Thermodynamics and Electrochemistry
Part 2 of 4: Gibbs Free Energy. This unit runs past the 40-minute ceiling, so it ships in 4 lectures.
The narration between entries paraphrases; it never reads an entry aloud before the entry plays. Every entry in this lecture plays exactly as it does on its own page — same file, same speed, nothing laid over it.
What you can do after this
- Say what ΔG folds together and what its sign decides
- Say which combinations of ΔH and ΔS make spontaneity depend on temperature
- Compute a crossover temperature without the unit error that spoils it
- Say why two different routes to ΔG° have to agree
- Say how ΔG° and K state the same fact two ways
What it covers
12 entries in 3 topics. Each number links to that entry on its own.
-
1. Gibbs free energy
-
2. Free energy from tables
-
3. Free energy, Q, and K
Check yourself
What does ΔG fold together, and what does its sign decide?
It folds a reaction's enthalpy change and its entropy change into a single number at a stated temperature: ΔG = ΔH − TΔS. The point of combining them is that neither alone decides anything. An exothermic reaction can still fail to proceed if its entropy change is unfavourable enough, and an endothermic one can proceed if its entropy change is favourable enough — reading either term on its own misses what actually settles the question. The sign of the combined number decides direction, at constant temperature and pressure: negative means the process is spontaneous as written, positive means it is nonspontaneous as written, and exactly zero means the system is at equilibrium. What the sign does not decide is rate. A reaction with a large negative ΔG may proceed immeasurably slowly, because spontaneity names direction and speed remains a kinetic question belonging to Unit 9.
Which combinations of ΔH and ΔS make spontaneity depend on temperature?
The two in which ΔH and ΔS share a sign. Where ΔH is negative and ΔS positive, both terms in ΔG = ΔH − TΔS work in the same favourable direction and the reaction is spontaneous at every temperature; where ΔH is positive and ΔS negative, both work against it and it is spontaneous at none. The other two combinations set the terms against each other, so which one wins depends on how large T is — and there is a specific temperature at which they balance exactly, the crossover. A reaction with both ΔH and ΔS positive is nonspontaneous below its crossover and spontaneous above it; a reaction with both negative behaves the other way round, spontaneous when cold and not when hot. The practical consequence is that calling such a reaction "spontaneous" without naming a temperature ignores its own crossover, since the word is only true on one side of it.
How do you compute a crossover temperature, and what spoils the answer?
Set ΔG to zero in ΔG = ΔH − TΔS, which is the definition of the crossover, and solve for temperature: T = ΔH/ΔS. Both ΔH and ΔS are taken as roughly constant over the temperature range of interest, an approximation that holds well enough for modest spans. For a reaction with ΔH = 178 kJ and ΔS = 160.5 J/K, the answer is about 1109 K, above which it becomes spontaneous. What spoils it is units. Enthalpies are tabulated in kilojoules and entropies in joules per kelvin, so dividing them as they stand gives a temperature off by a factor of exactly one thousand — the enthalpy has to be converted to joules first. The saving grace is that the error is not subtle in its consequences: a crossover temperature coming out near 1.1 K or near a million kelvin is implausible on its face, so checking the answer's magnitude against physical sense catches it. The same mismatch recurs later when ΔG° in kilojoules meets R in joules.
Why do two different routes to ΔG° have to agree?
Because free energy is a state function, and a state function depends only on a system's initial and final states rather than on the path taken between them. There are two routes available: assemble ΔG° from ΔH° and ΔS° through ΔG° = ΔH° − TΔS°, or sum tabulated ΔG°f values for the products, each times its coefficient, and subtract the same sum for the reactants. Both describe the same change between the same two states, so both must return the same number. In practice they return nearly the same number, because the tabulated values feeding each route carry their own rounding, and that rounding propagates differently through the two calculations. The useful consequence is diagnostic: a small discrepancy between the two routes is expected and means nothing, while a large one is a signal that something is wrong — an arithmetic slip, a missed coefficient, or the unit mismatch that turns up repeatedly in this material.
How do ΔG° and K state the same fact?
Through the nonstandard free energy expression, ΔG = ΔG° + RT ln Q, evaluated at equilibrium. At equilibrium two things become true at once: ΔG is zero, because there is no remaining drive in either direction, and Q equals K, because the reaction quotient has reached the equilibrium constant. Substituting both collapses the expression to ΔG° = −RT ln K. So the standard free energy change and the equilibrium constant are not two independent facts about a reaction that happen to correlate — they are one fact written two ways, and either can be computed from the other given the temperature. A strongly negative ΔG° and a large K are the same claim: the equilibrium mixture heavily favours products. Two cautions travel with it. The equation takes K specifically, not a nonequilibrium Q, and both quantities are tied to the temperature at which they were found, so carrying a K computed at one temperature over to another discards the dependence that produced it.
Lecture transcript
The narration as recorded, in order. Where an entry plays, the transcript links to it rather than reprinting it — the entry plays from its own file, unaltered, and says the same thing on its own page.
This is Unit Eleven of Chemistry, Part Two: Gibbs free energy. Part One built entropy — what it counts, what raises it, and the two laws that bound it and give it a zero. It ended on a calculation of a reaction's standard entropy change, and stopped there deliberately, because entropy alone does not decide whether a reaction goes. This part supplies the equation that does. Gibbs free energy folds Part One's entropy together with Unit Four's enthalpy into one number, ΔG equals ΔH minus T ΔS, and the sign of that number decides direction. Everything in this part is downstream of that one line. The second half then connects it back to Unit Ten: away from standard conditions free energy depends on the reaction quotient, and at equilibrium the relationship collapses into a direct link with the equilibrium constant itself. By the end you will be able to say what ΔG folds together and what its sign decides; say which combinations of ΔH and ΔS make spontaneity depend on temperature; compute a crossover temperature without the unit error that spoils it; say why two routes to ΔG° have to agree; and say how ΔG° and K state the same fact two ways.
The hinge of the whole unit arrives first. One equation folds Part One's entropy together with Unit Four's enthalpy and returns a single number whose sign decides direction. Then four sign combinations, two of which settle the question at every temperature and two of which hand it to temperature — and the calculation that finds where the answer changes.
Both terms, one number.
Chem 2103 Gibbs Free Energy Defined
Gibbs free energy change combines a reaction's enthalpy and entropy changes at a given temperature: ΔG equals ΔH minus T ΔS. A reaction with negative ΔH and positive ΔS has both terms favourable and a negative ΔG at every temperature. Reading either term alone misses what decides spontaneity.
Negative goes.
Chem 2106 Sign of ΔG Predicts Spontaneity
At constant temperature and pressure, a process is spontaneous where ΔG is negative, nonspontaneous where it is positive, and at equilibrium where it is exactly zero. A reaction at minus a hundred twenty kilojoules proceeds forward as written. The sign still says nothing about speed — that question stays with Unit Nine.
Two settled, two that flip.
Chem 2109 The Four ΔH/ΔS Sign Combinations
Negative ΔH with positive ΔS is spontaneous at every temperature; positive ΔH with negative ΔS is nonspontaneous at every temperature. The other two combinations flip with temperature. Hydrogen peroxide decomposing is exothermic and gains gas moles, so it belongs to the first case. A negative ΔH alone guarantees nothing — an unfavourable ΔS can still push ΔG positive.
Where the sign changes.
Chem 2112 Temperature-Dependent Spontaneity and the Crossover Temperature
Where ΔH and ΔS share a sign, ΔG's sign depends on temperature and switches at the point where ΔH equals T ΔS exactly. A reaction with both positive is nonspontaneous when cold and spontaneous past its crossover. Calling such a reaction spontaneous without naming the temperature ignores its own crossover point.
Set ΔG to zero.
Chem 2115 Calculating the Crossover Temperature
Set ΔG to zero in the free energy equation and solve: T equals ΔH over ΔS, with both values assumed roughly constant over the temperature range of interest. For ΔH of a hundred seventy-eight kilojoules and ΔS of a hundred sixty point five joules per kelvin, convert the enthalpy to joules first, and T is about eleven hundred nine kelvin. Mixing kilojoules against joules before dividing gives a crossover off by a factor of a thousand — an error the answer's implausible size should catch.
Three entries on the second route to a free energy change. Rather than assembling ΔG from an enthalpy and an entropy, it can be read from tabulated formation values in the same products-minus-reactants pattern already used for enthalpy in Unit Four and for entropy in Part One. The third entry checks that both routes agree — which, because free energy is a state function, they must.
Formed from its elements.
Chem 2118 Standard Free Energy of Formation ΔG°f
Standard free energy of formation is a compound's free energy change on forming from its elements in their standard states, with every element's own value exactly zero. Liquid water has a negative tabulated ΔG°f. The zero-for-elements convention mirrors ΔH°f, but this is a distinct quantity — it carries entropy's contribution too.
The familiar pattern.
Chem 2121 Calculating ΔG°rxn from ΔG°f Values
Sum the products' ΔG°f values, each times its coefficient, and subtract the same sum for reactants — the identical pattern already used for ΔH°rxn and for ΔS°rxn. Methane combustion works out to minus eight hundred eighteen kilojoules, strongly negative and spontaneous, consistent with methane burning readily once ignited. Oxygen contributes nothing to the reactant sum despite appearing in the equation, because an element in its standard state is exactly zero.
Two routes, one answer.
Chem 2124 ΔG as a Path-Independent State Function
Free energy is a state function, so ΔG°rxn found from formation values and from ΔH° minus T ΔS° agree within the rounding of the tabulated numbers — both depend only on the initial and final states. A small gap between the two routes reflects that rounding; only a large gap signals an arithmetic or unit error.
Four entries close the thermodynamics half, and they reach back to Unit Ten. Free energy away from standard conditions depends on the reaction quotient Q; at equilibrium, where Q becomes K and ΔG becomes zero, the relationship collapses into a direct link between the standard free energy change and the equilibrium constant. The capstone then reads one reaction four different ways off a single pair of numbers, and every reading has to agree with the others.
Away from standard.
Chem 2127 ΔG Under Nonstandard Conditions
Away from standard conditions, ΔG equals ΔG° plus R T times the natural log of Q, adjusting the standard value by the reaction quotient. A reaction with slightly positive ΔG° can still run forward if Q is small enough. ΔG° describes standard conditions; ΔG describes the mixture as it stands, and the two need not share a sign.
At equilibrium.
Chem 2130 ΔG° Relates to K
At equilibrium ΔG is zero and Q equals K, which reduces the expression to ΔG° equals minus R T times the natural log of K — a direct link between standard free energy change and the equilibrium constant. A large negative ΔG° corresponds to a large K. The equation takes K, never Q; a nonequilibrium value gives a meaningless result.
Solve, then exponentiate.
Chem 2133 Calculating K from ΔG°
Substitute ΔG° in joules and the absolute temperature into that relationship, isolate the natural log of K, and exponentiate. A ΔG° of minus twenty-nine kilojoules at two ninety-eight kelvin gives a log of about eleven point seven, so K is around one point two times ten to the fifth — large, as a strongly negative ΔG° predicts. Leaving ΔG° in kilojoules against R in joules scales the result by a thousand.
One pair of numbers, four readings.
Chem 2136 Capstone: Full Thermodynamic Profile of a Reaction
Dinitrogen tetroxide dissociating to two nitrogen dioxide has ΔH° of fifty-five point three kilojoules and ΔS° of a hundred seventy-five point eight joules per kelvin. At two ninety-eight kelvin those give ΔG° of about two point nine kilojoules — small and positive, so nonspontaneous as written — and summing tabulated formation values returns two point eight, agreeing within rounding. Because ΔH° and ΔS° share a sign, a crossover exists: ΔH° over ΔS° is about three hundred fifteen kelvin, above which the reaction turns spontaneous. That same small positive ΔG° converts to K of about zero point three one, under one, matching the published value. One ΔH° and ΔS° pair, read four ways, and the four answers agree by construction. Note that both ΔG° and K are tied to the temperature used — carrying a K to a different temperature discards that.
Five things to carry out of this part. Gibbs free energy folds enthalpy and entropy into one number at a stated temperature, ΔG equals ΔH minus T ΔS, and its sign decides direction: negative is spontaneous, positive nonspontaneous, exactly zero is equilibrium — and none of that says anything about speed. Of the four ΔH and ΔS sign combinations, negative-and-positive is spontaneous at every temperature and positive-and-negative at none, while the two matching-sign combinations hand the decision to temperature. Those two have a crossover, found by setting ΔG to zero and solving T equals ΔH over ΔS — and the standing trap there is units, because ΔH is tabulated in kilojoules while ΔS is in joules per kelvin, and dividing without converting gives an answer off by a factor of a thousand. Two routes to ΔG° have to agree because free energy is a state function: whether you assemble it from ΔH° and ΔS° or sum tabulated formation values, both depend only on the initial and final states, so a small gap is rounding and only a large one is an error. And ΔG° and K state one fact two ways: at equilibrium the nonstandard expression collapses to ΔG° equals minus R T log K, so a strongly negative standard free energy change and a large equilibrium constant are the same claim about a reaction. Part Three leaves thermodynamics for redox chemistry and the galvanic cell — where this same free energy turns up again, measured as a voltage.
This is free, and stays free. If it helped, here is what keeps it running.