Gibbs free energy ΔG = ΔH − TΔS; spontaneity criteria: Question bank

Before we start, let us build — on this page, from zero — every symbol these traps use. Nothing is borrowed on faith.

0. Building the symbols (so nothing is used before it is earned)

0.1 The three raw ingredients

0.2 Where $Δ G = Δ H - T Δ S$ actually comes from

We do not take this formula on trust. Here is the whole chain, self-contained.

Step 1 — heat into the surroundings. At constant pressure the system's heat is its enthalpy change, $q_{sy s} = Δ H$ . The surroundings get the opposite heat: $q_{s u r r} = - Δ H$ . Look at s01 — the arrow of heat leaving an exothermic beaker turns around and pours into the surroundings.

Step 2 — that heat becomes surroundings-entropy. The surroundings are an enormous reservoir at fixed $T$ , so $Δ S_{s u r r} = q_{s u r r} / T = - Δ H / T$ .

Step 3 — substitute into the second law. $Δ S_{u ni v} = Δ S_{sy s} - \frac{Δ H}{T}$

Step 4 — multiply through by $- T$ (a positive number times $- 1$ , so the inequality direction flips): $- T Δ S_{u ni v} = Δ H - T Δ S_{sy s}$

Step 5 — name the right-hand side. We define it as $Δ G$ :

0.3 The standard change $Δ G^{\circ}$ , the quotient $Q$ , and the constant $K$

Several traps below use $Δ G^{\circ}$ , $Q$ , and $K$ . Define them now.

0.4 The four sign-quadrants and the ΔG-vs-T lines

s02 is the quadrant map: which of the four sign combinations of $(Δ H, Δ S)$ is spontaneous, and when. Since $Δ G = Δ H - T Δ S$ , plotting $Δ G$ against $T$ gives a straight line with intercept $Δ H$ and slope $- Δ S$ — s04 draws all four lines and where each crosses zero (the crossover temperature $T = Δ H /Δ S$ ).

1. True or false — justify

Every "T/F" answer must say why, not just "true".

Δ G < 0

guarantees the reaction happens quickly.

False.

Δ G

measures feasibility, not speed. Diamond→graphite has

Δ G < 0

yet takes eons because of a huge activation-energy barrier — that is kinetics, a separate story.

A reaction with

Δ G > 0

can never occur in that direction.

False. It is non-spontaneous on its own, but it can be driven by coupling to another process (e.g. ATP hydrolysis, or an external voltage as in electrolysis). Also a tiny amount still occurs until equilibrium.

Δ H < 0

and

Δ S > 0

, the reaction is spontaneous at every temperature.

True.

Δ G = Δ H - T Δ S

= (negative) − (positive)(positive) = negative + negative, which stays below zero for all

T > 0

. Both driving forces pull the same way.

An exothermic reaction is always spontaneous.

False. Exothermic only means

Δ H < 0

. If

Δ S

is also negative, then at high enough

T

the

- T Δ S

term (positive and growing) overwhelms

Δ H

and

Δ G

turns positive.

For any reaction, raising

T

always makes

Δ G

more negative.

False.

Δ G = Δ H - T Δ S

has slope

- Δ S

against

T

. If

Δ S > 0

, raising

T

lowers

Δ G

; if

Δ S < 0

, raising

T

raises

Δ G

. The sign of

Δ S

decides.

At the melting point of ice,

Δ G = 0

for the solid→liquid change.

True. At a phase-transition temperature the two phases coexist in equilibrium, and equilibrium means

Δ G = 0

exactly (see Phase Transitions).

A large negative

Δ G^{\circ}

means the reaction is nearly complete at equilibrium.

True. Since

Δ G^{\circ} = - R T ln K

, a large negative

Δ G^{\circ}

gives a large positive

ln K

, so

K ≫ 1

and products dominate — see Chemical Equilibrium and K.

Δ G

and

Δ G^{\circ}

always have the same sign.

False.

Δ G = Δ G^{\circ} + R T ln Q

. If

Q

is small (few products yet),

R T ln Q

is very negative and can make

Δ G < 0

even when

Δ G^{\circ} > 0

. They agree only at standard state (

Q = 1

, so

ln Q = 0

At equilibrium,

Δ G^{\circ} = 0

False. At equilibrium the actual

Δ G = 0

, but

Δ G^{\circ} = - R T ln K

, which is nonzero whenever

K \neq = 1

. Do not confuse the standard benchmark with the current value.

Δ S_{u ni v} > 0

then

Δ G < 0

for the system.

True. By construction

Δ G = - T Δ S_{u ni v}

, and

T

is positive, so the two inequalities are equivalent (just multiplied by a positive number that flips the sign).

2. Spot the error

Each line quotes a flawed statement; the answer names the mistake and fixes it.

Δ G = Δ H - T Δ S = - 50 kJ - (300) (120) = - 36050 kJ

, where

Δ S = 120 J/K

Unit mismatch.

Δ S

is in J/K but

Δ H

is in kJ. Convert:

120 J/K = 0.120 kJ/K

, giving

- 50 - 36 = - 86 kJ

. The unconverted version is off by a factor of 1000.

"Since

Δ S_{s u r r} = q_{s u r r} / T

and heat flows out of the system,

Δ S_{s u r r} = + Δ H_{sy s} / T

Sign error. The surroundings absorb the opposite heat of the system:

q_{s u r r} = - Δ H_{sy s}

, so

Δ S_{s u r r} = - Δ H_{sy s} / T

. For an exothermic reaction (

Δ H < 0

) this is positive — surroundings gain disorder, correct.

"The reaction has

Δ H > 0

and

Δ S > 0

, so it is non-spontaneous because

Δ H

is positive."

Ignored the entropy term. With

Δ S > 0

, at high enough

T

the term

- T Δ S

becomes more negative than

Δ H

is positive, so

Δ G < 0

. It is spontaneous above

T = Δ H /Δ S

"To find the crossover temperature, set

Δ H = 0

Wrong condition. The crossover (spontaneous ↔ not) is where

Δ G = 0

, i.e.

Δ H = T Δ S

, giving

T = Δ H /Δ S

. Setting

Δ H = 0

has no physical meaning here.

"Diamond is stable forever because turning into graphite is non-spontaneous."

Confused thermo with kinetics. Graphite→ is actually spontaneous (

Δ G < 0

); diamond persists only because the rate is unimaginably slow (high activation energy), not because of

Δ G

Δ G^{\circ} = - R T ln K

, so if

K = 1

the reaction cannot proceed."

Misread.

K = 1

gives

Δ G^{\circ} = 0

, meaning reactants and products are equally favoured at standard state — it still proceeds toward that balanced equilibrium, just not far.

"Since

T

appears in

Δ G = Δ H - T Δ S

, cooling to

0 K

makes

Δ G = Δ H

and every exothermic reaction becomes spontaneous."

Careless at the limit. True that at

T \to 0

Δ G \to Δ H

, so an exothermic reaction (

Δ H < 0

) does give

Δ G < 0

. But

0 K

is unreachable, and rates vanish there too — a mathematical limit, not a practical route.

3. Why questions

Δ G

was invented to track only the system. Why is that useful?

Measuring the surroundings' entropy directly is impractical. Gibbs folds the surroundings' contribution (

- Δ H / T

) into system-only quantities, so we predict spontaneity from the beaker alone without ever touching the surroundings.

Why does temperature act as the "weight" on the entropy term but not on enthalpy?

Because

Δ S_{s u r r} = - Δ H / T

— the entropy the heat produces in the surroundings shrinks as

T

rises (the same heat matters less to a hotter reservoir). Rearranged into

Δ G

, this puts

T

multiplying

Δ S

, giving entropy more clout at high

T

Why must we use

q_{P} = Δ H

rather than

q_{V}

in the derivation?

Gibbs free energy is defined at constant pressure. Enthalpy

Δ H

is the heat exchanged at constant pressure by definition, so it is the correct heat term for the surroundings' entropy calculation.

Why do we treat the heat flow to the surroundings as reversible even when the reaction is not?

The surroundings are so vast their temperature never budges. A heat transfer into an infinite reservoir at fixed

T

is thermodynamically reversible for the reservoir, so

Δ S_{s u r r} = q_{s u r r} / T

holds exactly regardless of how irreversible the system's process is.

Why does multiplying

Δ S_{u ni v} > 0

- T

flip the inequality?

T

is a positive number; multiplying an inequality by any negative quantity (

- T

) reverses its direction. That is what turns "

Δ S_{u ni v} > 0

" into "

Δ G < 0

Why does industry run ammonia synthesis at moderate rather than very high temperature?

N_{2} + 3 H_{2} \to 2 N H_{3}

has

Δ H < 0

and

Δ S < 0

. Raising

T

makes

- T Δ S

increasingly positive, pushing

Δ G

toward zero and lowering yield. Moderate

T

balances a workable rate against still-favourable thermodynamics.

Why can a reaction be spontaneous yet have positive

Δ H

(endothermic)?

Δ S > 0

and

T

is large, the

- T Δ S

term is negative and larger in size than

Δ H

, so

Δ G < 0

. Disorder gain overrides the energy cost — e.g. ice melting above

0^{\circ}

Why does

Δ G

differ from

Δ G^{\circ}

as a reaction runs?

Δ G^{\circ}

is fixed at standard state, but the real mixture keeps changing, so its quotient

Q

changes, and

Δ G = Δ G^{\circ} + R T ln Q

drifts.

Δ G

slides toward zero as

Q

climbs toward

K

4. Edge cases

What is

Δ G

for a reaction with

Δ S = 0

Then

Δ G = Δ H

at all temperatures — the sign of the enthalpy alone decides spontaneity, and temperature has no effect. (In s04 this is a horizontal line.)

What happens to the crossover temperature

T = Δ H /Δ S

if both

Δ H

and

Δ S

are negative?

The ratio is positive (negative÷negative), giving a real crossover: spontaneous below it (enthalpy wins) and non-spontaneous above it (entropy penalty wins).

Δ H > 0

and

Δ S < 0

, does a crossover temperature exist?

No physically meaningful one.

Δ G = (+) - T (-) = (+) + (positive)

is positive for every

T > 0

; the formula

T = Δ H /Δ S

gives a negative "temperature", which is impossible. It is never spontaneous.

T \to 0 K

, what does

Δ G

approach and what does it mean?

Δ G \to Δ H

, so only enthalpy matters near absolute zero — nature's "prefer-low-energy" want fully dominates and the "prefer-disorder" want is switched off.

T \to \infty

, which term dominates

Δ G

The

- T Δ S

term grows without bound, so its sign controls

Δ G

: if

Δ S > 0

Δ G \to - \infty

(spontaneous); if

Δ S < 0

Δ G \to + \infty

(non-spontaneous). Enthalpy becomes negligible.

At the exact crossover temperature, what is the state of the system?

Δ G = 0

, so the forward and reverse processes are equally favoured — the system sits at equilibrium (like ice and water coexisting at

273 K

What does

Δ G = 0

tell you about

Q

and

K

Setting

Δ G = Δ G^{\circ} + R T ln Q = 0

gives

ln Q = - Δ G^{\circ} / R T = ln K

, so

Q = K

. Zero free-energy change means the reaction quotient has reached the equilibrium constant.

If a reaction has

Δ H = 0

exactly, when is it spontaneous?

Δ G = - T Δ S

, so spontaneity depends purely on entropy: spontaneous whenever

Δ S > 0

(for any

T > 0

), driven entirely by disorder — e.g. free mixing of two ideal gases.

Connections

Parent: Gibbs free energy — the full derivation and worked numbers.
Second Law of Thermodynamics — source of $Δ S_{u ni v} > 0$ .
Enthalpy and ΔH · Entropy and ΔS — the two competing wants.
Chemical Equilibrium and K — via $Δ G^{\circ} = - R T ln K$ .
Phase Transitions — the $Δ G = 0$ boundary.
Electrochemistry ΔG = −nFE — driving non-spontaneous reactions with voltage.

Figures on this page

s01 — deriving $Δ G = - T Δ S_{u ni v}$ from heat flow.
s02 — the four sign-quadrant map.
s03 — $Δ G$ sliding to the valley where $Q = K$ .
s04 — $Δ G$ vs $T$ straight lines for all four cases.

Figure — Gibbs free energy ΔG = ΔH − TΔS; spontaneity criteria