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Ch.5 THERMOCHEMISTRY

Ch.5 THERMOCHEMISTRY. Energy, E work, w 1 st Law Thermo Calorimetry. Enthalphy ,  H, 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, H, 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 or force that 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?

  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) system 2 mol O2 1 mol CH4 ∆PE E released to surroundings as Heat 2 mol H2O 1 mol CO2 E

  8. ∑PE(NO) > ∑PE(O2+ N2) system 2 mol NO ∆PE 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/ T diff w: work, other form E transfer mechanical, electrical, ∆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 = (1.3*108 J) + (-5.1*107 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 ? K or oC?  K = oC T? 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 = - 635.1 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”?? Enthalpies of Formation Ho?

  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 ∆Hf0 values from book table, appendix C, pg1100 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? 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? 2 SO2 (g) + O2 (g) -----> 2 SO3 (g)

  25. 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 = (4.5*105 J)/[(0.82 J/g-K)*(5.0*104 g)] = 11O decrease

  26. 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 Exothermic, -H What if rxn were reversed?????

  27. Write balanced eqn for the formation of 1 mol of NO2 gas from nitrogen monoxide gas and oxygen gas. Calculate DHOrxn 1 NO(g) + .5 O2(g) ---> NO2(g) DHOf : 90.3 kJ + .5(0) kJ ---> 33.2 kJ (33.2 kJ) - (90.3 + 0)kJ = -57.1 kJ

  28. Find the overall rxn, CH3OH(l) + H2O(l) ---> CO2(g) + 3 H2(g), from the given steps: H2(g) + CO(g) ---> CH3OH(l) CO(g) + H2O(l) ---> CO2(g) + H2(g) Calculate DHrxn for each step and find the overall DHrxn 1 CH3OH(l) ---> 2 H2(g) + 1 CO(g) 1 CO(g) + 1 H2O(l) ---> 1 CO2(g) + 1 H2(g) DHf: -238.6 0 -110.5 DHrxn = 128.1 kJ DHf: -110.5 -285.8 -393.5 0 DHrxn = 2.8 kJ 1 CH3OH(l) + H2O(l) ---> CO2(g) + 3 H2(g) DHrxn = 130.9 kJ

  29. Toss a ball upward. a. Does KE increase or decrease b. As ball goes higher, want effect to PE Decrease, KE converts to PE Increases Define a. System b. Closed system c. Not part of system Region of study w/ E changes exchange E not mass surroundings Explain a. 1st Law b. Internal E c. How internal E of closed system increase E not created nor destroyed, changes form Total E of system, KE + PE System absorbs heat or work done on system

  30. Calculate E of system, is endo- or exo- thermic a. Balloon cooled, remove 0.655 kJ heat, shrinks, & atmosphere does 382 J work on b. 100 g metal bar gains 25oC, absorbs 322 J of heat. Vol is constant c. Surroundings do 1.44 kJ work compressing gas in perfectly insulated container q “-” w “+” E =-0.655 kJ + 0.382 kJ = -0.273 kJ EXO q “+” w = 0 E = +322 J ENDO q = 0 (perfectly insulated) w “+” E = +1.44 kJ ENDO

  31. Ca(OH)2(s) ----- CaO(s) + H2O(g) Requires addition of 109 kJ of heat per mol of Ca(OH)2 a. Write balanced thermochemistry equation b. Draw enthalpy diagram Ca(OH)2(s) ----- CaO(s) + H2O(g) H = 109 kJ CaO(s) + H2O(g) H = 109 kJ Ca(OH)2(s)

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