CHEM 480 1nd Edition Lecture 7 Outline of Last Lecture I. Guest SpeakerOutline of Current Lecture I. Chemical ChangesII. Energy from biological fuelsIII. Energy is DispersedIV. Heat of formation, ΔfHϴV. Calculate average C-H bond enthalpy in CH4 given heats of formationVI. Potassium metal added to waterCurrent LectureI. Chemical Changesa. Chemical reaction, rearrangement of atoms in reactants to products, accompanied by enthalpy changesb. Bonds broken and new bonds formed c. Use bond enthalpy = molar quantityd. Bond enthalpy = endothermic quantityi. Bond enthalpy depends on moleculeII. Energy from biological fuelsa. How to calculate energy content from fuels – the body’s ability to use fuel to do workb. We use enthalpy of combustion, ΔcHϴ (molar quantity)c. May be standardized to: i. Specific enthalpy, ΔcHϴ (kJ/g)ii. Enthalpy density, ΔcHϴ (kJ/dm3)d. Remember that combustion reactions are exothermic e. Heat of combustion at constant pressure is heat associated w/ that changef. ΔcH = qP while ΔcU = qVg. ΔcH = ΔcU + p ΔcVgas = ΔcU + Δcνgas RTh. Combustion of glucose:i. C6H12O6 (s) + 6O2 (g) => 6CO2 (g) + 6H2O (l)i. ΔVgas = 0ii. ΔcH = ΔcU iii. If instead product was H2O (g), ΔVgas ≠ 0, ΔcH ≠ ΔcU These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.iv. Change in internal energy is more negative (absolute value ΔcU > ΔcH)j. Factors that influence fuel valuei. Extent of carbon atoms’ oxidationii. Number of carbon atomsk. Fuel valuesi. Simple sugar (glucose) provides ~ 16 kJ/gii. Complex carbohydrate ~ 17 kJ/giii. Proteins ~ 17 kJ/giv. Fats ~ 38 kJ/gv. *More C-C bonds greater oxidizing potentialvi. => Greater energy upon combustionIII. Energy is Disperseda. Not al energy released by oxidation of carbon results in workb. Some energy used to maintain body temperaturec. Way s of dissipating heati. Evaporation of water from skinii. Radiation of heat via blood flow d. Drawing on the fact that enthalpy is a state function:i. Hess’s law (the standard enthalpy of reaction is the sum of the standard enthalpies of the reactions that combine to equal the overall reaction)ii. Heats of formationIV. Heat of formation, ΔfHϴa. ΔfHϴ = enthalpy change associated with forming one mole of a substance from its elements in their reference state, where reference state = most stable form under prevailing conditions (usually 298 K and 1 bar)b. ΔfHϴ (element in reference state) = 0c. Exothermic compounds ΔfHϴ < 0d. Endothermic compounds ΔfHϴ > 0e. Skip 1.12 to 1.13V. Calculate average C-H bond enthalpy in CH4 given heats of formationa. CH4 = -74.8b. C = 719c. H = 218d. ΔfH = ΔfH (C) + 4ΔfH (H) - ΔfH (CH4) = 1670 kJ / 4 because 4 bondse. Answer: 416 kJ VI. Potassium metal added to water:a. 2K (s) + 2H2O (l) => 2KOH (aq) + H2 (g)b. Predict signs of w, q, ΔU and ΔH c. Equation causes an explosioni. ΔH = negative because exothermicii. Q = negativeiii. W = negative (doing work on surroundings) iv. ΔU =