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Ch.5 THERMOCHEMISTRY

Ch.5 THERMOCHEMISTRY. Energy, E work, w 1 st Law Thermo Calorimetry. Enthalphy, heat, q heat of rxn; enthalphyies of formation Hess’ Law. ∆H RXN. ∆H, Enthalpy. Specific heat. Thermochemistry. relationship bet chem rxns & E  es due to heat.

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Ch.5 THERMOCHEMISTRY

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  1. Ch.5 THERMOCHEMISTRY Energy, E work, w 1st Law Thermo Calorimetry Enthalphy, heat, q heat of rxn; enthalphyies of formation Hess’ Law ∆HRXN ∆H, Enthalpy Specific heat

  2. Thermochemistry relationship bet chem rxns & E es due to heat capacity to do work or transfer heat Energy, E Work, w E causes a  in direction or position of an object w = F*d Heat, q E to cause increase of temp of an object hotter -----> colder sys ----> surr exothermic, sys losses q surr ----> sys endothermic, sys gains q

  3. ENERGY PE: potential E stored E, amt E sys has available KE: kinetic E E in motion, Ek = 0.5 mv2 2 objects mass1> mass2 @ same speed which more Ek? 1 object v1< v2 @ same mass which more Ek? Internal E total KE + PE of system ∆Ealways => system surr ∆E = ∑Ef - ∑Ei = ∑Epdt - ∑Ereact ∆Esys = -∆Esurr Transfer of E results in work &/or heat

  4. H2 O2 NOW, think of atoms & molecules in random motion colliding!!!!!! What kind of Energies would be involved? motion thru space rotational, vibrational nuclei, e-

  5. E UNITS Joule, J EK = 0.5(2 Kg)(1 m/s)2 = 1 Kg-m2/s2 = 1 J calorie, cal E needed to raise 1 g H2O by 1oC 1 cal = 4.184 J 1 Cal (food) = 1000 cal= 1 kcal Transfer of E results in work &/or heat

  6. System – Surroundings Defined as ………….? What??? System: a defined region Surroundings: everything that will ∆ by influences of the system OPEN: matter & E ex w/ surr CLOSED: ex E, not matter w/ surr

  7. ∑PE(H2O+ CO2) < ∑PE(O2+ CH4) 2 mol O2 1 mol CH4 system ∆PE E released to surroundings as Heat 2 mol H2O 1 mol CO2 E

  8. ∑PE(NO) > ∑PE(O2+ N2) system 2 mol NO Heat absord from surroundings 1 mol N2 1 mol O2 E

  9. Determine the sign of DH in each process under 1 atm; eno or exo? 1. ice cube melts 2. 1 g butane gas burned to give CO2 & H2O must predict if heat absorbed or released 1. ice is the sys, ice absorbs heat to melt, DH “+”, ENDO 2. butane + O2 is the sys, combustion gives off heat, DH “-”, EXO

  10. Conservation of E 1st Law of Thermodynamics: total E of universe is constant - E is neither created nor destroyed but es form q: Heat, Internal H E transfer bet sys & surr w/ Temp diff w: work, other form E transfer mechanical, electrical, chemical ∆E = q + w sum of E transfer as heat &/or work ∆E = q + w + + + - - - + - + : sys gain E; w > q - + - : sys lost E; w > q

  11. What is ∆E when a process in which 15.6 kJ of heat and 1.4 kJ of work is done on the system? ∆E = q + w 15.6 + 1.4 kJ = 17.0 kJ

  12. State Functions Property of variable depends on current state; not how that state was obtained T, H, E, V, P use CAPITAL letters to indicate state fcts ∆ : state fcts depend on initial & final states ∆H Enthalpy Must measure q & w 2 types: electrical, PV - movement of charged particles - w of expanding gas w = -P∆V @ constant P ∆H = ∆E + P∆V q + w

  13. 3 Chemical Systems #1 no gas involved s, l, ppt, aq phases have little or not V change; P∆V ≈ 0, then ∆H ≈ ∆E #2 amt of gas no change H2 (g) + I2 (g) -- 2 HI (g) P∆V = 0, then ∆H = ∆E #3 amt of gas does change N2 (g) + 3 H2 (g) -- 2 NH3 (g) P∆V ≠ 0, then ∆H ≈ ∆E ∆E mostly transfer as Heat ∆H Enthalpy Changes ∆Hcomb ∆Hf ∆Hfus ∆Hvap combines w/ O2 cmpd formed subst melts subst vaporizes s -- l l -- g

  14. PV Work Calculate the work associated with the expansion of a gas from 46 L to 64L @ 15 atm. w = -P∆V w = -(15 atm)(18 L) = -270 L-atm NOTE: “PV” work - P in P∆V always refers to external P - P that causes compression or resists expansion

  15. A balloon is inflated by heating the air inside. The vol changes from 4.00*106 L to 4.5*106 L by the addition of 1.3*108 J of heat. Find ∆E, assuming const P = 1.0 atm Heat added, q = + P = 1.0 atm 1 L-atm = 101.3 J ∆V = 5.0*105 L ∆E = q + w w = -P∆V w = -(1.0 atm)(5.0*105 L) = -5.0*105 L-atm (-5.0*105 L-atm)(101.3 J / 1 L-atm) = -5.1*107 J ∆E = q + w = (-5.1*107 J) + (1.3*108 J) = 8*107 J More E added by heating than gas expanding, net increase in q, ∆E “+”

  16. REVIEW PE - KE J - cal 1st Law Enthalpy sys - surr PV STATE Fcts

  17. CALORIMETRY Heats of Reaction Measure of Heat flow, released or absorded, @ const P & V Not as simple as: ∆Hfinal - ∆Hinitial Solar-heated homes use rocks to store heat. An increase of 120C in temp of 50.0 Kg of rocks, will absorb what quantity of heat? Assume Cs = 0.82 J/Kg-K. What T would result in a release of 450 kJ? Heat Capacity, CSpecific Heat, Cs T when object absorbs heat C of 1 g of subst +q or -q? gains loss endo exo q = Cs*m*T

  18. How much Heat is transferred when 720 g of antifreeze cools 25.5 oC? Cs = 2.42 J/g-K THN IK!!!! q = Cs * mass * ∆T ∆ T = -25.5 ? T? K or oC?  K = oC q = (2.42 J/g-K) * (720 g) * (-25.5K) = -44400 J or -44.4 kJ

  19. HESS’S LAW Heat Summation Hess’ states: overall H is sum of individual steps Rxn are multi-step processes Calculate H from tabulated values REACTS ======> PDTS

  20. THN IK!!!! What effect H if - reverse rxn? - had 3X many moles?

  21. Calculate HRXN for Ca(s) + 0.5 O2(g) + CO2 (g) --------> CaCO3(s) HRXN = ? given the following steps: Ca (s) + 0.5 O2 (g) -----> CaO (s) Hof = - 635.1 kJ CaCO3 (s) -----> CaO (s) + CO2 (g) Hof = + 178.3 kJ Note: to obtain overall rxn ==> (1st rxn) + (-2nd rxn) Ca (s) + 0.5 O2 (g) -----> CaO (s) Hof = -635.5 kJ CaO (s) + CO2 (g) -----> CaCO3 (s) Hof = -178.3 kJ Ca(s) + 0.5 O2(g) + CO2 (g) ------> CaCO3(s) Horxn = -813.4 kJ “o”?? Ho? Enthalpies of Formation

  22. STANDARD STATES Set of specific conditions - gas: 1 atm, ideal behavior - aq solution: 1 M (mol/L) - pure subst: most stable form @ 1 atm & Temp T usually 25oC - forms 1 mole cmpd; kJ/mol Use superscript “o” indicates Std States Individual ∆Hf 0 values from book table, appendix C, pg. 1059 or 221 NOTE: look at state

  23. Ca (s) + 0.5 O2 (g) -----> CaO (s) H = -635.5 kJ -635.5 kJ - ( 0.0 + 0.0 )kJ = -635.5 kJ

  24. Solar-heated homes use rocks to store heat. An increase of 120C in temp of 50.0 Kg of rocks, will absorb what quantity of heat? Assume Cs = 0.82 J/Kg-K. What T would result in a release of 450 kJ? q = Cs*m*T q = (0.82 J/g-K)*(5.0*104 g)*(12.0 K) = 4.9 * 105J T = q/[Cs*m] T = (4.9*105 J)/[(0.82 J/g-K)*(5.0*104 g)] = 11O decrease What is the change in enthalpy for the reaction of sulfur dioxide and oxygen to form sulfur trioxide. All in gas form. Is this endo- or exo-thermic? Eqn: 2 SO2 (g) + O2 (g) -----> 2 SO3 (g)

  25. Find Hof per mole in tables (kJ/mol) SO2 = -296.8 SO3 = -396.0 O2 = 0.0 free element Sum Hf reactants using stoich coeff & also pdts H = Hf Pdts - Hf reacts H = (-792.0 kJ) - (-593.6 kJ) = -198.4 kJ H? What if rxn were reversed????? Exothermic, -H

  26. SUMMARY HEAT CAPACITY: amt of heat needed to D sys by 10C MOLAR HEAT CAPACITY: amt of heat needed to D mole of subst by 10C SPECIFIC HEAT: amt of heat needed to D 1 g of subst by 10C THERMO 1st LAW: if a sys undergoes series of Des that brings it back to its original state, net DE = 0 E is constant 2nd LAW: Gibbs E, DG, theoretical max work done by sys a rxn occurs spontaneously in direction that increases DS, disorder

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