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STATE FUNCTIONS and ENTHALPY. heat energy transferred. work done by the system. energy change. FIRST LAW OF THERMODYNAMICS. ∆E = q + w. Energy is conserved!. Chemical Reactivity. What drives chemical reactions? How do they occur?
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heat energy transferred work done by the system energy change FIRST LAW OF THERMODYNAMICS ∆E = q + w Energy is conserved!
Chemical Reactivity What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS. Have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED. • formation of a precipitate • gas formation • H2O formation (acid-base reaction) • electron transfer in a battery
Chemical Reactivity Energy transfer allows us to predict reactivity. In general, reactions that transfer energy to their surroundings are product-favored. So, let us consider heat transfer in chemical processes.
ENTHALPY: ∆H ∆H = Hfinal - Hinitial If Hfinal > Hinitial then ∆H is positive Process is ENDOTHERMIC If Hfinal < Hinitial then ∆H is negative Process is EXOTHERMIC
∆H along one path = ∆H along another path This equation is valid because ∆H is a STATE FUNCTION Depend ONLY on the state of the system NOT how it got there. Can NOT measure absolute H, ONLY measure ∆H.
Standard Enthalpy Values Most ∆H values are labeled ∆Ho Measured under standard conditions P = 1 bar Concentration = 1 mol/L T = usually 25 oC with all species in standard states e.g., C = graphite and O2 = gas
Standard Enthalpy Values ∆Hfo = standard molar enthalpy of formation — 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 formation H2(g) + 1/2 O2(g) --> H2O(g) ∆Hfo (H2O, g)= -241.8 kJ/mol By definition, ∆Hfo= 0 for elements in their standard states.
HESS’S LAW In general, when ALL enthalpies of formation are known, Calculate ∆H of reaction? ∆Horxn = ∆Hfo(products) - ∆Hfo(reactants) Remember that ∆ always = final – initial
Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆Horxn for CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Horxn = ∆Hfo(prod) - ∆Hfo(react)
Using Standard Enthalpy Values CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Horxn = ∆Hfo(prod) - ∆Hfo(react) ∆Horxn = ∆Hfo(CO2) + 2 ∆Hfo(H2O) - {3/2 ∆Hfo(O2) + ∆Hfo(CH3OH)} = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} ∆Horxn = -675.6 kJ per mol of methanol