Thermo_2 - CHEMICAL REACTIVITY What drives chemical...

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Unformatted text preview: CHEMICAL REACTIVITY What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS The THERMODYNAMICS and the second by KINETICS. and KINETICS Have already seen a number of “driving forces” for reactions that are PRODUCTforces” PRODUCTFAVORED. • formation of a precipitate • gas formation • H2O formation (acid-base reaction) • electron transfer in a battery Heat Energy Transfer in a Physical Process • CO2 (s, -78 oC) ---> (s, CO2 (g, -78 oC) (g, • A regular array of molecules in a solid -----> gas phase molecules. • Gas molecules have higher kinetic energy. CHEMICAL REACTIVITY But energy transfer also allows us to predict reactivity. In general, reactions that transfer energy to their surroundings are productfavored. Heat Energy Transfer in a Physical Process CO2 (s, -78 oC) ---> CO2 (g, -78 oC) (s, (g, So, let us consider heat transfer in chemical processes. Heat Energy Transfer in a Physical Process CO2 (s, -78 oC) ---> CO2 (g, -78 oC) (s, (g, Heat flows into the SYSTEM (solid CO 2) Heat SYSTEM (solid from the SURROUNDINGS in an from SURROUNDINGS in ENDOTHERMIC process. process. Heat Energy Transfer in a Physical Process CO2 gas ΔE = E(final) - E(initial) = E(gas) - E(solid) Surroundings System heat Page 1 CO2 solid Heat Energy Transfer in Physical Change CO2 (s, -78 oC) ---> CO2 (g, -78 oC) (s, (g, Two things have happened! • Gas molecules have higher kinetic energy. • Also, WORK is done by the Also, WORK is system in pushing aside the atmosphere. FIRST LAW OF THERMODYNAMICS heat Surroundings System heat SYSTEM ΔE = q + w energy change qsystem < 0 work done by the by system system Energy is conserved! ENTHALPY T(system) goes down T(surr) goes up ENDOTHERMIC w transfer in (+w) w transfer out (-w) ENTHALPY ΔH = Hfinal - Hinitial Heat transferred at constant P = qp If Hfinal > Hinitial tthen ΔH is positive If Hfinal > Hinitial hen ΔH is positive ΔH where H = enthalpy enthalpy Process is ENDOTHERMIC Process is ENDOTHERMIC Process ENDOTHERMIC and so ΔE = ΔH + w (and w is usually small) ΔH = heat transferred at constant P ≈ ΔE ΔH = change in heat content of the system of T(system) goes up T (surr) goes down ΔE = q + w Most chemical reactions occur at constant P, so qp = qsystem > 0 heat transfer out (exothermic), -q heat energy transferred Endo- and Exothermic Endo- and Exothermic Surroundings System heat transfer in (endothermic), +q ΔH = Hfinal - Hinitial EXOTHERMIC Page 2 If Hfinal < Hinitial tthen ΔH is negative If Hfinal < Hinitial hen ΔH is negative Process is EXOTHERMIC Process is EXOTHERMIC Process EXOTHERMIC USING ENTHALPY Consider the decomposition of water H2O(g) + 242 kJ O(g) 242 ---> H2(g) + 1/2 O2(g) Endothermic reaction — heat is a “reactant” ΔH = + 242 kJ USING ENTHALPY USING ENTHALPY Making H2 from H2O involves two steps. Making H2 from H2O involves two steps. Each step requires energy. H2O(liq) + 44 kJ ---> H2O(g) H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g) ----------------------------------------------------------------------- Liquid H 2O H2 + O2 gas H2O vapor USING ENTHALPY Calc. ΔH for S(s) + 3/2 O2(g) --> SO3(g) knowing that S(s) + O2(g) --> SO2(g) ΔH1 = -320.5 kJ SO2(g) + 1/2 O2(g) --> SO3(g) ΔH2 = -75.2 kJ The two equations add up to give the desired equation, so ΔHnet = ΔH1 + ΔH2 = -395.7 kJ energy S solid direct path + 3/2 O2 +O2 ΔH1 = -320.5 kJ ΔH = -395.7 kJ SO3 gas SO2 gas + 1/2 O2 ΔH2 = -75.2 kJ Σ ΔH along one path = Σ ΔH along one path = Σ ΔH along another path Σ ΔH along another path Page 3 H2O(liq) + 286 kJ --> H2(g) + 1/2 O2(g) Example of HESS’S LAW— Example HESS’S If a rxn. is the sum of 2 or more others, the net ΔH is the sum of the ΔH’s of the other rxns. Σ ΔH along one path = Σ ΔH along one path = Σ ΔH along another path Σ ΔH along another path • This equation is valid because ΔH is a STATE FUNCTION STATE • These depend only on the state of the system and not how it got there. • V, T, P, energy — and your bank account! • Unlike V, T, and P, one cannot measure absolute H. Can only measure ΔH. Standard Enthalpy Values Most ΔH values are labeled ΔH o Most Measured under standard conditions Measured standard P = 1 bar Concentration = 1 mol/L T = usually 25 oC with all species in standard states e.g., C = graphite and O2 = gas Enthalpy Values Depend on how the reaction is written and on phases of reactants and products H2(g) + 1/2 O 2(g) --> H2O(g) ΔH˚ = -242 kJ 2 H2(g) + O 2(g) --> 2 H2O(g) ΔH˚ = -484 kJ H2O(g) ---> H 2(g) + 1/2 O2(g) (g) ΔH˚ = +242 kJ H2(g) + 1/2 O 2(g) --> H2O(liquid) Standard Enthalpy Values NIST (Nat’l Institute for Standards and Technology) gives values of ΔHfo = standard molar enthalpy of formation This is the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 6.2 and Appendix L ΔH˚ = -286 kJ ΔHfo, standard molar enthalpy of formation H2(g) + 1/2 O2(g) --> H2O(g) Using Standard Enthalpy Values Use ΔH˚’s to calculate enthalpy change for H2O(g) + C(graphite) --> H2(g) + CO(g) Using Standard Enthalpy Values H2O(g) + C(graphite) --> H2(g) + CO(g) From reference books we find • H2(g) + 1/2 O2(g) --> H2O(g) ΔHfo (H2O, g)= -241.8 kJ/mol By definition, ΔHfo = 0 for By for elements in their standard states. ΔHf˚ of H2O vapor = - 242 kJ/mol • C(s) + 1/2 O2(g) --> CO(g) (product is called “water gas”) Page 4 ΔHf˚ of CO = - 111 kJ/mol Using Standard Enthalpy Values Using Standard Enthalpy Values H2O(g) --> H2(g) + 1/2 O2(g) ΔHo = +242 kJ ΔHo C(s) + 1/2 O2(g) --> CO(g) ΔH = -111 kJ -------------------------------------------------------------------------------- H2O(g) + C(graphite) --> H2(g) + CO(g) ΔHonet = +131 kJ To convert 1 mol of water to 1 mol each of H2 and CO requires 131 kJ of energy. requires Calculate Δ H of reaction? Using Standard Enthalpy Values In general, when ALL In ALL enthalpies of formation are known, Calculate the heat of combustion of methanol, i.e., ΔHorxn for ΔHorxn = ΔHorxn = Σ ΔHfo (prod) - Σ ΔHfo (react) (react) CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) Σ ΔHfo (products) (products) - Σ ΔHfo (reactants) (reactants) The “water gas” reaction is ENDOthermic. ENDO Using Standard Enthalpy Values CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ΔHorxn = Σ ΔHfo (prod) - Σ ΔHfo (react) (react) ΔHorxn = ΔHfo (CO2) + 2 ΔHf (H2O) (CO (H - {3/2 ΔHfo (O2) + ΔHfo (CH3OH)} (O (CH = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} o ΔH rxn = -675.6 kJ per mol of methanol o CALORIMETRY Measuring Heats of Reaction • E transferred from system = E transferred to surroundings • System = reaction • Surround = water + “bomb” Page 5 Measuring Heats of Reaction Measuring Heats of Reaction CALORIMETRY CALORIMETRY Calculate heat of combustion of octane. C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O • Burn 1.00 g of octane • Temp rises from 25.00 to 33.20 oC • Calorimeter contains 1200 g water • Heat capacity of bomb = 837 J/K Measuring Heats of Reaction Measuring Heats of Reaction CALORIMETRY CALORIMETRY Step 1 Calc. heat transferred from reaction to water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)( ΔT) = (837 J/K)(8.20 K) = 6860 J Step 3 Total heat evolved 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ Page 6 ...
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This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Reynolds during the Fall '01 term at University of Florida.

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