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Chemical Thermodynamics

Ch. 4Std 12

Easy Overview

Why does ice melt above 0°C but not below? Why does gasoline burn spontaneously but diamond does not turn into graphite at room temperature (even though it is thermodynamically more stable)? Thermodynamics answers these questions. It is the energy accounting of chemical reactions — telling us which reactions can happen on their own, how much energy they release or absorb, and when they reach equilibrium. Every chemical reaction involves energy. A system is the part of the universe you are studying. The surroundings is everything else. Systems can be open (exchanges matter and energy), closed (exchanges only energy), or isolated (exchanges nothing). State functions (P, V, T, U, H, S, G) depend only on the current state, not how you got there. Path functions (heat q and work w) depend on the route taken. The First Law is energy conservation: ΔU = q + w. The most useful form for chemistry is enthalpy ΔH — heat absorbed or released at constant pressure. Exothermic reactions (ΔH < 0) release heat; endothermic (ΔH > 0) absorb heat. Hess's Law says ΔH is the same regardless of the path — a powerful tool for calculating unknown enthalpy changes. But enthalpy alone does not determine spontaneity. The Second Law says the entropy of the universe always increases in a spontaneous process. Entropy (S) is a measure of disorder. When a solid dissolves, a liquid evaporates, or a gas forms, entropy increases. Gibbs free energy ΔG = ΔH - TΔS combines both factors — a process is spontaneous when ΔG < 0. The relationship ΔG° = -RT ln K ties thermodynamics to the equilibrium constant. This chapter gives you the tools to predict whether a reaction will happen, how temperature affects it, and what conditions favor the products.

System, surroundings, and types of systems

The system is the specific part of the universe you are studying (the reaction mixture). The surroundings is everything else. The boundary separates them. Three types: (1) Open system — exchanges BOTH matter and energy. Example: a boiling beaker — water vapor escapes (matter), heat flows in (energy). (2) Closed system — exchanges ONLY energy, not matter. Example: a sealed flask on a hot plate. Most chemical reactions in the lab are closed. (3) Isolated system — exchanges NEITHER matter nor energy. Example: a thermos flask (ideally). The universe is the only truly isolated system. Understanding these is crucial because the mathematical treatment differs. ΔU = q_v (heat at constant volume) applies to closed systems with no PV work. ΔH = q_p (heat at constant pressure) is for closed systems open to the atmosphere. Board exam: 'Classify a living plant, a sealed soda bottle, and a thermos of coffee.' Plant: open. Soda bottle: closed. Thermos: approximately isolated.

State functions vs path functions

State functions depend only on the current state of the system — P, V, T, U, H, S, G. The change in a state function between two states is path-independent. Think of altitude: if you hike to the top of a mountain, your altitude depends only on where you are, not on the trail you took. Path functions (q and w) depend on the ROUTE taken. Climbing the same mountain via different routes involves different amounts of work and heat loss. The First Law: ΔU = q + w. ΔU is a state function (path-independent), but q and w individually are path functions — however, their sum (ΔU) is always the same. A common example: go from state A to state B by constant volume then constant pressure, or by a different path. ΔU is the same, but q and w differ. This is why we define thermodynamic quantities that are state functions — we can calculate their changes without knowing the path. Board exam question: 'Is work a state function?' NO.

First Law — ΔU = q + w

Energy cannot be created or destroyed, only converted. For a closed system: ΔU = q + w. q = heat absorbed BY the system (positive = heat flows in, negative = heat flows out). w = work done ON the system (positive = work done on system like compression, negative = system does work like expansion). For expansion work: w = -PΔV. Two important special cases: (1) Constant volume (ΔV = 0): no PV work, so ΔU = q_v — heat measured at constant volume equals internal energy change. Bomb calorimeters measure q_v. (2) Constant pressure (common in chemistry): ΔH = q_p — heat at constant pressure equals enthalpy change. For reactions: ΔH ≈ ΔU + Δn_g RT, where Δn_g = change in moles of gas. If no gases are involved, ΔH ≈ ΔU. Sign convention trap: when the system RELEASES heat (exothermic), q is NEGATIVE (heat leaving). When the system EXPANDS, w is NEGATIVE (system does work on surroundings). Build the habit of writing signs carefully.

Enthalpy (H) and enthalpy changes

Enthalpy H = U + PV. Since U, P, V are state functions, H is also a state function. At constant pressure, ΔH = ΔU + PΔV = q_p. Exothermic: ΔH < 0 — system loses heat to surroundings. Example: CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol. Endothermic: ΔH > 0 — system absorbs heat. Example: CaCO₃ → CaO + CO₂, ΔH = +178 kJ/mol. Thermochemical equations include states: H₂(g) + ½O₂(g) → H₂O(l), ΔH = -285.8 kJ/mol. The magnitude of ΔH depends on: (1) physical states (H₂O(l) vs H₂O(g) — difference of 44 kJ/mol), (2) amounts (proportional — double reaction, double ΔH), (3) temperature and pressure. Standard enthalpy change ΔH° refers to 1 bar pressure and usually 298 K. The enthalpy of reaction can be calculated from standard enthalpies of formation: ΔH° = ΣΔfH°(products) - ΣΔfH°(reactants). This is one of the most used formulas in thermodynamics.

Standard enthalpy of formation (ΔfH°)

Standard enthalpy of formation ΔfH° is the enthalpy change when ONE MOLE of a compound is formed from its elements in their standard states at 1 bar and 298 K. By convention: ΔfH° of an element in its standard state is ZERO. So ΔfH° of O₂(g) = 0, C(graphite) = 0, H₂(g) = 0, but C(diamond) ≠ 0 (graphite is the standard state for carbon at 298 K). Examples: ΔfH° of CO₂(g) = -393.5 kJ/mol, H₂O(l) = -285.8 kJ/mol, CH₄(g) = -74.8 kJ/mol. Most ΔfH° values are negative (exothermic), but a few are positive — C₂H₂(g) (+227 kJ/mol), NO(g) (+90.25 kJ/mol). More negative ΔfH° = more stable compound. Using ΔfH° values: ΔH° = ΣΔfH°(products) - ΣΔfH°(reactants). Example: for C₂H₄ + H₂ → C₂H₆: ΔfH° C₂H₄ = +52.3, H₂ = 0, C₂H₆ = -84.7 kJ/mol. ΔH° = (-84.7) - (+52.3 + 0) = -137.0 kJ/mol. Enthalpy of combustion (ΔcH°): heat released when one mole burns completely in excess O₂. Used to compare fuels.

Hess's Law of Constant Heat Summation

Hess's Law: the total enthalpy change for a reaction is the same regardless of the number of steps or the route. Enthalpy is a state function — path does not matter. This is incredibly useful because many reactions cannot be measured directly. Classic example: formation of CO from C and O₂. You cannot burn carbon in limited oxygen without forming some CO₂. But: (1) C(s) + O₂(g) → CO₂(g), ΔH₁ = -393.5 kJ. (2) CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = -283.0 kJ. Reverse (2): CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ. Add to (1): C(s) + ½O₂(g) → CO(g), ΔH = -393.5 + 283.0 = -110.5 kJ/mol. Another use: calculate reaction enthalpy from ΔfH°: ΔH° = ΣΔfH°(products) - ΣΔfH°(reactants). Hess's Law also explains why you can add/subtract chemical equations like algebraic equations — just remember to do the same with ΔH values. Board exam tip: when solving Hess's Law problems, write each equation with states, orient them correctly (you may need to reverse or multiply), and cancel species appearing on both sides.

Bond enthalpies — estimating ΔH from bonds

Bond dissociation enthalpy is the energy to break one mole of a specific bond in the gas phase. H₂(g) → 2H(g), ΔH = +436 kJ/mol (H-H bond). For polyatomic molecules, mean bond enthalpies are used (averaged over many compounds). To estimate ΔH: ΔH = Σ(bond enthalpies of bonds BROKEN) - Σ(bond enthalpies of bonds FORMED). Breaking bonds absorbs energy (+), forming bonds releases energy (-). Example: CH₄ + 2O₂ → CO₂ + 2H₂O. Bonds broken: 4 C-H (4 × 413 = 1652), 2 O=O (2 × 498 = 996). Total = 2648 kJ. Bonds formed: 2 C=O (2 × 799 = 1598), 4 O-H (4 × 463 = 1852). Total = 3450 kJ. ΔH = 2648 - 3450 = -802 kJ. Compare with actual ΔcH° of CH₄ = -890 kJ/mol — the bond enthalpy method gives an approximation. It is useful when ΔfH° data is unavailable. Common mistake: forgetting to count the NUMBER of each bond type (like 4 C-H bonds in CH₄, 2 O=O in 2O₂).

Second Law — entropy and spontaneity

Spontaneous processes occur without outside intervention. Water flows downhill, ice melts above 0°C, iron rusts. The First Law (energy conservation) does not explain why some reactions happen and others do not. The Second Law provides the criterion: the entropy of the universe always increases in a spontaneous process. ΔS_universe = ΔS_system + ΔS_surroundings > 0. Entropy (S) is a measure of disorder or randomness. Think of a tidy room vs a messy room — the messy room has higher entropy. Natural processes tend toward higher entropy. A gas spreading into a vacuum, a drop of ink diffusing in water, ice melting — all increase entropy. The Second Law does not require entropy to increase in the system — it can decrease (like forming a crystal from solution). But the entropy increase in the surroundings MUST outweigh the system decrease, so the total entropy of the universe increases. This is why some endothermic reactions can be spontaneous — if the entropy increase is large enough and temperature is high enough, TΔS can overcome the positive ΔH. Entropy units: J K⁻¹ mol⁻¹. Standard molar entropies (S°) are tabulated — solids < liquids < gases.

Predicting entropy changes

ΔS° = ΣS°(products) - ΣS°(reactants). Entropy increases (ΔS° > 0) when: (1) More gas molecules are produced than consumed. N₂ + 3H₂ → 2NH₃: 4 gas → 2 gas, ΔS negative. CaCO₃ → CaO + CO₂: 0 → 1 gas, ΔS positive. (2) Phase change: solid → liquid → gas. Melting ice: +22 J/K·mol. Boiling water: +109 J/K·mol (gases are far more disordered). (3) Solids dissolve: NaCl(s) → Na⁺(aq) + Cl⁻(aq), ΔS ≈ +43 J/K·mol. (4) Temperature increases — more kinetic energy = more possible arrangements. (5) Complex molecules break into simpler ones — more particles = more disorder. The magnitude of ΔS° is largest for gas-producing reactions (often 100-500 J/K·mol). Solid-solid or solid-liquid reactions have smaller ΔS°. For board exams, predict the SIGN of ΔS°: if the number of gas molecules increases, ΔS° is positive. If it decreases, negative. If no gas is involved, look for phase changes or changes in the number of particles.

Gibbs free energy — the spontaneity decider

Gibbs free energy G = H - TS. At constant T and P: ΔG = ΔH - TΔS. ΔG < 0: spontaneous (exergonic). ΔG > 0: non-spontaneous (endergonic). ΔG = 0: equilibrium. This is the most important equation in chemical thermodynamics — it shows the tug-of-war between enthalpy (ΔH) and entropy (TΔS). At low temperatures, ΔH dominates. At high temperatures, TΔS becomes more significant. For a reaction with ΔH > 0 and ΔS > 0 (like ice melting), it is non-spontaneous at low T but spontaneous at high T. The crossover temperature T = ΔH/ΔS is where ΔG = 0. For ice: ΔH = +6.01 kJ/mol = 6010 J/mol, ΔS = +22.0 J/K·mol. T = 6010/22.0 = 273 K = 0°C. Below 0°C, ΔG > 0 (ice does not melt spontaneously). Above 0°C, ΔG < 0 (ice melts). At 0°C, ΔG = 0 (ice and water coexist at equilibrium). Nature uses this to set phase transition temperatures.

Four temperature-spontaneity cases

The signs of ΔH and ΔS create four scenarios. Case 1: ΔH < 0, ΔS > 0. ΔG always negative. Spontaneous at ALL temperatures. Example: 2H₂O₂ → 2H₂O + O₂ (exothermic, entropy increase from gas production). Case 2: ΔH > 0, ΔS < 0. ΔG always positive. NEVER spontaneous. Example: the reverse of Case 1. Case 3: ΔH < 0, ΔS < 0. Spontaneous at LOW temperatures (ΔH dominates), non-spontaneous at HIGH temperatures (TΔS dominates). Example: N₂ + 3H₂ → 2NH₃ — exothermic (-92.4 kJ), entropy decreases (4 gas → 2 gas, ΔS ≈ -198 J/K). Below T = ΔH/ΔS = 92400/198 = 467 K (194°C), the reaction is spontaneous — this is why the Haber process runs at moderate temperatures. Case 4: ΔH > 0, ΔS > 0. Non-spontaneous at LOW temperatures, spontaneous at HIGH temperatures. Example: CaCO₃ → CaO + CO₂ — endothermic (+178 kJ), ΔS ≈ +160 J/K. T = 178000/160 = 1112 K (839°C). Above this temperature, limestone decomposes — this is how lime is produced industrially. Board exam favorite: identify the case and find T.

ΔG° and the equilibrium constant

ΔG° = -RT ln K. This connects thermodynamics to equilibrium. At 298 K: ΔG° = -5.71 × log₁₀ K kJ/mol. If ΔG° < 0 → K > 1 → products favored. If ΔG° > 0 → K < 1 → reactants favored. If ΔG° = 0 → K = 1. Combining with ΔG° = ΔH° - TΔS°: ln K = -ΔH°/(RT) + ΔS°/R (van't Hoff equation). This shows ln K varies linearly with 1/T. Slope = -ΔH°/R. If ΔH° > 0 (endothermic), as T increases, ln K increases — higher temperature favors products (Le Chatelier confirmed!). If ΔH° < 0 (exothermic), higher temperature favors reactants. Two-point van't Hoff equation: ln(K₂/K₁) = (ΔH°/R)(1/T₁ - 1/T₂). This allows calculation of ΔH° from equilibrium constants at two temperatures. Example: ΔG° = -50 kJ/mol at 298 K. ln K = -ΔG°/RT = 50000/(8.314 × 298) = 20.18. K = e²⁰·¹⁸ ≈ 5.8 × 10⁸ — products heavily favored.

Third Law of Thermodynamics

The Third Law: the entropy of a perfectly crystalline substance is ZERO at absolute zero (0 K). At 0 K, a perfect crystal has only one possible arrangement — no disorder, S = 0. Only PERFECT crystals have S = 0 at 0 K. Imperfect crystals or glasses have residual entropy. For example, CO cannot form a perfect crystal at 0 K because CO molecules can be oriented as CO or OC — this gives residual entropy. The Third Law allows calculation of absolute entropies (S°) by measuring heat capacity from near 0 K to the desired temperature: S°(T) = ∫₀ᵀ (C_p/T) dT. This is why standard molar entropies are absolute values (unlike enthalpy, where we can only measure changes). Tabulated S° values are always positive. Trends: S° increases with molecular mass (I₂ > F₂), molecular complexity (propane > ethane), and temperature. The Third Law also explains why approaching absolute zero is so difficult — as T → 0, ΔS → 0, meaning nearly all processes become impossible.

Coupled reactions — making the impossible happen

A non-spontaneous reaction (ΔG > 0) can be DRIVEN by coupling it with a spontaneous reaction (ΔG < 0) whose combined ΔG is negative. This is how living organisms function. Classic example: Fe₂O₃ reduction to Fe is non-spontaneous, but carbon oxidation (C + O₂ → CO₂, ΔG° = -394 kJ/mol) provides the driving force. The coupled reaction Fe₂O₃ + 3C → 2Fe + 3CO has negative overall ΔG. In biology, ATP hydrolysis (ΔG ≈ -30.5 kJ/mol under cellular conditions) drives countless non-spontaneous reactions — muscle contraction, protein synthesis, nerve impulse transmission, active transport. ATP is the energy currency of the cell — your body cycles through about 40 kg of ATP per day. Photosynthesis: the synthesis of glucose from CO₂ and H₂O (ΔG° = +2870 kJ/mol) is coupled with sunlight absorption through chlorophyll. The light reactions convert light into chemical energy (ATP and NADPH), which drives the Calvin cycle. In industry: the production of TiO₂ from TiCl₄ uses the highly exothermic reaction of TiCl₄ with oxygen. Board exam tip: 'Explain coupled reactions with ATP hydrolysis as an example.'

Thermodynamics vs kinetics — a crucial distinction

Thermodynamics tells you IF a reaction CAN occur (ΔG < 0 = spontaneous). Kinetics tells you HOW FAST it will occur (activation energy, E_a). They are independent. Diamond is thermodynamically unstable relative to graphite (ΔG = -2.9 kJ/mol for diamond → graphite). But it does not convert in your lifetime because the activation energy is enormous — the C-C bonds are incredibly strong and the entire crystal structure must rearrange. Diamond is kinetically stable even though it is thermodynamically unstable. Gasoline does not spontaneously react with O₂ at room temperature (thank goodness), even though ΔG for combustion is very negative. You need a spark to supply activation energy. Once started, the reaction releases enough energy to sustain itself. A mixture of H₂ and O₂ at room temperature is stable for years — but a tiny spark causes an explosive reaction. Always remember: ΔG predicts direction, not speed. A reaction can be spontaneous but incredibly slow (diamond → graphite) or non-spontaneous but potentially fast (the reverse reaction). This distinction shows up repeatedly in board exams — be prepared to explain it.

Key Points

  • •System: open (matter+energy), closed (energy only), isolated (neither)
  • •State functions: P, V, T, U, H, S, G — path independent
  • •Path functions: heat (q) and work (w) — depend on the route
  • •First Law: ΔU = q + w; at constant volume ΔU = q_v; at constant pressure ΔH = q_p
  • •Exothermic: ΔH < 0; Endothermic: ΔH > 0
  • •ΔfH° of element in standard state = 0; ΔH° = ΣΔfH°(products) - ΣΔfH°(reactants)
  • •Hess's Law: ΔH depends only on initial and final states, not path
  • •Enthalpy of neutralization (strong acid + strong base): ≈ -57.3 kJ/mol
  • •Bond enthalpy method: ΔH ≈ Σ(bonds broken) - Σ(bonds formed)
  • •Second Law: ΔS_universe > 0 for spontaneous processes
  • •Entropy (S): measure of disorder; S° values: solid < liquid < gas
  • •ΔS° = ΣS°(products) - ΣS°(reactants); increases with gas production, dissolution, T rise
  • •Third Law: S = 0 for perfect crystal at 0 K; allows calculation of absolute entropies
  • •Gibbs free energy: ΔG = ΔH - TΔS; ΔG < 0 = spontaneous; ΔG = 0 = equilibrium
  • •ΔG° = -RT ln K; connects thermodynamics to equilibrium constant
  • •Four cases: (ΔH-ΔS) signs — (--, ++, -+, +-) determine temperature dependence
  • •Coupled reactions: non-spontaneous driven by spontaneous — ATP hydrolysis example
  • •Thermodynamics ≠ Kinetics: diamond → graphite is spontaneous but extremely slow

Practice Questions

  • State the First Law. A gas expands from 2 L to 5 L against constant 2 atm pressure, absorbing 500 J of heat. Calculate ΔU. (1 L atm = 101.3 J)
  • Explain Hess's Law. Calculate ΔH for C + ½O₂ → CO from: C + O₂ → CO₂ ΔH = -393.5 kJ; CO + ½O₂ → CO₂ ΔH = -283 kJ.
  • For a reaction ΔH = +30 kJ, ΔS = +100 J/K. At what temperature does it become spontaneous? Is it spontaneous at 250 K? At 400 K?
  • Define Gibbs free energy. Derive ΔG = ΔH - TΔS. Explain why ΔG < 0 for spontaneous processes.
  • Derive the relationship between ΔG° and K. For a reaction with ΔG° = -20 kJ/mol at 298 K, calculate K. (R = 8.314 J/K·mol)
  • Predict the sign of ΔS for: (a) N₂ + 3H₂ → 2NH₃, (b) CaCO₃ → CaO + CO₂, (c) NaCl(s) → Na⁺(aq) + Cl⁻(aq), (d) H₂O(l) → H₂O(s). Explain each.
  • What is the Third Law? How does it allow calculation of absolute entropies? Why can't CO form a perfect crystal at 0 K?
  • Distinguish between thermodynamics and kinetics. Why does diamond not convert to graphite at room temperature even though ΔG is negative?