Section 17: Enthalpy

17.1 ●​ Enthalpy (ΔH) is a state function, meaning it depends only on the initial and final states, not on the path or number of steps taken. ●​ Hess’s Law: The enthalpy change of a reaction is the same regardless of the number of steps or use of a catalyst. ●​ Endothermic: Energy is absorbed; ΔH is positive. ●​ Exothermic: Energy is released; ΔH is negative. ●​ Breaking bonds requires energy ("takes to break"), forming bonds releases energy ("frees to form"). Energy Profile Diagrams: ●​ If products have lower energy than reactants → exothermic (ΔH < 0)​ ●​ If products have higher energy than reactants → endothermic (ΔH > 0) Two Main Rules for Using Hess’s Law: ●​ If a reaction is reversed, the sign of ΔH is also reversed.​ ●​ If the coefficients in a reaction are multiplied by an integer, then ΔH is also multiplied by that integer. Example 1: Graphite to Diamond ●​ Combustion reactions for graphite and diamond are known.​ ●​ Reverse the combustion of diamond to cancel CO₂ and isolate graphite → diamond.​ ●​ Add the reactions; ΔH for graphite to diamond = +2 kJ/mol (endothermic). Example 2: Synthesis of Diborane (B₂H₆) ●​ Combine multiple known reactions to derive the target equation:​ ○​ Equation (a): Use as is for boron.​ ○​ Equation (c): Multiply by 3 for H₂.​ ○​ Equation (b): Reverse to put B₂H₆ on the product side.​ ○​ Equation (d): Multiply by 3 to account for phase change in water (gas to liquid).​ ●​ Cancel common species on both sides.​ ●​ Total ΔH = +36 kJ/mol (endothermic). 17.2 ●​ Energy is transferred between the system and surroundings. ○​ q(system) = –q(surroundings) ●​ ΔH can be calculated via:​ ○​ Calorimetry​ ○​ Hess’s Law​ ○​ Standard enthalpies of formation​ ○​ Bond energies Standard Enthalpy of Formation (ΔHf°): Defined as the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (298 K, 1 atm). ●​ Standard State Definitions: ●​ Gases: 1 atm​ ●​ Pure substances (solid/liquid): pure form​ ●​ Solutions: 1 M concentration​ ●​ Elements: most stable form at standard conditions (e.g., O₂, H₂, graphite)​ ●​ Formation Reaction Characteristics: ●​ Reactants are pure elements in standard states​ ●​ Only 1 mole of product​ ●​ May include fractional coefficients Example: Combustion of Methane (CH₄) ●​ Visualize the reaction as:​ ○​ Step 1: Decompose CH₄ to elements (reverse of formation reaction, ΔH = +75 kJ/mol)​ ○​ Step 2: O₂ already in standard state (ΔH = 0)​ ○​ Step 3: Form CO₂ from C and O₂ (ΔH = –394 kJ/mol)​ ○​ Step 4: Form 2 H₂O from H₂ and O₂ (ΔH = –572 kJ/mol)​ ●​ Total ΔH = 75 + 0 – 394 – 572 = –891 kJ/mol (exothermic) ●​ General Equation: ΔH°rxn = ΣnΔHf°(products) – ΣnΔHf°(reactants)​ ●​ Multiply ΔHf° values by the number of moles (n) for each substance Example: Reaction of Ammonia with Oxygen ●​ 4 NH₃ + 7 O₂ → 4 NO₂ + 6 H₂O​ ●​ Use formation values:​ ○​ NO₂: 4 × 34 = 136 kJ​ ○​ H₂O: 6 × (–286) = –1716 kJ​ ○​ NH₃: 4 × (–46) = –184 kJ​ ○​ O₂: 0 kJ (element in standard state)​ ●​ ΔH = (136 – 1716) – (–184 + 0) = –1396 kJ (exothermic) Bond Energies: ●​ Breaking bonds requires energy (positive ΔH)​ ●​ Forming bonds releases energy (negative ΔH)​ ●​ Bond energy is the energy required to break 1 mole of a bond in the gas phase​ ●​ Stronger bonds have higher bond energies Example: Formation of Ammonia (2 NH₃) ●​ Break: ○​ 1 N≡N triple bond = 945 kJ​ ○​ 3 H–H bonds = 3 × 436 = 1308 kJ​ ○​ Total = 2253 kJ ●​ Form:​ ○​ 6 N–H bonds = 6 × 391 = 2346 kJ ●​ ΔH = 2253 – 2346 = –93 kJ (exothermic)