Thermo_2 - CHEMICAL REACTIVITY REACTIVITY What drives...

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Unformatted text preview: CHEMICAL REACTIVITY REACTIVITY What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS. KINETICS. Have already seen a number of “driving forces” for reactions that are PRODUCTFAVORED. FAVORED. • formation of a precipitate • gas formation • H2O formation (acid-base reaction) • electron transfer in a battery CHEMICAL REACTIVITY REACTIVITY But energy transfer also allows us to predict reactivity. Heat Energy Transfer in a Physical Process CO2 (s, -78 oC) ---> CO2 (g, -78 oC) (s, (g, In general, reactions that transfer energy to their surroundings are productfavored. So, let us consider heat transfer in chemical processes. 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. 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) SYSTEM (solid from the SURROUNDINGS in an SURROUNDINGS in Heat Energy Transfer in a Physical Process CO2 gas gas ENDOTHERMIC process. process. Surroundings System ∆E = E(final) - E(initial) = E(gas) - E(solid) heat CO2 solid solid Page 1 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 WORK is system in pushing aside the atmosphere. FIRST LAW OF THERMODYNAMICS heat energy transferred heat transfer in (endothermic), +q heat transfer out (exothermic), -q SYSTEM ∆E = q + w energy change work done by the system w transfer in (+w) ∆E = q + w Energy is conserved! w transfer out (-w) Endo- and Exothermic Endo- and Exothermic Surroundings System Surroundings System ENTHALPY Most chemical reactions occur at constant P, so Heat transferred at constant P = qp qp = ENTHALPY ∆H = Hfinal - Hinitial If Hfinal > Hinitial then ∆H is positive hen If Hfinal > Hinitial tthen ∆H is positive heat heat ∆H where H where = enthalpy enthalpy Process is ENDOTHERMIC Process is ENDOTHERMIC ENDOTHERMIC qsystem > 0 qsystem < 0 and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system If Hfinal < Hinitial then ∆H is negative hen If Hfinal < Hinitial tthen ∆H is negative Process is EXOTHERMIC Process is EXOTHERMIC EXOTHERMIC T(system) goes up T (surr) goes down (surr) T(system) goes down T(surr) goes up T(surr) ∆H = Hfinal - Hinitial ENDOTHERMIC EXOTHERMIC Page 2 USING ENTHALPY Consider the decomposition of water H2O(g) + 242 kJ 242 ---> H2(g) + 1/2 O2(g) ---> Endothermic reaction — heat is a “reactant” ∆H = + 242 kJ USING ENTHALPY Making H2 from H2O involves two steps. from Each step requires energy. USING ENTHALPY Making H2 from H2O involves two steps. from H2O(liq) + 44 kJ ---> H2O(g) O(liq) H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g) ----------------------------------------------------------------------- Liquid H 2O H2 + O2 gas H2O vapor H2O(liq) + 286 kJ --> H2(g) + 1/2 O2(g) O(liq) Example of HESS’S LAW— LAW— If a rxn. is the sum of 2 or more others, rxn. the net ∆H is the sum of the ∆H’s of the other rxns. rxns. USING ENTHALPY Calc. ∆H for S(s) + 3/2 O2(g) --> SO3(g) Calc. knowing that S(s) + O2(g) --> SO2(g) ∆H1 = -320.5 kJ -320.5 SO2(g) + 1/2 O2(g) --> SO3(g) ∆H2 = -75.2 kJ -75.2 The two equations add up to give the desired equation, so ∆Hnet = ∆H1 + ∆H2 = -395.7 kJ -395.7 energy S solid direct path + 3/2 O2 ∆H = -395.7 kJ SO3 gas +O2 ∆H1 = -320.5 kJ SO2 gas + 1/2 O2 ∆H2 = -75.2 kJ Σ ∆H along one path = Σ ∆H along one path = Σ ∆H along another path Σ ∆H along another path • This equation is valid because ∆H is a STATE FUNCTION • 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 ∆H. Σ ∆H along one path = Σ ∆H along one path = Σ ∆H along another path Σ ∆H along another path Page 3 Standard Enthalpy Values Most ∆H values are labeled ∆H o Measured under standard conditions P = 1 bar Concentration = 1 mol/L mol/L T = usually 25 oC with all species in standard states e.g., C = graphite and O2 = gas 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) ∆H˚ = -242 kJ 2 H2(g) + O 2(g) --> 2 H2O(g) ∆H˚ = -484 kJ H2O(g) ---> H 2(g) + 1/2 O2(g) ∆H˚ = +242 kJ H2(g) + 1/2 O 2(g) --> H2O(liquid) ∆H˚ = -286 kJ Standard Enthalpy Values NIST (Nat’l Institute for Standards and (Nat’l Technology) gives values 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 ∆Hfo = standard molar enthalpy of standard ∆Hfo, standard molar enthalpy of formation H2(g) + 1/2 O2(g) --> H2O(g) ∆Hfo (H2O, g)= -241.8 kJ/mol (H kJ/mol By definition, ∆Hfo = 0 for for elements in their standard states. Using Standard Enthalpy Values Use ∆H˚’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) ∆Hf˚ of H2O vapor = - 242 kJ/mol kJ/mol • C(s) + 1/2 O2(g) --> CO(g) (product is called “water gas”) “water gas”) ∆Hf˚ of CO = - 111 kJ/mol kJ/mol Page 4 Using Standard Enthalpy Values H2O(g) --> H2(g) + 1/2 O2(g) ∆Ho = +242 kJ +242 C(s) + 1/2 O2(g) --> CO(g) ∆H = -111 kJ -111 -------------------------------------------------------------------------------- Using Standard Enthalpy Values Standard In general, when ALL ALL enthalpies of formation are known, Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆Horxn for for CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Horxn = Σ ∆Hfo (prod) - Σ ∆Hfo (react) (prod) (react) ∆Ho Calculate ∆ H of reaction? H2O(g) + C(graphite) --> H2(g) + CO(g) ∆Honet = +131 kJ +131 To convert 1 mol of water to 1 mol each of H2 and CO requires 131 kJ of energy. and The “water gas” reaction is ENDOthermic. ENDOthermic. ∆Horxn = Σ ∆Hfo (products) (products) - Σ ∆Hfo (reactants) (reactants) Using Standard Enthalpy Values CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Horxn = Σ ∆Hfo (prod) - Σ ∆Hfo (react) (prod) (react) (CO2) + 2 ∆Hf (H2O) (CO (H - {3/2 ∆Hfo (O2) + ∆Hfo (CH3OH)} {3/2 (O (CH = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} o ∆H rxn = -675.6 kJ per mol of methanol -675.6 = o CALORIMETRY Measuring Heats of Reaction • E transferred from system = E transferred to surroundings • System = reaction • Surround = water + “bomb” Measuring Heats of Reaction Measuring Heats of Reaction CALORIMETRY CALORIMETRY Calculate heat of combustion of octane. C8H18 + 25/2 O2 --> 25/2 --> 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 ∆Horxn ∆Hfo Page 5 Measuring Heats of Reaction Measuring Heats of Reaction CALORIMETRY CALORIMETRY Step 1 Calc. heat transferred from Calc. reaction to water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. heat transferred from Calc. reaction to bomb. q = (bomb heat capacity)( ∆T) = (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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