Intuition What this deep-dive doesThe parent note gave you the equation Δ G = Δ H − T Δ S every possible situation that equation can face — all four sign patterns, the moment when Δ S = 0 Δ H = 0
Before anything, we fix the vocabulary so no symbol is used before it is earned.
Definition The three characters (plain words)
Δ H enthalpy change = the heat the reaction releases (negative) or soaks up (positive) at constant pressure. Think "downhill or uphill in energy." Built in Enthalpy and ΔH .Δ S entropy change = how much more "spread out / messy" the system becomes (positive) or how much tidier it gets (negative). Built in Entropy and ΔS .T kelvin (always positive, so it can never flip a sign by itself — it only sets the weight on Δ S Δ G goes down below zero, the change happens on its own. From the parent topic note .
Definition A unit rule we will use every timeΔ H kilojoules (kJ) but Δ S joules per kelvin (J/K) — a thousand-fold mismatch. Before adding Δ H − T Δ S same unit. We will always convert Δ S
Every problem this topic can throw is one of these cells. The rest of the page fills each one.
Cell
Δ H Δ S Extra condition
What decides
Example
A −
+
none
always spontaneous
Ex 1
B +
−
none
never spontaneous
Ex 2
C −
−
temperature
low-T
Ex 3
D +
+
temperature
high-T
Ex 4 (word problem)
E any
= 0 degenerate
Δ G = Δ H T Ex 5
F = 0 any
degenerate
pure entropy drives it
Ex 6
E′ = 0 = 0 double-zero
Δ G = 0 T Ex 5 (note)
G limiting
—
T → 0 T → ∞ which term survives
Ex 7
H standard vs actual
—
non-standard Q
Δ G Δ G ∘ Ex 8
I link to K
—
equilibrium
Δ G ∘ = − R T ln K Ex 9
Intuition How to read the master figure aboveThe horizontal axis is temperature T vertical axis is Δ G Δ G = 0 Δ G = Δ H − T Δ S straight : its intercept (where it meets T = 0 Δ H slope equals − Δ S lavender = Cell A , coral = Cell B , mint = Cell C , butter = Cell D , dashed slate = Cell E (Δ S = 0 , dotted grey = Cell F (Δ H = 0 . A line lying below the grey zero-line means the reaction is spontaneous there; the butter line (D) crosses zero at the marked crossover T = Δ H /Δ S
Worked example Ex 1 — Combustion-like reactionΔ H = − 200 kJ Δ S = + 150 J/K 250 K 2000 K
Forecast: Both terms help — energy falls and mess rises. Guess: spontaneous everywhere.
Match units. Δ S = 150 J/K = 0.150 kJ/K Why this step? Δ H − T Δ S At 250 K: Δ G = − 200 − ( 250 ) ( 0.150 ) = − 200 − 37.5 = − 237.5 kJ Why: plug into Δ G = Δ H − T Δ S At 2000 K: Δ G = − 200 − ( 2000 ) ( 0.150 ) = − 200 − 300 = − 500 kJ Why: raising T − T Δ S more negative here, since Δ S > 0
Verify: Both negative → spontaneous at every temperature. Line A (lavender) in the figure never crosses zero and slopes downward. ✓ Matches forecast.
Worked example Ex 2 — Forcing order while soaking heatΔ H = + 120 kJ Δ S = − 90 J/K T
Forecast: Both terms fight us — energy climbs and mess falls. Guess: never spontaneous.
Match units: Δ S = − 90 J/K = − 0.090 kJ/K Why this step? Same unit rule — Δ H Δ S T Δ S General expression: Δ G = 120 − T ( − 0.090 ) = 120 + 0.090 T Why this step? Writing Δ G T whole temperature behaviour at once, not just one point.Read it: for any T > 0 120 0.090 T Δ G > 0 Why: a positive constant plus a positive multiple of T Δ H Δ S
Verify: Try T = 1000 Δ G = 120 + 90 = 210 kJ > 0
Worked example Ex 3 — Freezing water,H 2 O ( l ) → H 2 O ( s ) Δ H = − 6.01 kJ/mol Δ S = − 22.0 J/mol⋅K T
Forecast: Energy wants it (heat released), mess hates it (liquid → orderly solid). Enthalpy and entropy disagree → temperature is the referee. Guess: spontaneous only when cold.
Match units: Δ S = − 22.0 J/mol⋅K = − 0.0220 kJ/mol⋅K Δ H = − 6.01 kJ/mol Why this step? We will divide Δ H Δ S Set Δ G = 0 to find the crossover.
Why this step? The two effects exactly cancel at the tie point; that temperature is the boundary between "go" and "no go."Solve: T = Δ S Δ H = − 0.0220 kJ/mol⋅K − 6.01 kJ/mol = 273.2 K Why: both negatives cancel, giving a real positive temperature (a physically valid crossover exists only when Δ H Δ S Pick a side and read the sign. At T = 270 K Δ G = − 6.01 − 270 ( − 0.0220 ) = − 6.01 + 5.94 = − 0.07 kJ < 0 T = 276 K Δ G = − 6.01 + 6.07 = + 0.06 kJ > 0 Why this step? Knowing the crossover is not enough — we must test one temperature on each side to learn which side is spontaneous, since Δ G = 0
Verify: Crossover is 273.2 K = 0 ∘ C Phase Transitions . ✓
Worked example Ex 4 — Word problem: baking limestoneA kiln decomposes limestone: CaCO 3 ( s ) → CaO ( s ) + CO 2 ( g ) Δ H = + 178 kJ Δ S = + 161 J/K What minimum kiln temperature makes this go? Will 1100 K
Forecast: Endothermic (needs heat), but a gas is born (mess shoots up). Guess: only hot enough.
Match units: Δ H = 178 kJ = 178000 J Δ S = 161 J/K Why this step? Here we convert Δ H up to joules (×1000) so it matches Δ S Crossover temperature: T = Δ S Δ H = 161 J/K 178000 J = 1105.6 K Why this step? Above this T T Δ S Δ H Δ G Test 1100 K and read the sign: Δ G = 178 − 1100 ( 0.161 ) = 178 − 177.1 = + 0.9 kJ > 0 Why this step? 1100 < 1105.6 Δ G Conclusion: need T > 1106 K 1100 K
Verify: At 1200 K Δ G = 178 − 1200 ( 0.161 ) = 178 − 193.2 = − 15.2 kJ < 0 1106 K
Worked example Ex 5 — WhenΔ S = 0 Δ H = − 15 kJ Δ S = 0
Forecast: No entropy term at all → the scoreboard is just the energy.
Substitute Δ S = 0 Δ G = Δ H − T ( 0 ) = Δ H = − 15 kJ Why this step? The T Δ S Δ G Δ H every temperature.Read: Δ G = − 15 kJ < 0 T
Verify: In the figure a Δ S = 0 horizontal (slope − Δ S = 0 Δ H < 0
Cell E′ — the double-zero corner: if both Δ H = 0 and Δ S = 0 Δ G = 0 − T ( 0 ) = 0 every temperature. There is neither an energy push nor an entropy push, so Δ G = 0 no spontaneous drive in either direction. This is the flat line lying exactly on the grey zero-axis. It completes the taxonomy: every combination of { − , 0 , + } Δ H Δ S
Worked example Ex 6 — Mixing two ideal gasesTwo ideal gases mix with Δ H = 0 Δ S = + 12 J/K
Forecast: No energy help or hindrance — only messiness talks. Mess goes up → should go.
Match units: Δ S = + 12 J/K = + 0.012 kJ/K Why this step? Even though Δ H = 0 Δ G Δ S Substitute Δ H = 0 Δ G = 0 − T Δ S = − T Δ S Why this step? With enthalpy gone, spontaneity rides entirely on the sign of − T Δ S Read the sign: since T > 0 Δ S = + 0.012 kJ/K > 0 − T ( 0.012 ) T > 0 Why this step? A negative times two positives is negative — so Δ G < 0 At 298 K: Δ G = − 298 ( 0.012 ) = − 3.58 kJ
Verify: In the figure a Δ H = 0 origin (intercept Δ H = 0 T > 0
Worked example Ex 7 — Which term survives at the extremes?Take the limestone reaction from Ex 4 (Δ H = + 178 kJ Δ S = + 0.161 kJ/K Δ G T → 0 T → ∞
Forecast: Cold → energy rules; hot → entropy rules (parent's "Cold loves H, Hot loves S").
T → 0 Δ G → Δ H − 0 = + 178 kJ > 0 Why this step? The T Δ S T → ∞ − T Δ S = − T ( 0.161 ) + 178 Δ G → − ∞ Why: the entropy term is linear in T Δ H T Δ S > 0 Numbers: at T = 10 K Δ G = 178 − 1.61 = 176.4 kJ T = 5000 K Δ G = 178 − 805 = − 627 kJ
Verify: These are the two ends of line D (butter) in the figure: it starts near its intercept + 178 T = 0 Second Law of Thermodynamics — at high T
Definition The reaction quotientQ a A + b B → c C + d D reaction quotient is
Q = [ A ] a [ B ] b [ C ] c [ D ] d Q how far along the reaction is right now : Q Q current, not-yet-settled amounts. More in Chemical Equilibrium and K .
Worked example Ex 8 — A "non-spontaneous" reaction that runsA reaction has Δ G ∘ = + 5.0 kJ/mol 298 K Q = 1.0 × 1 0 − 3 now ?
Forecast: Standard says no, but starved of product a reaction often wants to go forward. Guess: yes, spontaneous now.
Use the full form Δ G = Δ G ∘ + R T ln Q Why this step? Δ G ∘ actual Δ G Q Δ G Δ G ∘ Compute R T ln Q R = 8.314 J/mol⋅K R T = 8.314 × 298 = 2477.6 J/mol ln ( 1.0 × 1 0 − 3 ) = − 6.9078 = 2477.6 × ( − 6.9078 ) = − 17117 J/mol = − 17.1 kJ/mol Why: Q < 1 ln Q Δ G Add: Δ G = + 5.0 + ( − 17.1 ) = − 12.1 kJ/mol < 0
Verify: Negative → spontaneous right now , even though Δ G ∘ > 0 Q Chemical Equilibrium and K . ✓ Matches forecast.
Definition The equilibrium constantK Q equilibrium constant K Q K = [ A ] a [ B ] b [ C ] c [ D ] d equilibrium K T K ≫ 1 K ≪ 1 Δ G ∘ = − R T ln K Chemical Equilibrium and K .
Worked example Ex 9 — FromΔ G ∘ K Δ G ∘ = − 33.3 kJ 298 K K
Forecast: Strongly negative Δ G ∘ K
Rearrange Δ G ∘ = − R T ln K ln K = R T − Δ G ∘ Why this step? This bridge (derived in the parent framework) converts the energy scoreboard into a measurable equilibrium ratio.Numbers: ln K = 8.314 × 298 33300 = 2477.6 33300 = 13.44 Exponentiate: K = e 13.44 ≈ 6.9 × 1 0 5
Verify: K ≫ 1 Δ G ∘ < 0 Electrochemistry ΔG = −nFE , where the same Δ G ∘
Recall Which cell is which — quick self-testΔ H < 0 , Δ S > 0 Δ H > 0 , Δ S < 0 T = Δ H /Δ S Δ S = 0 Δ G Δ G = Δ H Δ H = 0 Δ G = − T Δ S Δ H = 0 Δ S = 0 Δ G 0 T T → ∞ Δ S > 0 Δ G → − ∞ T Δ G Δ G ∘ Δ G = Δ G ∘ + R T ln Q Δ G Q
Mnemonic Reading any line in the master figureIntercept = Δ H − Δ S A line dipping below zero = spontaneous there. Horizontal line = Δ S = 0 Δ H = 0
