occur at the boundary $x = a$ or $x = b$ even if interior stationary points exist. Always evaluate $f$ at both endpoints and compare.
Assuming f″ = 0 means an inflection point. If $f^{''} (x_{0}) = 0$ you must check that $f^{''}$ actually changes sign at $x_{0}$ — if it doesn't, the second-derivative test is inconclusive and the point could still be a maximum or minimum (e.g. $f (x) = x^{4}$ at $x = 0$ ).
Confusing the force with the potential minimum. In a potential-energy well, $F = - d V / d r = 0$ at the equilibrium position. This is not where $V$ is zero, but where the slope of $V$ is zero.

Stationary-point finder

Adjust the parameter — the red dot is the maximum (transition state analogue) and the green dot the minimum (stable intermediate).

Where f′ = 0, the curve has a peak or a valley

f(x) f′(x)

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

parameter a3.00

stationary points x±1.00

maximum atx = −1.00

minimum atx = +1.00

Where the dashed f′ crosses zero, f is flat. The second derivative f″ = 6x decides which: negative (left) gives the max, positive (right) gives the min — just like a transition state versus an intermediate.

Worked example 1Transition state and stable intermediate on an energy profile

A reaction-coordinate energy model gives $E (x) = x^{3} - 6 x^{2} + 9 x$ kJ mol⁻¹. Find all stationary points and classify each.

Worked example 2Most-probable speed from the Maxwell–Boltzmann distribution

The Maxwell–Boltzmann speed distribution for a gas of molar mass $M$ at temperature $T$ is $f (v) \propto v^{2} e^{- M v^{2} /2 RT}$ . Find the most-probable speed $v_{mp}$ by differentiating and setting equal to zero.

Worked examples

Optimization across kinetics, molecular structure and thermodynamics.

Worked example 3Lennard-Jones potential minimum

The Lennard-Jones potential is $V (r) = 4 ε [(\frac{σ}{r})^{12} - (\frac{σ}{r})^{6}]$ . Find the equilibrium separation $r_{0}$ where $V$ is a minimum.

Worked example 4Optimising the intermediate concentration in a consecutive reaction

In consecutive first-order reactions $A k_{1} B k_{2} C$ with $k_{1} = 0.10 s^{- 1}$ and $k_{2} = 0.040 s^{- 1}$ , the intermediate B reaches its maximum at time $t_{m a x} = \frac{ln ( k _{1} / k _{2} )}{k _{1} - k _{2}}$ . Evaluate this time.

Worked example 5Minimum free energy — equilibrium condition

For a simple reaction $A ⇌ B$ , the Gibbs free energy is $G (ξ) = G_{0} + (Δ G^{\circ} + RT ln Q) ξ$ near equilibrium, where $Q$ depends on $ξ$ . The minimum is where $d G / d ξ = 0$ . Show that this condition gives $Δ G^{\circ} = - RT ln K$ .

Worked example 6Inflection point in an autocatalytic concentration curve

An autocatalytic reaction gives the logistic concentration profile $[B] (t) = \frac{[ B ] _{\infty}}{1 + e ^{- r (t - t_{0})}}$ where $[B]_{\infty}$ is the maximum concentration, $r$ is a growth rate constant, and $t_{0}$ is a reference time. Find the inflection point and state what it represents physically.

Worked example 7Lennard-Jones force and the equilibrium distance

The Lennard-Jones pair potential is $V (r) = 4 ε [(\frac{σ}{r})^{12} - (\frac{σ}{r})^{6}]$ . The pair force is $F = - d V / d r$ . Show that $F = 0$ at $r_{0} = 2^{1/6} σ$ and confirm the second derivative confirms a minimum of $V$ there.

ChallengeMaximising intermediate yield at optimal quench time

In $A k_{1} B k_{2} C$ (consecutive, first order) with $[A]_{0} = 1$ , the intermediate concentration is:

[B] (t) = \frac{k _{1}}{k _{2} - k _{1}} (e^{- k_{1} t} - e^{- k_{2} t})

(a) Find the time $t_{m a x}$ at which $[B]$ is maximised.
(b) Show that $[B]_{m a x} = (k_{1} / k_{2})^{k_{2} / (k_{2} - k_{1})}$ .
(c) Evaluate both for $k_{1} = 0.10 s^{- 1}$ , $k_{2} = 0.40 s^{- 1}$ .

Check yourself

Four quick questions on stationary points and their chemical meaning.

Question 1 of 4 · Score 0

A stationary point of f(x) occurs where:

Choose an answer.