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PE from bonds and IMFS, called chemical energy (same as E) AT = Tf - Ti -If AT > O, System absorbed energy (endothermic) -If AT < O, system released energy (exothermic) work AT = a + W heat -Heat (): transfer of thermal energy due to temp difference - Work (W): force acting over a distance # of mols -Sign conventions: +q +W +AT - Formula: gain heat Work done on System MAH = net gain of energy net loss of energy - Heat flowing in/out results in change in state or temp of System = MCAT T mass -specific heelt = 9P •chanse -q -W in temp loses heat - AT heat - Enthalphy (H): total energy of a system at constant pressure •Can only measure change Extensive property Work done by system note: D = A • Units are kJ/mol - Energy change to a system that doesn't result in ot is measured as DH • Phase changes -Types of enthalpy change: • Enthalphy of reaction (DHixn): heat absorbed or released during a chemical reaction • Enthalpy of combustion (DH comb): heat absorbed or released by burning (usually O₂) in kJ/mol • Enthalpy of solution (DH soln); heat absorbed or released by a substance when it dissolves • Enthalpy of fusion (OHfus): heat absorbed to change I mol frm sollig at melting point Freezing is exothermic, so - DH Melting is endothermic. So + DH • Enthalpy of vaporization (DHvap): heelt absorbed to change I mol frm liggas at boiling point ▶ Condensing is exothermie, so - DH ▶ · Boiling is endothermic, so +DH - For a chemical reaction: Hreaction = DH products - OH reactants If Hiran is t, endothermic if olt is -, exothermic -Enthalphy diagrams Show if a reaction is endo or exo: ENDOTHERMIC EXOTHERMIC Reactants -DH Lactants Products - PE Diagrams: Activati Energy energ released 1xn Progress exothermic Products Reactants Reactants 1xn Progress endothermic Products + DH Products -Heat Capacity (C): quantity of heat required to change the temp by 1°C or IK - Specific Heat (CP): quantity of heat required to change the temp by 1°C or Ik for Ig of substance -heat capacity Spec. healt = mass q MAT Molar Heat Capacity is used when amt is I mol •Units are J/mol · K - Useful spec heat equation: 9 = mcAT - Calorimetry: measure of heat flow Instrument called a change in temp - 9 reaction = 9, calorimeter OR -9rxn=9cal + 9 soln Thermometer Glass stirrer -Thermal Equilibrium: occurs when system and surroundings reach same temp and heat transfer stops - Coffee Cup Calorimetry: Cork stopper Calorimeter used Two Styrofoam® cups nested together Reaction mixture ・9ix = 9soln 9 soln = Csoln M soin • (Tfinal - Tinitial) - Ex/ A 0.500-g sample of KCl is added to 50.0 g of water in a calorimeter. If the temperature decreases by 1.05 C, what's the approximate amount of heat involved in the dissolution of the KCl? Assume the heat capacity of the resulting solution is 4.18 J/g C. Is the reaction exothermic or endothermic? Given: m KCl = 0.500 9 m water = 50.0 g 4T = -1.05 °C C = 4.184 J/g °C 50.09 + 0.500, So. 500 9 of solution Asking for: 9rxn -9rxn= (50.500) - (4.184) (-1.05 ˚C) -9rxn=-221.8566 J 91xn=221.8566 J Keer 3 sig figs! 19rxn= 222 J -Thermochemical Equations: describe the ixn and energy changes. during said IxN •AH written in Products or off to the side as AH = ·PHYSICAL STATE OF REACTANTS & PRODUCTS ARE IMPORTANT - Thermochemical equations can be manipulated to help us Problems by: . • Flipping the equation Must also • Multiplying the entire equation by a factor I chounse alt - Can also use oH as a Conversion factor: 2 CH4O+ 30₂ 4 H₂O + 2 CO₂ AH = -1452 kJ/mol 1452 KJ 1452 kJ 1452 kJ 1452 kJ 2 mol CH₂O 4 mol H₂O 2 mol CO₂ 3 mol O₂ solve - Ex/ W/ the reaction above, how much heat is released if 14.8 g of CH40 is combusted? mm CH40 = 32.0419 9 14.81 mol CH₂O 1452 KJ 32.041992 al CHYⒸ -Hess's Law: b/c enthalpy is a state function, total oH depends on initial state (reactants) and final state (Products) •Allows us to calc all for hard to find out ixns - If rxn can be expressed as steps, then AH for target Ixn is Sum of AH of each 새 tars = Eat givens + Sum - Ex/ - 04 To solve Hess's Law Probs, algebraically manipulate equations and a to find all of target equation - Ex/ Target: C3H8 (9) + 5 0₂ (9) → 3 CO₂ (3) + 4 17₂0 (1) 4H= ? -3 step process to set to target 1. Decomp of propane: 2. Formation of CO₂: 3. Formation of H₂O 335 kJ is released Target: 2 C(grap) + 3 (3H₂ (3) → 3 (₁₂) + 4H₂ (3) AH= 103.89 KJ 3 C G + 3 0₂ (3) → 3 CO₂ (5) 4 H₂ (9) + 2 O₂ (s) → 4 H₂O (1) Calo & ΔΗ=? H₂ (9) → 20 Casian) + O2 (9) → COn () AH = -393.5 kJ = ( 2 Colto (9) + 7 0₂ (9)→ 4 CO₂ (9) + 6 H₂0 (1) AH = -3119.6 KJ) flip 33 H₂ (3) + 1⁄2 0₂ (9) → H₂0 (1) &H= -285.8 KJ 2 C (grap) + 22₂ (9) → a CO₂ (2) 20026) + 3H00 Co Ho (9) + 3 1₂ (9)+₂2 →3H₂(1) AH = -1181 kJ OH = -1143 kJ AH = -787 KJ OH = AH = 2 C (graphite) + 3 H₂ (9) → C6 H6 (0) AH = -84.6 KJ G 1559.8 kJ -857.4 KJ -Enthalpy of Formation (AHI): enthalpy change for the reaction where a Compound is made frm constituent elements in their elemental forms Standard Enthalpy of Formation (Alt°f): enthalpy change when I mol of a substance is formed from its component elements most stable states in their •AKA Standard Heat of Formation • Standard Conditions (25°C and 1.00 atm pressure) are denoted by 0 -AHF for any element in its most stable state is O Sign | Hixh = (Σn₂ & Hi (products)) - Σn₂ & Hi (xcocinas) (reactants) also - Bond Enthalpies : + used to calculate appx enthalpies for a reaction -To break a bond, energy is needed b/c of ators are at lowest PE and more stable (endothermic) -When bonds for, atoms must lose energy to lower PE, excess energy is heat (exothermic) - To use bond enthalpics, reactants and products must be in a gas state If there's liquids, you need enthalpy change of vaporization - the more energy needed to break a bond, the more stable the Compound is - Ex/ Calculate H: CH4 (9) + 2 O₂ (9) → 2 H₂O (9) + CO₂ (9) Bonds broken: 4 (H+- C-H) Bonds formed: 4 (₁) and 2 (0=C=0) AH = (4 (413) + 2 (498)) - (4 (464) + 2 (805)) AH = 2658 kJ 3466 kJ H = -808 KJ and 2 (0 = 0)