1. 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, Enthalpy ∆H RXN Specific heat

2. Thermochemistry Energy, E capacity to do work or transfer heat Work, w E or force that causes a  in direction or position of an object w = F*d relationship bet chem rxns & E  es due to heat Heat, qE 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 Estored E, amt E sys has available E in motion, E k = 0.5 mv 2 KE: kinetic E 2 objects mass 1 > mass 2 @ same speed which more E k ? 1 object v 1 < v 2 @ same mass which more E k ? Internal E total KE + PE of system ∆E = ∑E f - ∑E i = ∑E pdt - ∑E react ∆E sys = -∆E surr ∆E always => system surr Transfer of E results in work &/or heat

4. H 2 O 2 NOW, think of atoms & molecules in random motion colliding!!!!!! What kind of Energies would be involved?

5. E UNITS Joule, J E K = 0.5(2 Kg)(1 m / s) 2 = 1 Kg-m 2 / s 2 = 1 J calorie, cal E needed to raise 1 g H 2 O by 1 o C 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. E 2 mol O 2 1 mol CH 4 ∆PE 2 mol H 2 O 1 mol CO 2 E released to surroundings as Heat ∑PE (H2O + CO2) < ∑PE (O2 + CH4) system

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

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

10. 1 st Law of Thermodynamics: total E of universe is constant - E is neither created nor destroyed but  es form Conservation of E 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 H 2 (g) + I 2 (g) --  2 HI (g) P∆V ≠ 0, then ∆H ≈ ∆E ∆E mostly transfer as Heat P∆V = 0, then ∆H = ∆E #3 amt of gas does change N 2 (g) + 3 H 2 (g) --  2 NH 3 (g) ∆H Enthalpy Changes ∆H comb ∆H f ∆H fus ∆H vap combines w/ O 2 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*10 6 L to 4.5*10 6 L by the addition of 1.3*10 8 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*10 5 L ∆E = q + w w = -P∆V w = -(1.0 atm)(5.0*10 5 L) = -5.0*10 5 L-atm (-5.0*10 5 L-atm)(101.3 J / 1 L-atm) = -5.1*10 7 J ∆E = q + w = (1.3*10 8 J) + (-5.1*10 7 J) = 8*10 7 J More E added by heating than gas expanding, net increase in q, ∆E “+”

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

17. Measure of Heat flow, released or absorded, @ const P & V CALORIMETRY Heats of Reaction Heat Capacity, C Specific Heat, C s  T when object absorbs heat C of 1 g of subst Solar-heated homes use rocks to store heat. An increase of 12 0 C in temp of 50.0 Kg of rocks, will absorb what quantity of heat? Assume C s = 0.82 J / Kg-K. What  T would result in a release of 450 kJ? Not as simple as: ∆H final - ∆H initial q = C s *m*  T +q or -q? gains loss endo exo

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

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

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

21. Calculate  H RXN for Ca(s) + 0.5 O 2 (g) + CO 2 (g) --------> CaCO 3 (s)  H RXN = ? given the following steps: Ca (s) + 0.5 O 2 (g) -----> CaO (s)  H o f = - 635.1 kJ CaCO 3 (s) -----> CaO (s) + CO 2 (g)  H o f = + 178.3 kJ Note: to obtain overall rxn ==>(1st rxn) + (-2nd rxn) Ca (s) + 0.5 O 2 (g) -----> CaO (s)  H o f = - 635.1 kJ CaO (s) + CO 2 (g) -----> CaCO 3 (s)  H o f = - 178.3 kJ Ca(s) + 0.5 O 2 (g) + CO 2 (g) ------> CaCO 3 (s)  H o rxn = - 813.4 kJ Ho?Ho? “o”?? 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 25 o C - forms 1 mole cmpd; kJ / mol Use superscript “o” indicates Std States Individual ∆H f 0 values from book table, appendix C, pg 1100 NOTE: look at state

23. Ca (s) + 0.5 O 2 (g) -----> CaO (s)  H = - 635.5 kJ -635.5 kJ - (0.0 + 0.0)kJ = -635.5 kJ

24. 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 SO 2 (g) + O 2 (g) -----> 2 SO 3 (g) Solar-heated homes use rocks to store heat. An increase of 12 0 C in temp of 50.0 Kg of rocks, will absorb what quantity of heat? Assume C s = 0.82 J / Kg-K. What  T would result in a release of 450 kJ?

25. q = (0.82 J / g-K)*(5.0*10 4 g)*(12.0 K) = 4.9 * 10 5 J q = C s *m*  T  T = q / [C s *m]  T = (4.5*10 5 J)/[(0.82 J / g-K)*(5.0*10 4 g)] = 11 O decrease

26.  H =  H f Pdts -  H f reacts  H = (-792.0 kJ) - (-593.6 kJ) = -198.4 kJ Exothermic, -  H What if rxn were reversed????? Find  H o f per mole in tables (kJ/mol) SO 2 = -296.8 SO 3 = -396.0 O 2 = 0.0 free element Sum  H f reactants using stoich coeff & also pdts

27. Write balanced eqn for the formation of 1 mol of NO 2 gas from nitrogen monoxide gas and oxygen gas. Calculate  H O rxn 1 NO (g) +.5 O 2 (g) ---> NO 2 (g)  H O f : 90.3 kJ +.5(0) kJ ---> 33.2 kJ (33.2 kJ) - (90.3 + 0)kJ = -57.1 kJ

28. Find the overall rxn, CH 3 OH (l) + H 2 O (l) ---> CO 2 (g) + 3 H 2 (g), from the given steps: H 2 (g) + CO (g) ---> CH 3 OH (l) CO (g) + H 2 O (l) ---> CO 2 (g) + H 2 (g) Calculate  H rxn for each step and find the overall  H rxn 1 CH 3 OH (l) ---> 2 H 2 (g) + 1 CO (g) 1 CO (g) + 1 H 2 O (l) ---> 1 CO 2 (g) + 1 H 2 (g)  H f : -238.6 0 -110.5  H rxn = 128.1 kJ  H f : -110.5 -285.8 -393.5 0  H rxn = 2.8 kJ 1 CH 3 OH (l) + H 2 O (l) ---> CO 2 (g) + 3 H 2 (g)  H rxn = 130.9 kJ

29. Toss a ball upward. a. Does KE increase or decrease b. As ball goes higher, want effect to PE Define a. System b. Closed system c. Not part of system Explain a. 1 st Law b. Internal E c. How internal E of closed system increase Decrease, KE converts to PE Increases Region of study w/ E changes exchange E not mass surroundings 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 25 o C, 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) + H 2 O (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) + H 2 O (g)  H = 109 kJ CaO (s) + H 2 O (g) Ca(OH) 2 (s)  H = 109 kJ