Energetics is one of the most calculation-heavy and conceptually demanding topics in IB Chemistry. It spans from basic enthalpy changes in Topic 5 (SL/HL) all the way to Born-Haber cycles and entropy-driven spontaneity in Topic 15 (HL only). Whether you are aiming for a solid 5 or pushing for that elusive 7, a systematic understanding of energetics is non-negotiable. This comprehensive guide breaks down every core concept — from exothermic endothermic classification, through calorimetry calculations and Hess's Law, to the subtleties of Gibbs free energy and reaction spontaneity — all supplemented with worked examples, common pitfalls, and proven exam strategies.
能量学是IB化学中计算量最大、概念密度最高的章节之一。它横跨Topic 5的基础焓变（SL/HL共用）和Topic 15的玻恩-哈伯循环与熵驱动自发性（仅HL）。无论你目标是稳稳拿5分还是冲刺7分，系统掌握能量学都是绕不开的硬仗。本文逐项拆解所有核心概念——从放热吸热的分类，到量热法计算与赫斯定律，再到吉布斯自由能和反应自发性的精细判断——全部配有例题解析、常见错误提醒和高分考试策略。
--- ### 1. Exothermic vs Endothermic Reactions 放热与吸热反应 The very first distinction every IB Chemistry student must internalize is whether a reaction releases energy to the surroundings or absorbs energy from them. This classification underpins everything that follows — enthalpy sign conventions, energy profile diagram labels, and even the qualitative predictions you make about spontaneity. At the molecular level, this distinction comes down to bond energetics. In an exothermic reaction, the bonds formed in the products are collectively stronger (more stable, lower in energy) than the bonds broken in the reactants. The excess energy is released as heat. Conversely, in an endothermic reaction, the bonds broken are collectively stronger than those formed, so the system must absorb energy from the surroundings to proceed.Exothermic Reactions 放热反应
- Energy is released to the surroundings
- ΔH is negative (ΔH < 0)
- Products are more stable (lower energy) than reactants
- Temperature of surroundings increases
- Classic examples: combustion of fuels, neutralization of acids and bases, respiration (C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O), the thermite reaction (Fe₂O₃ + 2Al → 2Fe + Al₂O₃), and condensation of steam
- In an energy profile diagram, the products sit at a lower energy level than the reactants
Endothermic Reactions 吸热反应
- Energy is absorbed from the surroundings
- ΔH is positive (ΔH > 0)
- Products are less stable (higher energy) than reactants
- Temperature of surroundings decreases
- Classic examples: photosynthesis (6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂), thermal decomposition of calcium carbonate (CaCO₃ → CaO + CO₂), dissolving ammonium chloride in water, melting of ice, and the reaction of barium hydroxide with ammonium thiocyanate
- In the energy profile, products end up at a higher energy level than reactants
Exam Precision 考试精确表达
When labelling an energy profile diagram in Paper 2, you must include: (a) the y-axis labelled "Potential Energy" or "Enthalpy H", (b) clear horizontal lines for reactants and products with their relative heights correctly showing exothermic or endothermic, (c) ΔH clearly marked as the vertical difference between products and reactants, (d) the activation energy Ea shown as the peak height measured from the reactants' energy level, and (e) if it is a multi-step reaction, the transition state(s) and any intermediates at local minima. Students lose marks every year for sloppy or incomplete labelling. The IB mark scheme rewards precision.
--- ### 2. Enthalpy Change ΔH 焓变 Enthalpy (symbol H) is formally defined as the total heat content of a system at constant pressure. Since absolute enthalpy cannot be measured directly, we always work with enthalpy changes. The defining equation is deceptively simple but must be applied with care:ΔH = H(products) - H(reactants)
When products are at a lower enthalpy than reactants, ΔH is negative (exothermic). When products are at a higher enthalpy, ΔH is positive (endothermic). The magnitude of ΔH tells us how much energy is transferred as heat at constant pressure. **Standard conditions** are critical because enthalpy changes vary with temperature, pressure, and concentration. The IB defines standard conditions as: temperature = 298 K (exactly 25°C), pressure = 100 kPa (1 bar). All substances must be in their standard states — the most stable physical state under these conditions. For solutions, the standard concentration is 1 mol dm⁻³. When an enthalpy change is measured or calculated under standard conditions, we append the plimsoll symbol (⦵):Standard enthalpy change = ΔH⦵
The plimsoll is an essential notation — omitting it in a definition answer can cost you the mark. Always check: if the question specifies standard conditions, include the ⦵ symbol in your ΔH notation. --- ### 3. Types of Standard Enthalpy Changes 标准焓变类型详解 The IB syllabus (Topics 5 and 15) requires you to define, recognize, and apply several distinct types of standard enthalpy change. Each has a precise definition that must include specific wording — "one mole" is almost always present, as is the specification of standard conditions. Here is the complete taxonomy:ΔH⦵f (Standard Enthalpy of Formation 标准生成焓): The enthalpy change when ONE MOLE of a compound is formed from its constituent elements in their standard states under standard conditions. Example: C(s, graphite) + O₂(g) → CO₂(g). Note well: ΔH⦵f of any element in its standard state is exactly zero by definition. This is tested every year without fail.
ΔH⦵c (Standard Enthalpy of Combustion 标准燃烧焓): The enthalpy change when ONE MOLE of a substance is completely burned in excess oxygen under standard conditions. Example: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Combustion is always exothermic, so all ΔH⦵c values are negative.
ΔH⦵neut (Standard Enthalpy of Neutralization 标准中和焓): The enthalpy change when an acid and a base react to form ONE MOLE of water under standard conditions. For strong acid with strong base (e.g. HCl + NaOH), ΔH⦵neut is approximately -57 kJ mol⁻¹. For weak acids or bases, the magnitude is smaller because some energy is used to ionize the weak electrolyte.
ΔH⦵sol (Standard Enthalpy of Solution 标准溶解焓): The enthalpy change when ONE MOLE of solute dissolves in an excess of solvent to form a solution of infinite dilution under standard conditions. Example: NaCl(s) → Na⁺(aq) + Cl⁻(aq). Can be endothermic (NaCl, +3.9 kJ mol⁻¹) or exothermic (NaOH, -44.5 kJ mol⁻¹).
ΔH⦵hyd (Standard Enthalpy of Hydration 标准水合焓): The enthalpy change when ONE MOLE of gaseous ions is dissolved in water to form an infinitely dilute solution. Always exothermic because ion-dipole attractions release energy. Example: Na⁺(g) → Na⁺(aq). This is a core concept for HL students working with Born-Haber cycles.
ΔH⦵at (Standard Enthalpy of Atomization 标准原子化焓): The enthalpy change when ONE MOLE of gaseous atoms is formed from the element in its standard state. Always endothermic (bonds must be broken). Example: ½Cl₂(g) → Cl(g). For metals: Na(s) → Na(g). This is the sublimation or vaporization step in Born-Haber cycles.
Top 3 Definition Mistakes 最常见的三个定义错误
- Forgetting "one mole" — all standard enthalpy definitions specify the amount. Saying "when a substance is burned" loses the mark; you must say "when ONE MOLE of a substance is completely burned."
- Omitting "under standard conditions" — if the definition includes the plimsoll (⦵), standard conditions must be stated explicitly.
- Confusing formation with combustion — formation starts from elements and makes one mole of compound; combustion starts from one mole of substance and burns it completely.
q = mcΔT
Where q is the heat energy transferred (in joules, J), m is the mass of the substance being heated (typically water or a dilute aqueous solution, in grams), c is the specific heat capacity (for water, c = 4.18 J g⁻¹ K⁻¹), and ΔT is the temperature change (in Kelvin or Celsius; the interval is the same). Once q is calculated, the enthalpy change per mole is found by:ΔH = -q / n
The negative sign is essential and frequently forgotten. In a calorimetry experiment, q is the heat absorbed by the water (positive if the water warms up). Since the reaction released that heat (exothermic), the reaction's ΔH must be negative. This sign convention trips up students every single year.Worked Example 1: Neutralization Calorimetry 中和量热
50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid is mixed with 50.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide solution in a polystyrene cup. The temperature rises from 21.0°C to 27.5°C. Calculate the standard enthalpy of neutralization ΔH⦵neut.
Step 1: Determine the total mass. Total volume = 50.0 + 50.0 = 100.0 cm³. Assuming the density of dilute aqueous solution ≈ 1.00 g cm⁻³, mass m = 100.0 g.
Step 2: Calculate temperature change. ΔT = 27.5 - 21.0 = 6.5°C (or 6.5 K).
Step 3: Calculate q. q = mcΔT = 100.0 × 4.18 × 6.5 = 2717 J.
Step 4: Find moles of water formed. HCl + NaOH → NaCl + H₂O. n(HCl) = 0.0500 × 1.00 = 0.0500 mol. Since both reactants are present in equal stoichiometric amounts, n(H₂O) formed = 0.0500 mol.
Step 5: Calculate ΔH. ΔH = -q / n = -2717 / 0.0500 = -54340 J mol⁻¹ = -54.3 kJ mol⁻¹ (3 significant figures).
Evaluation: The literature value for strong acid-strong base neutralization is approximately -57 kJ mol⁻¹. Our value of -54.3 kJ mol⁻¹ is slightly less exothermic, likely due to heat loss to the surroundings, the heat capacity of the polystyrene cup not being accounted for, and the assumption that the solution has the same specific heat capacity as pure water.
Worked Example 2: Combustion Calorimetry 燃烧量热
0.500 g of ethanol (C₂H₅OH, Mr = 46.0) is burned completely in a spirit burner. The flame heats 200 g of water in a copper calorimeter. The water temperature rises from 22.0°C to 36.5°C. Calculate ΔH⦵c of ethanol.
Step 1: ΔT = 36.5 - 22.0 = 14.5°C.
Step 2: q = 200 × 4.18 × 14.5 = 12122 J = 12.12 kJ.
Step 3: n(ethanol) = 0.500 / 46.0 = 0.01087 mol.
Step 4: ΔH = -12.12 / 0.01087 = -1115 kJ mol⁻¹.
Comparison: The Data Booklet value for ethanol combustion is -1367 kJ mol⁻¹. Our experimental value is significantly less exothermic. Major sources of error include: incomplete combustion (producing CO and soot instead of CO₂), heat loss to the surroundings and the copper calorimeter, evaporation of ethanol from the wick, and the fact that not all the heat from the flame is transferred to the water. This is a classic evaluation question in Paper 2.
Improving Calorimetry Accuracy 提高量热精度
- Use a draught shield to minimize convective heat loss
- Stir continuously to ensure uniform temperature distribution
- Use a calibrated thermometer with 0.1°C precision
- For combustion: use a bomb calorimeter instead of a spirit burner — it ensures complete combustion in excess oxygen and minimizes heat loss
- Plot a temperature-time cooling curve and extrapolate to find the theoretical maximum ΔT, correcting for heat loss during the measurement period
- Account for the heat capacity of the calorimeter itself by calibrating with a reaction of known ΔH, such as the combustion of benzoic acid
ΔH(direct) = ΔH(route 1) + ΔH(route 2) + ... + ΔH(route n)
There are three main computational strategies for applying Hess's Law:Strategy 1: Enthalpy Cycle Diagram 焓循环图
Draw a cycle with reactants and products at the top. Connect both to a common intermediate below (usually the constituent elements or the combustion products CO₂ + H₂O). Apply the rule: the sum of clockwise enthalpy changes equals the sum of anticlockwise enthalpy changes. This visual method minimizes sign errors because you can see the direction of each arrow.
Strategy 2: ΔH⦵f Formula 生成焓法
ΔH⦵ = Σ [coeff × ΔH⦵f(product)] - Σ [coeff × ΔH⦵f(reactant)]. Each standard enthalpy of formation is multiplied by the corresponding stoichiometric coefficient. This is the fastest method when ΔH⦵f data is provided.
Strategy 3: ΔH⦵c Formula 燃烧焓法
ΔH⦵ = Σ [coeff × ΔH⦵c(reactant)] - Σ [coeff × ΔH⦵c(product)]. Note the order is reversed compared to the formation formula: reactants minus products, not products minus reactants.
Worked Example 3: Hess's Law via Combustion Data 燃烧数据法
Problem: Calculate ΔH⦵f for ethene (C₂H₄) given the following standard enthalpies of combustion:
ΔH⦵c[C(s, graphite)] = -394 kJ mol⁻¹
ΔH⦵c[H₂(g)] = -286 kJ mol⁻¹
ΔH⦵c[C₂H₄(g)] = -1411 kJ mol⁻¹
Target reaction (formation of ethene): 2C(s) + 2H₂(g) → C₂H₄(g)
Applying the combustion formula: ΔH⦵f = Σ ΔH⦵c(reactants) - Σ ΔH⦵c(products)
ΔH⦵f = [2(-394) + 2(-286)] - (-1411)
ΔH⦵f = [-788 + (-572)] + 1411 = -1360 + 1411 = +51 kJ mol⁻¹
Interpretation: The formation of ethene from its elements is endothermic (+51 kJ mol⁻¹). This makes sense because ethene contains a C=C double bond which is weaker than two C-C single bonds; forming it from graphite (which has strong C-C bonds in its layered structure) requires energy input.
Worked Example 4: Hess's Law with an Enthalpy Cycle 焓循环法
Problem: Calculate ΔH for the reaction: 2NO₂(g) → N₂O₄(g)
Given: (1) ½N₂(g) + O₂(g) → NO₂(g) ΔH = +33.2 kJ mol⁻¹
(2) N₂(g) + 2O₂(g) → N₂O₄(g) ΔH = +9.2 kJ mol⁻¹
Constructing the cycle: Route through the elements N₂ + 2O₂.
2NO₂ → N₂ + 2O₂ (reverse of equation (1) × 2): ΔH = -2 × 33.2 = -66.4 kJ
N₂ + 2O₂ → N₂O₄ (equation (2)): ΔH = +9.2 kJ
ΔH = -66.4 + 9.2 = -57.2 kJ mol⁻¹
The dimerization of NO₂ to N₂O₄ is exothermic, which is why N₂O₄ is the favored species at lower temperatures.
--- ### 6. Bond Enthalpies 键能 Bond enthalpy is the energy required to break one mole of a particular covalent bond in the gaseous state. It provides an alternative route for estimating ΔH that does not require the standard enthalpy data used in Hess's Law. The core equation is:ΔH = Σ (bond enthalpies of bonds broken) - Σ (bond enthalpies of bonds formed)
The logic is straightforward. Breaking bonds requires energy input (endothermic, positive contribution to ΔH). Forming bonds releases energy (exothermic, negative contribution to ΔH). Therefore, if the bonds formed are stronger overall than the bonds broken, the reaction is exothermic.Critical Distinction 关键区分
Average bond enthalpy is the average energy required to break a given bond type across many different compounds. For example, the C-H bond enthalpy of 414 kJ mol⁻¹ is the average for C-H bonds in methane, ethane, ethene, and many other molecules.
Bond dissociation enthalpy is the exact energy required to break a specific bond in a specific molecule. For methane, breaking the first C-H bond requires 439 kJ mol⁻¹, the second requires 460 kJ mol⁻¹, and so on. They differ because the molecular environment changes after each bond is broken.
For IB exam purposes, you use average bond enthalpies from Section 11 of the Data Booklet. This means bond enthalpy calculations give approximate ΔH values. You must state this limitation explicitly when evaluating results.
Worked Example 5: Bond Enthalpy Calculation 键能计算
Problem: Use average bond enthalpies to estimate ΔH for the hydrogenation of ethene: C₂H₄(g) + H₂(g) → C₂H₆(g)
Step 1: Identify bonds broken and formed. Draw the structural formulas:
C₂H₄: H₂C=CH₂ (4 × C-H, 1 × C=C) + H₂: (1 × H-H)
C₂H₆: H₃C-CH₃ (6 × C-H, 1 × C-C)
Step 2: Bonds broken: 4 × C-H (414) + 1 × C=C (614) + 1 × H-H (436) = 1656 + 614 + 436 = 2706 kJ
Step 3: Bonds formed: 6 × C-H (414) + 1 × C-C (346) = 2484 + 346 = 2830 kJ
Step 4: ΔH = 2706 - 2830 = -124 kJ mol⁻¹
Note: This is an estimate because average bond enthalpies are used. The actual ΔH⦵ for this reaction is -137 kJ mol⁻¹. The discrepancy arises because the C=C bond in ethene and the C-C and C-H bonds in ethane have bond enthalpies that differ from the average values.
Bond Enthalpy Limitations 键能法的局限性
- Only valid for gaseous reactions — bond enthalpies are defined in the gas phase. If liquids or solids are involved, enthalpy changes for vaporization or fusion must also be considered.
- Average bond enthalpies do not account for the specific molecular environment.
- The method ignores intermolecular forces, which are present in condensed phases.
- For resonance-stabilized molecules (e.g. benzene, ozone), the actual bond energies differ significantly from the tabulated averages.
Entropy Increases (ΔS > 0)
- Solid melts to liquid
- Liquid vaporizes to gas
- Solid sublimes directly to gas
- Number of gas moles increases in reaction
- Solid dissolves in solvent (ions become dispersed)
- Temperature increases (more kinetic energy, more microstates)
- A larger molecule breaks into smaller fragments
Entropy Decreases (ΔS < 0)
- Gas condenses to liquid
- Liquid freezes to solid
- Gas deposits directly to solid
- Number of gas moles decreases in reaction
- Ions crystallize from solution
- Small molecules polymerize into a large chain
- Temperature decreases
Quick Prediction Rule 快速判断法则
The single most reliable predictor of ΔS sign is the change in the number of moles of gas: Δn(gas) = n(gas, products) - n(gas, reactants). If Δn(gas) > 0, entropy almost certainly increases. If Δn(gas) < 0, entropy almost certainly decreases. If Δn(gas) = 0, you need to look at other factors like increased molecular complexity or changes of state.
--- ### 8. Gibbs Free Energy ΔG 吉布斯自由能 (HL only) The Gibbs free energy equation is arguably the single most important equation in chemical thermodynamics. It unifies enthalpy (heat), entropy (disorder), and temperature into a single criterion for spontaneity:ΔG = ΔH - TΔS
Where ΔG is in kJ mol⁻¹, ΔH in kJ mol⁻¹, T in Kelvin (K), and ΔS in kJ K⁻¹ mol⁻¹. The unit conversion for ΔS is the single most common source of error: the Data Booklet lists ΔS values in J K⁻¹ mol⁻¹, so you must divide by 1000 before inserting into the equation. The spontaneity criterion is elegantly simple:ΔG < 0: The reaction is spontaneous (thermodynamically feasible) in the forward direction under the given conditions.
ΔG = 0: The system is at equilibrium. No net change occurs. This is the condition from which we derive the relationship between ΔG⦵ and the equilibrium constant K: ΔG⦵ = -RT ln K.
ΔG > 0: The forward reaction is non-spontaneous under the given conditions. The reverse reaction is spontaneous.
The four combinations of ΔH and ΔS signs determine how temperature affects spontaneity:Case 1: ΔH < 0, ΔS > 0
ΔG = negative - (positive × positive) = always negative. Spontaneous at all temperatures. Example: 2H₂O₂(l) → 2H₂O(l) + O₂(g). Exothermic AND produces gas — both factors favor spontaneity.
Case 2: ΔH < 0, ΔS < 0
ΔG = negative - (positive × negative) = negative + positive. Spontaneous only when |ΔH| > |TΔS|. Spontaneous at low T. Example: NH₃(g) + HCl(g) → NH₄Cl(s). Exothermic but entropy decreases (gas → solid).
Case 3: ΔH > 0, ΔS > 0
ΔG = positive - (positive × positive). Spontaneous only when TΔS > ΔH. Spontaneous at high T. Example: CaCO₃(s) → CaO(s) + CO₂(g). Endothermic but entropy increases (gas produced).
Case 4: ΔH > 0, ΔS < 0
ΔG = positive - (positive × negative) = positive + positive = always positive. Never spontaneous. Example: 3O₂(g) → 2O₃(g). Endothermic AND entropy decreases — both factors oppose spontaneity.
--- ### 9. Temperature at Which Spontaneity Changes 自发性的温度转折点 For reactions where ΔH and ΔS have the same sign (Cases 2 and 3 above), there is a threshold temperature at which the reaction transitions from non-spontaneous to spontaneous (or vice versa). At this temperature, ΔG = 0:T = ΔH / ΔS (when ΔG = 0)
Worked Example 6: Temperature Crossover 温度转折
Problem: For the thermal decomposition of calcium carbonate: CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. Above what temperature does this reaction become spontaneous?
Step 1: Convert ΔS to kJ K⁻¹ mol⁻¹: ΔS = 161 / 1000 = 0.161 kJ K⁻¹ mol⁻¹.
Step 2: Set ΔG = 0: 0 = ΔH - TΔS → T = ΔH / ΔS = 178 / 0.161 = 1106 K.
Step 3: At T > 1106 K (833°C), the reaction is spontaneous. This explains why limestone must be heated to very high temperatures in a kiln to produce quicklime (CaO) — an industrial process that has been used for millennia.
--- ### 10. Comprehensive Exam Strategy 综合考试策略Paper 1 Mastery 选择题高分策略
- When you see a Hess's Law MCQ, scan the given equations for how they can be added or reversed to yield the target. Often you do not even need to calculate the full answer — eliminating implausible options by sign is faster.
- For calorimetry MCQs, check the limiting reagent before calculating n. If one reactant is in excess, the moles of water (or product) formed are determined by the limiting reagent only.
- In ΔG questions, always verify the units of ΔS before applying ΔG = ΔH - TΔS. If ΔS is given in J, convert to kJ.
- Remember the sign of ΔH⦵f for elements is zero. This single fact is tested in some form almost every examination session.
- For bond enthalpy MCQs: if all reactants and products are gases, proceed with confidence. If liquids or solids are present, the result is approximate — and the question may test your awareness of this limitation.
Paper 2 Structured Questions 大题策略
- Always show your working. Hess's Law and calorimetry questions have method marks. Even if your final answer is wrong, a clearly laid-out method with correct intermediate steps will earn partial credit.
- When drawing an enthalpy cycle, label every arrow with its ΔH value and the direction (up = endothermic, down = exothermic). This visual clarity prevents sign confusion and impresses examiners.
- For calorimetry evaluation questions, state at least three sources of error AND suggest a specific improvement for each. Generic statements like "heat loss" without a remedy only earn partial marks. Pair each error with a fix: "Heat loss to surroundings — use a draught shield and insulate the calorimeter."
- Definitions must be PRECISE. Practice writing out each standard enthalpy definition word for word as it appears in the syllabus. The phrase "one mole" appears in every single one of them.
- When comparing your experimental ΔH with the literature value, calculate the percentage error and suggest chemical reasons for the discrepancy (incomplete combustion, heat capacity of the apparatus, evaporation of reactants) rather than just saying "experimental error."
- For entropy questions, always count gas moles first. The Δn(gas) sign determines ΔS sign in the vast majority of reactions.
The Fatal Five: Most Common Mistakes 五大致命错误
- Sign error in ΔH = -q/n: Heat absorbed by water = +q. Heat released by reaction = -ΔH. When water warms up, the reaction is exothermic, so ΔH MUST be negative. Always apply the negative sign.
- Units mismatch in ΔG = ΔH - TΔS: ΔS is almost always provided in J K⁻¹ mol⁻¹. You must divide by 1000 to convert to kJ K⁻¹ mol⁻¹ before multiplying by T. Forgetting this converts a small entropy contribution into one ten times too large or small.
- Stoichiometry in formation/combustion formulas: When using Σ ΔH⦵f(products) - Σ ΔH⦵f(reactants), every ΔH⦵f is multiplied by its coefficient from the balanced equation. Forgetting to multiply by the coefficient is the most common Hess's Law error.
- Limiting reagent in calorimetry: The value of n in ΔH = -q/n must be the moles of the limiting reactant, not the total moles of both reactants. If 50 cm³ of 2M HCl reacts with 50 cm³ of 1M NaOH, the NaOH is limiting (0.050 mol), not the HCl (0.100 mol).
- Temperature in Celsius vs Kelvin: In ΔG = ΔH - TΔS, T must be in Kelvin. Add 273 (or more precisely 273.15) to Celsius temperatures. In q = mcΔT, however, ΔT in Celsius is identical to ΔT in Kelvin, so either unit works for temperature differences.
Key Data Booklet Reference (Section 11) 数据手册参考
Average Bond Enthalpies (kJ mol⁻¹): C-H 414, C-C 346, C=C 614, C≡C 839, C-O 358, C=O 804 (in CO₂), O=O 498, H-H 436, O-H 463, N≡N 945, N-H 391, C-Cl 324, H-Cl 431
Standard enthalpy of vaporization of water: +44.0 kJ mol⁻¹ at 298 K
Standard enthalpy of fusion of water: +6.01 kJ mol⁻¹ at 273 K
Specific heat capacity of water: 4.18 kJ kg⁻¹ K⁻¹ (equivalent to 4.18 J g⁻¹ K⁻¹)
Note: These values are provided in the IB Chemistry Data Booklet. You do NOT need to memorize them — they will be given to you in the exam. What you need to know is how and when to use each one.
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