System, state functions, work and the first law
Link to Std 9–10 : You know that energy changes from one form to another and is never lost; thermodynamics counts exactly how much heat and work move in a chemical change.
Think of a bank account. The balance is like internal energy; deposits and withdrawals are like heat and work. The balance depends only on where you stand now, not on how you got there.
The system is the part of the universe we study; everything else is the surroundings. The first law says energy is conserved: ΔU = q + w, where q is heat absorbed and w is work done on the system.
In the figure : Sign convention: heat entering the system is positive; heat leaving is negative. Work done on the system is positive; work done by the system is negative.
Rule 1: Types of systems and state functions — An open system exchanges matter and energy, a closed system exchanges only energy, and an isolated system exchanges neither. A state function depends only on the present state (U, H, S, G, p, V, T); q and w depend on the path.
- Easy : A sealed flask in a water bath is a closed system.
- Medium : A thermos flask is nearly an isolated system.
- Exam-type : Heat (q) and work (w) are not state functions, but their sum ΔU is.
Rule 2: Work of expansion — Work done on the system w = −p(ext) × ΔV. In free expansion into vacuum p(ext) = 0, so w = 0. 1 L atm = 101.325 J.
- Easy : A gas expands from 2.0 L to 5.0 L against 1.0 atm: w = −1.0 × 3.0 = −3.0 L atm = −304 J.
- Medium : For free expansion into vacuum, w = 0.
- Exam-type : Isothermal reversible expansion of 1 mol of ideal gas at 300 K to double its volume: w = −2.303 nRT log(V₂/V₁) = −2.303 × 1 × 8.314 × 300 × 0.301 ≈ −1729 J.
Rule 3: First law in use — ΔU = q + w. For an isothermal change of an ideal gas ΔU = 0, so q = −w.
- Easy : A system absorbs 500 J of heat and does 200 J of work: ΔU = 500 − 200 = +300 J.
- Medium : Surroundings do 150 J of work on a gas while it loses 400 J of heat: ΔU = 150 − 400 = −250 J.
- Exam-type : In the 1 mol gas expansion above, ΔU = 0, so q = +1729 J; the heat taken in is converted completely to work.
Rule 4: Internal energy — U is the total of all kinetic and potential energies of particles in the system. Only the change ΔU can be measured.
- Easy : Heating a gas at constant volume raises its U.
- Medium : For an ideal gas U depends only on temperature.
- Exam-type : An adiabatic process has q = 0, so ΔU = w.
A state function has the same change however the system gets from the start to the end.
Getting the sign of w wrong. Work done by the system on the surroundings is negative.
Memory clue: "Into the system is plus" — heat in or work in is positive, heat out or work out is negative.
Spot the difference
| Point | State function | Path function |
|---|---|---|
| Depends on | Initial and final state only | Path taken |
| Examples | U, H, S, G | q, w |
| Change | Always the same | Can differ |
Practice questions
- 1Define an isolated system.
- 2Give two examples of state functions.
- 3What is w in free expansion into a vacuum?
- 4A gas expands from 2.0 L to 5.0 L against 1.0 atm. Find w in joules.
- 5A system absorbs 500 J of heat and does 200 J of work. Find ΔU.
- 6A gas is compressed with 150 J of work and loses 400 J of heat. Find ΔU.
- 7Find w and q for isothermal reversible doubling of the volume of 1 mol of ideal gas at 300 K.
- 8Why is ΔU zero for an isothermal change of an ideal gas?
Show answers (try them yourself first)
- 1A system that exchanges neither matter nor energy with its surroundings.
- 2Internal energy and enthalpy (also entropy, pressure, volume).
- 3Zero, since the external pressure is zero.
- 4w = −1.0 × (5.0 − 2.0) = −3.0 L atm = −3.0 × 101.325 ≈ −304 J.
- 5ΔU = 500 − 200 = +300 J.
- 6ΔU = +150 − 400 = −250 J.
- 7w = −2.303 × 1 × 8.314 × 300 × log 2 ≈ −1729 J and q = +1729 J.
- 8U of an ideal gas depends only on temperature; the temperature does not change, so ΔU = 0.
Enthalpy: heat at constant pressure
Link to Std 9–10 : You saw that some reactions give out heat (burning) and others absorb it (dissolving some salts); enthalpy puts a number on this.
A hot-pack warms your hands because a reaction releases heat; an instant cold-pack cools because dissolving absorbs heat.
Enthalpy H = U + pV. The heat taken in at constant pressure equals the change in enthalpy, ΔH = q(p). For reactions of gases: ΔH = ΔU + Δn(g)RT.
Enthalpy: heat at constant pressure
In the app you can move this figure with your finger.
Open in the appIn the figure : In an exothermic reaction the products have less enthalpy than the reactants (ΔH < 0). In an endothermic reaction the products have more (ΔH > 0). The calculator below finds ΔH − ΔU for any reaction.
Rule 1: Sign of ΔH — ΔH is negative when heat is released and positive when heat is absorbed.
- Easy : Burning methane has ΔH < 0 (exothermic).
- Medium : Melting of ice has ΔH > 0 (endothermic).
- Exam-type : If ΔH = −50 kJ, the reverse reaction has ΔH = +50 kJ.
Rule 2: ΔH and ΔU — Δn(g) = moles of gaseous products − moles of gaseous reactants. Solids and liquids are ignored. R = 8.314 J K⁻¹ mol⁻¹.
- Easy : For 2H₂(g) + O₂(g) → 2H₂O(l), Δn(g) = 0 − 3 = −3, so ΔH − ΔU = −3 × 8.314 × 298 ÷ 1000 = −7.43 kJ.
- Medium : For N₂O₄(g) → 2NO₂(g), Δn(g) = +1, so ΔH − ΔU = +8.314 × 298 ÷ 1000 = +2.48 kJ.
- Exam-type : For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890.3 kJ and Δn(g) = 1 − 3 = −2. ΔU = ΔH − Δn RT = −890.3 + 4.96 = −885.3 kJ.
Rule 3: Standard conditions — Standard enthalpy change ΔH° is for substances in their standard states at 1 bar pressure, usually quoted at 298 K.
- Easy : The standard state of oxygen is O₂ gas at 1 bar.
- Medium : The standard state of carbon is graphite.
- Exam-type : The standard enthalpy of formation of an element in its standard state is zero.
Rule 4: Thermochemical equations — The enthalpy change goes with the balanced equation. Doubling the equation doubles ΔH, and reversing it changes the sign. State symbols are needed.
- Easy : If H₂ + ½O₂ → H₂O(l) has ΔH = −285.8 kJ, then 2H₂ + O₂ → 2H₂O(l) has ΔH = −571.6 kJ.
- Medium : H₂O(l) → H₂ + ½O₂ has ΔH = +285.8 kJ.
- Exam-type : H₂O(g) and H₂O(l) have different ΔH values for the same reaction because condensation releases heat.
At constant pressure, heat exchanged is the enthalpy change; at constant volume it is the internal energy change.
Counting solids and liquids in Δn(g). Only gases count.
Memory clue: "ΔH = ΔU + ΔnRT; only gas moles count".
Spot the difference
| Point | Exothermic | Endothermic |
|---|---|---|
| ΔH | Negative | Positive |
| Surroundings | Warm up | Cool down |
| Example | Combustion | Dissolving NH₄NO₃ |
Practice questions
- 1Define enthalpy.
- 2What is the sign of ΔH for an exothermic reaction?
- 3What is the standard enthalpy of formation of O₂(g)?
- 4Find ΔH − ΔU at 298 K for 2H₂(g) + O₂(g) → 2H₂O(l).
- 5Find ΔH − ΔU at 298 K for N₂O₄(g) → 2NO₂(g).
- 6If H₂ + ½O₂ → H₂O(l) has ΔH = −285.8 kJ, what is ΔH for 2H₂O(l) → 2H₂ + O₂?
- 7CH₄ + 2O₂ → CO₂ + 2H₂O(l) has ΔH = −890.3 kJ at 298 K. Find ΔU.
- 8Why is Δn(g) taken only for gases?
Show answers (try them yourself first)
- 1H = U + pV; ΔH is the heat exchanged at constant pressure.
- 2Negative.
- 3Zero.
- 4Δn(g) = −3, so −3 × 8.314 × 298 ÷ 1000 = −7.43 kJ.
- 5Δn(g) = +1, so +2.48 kJ.
- 6+571.6 kJ.
- 7Δn(g) = −2, so ΔU = −890.3 + 2 × 8.314 × 298 ÷ 1000 = −885.3 kJ.
- 8The volume change, and so the pV work, comes almost entirely from gases; the volume of solids and liquids is negligible.
Heat capacity and calorimetry
Link to Std 9–10 : You know that different materials heat up at different speeds; here we measure the heat from the temperature rise.
A metal spoon in hot tea warms quickly, but the tea itself takes much longer to cool. Water needs a lot of heat for a small temperature rise.
Heat capacity (C) is the heat needed to raise the temperature of a body by 1 K. Specific heat (c) is the heat needed for 1 g (or 1 kg). Heat absorbed is q = m × c × ΔT.
In the figure : A sample burns in oxygen inside a sealed steel vessel (bomb) placed in an insulated water bath. The rise in temperature of the bath gives the heat of the reaction.
Rule 1: Heat from temperature change — q = m c ΔT, with c of water = 4.184 J g⁻¹ K⁻¹.
- Easy : Heating 250 g of water by 4.0 K needs 250 × 4.184 × 4.0 = 4184 J.
- Medium : The same heat given to 500 g of water raises the temperature by only 2.0 K.
- Exam-type : When a hot metal is dropped in water at constant total energy, heat lost by the metal equals heat gained by the water.
Rule 2: Calorimeter — A bomb calorimeter at constant volume measures ΔU. Heat released = calorimeter constant (C) × temperature rise. A coffee-cup calorimeter at constant pressure measures ΔH.
- Easy : A calorimeter constant is 8.0 kJ K⁻¹ and the temperature rises by 3.0 K: heat = 24 kJ.
- Medium : If 2.0 g of a substance (M = 40 g mol⁻¹) gives that 24 kJ, then 0.050 mol gives 24 kJ, so ΔU = −24 ÷ 0.050 = −480 kJ mol⁻¹.
- Exam-type : A bomb calorimeter gives ΔU. To find ΔH convert with ΔH = ΔU + Δn(g)RT.
Rule 3: Cp and Cv — For an ideal gas Cp − Cv = R per mole. For a monatomic gas Cv = (3/2)R and Cp = (5/2)R.
- Easy : For a monatomic gas Cv = 1.5 × 8.314 = 12.47 J K⁻¹ mol⁻¹ and Cp = 20.79 J K⁻¹ mol⁻¹.
- Medium : 2 mol of gas with Cv = 12.5 J K⁻¹ mol⁻¹ heated by 40 K at constant volume needs q = 2 × 12.5 × 40 = 1000 J.
- Exam-type : At constant pressure the same heating needs q = 2 × (12.5 + 8.314) × 40 = 1665 J, more than at constant volume because the gas also does expansion work.
Rule 4: Water's large specific heat — Because of its high c, water stores much heat, which keeps coastal climates mild and makes water a good coolant.
- Easy : Sea breeze cools the coast in summer because water warms slowly.
- Medium : Water in a car radiator carries away engine heat.
- Exam-type : Sand heats faster than water in the Sun because its specific heat is much lower.
At constant volume no pV work is done, so the heat equals ΔU.
Mixing up heat capacity (for the whole body) and specific heat (per gram).
Memory clue: "q = m c ΔT, then ÷ moles for molar".
Spot the difference
| Point | Bomb calorimeter | Coffee-cup calorimeter |
|---|---|---|
| Condition | Constant volume | Constant pressure |
| Measures | ΔU | ΔH |
| Used for | Combustion | Reactions in solution |
Practice questions
- 1Define specific heat.
- 2How much heat raises 250 g of water by 4.0 K?
- 3Which calorimeter measures ΔU?
- 4A calorimeter constant is 8.0 kJ K⁻¹ and the temperature rise is 3.0 K. How much heat is released?
- 5If 2.0 g of a substance (M = 40 g mol⁻¹) gives that heat, find ΔU per mole.
- 6Find Cv and Cp for a monatomic ideal gas.
- 7How much heat is needed to heat 2 mol of gas (Cv = 12.5 J K⁻¹ mol⁻¹) by 40 K at constant volume, and at constant pressure?
- 8Why is Cp greater than Cv?
Show answers (try them yourself first)
- 1The heat needed to raise the temperature of 1 g of a substance by 1 K.
- 2250 × 4.184 × 4.0 = 4184 J.
- 3The bomb calorimeter.
- 48.0 × 3.0 = 24 kJ.
- 52.0 ÷ 40 = 0.050 mol, so ΔU = −24 ÷ 0.050 = −480 kJ mol⁻¹.
- 6Cv = (3/2)R = 12.47 and Cp = (5/2)R = 20.79 J K⁻¹ mol⁻¹.
- 7At constant volume q = 2 × 12.5 × 40 = 1000 J; at constant pressure q = 2 × 20.81 × 40 ≈ 1665 J.
- 8At constant pressure the gas expands and does work, so more heat is needed for the same temperature rise.
Hess's law and enthalpies of formation, combustion and bond
Link to Std 9–10 : You know that reactions can be added; here we add their heats in the same way to find values we cannot measure directly.
Climbing from the ground to the roof gives the same height gain whether you take the stairs or the lift. Enthalpy change too depends only on start and end.
Hess's law: the total enthalpy change of a reaction is the same whether it takes place in one step or in several. Therefore, ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants).
In the figure : Burning graphite directly to CO₂ gives −393.5 kJ. The two-step route (to CO, then CO to CO₂) gives −110.5 and −283.0, which add up to the same value.
Rule 1: Adding reactions — Reverse a reaction and its ΔH changes sign; multiply a reaction and ΔH is multiplied.
- Easy : C + O₂ → CO₂ (−393.5) and CO + ½O₂ → CO₂ (−283.0). Subtracting gives C + ½O₂ → CO with ΔH = −393.5 + 283.0 = −110.5 kJ.
- Medium : If A → B has ΔH = −40 and B → C has ΔH = −25, then A → C has ΔH = −65 kJ.
- Exam-type : For 2A → 2B with ΔH(A → B) = −40, ΔH = −80 kJ; the reverse 2B → 2A has +80 kJ.
Rule 2: Using enthalpies of formation — Use the formula with the stoichiometric numbers.
- Easy : CH₄ + 2O₂ → CO₂ + 2H₂O(l): ΔfH° (CO₂) = −393.5, (H₂O, l) = −285.8, (CH₄) = −74.8. ΔrH° = −393.5 + 2(−285.8) − (−74.8) = −890.3 kJ.
- Medium : CaCO₃(s) → CaO(s) + CO₂(g): with −1206.9 for CaCO₃, −635.1 for CaO and −393.5 for CO₂, ΔrH° = −635.1 − 393.5 + 1206.9 = +178.3 kJ.
- Exam-type : From the CH₄ combustion above, the heat released on burning 8 g of CH₄ (0.5 mol) is 0.5 × 890.3 = 445 kJ.
Rule 3: Bond enthalpies — ΔrH° ≈ Σ (bonds broken) − Σ (bonds formed), using average bond enthalpies of gaseous species.
- Easy : H₂ + Cl₂ → 2HCl: broken 436 + 242 = 678 kJ; formed 2 × 431 = 862 kJ; ΔH = 678 − 862 = −184 kJ.
- Medium : Bond enthalpies give only approximate answers because they are averages.
- Exam-type : A reaction is exothermic when the bonds formed are stronger in total than the bonds broken.
Rule 4: Enthalpies of phase change — Fusion for melting and vaporisation for boiling. For water ΔfusH = 6.01 kJ mol⁻¹ and ΔvapH = 40.7 kJ mol⁻¹.
- Easy : Melting 36 g (2 mol) of ice needs 2 × 6.01 = 12.02 kJ.
- Medium : Vaporising 9 g (0.5 mol) of water needs 0.5 × 40.7 = 20.35 kJ.
- Exam-type : ΔvapH is much larger than ΔfusH because boiling separates molecules completely, while melting only loosens them.
ΔH for the same overall change is the same for any route; only the start and the end matter.
Forgetting to multiply ΔfH by the stoichiometric coefficient.
Memory clue: "Products minus reactants" for formation data, and "broken minus formed" for bond data.
Spot the difference
| Point | Using ΔfH° | Using bond enthalpy |
|---|---|---|
| Formula | Σ products − Σ reactants | Σ broken − Σ formed |
| Accuracy | Good | Approximate |
| Data | Formation values | Average bond values |
Practice questions
- 1State Hess's law.
- 2What is ΔfH° of an element in its standard state?
- 3Which is larger for water, ΔfusH or ΔvapH?
- 4Find ΔH for C + ½O₂ → CO from C + O₂ → CO₂ (−393.5) and CO + ½O₂ → CO₂ (−283.0).
- 5Find ΔH for H₂ + Cl₂ → 2HCl from the bond enthalpies H–H 436, Cl–Cl 242 and H–Cl 431 kJ mol⁻¹.
- 6How much heat is needed to melt 36 g of ice?
- 7Find ΔrH° for CaCO₃(s) → CaO(s) + CO₂(g) using ΔfH° of CaCO₃ = −1206.9, CaO = −635.1 and CO₂ = −393.5 kJ mol⁻¹.
- 8Why do bond enthalpy calculations give only approximate results?
Show answers (try them yourself first)
- 1The total enthalpy change of a reaction is the same whether it occurs in one step or many.
- 2Zero.
- 3ΔvapH (40.7 against 6.01 kJ mol⁻¹).
- 4−393.5 − (−283.0) = −110.5 kJ.
- 5436 + 242 − 2 × 431 = −184 kJ.
- 636 g = 2 mol, so 2 × 6.01 = 12.02 kJ.
- 7−635.1 − 393.5 − (−1206.9) = +178.3 kJ.
- 8They use average bond enthalpies, but the actual value depends on the molecule.
Spontaneity and entropy
Link to Std 9–10 : You know that heat flows from hot to cold and a gas spreads to fill a room; entropy explains why these changes happen by themselves.
A tidy desk becomes messy without effort, but a messy desk does not tidy itself. Nature moves towards more disorder.
A spontaneous process occurs on its own without outside help. Entropy (S) is a measure of the disorder (the number of ways to arrange the particles). For a reversible change at temperature T, ΔS = q(rev) ÷ T.
In the figure : Particles are ordered in a solid, more free in a liquid and fully random in a gas. Entropy rises in the order solid < liquid < gas.
Rule 1: Predicting the sign of ΔS — ΔS is positive when disorder increases: melting, vaporising, dissolving, a reaction that increases moles of gas.
- Easy : Melting of ice has ΔS > 0.
- Medium : N₂ + 3H₂ → 2NH₃ has ΔS < 0 because 4 moles of gas become 2.
- Exam-type : CaCO₃(s) → CaO(s) + CO₂(g) has ΔS > 0 because a gas is formed.
Rule 2: Calculating ΔS of phase change — At the transition temperature, ΔS = ΔH(transition) ÷ T.
- Easy : Fusion of ice at 273.15 K: ΔS = 6010 ÷ 273.15 = 22.0 J K⁻¹ mol⁻¹.
- Medium : Vaporisation of water at 373.15 K: ΔS = 40 700 ÷ 373.15 = 109.1 J K⁻¹ mol⁻¹.
- Exam-type : Vaporisation has the larger ΔS because the gas has far more disorder than the liquid.
Rule 3: Second law — A process is spontaneous if the total entropy of the system and the surroundings increases: ΔS(total) = ΔS(system) + ΔS(surroundings) > 0. For surroundings ΔS = −ΔH(system) ÷ T.
- Easy : An exothermic reaction with ΔH = −60 kJ at 300 K gives ΔS(surr) = +60 000 ÷ 300 = +200 J K⁻¹.
- Medium : If ΔS(system) = −50 J K⁻¹, then ΔS(total) = −50 + 200 = +150 J K⁻¹ > 0, so it is spontaneous.
- Exam-type : The entropy of the universe never decreases in a spontaneous process.
Rule 4: Third law — The entropy of a perfect crystal at 0 K is zero, so absolute entropies can be found.
- Easy : Entropy of a pure crystal is least at 0 K.
- Medium : Entropy rises when the temperature of a substance is raised.
- Exam-type : Because of the third law, standard molar entropies S° are always positive.
A process can be spontaneous even if the entropy of the system falls, when the surroundings gain more.
Using ΔS(system) alone to judge spontaneity.
Memory clue: "Solid < liquid < gas; more gas moles means more S".
Spot the difference
| Point | Spontaneous | Non-spontaneous |
|---|---|---|
| Needs outside help | No | Yes |
| ΔS(total) | Positive | Negative |
| Example | Ice melting at 25 °C | Water freezing at 25 °C |
Practice questions
- 1Define entropy.
- 2Which has higher entropy, ice or steam?
- 3State the third law of thermodynamics.
- 4Predict the sign of ΔS for N₂ + 3H₂ → 2NH₃.
- 5Find ΔS for fusion of ice at 273.15 K (ΔH = 6.01 kJ mol⁻¹).
- 6Find ΔS(surroundings) when a reaction with ΔH = −60 kJ occurs at 300 K.
- 7If ΔS(system) = −50 J K⁻¹ in the reaction of question 6, is it spontaneous?
- 8Why does a gas expand into a vacuum spontaneously although no heat is exchanged?
Show answers (try them yourself first)
- 1A measure of the disorder of a system.
- 2Steam.
- 3The entropy of a perfect crystal at 0 K is zero.
- 4Negative, because 4 moles of gas change to 2.
- 56010 ÷ 273.15 = 22.0 J K⁻¹ mol⁻¹.
- 6+60 000 ÷ 300 = +200 J K⁻¹.
- 7ΔS(total) = −50 + 200 = +150 J K⁻¹ > 0, so it is spontaneous.
- 8The gas can occupy a much larger volume in many more ways, so the entropy of the system rises while the surroundings do not change.
Gibbs energy and equilibrium
Link to Std 9–10 : You know that some changes need heating and some do not; Gibbs energy combines energy and disorder into a single check for spontaneity.
Ice melts above 0 °C but freezes below it. Neither energy nor disorder alone decides this; the temperature decides which one wins.
Gibbs energy G = H − TS. At constant temperature and pressure, ΔG = ΔH − TΔS. A process is spontaneous when ΔG < 0, at equilibrium when ΔG = 0.
Gibbs energy and equilibrium
In the app you can move this figure with your finger.
Open in the appIn the figure : Lines of ΔG against T. A reaction with ΔH > 0 and ΔS > 0 becomes spontaneous at high temperature; one with ΔH < 0 and ΔS < 0 is spontaneous at low temperature. The calculator below gives ΔG and the crossover temperature.
Rule 1: Four sign cases — ΔH − and ΔS + : always spontaneous. ΔH + and ΔS − : never spontaneous. ΔH − and ΔS − : spontaneous at low T. ΔH + and ΔS + : spontaneous at high T.
- Easy : Burning of fuel (ΔH < 0, ΔS > 0) is spontaneous at all temperatures.
- Medium : Melting of ice (ΔH > 0, ΔS > 0) is spontaneous above 273 K.
- Exam-type : A reaction with ΔH > 0 and ΔS < 0 can never become spontaneous at any temperature.
Rule 2: Calculating ΔG and the crossover temperature — T(eq) = ΔH ÷ ΔS when ΔG = 0.
- Easy : ΔH = +30 kJ, ΔS = +100 J K⁻¹. At 400 K, ΔG = 30 − 400 × 0.100 = −10 kJ: spontaneous.
- Medium : At 200 K the same reaction has ΔG = 30 − 20 = +10 kJ: non-spontaneous.
- Exam-type : The crossover temperature is 30 000 ÷ 100 = 300 K.
Rule 3: ΔG° and equilibrium constant — ΔG° = −2.303 RT log K. A negative ΔG° means K > 1.
- Easy : At 298 K, ΔG° = −11.4 kJ gives log K = 11 400 ÷ (2.303 × 8.314 × 298) ≈ 2.0, so K ≈ 100.
- Medium : K = 10 corresponds to ΔG° = −2.303 × 8.314 × 298 × 1 ÷ 1000 = −5.71 kJ.
- Exam-type : ΔG° = 0 means K = 1.
Rule 4: Free energy and useful work — The decrease in Gibbs energy is the maximum non-expansion (useful) work obtainable.
- Easy : The work from a fuel cell comes from the fall in Gibbs energy.
- Medium : A cell reaction with a large negative ΔG° gives a high cell voltage.
- Exam-type : ΔG is a state function and its sign decides the direction at the given temperature.
ΔG < 0 spontaneous, ΔG = 0 equilibrium, ΔG > 0 non-spontaneous in the forward direction.
Mixing kJ and J. ΔH is in kJ and ΔS in J K⁻¹; convert before using ΔH − TΔS.
Memory clue: "Heat out + more mess = always; heat in + more mess = hot" — ΔH negative with ΔS positive always works; both positive needs high T.
Spot the difference
| Point | ΔG < 0 | ΔG > 0 |
|---|---|---|
| Direction | Forward spontaneous | Backward spontaneous |
| K | Greater than 1 | Less than 1 |
| Example | Combustion | Reverse of combustion |
Practice questions
- 1Write the Gibbs equation.
- 2What is the sign of ΔG for a spontaneous process?
- 3What does ΔG° = 0 say about K?
- 4ΔH = +30 kJ, ΔS = +100 J K⁻¹. Find ΔG at 400 K.
- 5Find the temperature at which the reaction in question 4 is at equilibrium.
- 6Find K when ΔG° = −11.4 kJ at 298 K.
- 7For which sign of ΔH and ΔS is a reaction spontaneous only at low temperature?
- 8Why can a reaction with ΔH > 0 still be spontaneous?
Show answers (try them yourself first)
- 1ΔG = ΔH − TΔS.
- 2Negative.
- 3K = 1.
- 4ΔG = 30 − 400 × 0.100 = −10 kJ.
- 5T = 30 000 ÷ 100 = 300 K.
- 6log K ≈ 2.0, so K ≈ 100.
- 7ΔH negative and ΔS negative.
- 8If ΔS is positive and T is high enough that TΔS is larger than ΔH, ΔG becomes negative.