Home›Chemistry›Thermochemistry
⚗️ Chemistry · Thermochemistry

Memory tricks for enthalpy, entropy & Gibbs energy

From ΔH to ΔG to Hess's law — thermochemistry is all about signs and direction. These memory tricks lock in when reactions are spontaneous, how to use Hess's law, and how to interpret every energy quantity on the exam.

Study with others
📋
Thermochemistry Forum
Ask questions · Share techniques
 💬
Live Study Room
Study together now
🌡️ Thermochemistry — 9 Memory Tricks  ·  Click any card to expand · Save favorites · Switch to Flashcard or Quiz mode below
🎥 How Flashcards Work
A quick walkthrough of tap-to-flip, rating, and how card colors track what you're struggling with.
← Back Next →
Thermochemistry deck1 of 9
Tap to flip
← →
How well do YOU think you know this?
Easy Medium Hard Harder
Tap to flip back
Thermochemistry deck
Easy0
Medium0
Hard0
Harder0
Enthalpy
EXO exits · ENDO enters
Exothermic ΔH < 0 (heat released) · Endothermic ΔH > 0 (heat absorbed)
EXO exits — heat leaves the system, surroundings warm up, ΔH is negative. ENDO enters — heat flows into the system, surroundings cool down, ΔH is positive. Hand warmers = exo. Ice packs = endo. Combustion is always exothermic. Photosynthesis is endothermic.
📖 Full Lesson → Difficulty: Beginner
Energy diagram — exothermic
Products are LOWER than reactants. Energy is released. ΔH = H(products) - H(reactants) = negative value.
Energy diagram — endothermic
Products are HIGHER than reactants. Energy is absorbed. ΔH = positive value. System gains potential energy.
Bond energy connection
Breaking bonds = endothermic (requires energy). Forming bonds = exothermic (releases energy). ΔH_rxn = Σ(bonds broken) - Σ(bonds formed).
Standard enthalpy
ΔH° measured at 25°C and 1 atm. ΔH°_rxn = Σ ΔH°f(products) - Σ ΔH°f(reactants). Elements in standard state have ΔH°f = 0.
🎥 Watch Instead
▶
Video coming soon
This lesson's animated video hasn't been made yet — check back soon.
Flashcard
🃏 Enthalpy
Exothermic vs endothermic — which way does heat flow?
Tap to flip
🃏 Answer
EXO exits · ENDO enters
Energy diagram — exothermic — Products are LOWER than reactants. Energy is released. ΔH = H(products) - H(reactants) = negative value.
Energy diagram — endothermic — Products are HIGHER than reactants. Energy is absorbed. ΔH = positive value. System gains potential energy.
Bond energy connection — Breaking bonds = endothermic (requires energy). Forming bonds = exothermic (releases energy). ΔH_rxn = Σ(bonds broken) - Σ(bonds formed).
Standard enthalpy — ΔH° measured at 25°C and 1 atm. ΔH°_rxn = Σ ΔH°f(products) - Σ ΔH°f(reactants). Elements in standard state have ΔH°f = 0.
Tap to flip back
Hess's Law
Path doesn't matter — only start and end
ΔH is a state function — add or reverse steps to get target reaction
Hess's Law: ΔH is independent of the path — only the initial and final states matter. To use it: flip equations to get reactants/products where needed (flip = change sign of ΔH), multiply by coefficients (multiply = multiply ΔH), then add all steps. The ΔH values add up to give the overall ΔH.
📖 Full Lesson → Difficulty: Intermediate
Flipping a reaction
If you reverse an equation, multiply its ΔH by -1. Exothermic forward = endothermic reverse and vice versa.
Multiplying a reaction
If you double all coefficients, double ΔH. If you halve, halve ΔH. ΔH scales proportionally with moles.
Cancellation strategy
Arrange steps so intermediates appear on opposite sides and cancel. What's left should match your target equation exactly.
State function meaning
H is a state function like altitude — it doesn't matter how you climbed the mountain, only how high you went. Same start and end = same ΔH regardless of route.
🎥 Watch Instead
▶
Video coming soon
This lesson's animated video hasn't been made yet — check back soon.
Flashcard
🃏 Hess's Law
Hess's law — what does it say?
Tap to flip
🃏 Answer
Path doesn't matter — only start and end
Flipping a reaction — If you reverse an equation, multiply its ΔH by -1. Exothermic forward = endothermic reverse and vice versa.
Multiplying a reaction — If you double all coefficients, double ΔH. If you halve, halve ΔH. ΔH scales proportionally with moles.
Cancellation strategy — Arrange steps so intermediates appear on opposite sides and cancel. What's left should match your target equation exactly.
State function meaning — H is a state function like altitude — it doesn't matter how you climbed the mountain, only how high you went. Same start and end = same ΔH regardless of route.
Tap to flip back
Entropy
Entropy = Disorder — more ways to arrange = higher S
ΔS > 0 = disorder increases · ΔS < 0 = disorder decreases
Entropy (S) measures the number of possible microstates — how many ways can the system be arranged? More disorder = higher entropy. Gas has far more entropy than liquid, which has more than solid. Mixing, dissolving, and expanding into more space all increase entropy. The universe tends toward maximum entropy (Second Law of Thermodynamics).
📖 Full Lesson → Difficulty: Intermediate
Predicting ΔS sign
ΔS positive: solid→liquid→gas, dissolution, more moles of gas produced, mixing. ΔS negative: condensation, crystallization, fewer moles of gas, purification.
Third Law
A perfect crystal at absolute zero (0 K) has S = 0. This is the reference point for all absolute entropy values. Pure substances have positive absolute entropy at any T above 0 K.
Standard entropy
ΔS°_rxn = Σ S°(products) - Σ S°(reactants). Unlike ΔH°f, standard entropy of pure elements ≠ 0 — they have real entropy values.
Boltzmann equation
S = k·ln(W) where W = number of microstates and k = Boltzmann constant. More microstates = higher entropy. Basis of statistical thermodynamics.
🎥 Watch Instead
▶
Video coming soon
This lesson's animated video hasn't been made yet — check back soon.
Flashcard
🃏 Entropy
Entropy — what is it?
Tap to flip
🃏 Answer
Entropy = Disorder — more ways to arrange = higher S
Predicting ΔS sign — ΔS positive: solid→liquid→gas, dissolution, more moles of gas produced, mixing. ΔS negative: condensation, crystallization, fewer moles of gas, purification.
Third Law — A perfect crystal at absolute zero (0 K) has S = 0. This is the reference point for all absolute entropy values. Pure substances have positive absolute entropy at any T above 0 K.
Standard entropy — ΔS°_rxn = Σ S°(products) - Σ S°(reactants). Unlike ΔH°f, standard entropy of pure elements ≠ 0 — they have real entropy values.
Boltzmann equation — S = k·ln(W) where W = number of microstates and k = Boltzmann constant. More microstates = higher entropy. Basis of statistical thermodynamics.
Tap to flip back
Gibbs Energy
ΔG = ΔH − TΔS · Negative = Spontaneous
ΔG < 0 = spontaneous · ΔG = 0 = equilibrium · ΔG > 0 = non-spontaneous
Gibbs free energy (ΔG) tells you if a reaction happens on its own. ΔG = ΔH − TΔS. Negative ΔG = spontaneous (favorable). Positive = non-spontaneous (needs energy input). Zero = system is at equilibrium. Temperature is the key variable — it can flip a reaction from non-spontaneous to spontaneous.
📖 Full Lesson → Difficulty: Intermediate
4 ΔH/ΔS combinations
(-H,+S)=always spontaneous · (+H,-S)=never spontaneous · (-H,-S)=spontaneous at low T · (+H,+S)=spontaneous at high T. Temperature determines the last two cases.
ΔG and equilibrium
ΔG° = -RT·ln(K). Large negative ΔG° → large K → products favored. Large positive ΔG° → small K → reactants favored. ΔG = 0 when Q = K (equilibrium).
Maximum work
-ΔG = maximum non-expansion work available from a process. Fuel cells convert chemical ΔG directly to electrical work. More negative ΔG = more work available.
Spontaneous ≠ fast
ΔG tells you IF a reaction can occur, not how fast. Diamond → graphite is spontaneous (ΔG negative) but takes millions of years. Kinetics determines rate; thermodynamics determines direction.
🎥 Watch Instead
▶
Video coming soon
This lesson's animated video hasn't been made yet — check back soon.
Flashcard
🃏 Gibbs Energy
Gibbs free energy — the equation, and which sign means spontaneous?
Tap to flip
🃏 Answer
ΔG = ΔH − TΔS · Negative = Spontaneous
4 ΔH/ΔS combinations — (-H,+S)=always spontaneous · (+H,-S)=never spontaneous · (-H,-S)=spontaneous at low T · (+H,+S)=spontaneous at high T. Temperature determines the last two cases.
ΔG and equilibrium — ΔG° = -RT·ln(K). Large negative ΔG° → large K → products favored. Large positive ΔG° → small K → reactants favored. ΔG = 0 when Q = K (equilibrium).
Maximum work — -ΔG = maximum non-expansion work available from a process. Fuel cells convert chemical ΔG directly to electrical work. More negative ΔG = more work available.
Spontaneous ≠ fast — ΔG tells you IF a reaction can occur, not how fast. Diamond → graphite is spontaneous (ΔG negative) but takes millions of years. Kinetics determines rate; thermodynamics determines direction.
Tap to flip back
Calorimetry
q = mcΔT — "Queue My Chemistry, ΔT!"
q = heat · m = mass · c = specific heat · ΔT = temperature change
q = mcΔT is the fundamental calorimetry equation. Queue (q) = heat transferred in joules. m = mass in grams. c = specific heat capacity in J/(g·°C). ΔT = T_final − T_initial. Water's specific heat = 4.184 J/(g·°C) — the highest of any common liquid, which is why it's such an effective coolant and why oceans moderate climate.
📖 Full Lesson → Difficulty: Beginner
Sign convention
q_system = -q_surroundings. If water heats up (+ΔT), reaction is exothermic → q_rxn is negative. If water cools down, reaction is endothermic → q_rxn positive.
Coffee cup calorimeter
Constant pressure (open to atmosphere) → measures ΔH directly. q_rxn = -q_solution = -mcΔT. Assumes no heat loss to surroundings or to the cup itself.
Bomb calorimeter
Constant volume → measures ΔE (internal energy), not ΔH directly. q_rxn = -C_cal × ΔT where C_cal = heat capacity of the calorimeter. Used for combustion reactions.
Specific heat values
Water = 4.184 J/(g·°C) · Aluminum = 0.900 · Iron = 0.449 · Copper = 0.385. Higher specific heat = harder to change temperature = better thermal reservoir.
🎥 Watch Instead
▶
Video coming soon
This lesson's animated video hasn't been made yet — check back soon.
Flashcard
🃏 Calorimetry
Calorimetry — the equation?
Tap to flip
🃏 Answer
q = mcΔT — "Queue My Chemistry, ΔT!"
Sign convention — q_system = -q_surroundings. If water heats up (+ΔT), reaction is exothermic → q_rxn is negative. If water cools down, reaction is endothermic → q_rxn positive.
Coffee cup calorimeter — Constant pressure (open to atmosphere) → measures ΔH directly. q_rxn = -q_solution = -mcΔT. Assumes no heat loss to surroundings or to the cup itself.
Bomb calorimeter — Constant volume → measures ΔE (internal energy), not ΔH directly. q_rxn = -C_cal × ΔT where C_cal = heat capacity of the calorimeter. Used for combustion reactions.
Specific heat values — Water = 4.184 J/(g·°C) · Aluminum = 0.900 · Iron = 0.449 · Copper = 0.385. Higher specific heat = harder to change temperature = better thermal reservoir.
Tap to flip back
Laws of Thermo
You can't win, you can't break even, you can't quit
1st: energy conserved · 2nd: entropy increases · 3rd: can't reach absolute zero
The three laws of thermodynamics summarized as a game you can't win: 1st Law — you can't create energy (can't win). 2nd Law — you always lose some to entropy (can't break even). 3rd Law — you can't reach absolute zero to start over (can't quit the game). Every energy conversion loses some to unusable heat — that's the 2nd Law in action.
📖 Full Lesson → Difficulty: Beginner
First Law (ΔE = q + w)
Energy of the universe is constant. ΔE = q + w. q = heat absorbed by system. w = work done ON system. Conservation of energy — total energy never created or destroyed.
Second Law
Total entropy of universe always increases in any spontaneous process. ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0. Explains why heat flows hot→cold, not cold→hot.
Third Law
Entropy of a perfect crystal at 0 K = 0. In practice, absolute zero is unattainable — you can get arbitrarily close but never reach it exactly.
Zeroth Law
Often forgotten: if A is in thermal equilibrium with B, and B with C, then A and C are in equilibrium with each other. Basis for temperature measurement.
🎥 Watch Instead
▶
Video coming soon
This lesson's animated video hasn't been made yet — check back soon.
Flashcard
🃏 Laws of Thermo
You can't win, you can't break even, you can't quit
Tap to flip
🃏 Answer
First Law (ΔE = q + w) — Energy of the universe is constant. ΔE = q + w. q = heat absorbed by system. w = work done ON system. Conservation of energy — total energy never created or destroyed.
Second Law — Total entropy of universe always increases in any spontaneous process. ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0. Explains why heat flows hot→cold, not cold→hot.
Third Law — Entropy of a perfect crystal at 0 K = 0. In practice, absolute zero is unattainable — you can get arbitrarily close but never reach it exactly.
Zeroth Law — Often forgotten: if A is in thermal equilibrium with B, and B with C, then A and C are in equilibrium with each other. Basis for temperature measurement.
Tap to flip back
Phase Changes
Heating curve flat lines = phase changes — ΔT = 0 during change
Temperature constant while solid melts or liquid boils — all energy goes to changing phase
During a phase change (melting, boiling), temperature stays constant even though heat is added. All energy goes into breaking intermolecular forces — none into raising temperature. On a heating curve, the flat horizontal lines are melting and boiling points. The slope of rising sections depends on specific heat; the length of flat sections depends on ΔH_fus or ΔH_vap.
📖 Full Lesson → Difficulty: Intermediate
ΔH_fus vs ΔH_vap
For water: ΔH_fus = 6.01 kJ/mol (melting) · ΔH_vap = 40.7 kJ/mol (boiling). Vaporization requires much more energy — all intermolecular forces must be overcome to separate molecules into gas.
Heating curve calculation
Heat ice from -20°C to steam at 120°C = 5 steps: heat ice (q=mcΔT) + melt ice (q=n·ΔH_fus) + heat water (q=mcΔT) + boil water (q=n·ΔH_vap) + heat steam (q=mcΔT).
Clausius-Clapeyron
ln(P₂/P₁) = -ΔH_vap/R × (1/T₂ - 1/T₁). Relates vapor pressure to temperature. Higher ΔH_vap = vapor pressure changes more steeply with temperature.
Sublimation
Solid → gas directly. ΔH_sub = ΔH_fus + ΔH_vap (by Hess's Law). Dry ice (CO₂) and iodine sublime at atmospheric pressure.
🎥 Watch Instead
▶
Video coming soon
This lesson's animated video hasn't been made yet — check back soon.
Flashcard
🃏 Phase Changes
Heating curves — what do the flat sections mean?
Tap to flip
🃏 Answer
Heating curve flat lines = phase changes — ΔT = 0 during change
ΔH_fus vs ΔH_vap — For water: ΔH_fus = 6.01 kJ/mol (melting) · ΔH_vap = 40.7 kJ/mol (boiling). Vaporization requires much more energy — all intermolecular forces must be overcome to separate molecules into gas.
Heating curve calculation — Heat ice from -20°C to steam at 120°C = 5 steps: heat ice (q=mcΔT) + melt ice (q=n·ΔH_fus) + heat water (q=mcΔT) + boil water (q=n·ΔH_vap) + heat steam (q=mcΔT).
Clausius-Clapeyron — ln(P₂/P₁) = -ΔH_vap/R × (1/T₂ - 1/T₁). Relates vapor pressure to temperature. Higher ΔH_vap = vapor pressure changes more steeply with temperature.
Sublimation — Solid → gas directly. ΔH_sub = ΔH_fus + ΔH_vap (by Hess's Law). Dry ice (CO₂) and iodine sublime at atmospheric pressure.
Tap to flip back
Bond Enthalpy
Break bonds IN · Form bonds OUT
ΔH_rxn ≈ Σ(bonds broken) − Σ(bonds formed) using average bond enthalpies
Break bonds IN (endothermic, add energy) · Form bonds OUT (exothermic, release energy). ΔH_rxn ≈ sum of bonds broken minus sum of bonds formed. If stronger bonds form than break → exothermic overall. Note: bond enthalpy method gives approximate values — standard enthalpies of formation give exact values. Use bond enthalpies when no ΔH°f data is given.
📖 Full Lesson → Difficulty: Intermediate
Bond strength order
Triple bonds > double bonds > single bonds in strength. C≡C (835 kJ/mol) > C=C (614) > C-C (347). Stronger bond = more energy to break = more stable molecule.
Example: H₂ + Cl₂ → 2HCl
Break: H-H (436) + Cl-Cl (243) = 679 kJ in. Form: 2 × H-Cl (432) = 864 kJ out. ΔH ≈ 679 - 864 = -185 kJ. Exothermic — stronger H-Cl bonds form.
Why approximate?
Bond enthalpies are averages across many molecules. A C-H bond in methane vs. in ethanol has slightly different energy. ΔH°f method uses actual measured values for specific compounds.
Resonance structures
For molecules with resonance (like benzene), use the average bond order when looking up bond enthalpies. C-C in benzene is between single and double (≈507 kJ/mol).
🎥 Watch Instead
▶
Video coming soon
This lesson's animated video hasn't been made yet — check back soon.
Flashcard
🃏 Bond Enthalpy
Bond enthalpy — does breaking or forming bonds release energy?
Tap to flip
🃏 Answer
Break bonds IN · Form bonds OUT
Bond strength order — Triple bonds > double bonds > single bonds in strength. C≡C (835 kJ/mol) > C=C (614) > C-C (347). Stronger bond = more energy to break = more stable molecule.
Example: H₂ + Cl₂ → 2HCl — Break: H-H (436) + Cl-Cl (243) = 679 kJ in. Form: 2 × H-Cl (432) = 864 kJ out. ΔH ≈ 679 - 864 = -185 kJ. Exothermic — stronger H-Cl bonds form.
Why approximate? — Bond enthalpies are averages across many molecules. A C-H bond in methane vs. in ethanol has slightly different energy. ΔH°f method uses actual measured values for specific compounds.
Resonance structures — For molecules with resonance (like benzene), use the average bond order when looking up bond enthalpies. C-C in benzene is between single and double (≈507 kJ/mol).
Tap to flip back
Spontaneity
ΔG table: TΔS wins at high T, ΔH wins at low T
4 combinations of ΔH and ΔS signs determine when reactions are spontaneous
ΔG = ΔH − TΔS. At low temperature, ΔH dominates (T is small so TΔS is small). At high temperature, TΔS dominates. The crossover point is where ΔG = 0: T = ΔH/ΔS. Below this T one phase is stable; above it the other is. Ice melting at exactly 0°C is the classic example — ΔH and TΔS exactly balance at 273 K.
📖 Full Lesson → Difficulty: Advanced
Case 1: ΔH < 0, ΔS > 0
Always spontaneous at all temperatures. Both enthalpy and entropy favor it. ΔG always negative. Example: combustion reactions.
Case 2: ΔH > 0, ΔS < 0
Never spontaneous at any temperature. Both factors oppose it. ΔG always positive. Reverse reaction is always favored instead.
Case 3: ΔH < 0, ΔS < 0
Spontaneous at LOW temperature (ΔH term dominates). Example: water freezing — exothermic but decreases entropy. Spontaneous below 0°C.
Case 4: ΔH > 0, ΔS > 0
Spontaneous at HIGH temperature (TΔS term dominates). Example: ice melting — endothermic but increases entropy. Spontaneous above 0°C.
🎥 Watch Instead
▶
Video coming soon
This lesson's animated video hasn't been made yet — check back soon.
Flashcard
🃏 Spontaneity
Spontaneity — how do temperature, ΔH and ΔS interact?
Tap to flip
🃏 Answer
ΔG table: TΔS wins at high T, ΔH wins at low T
Case 1: ΔH < 0, ΔS > 0 — Always spontaneous at all temperatures. Both enthalpy and entropy favor it. ΔG always negative. Example: combustion reactions.
Case 2: ΔH > 0, ΔS < 0 — Never spontaneous at any temperature. Both factors oppose it. ΔG always positive. Reverse reaction is always favored instead.
Case 3: ΔH < 0, ΔS < 0 — Spontaneous at LOW temperature (ΔH term dominates). Example: water freezing — exothermic but decreases entropy. Spontaneous below 0°C.
Case 4: ΔH > 0, ΔS > 0 — Spontaneous at HIGH temperature (TΔS term dominates). Example: ice melting — endothermic but increases entropy. Spontaneous above 0°C.
Tap to flip back
🎓 Common Exam Questions
0
correct
0
wrong
5
remaining
questions correct  · 
No saved cards yet.
Click ☆ Save on any memory trick to save it here.