Ch.5 Energy, E work, w 1st Law Thermo Calorimetry ∆H, Enthalpy THERMOCHEMISTRY Enthalphy, H, heat, q heat of rxn; enthalphyies of formation Hess’ Law ∆HRXN Specific heat Thermochemistry relationship bet chem rxns & E es due to heat 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 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 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 1 object v1 < v2 @ same mass Internal E which more Ek? which more Ek? total KE + PE of system ∆E = ∑Ef - ∑Ei = ∑Epdt - ∑Ereact ∆Esys = -∆Esurr ∆E always => system surr Transfer of E results in work &/or heat NOW, think of atoms & molecules in random motion colliding!!!!!! What kind of Energies would be involved? H2 O2 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 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 ∑PE(H2O + CO2) < ∑PE(O2 + CH4) 2 mol O2 1 mol CH4 system ∆PE E released to surroundings as Heat E 2 mol H2O 1 mol CO2 ∑PE(NO) > ∑PE(O2 + N2) system 2 mol NO ∆PE Heat absord from surroundings E 1 mol N2 1 mol O2 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 CO2 & H2O must predict if heat absorbed or released 1. ice is the sys, ice absorbs heat to melt, H “+”, ENDO 2. butane + O2 is the sys, combustion gives off heat, H “-”, EXO 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| 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 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 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) ∆H ∆Hcomb combines w/ O2 P∆V ≠ 0, then ∆H ≈ ∆E ∆E mostly transfer as Heat Enthalpy Changes ∆Hf cmpd formed ∆Hfus subst melts s -- l ∆Hvap subst vaporizes l -- g 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 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 = + 1 L-atm = 101.3 J ∆V = 5.0*105 L P = 1.0 atm ∆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 “+” REVIEW PE - KE Enthalpy sys - surr J - cal 1st Law PV STATE Fcts 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, C T when object absorbs heat +q or -q? gains loss endo exo Specific Heat, Cs C of 1 g of subst heat trans q C J gK s g * (Tf - Ti ) m * T q = Cs*m*T How much Heat is transferred when 720 g of antifreeze cools 25.5 oC? Cs = 2.42 J/g-K q = Cs * mass * ∆T ∆ T = -25.5 ? oC? K or K = oC THN IK!!!! T? q = (2.42 J/g-K) * (720 g) * (-25.5K) = -44400 J or -44.4 kJ 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 Figure 05.22 THN IK!!!! What effect H if - reverse rxn? - had 3X many moles? Figure 05.22 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) CaO (s) + CO2 (g) -----> CaCO3 (s) Hof = - 635.1 kJ Hof = - 178.3 kJ Ca(s) + 0.5 O2(g) + CO2 (g) ------> CaCO3(s) “o”?? Ho? Horxn = - 813.4 kJ Enthalpies of Formation 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 H 0 f H 0 rxn formation of subst from its elemental parts Individual ∆Hf 0 values from book table, appendix C, pg 1100 NOTE: look at state H = - 635.5 kJ Ca (s) + 0.5 O2 (g) -----> CaO (s) 0 f CaO 0.5 O H 0.0 Ca (s) H 0.0 2 (g) (s) H - 635.5 0 f 0 f H (Pdts) - H (Reacts) H O f O f -635.5 kJ - (0.0 + 0.0)kJ = -635.5 kJ O rxn 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) q = Cs*m*T q = (0.82 J/g-K)*(5.0*104 g)*(12.0 K) = 4.9 * 105 J T = q/[Cs*m] T = (4.5*105 J)/[(0.82 J/g-K)*(5.0*104 g)] = 11O decrease 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 2 mol SO 2 296.8 kJ 1 mol O 2 0.0 kJ - 593.6 kJ 1 mol 1 mol 396.0 kJ 2 mol SO 3 - 792.0 kJ 1 mol H = Hf Pdts - Hf reacts H = (-792.0 kJ) - (-593.6 kJ) = -198.4 kJ Exothermic, -H What if rxn were reversed????? Write balanced eqn for the formation of 1 mol of NO2 gas from nitrogen monoxide gas and oxygen gas. Calculate HOrxn 1 NO(g) + .5 O2(g) ---> NO2(g) HOf : 90.3 kJ + .5(0) kJ ---> 33.2 kJ (33.2 kJ) - (90.3 + 0)kJ = -57.1 kJ 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 Hrxn for each step and find the overall Hrxn 1 CH3OH(l) ---> 2 H2(g) + 1 CO(g) Hf: -238.6 0 -110.5 Hrxn = 128.1 kJ 1 CO(g) + 1 H2O(l) ---> 1 CO2(g) + 1 H2(g) Hf: -110.5 -285.8 -393.5 0 Hrxn = 2.8 kJ 1 CH3OH(l) + H2O(l) ---> CO2(g) + 3 H2(g) Hrxn = 130.9 kJ Toss a ball upward. a. Does KE increase or decrease Decrease, KE converts to PE b. As ball goes higher, want effect 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 Calculate E of system, is endo- or exo- thermic a. Balloon cooled, remove 0.655 kJ heat, shrinks, & atmosphere does 382 J work on q “-” w “+” E =-0.655 kJ + 0.382 kJ = -0.273 kJ EXO b. 100 g metal bar gains 25oC, absorbs 322 J of heat. Vol is constant q “+” w = 0 E = +322 J ENDO c. Surroundings do 1.44 kJ work compressing gas in perfectly insulated container q = 0 (perfectly insulated) w “+” E = +1.44 kJ ENDO 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 Ca(OH)2(s) ----- CaO(s) + H2O(g) H = 109 kJ b. Draw enthalpy diagram CaO(s) + H2O(g) H = 109 kJ Ca(OH)2(s)