State Functions -depend only on the current state of the system.

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1 #47 Notes Ch. 6 Thermochemistry (Thermodynamics) -is the study of energy. I. Energy is the capacity to do work or produce heat. Law of Conservation of Energy Energy cannot be created or destroyed, just transferred. 1 st Law of Thermodynamics The energy in the universe is constant. State Functions -depend only on the current state of the system. P, V, T, energy Heat Heat flows from hot to cold: high KE, high velocity particles collide with cold/slower particles, making them speed up, increasing their KE. Exothermic Heat energy flows out of a system into its surroundings. (hot pack) **burning paper Endothermic Heat energy flows into a system from its surroundings. (cold pack) **ice cube melting ΔE = q + w ΔE = change in internal energy of a system (joules) q = heat, w = work +q = heat flows into the system (endothermic) -q = heat flows out of the system (exothermic) -w = the system does work (loses energy) +w = the surroundings do work on the system w = -PΔV P = pressure ΔV = V f - V i (change in volume) ** 1 L atm = J Ex. 1) A system gives off 196 kj of heat to its surroundings and the surroundings do 4.20 X10 5 j of work on the system. Find ΔE of the system X10 5 J 1 kj = 420 kj 1 X10 3 J ΔE = q + w = -196 kj kj = 224 kj (-) q, since heat is lost, (+) w, since work gained by system (work done on it)

2 Ex. 2) A balloon is being inflated. The volume of the balloon changes from 400 L to 750 L by the addition of 2.40 X10 6 j of heat. Assume the balloon expands against a constant pressure of 1.00 atm. Find ΔE of the balloon. ΔE = q + w w = -PΔV ΔV = V f - V i w = -(1.00 atm) (750 L 400 L) w = -350 L atm w = -350 L atm J = X10 4 J 1 L atm ΔE = q + w = 2.40 X10 6 J 3.54 X10 4 J ΔE = 2.36 X10 6 J

3 #48 Notes III. Calorimetry -is the science of measuring heat. ΔE = q + w w = -PΔV q = smδt where s = specific heat capacity, m = mass, and ΔT = T f - T i Specific heat capacity (s) -is the energy required to raise the temperature one degree Celsius for one gram of the substance. s for H 2 O = J/(g oc) = 1 calorie, s for Fe = 0.45 J/(g oc) Molar heat capacity -is the same except for 1 mol of the substance. J/(mol oc) **so the q equation would have mols for m, instead of mass Ex. 1a) How much heat is required to raise 8.77 g CCl 4 from 37.1 o C to 56.4 o C? s for CCl 4 = J/(g oc) q = smδt = (0.856 J/(g oc)) ( 8.77 g) ( 56.4 o C 37.1 o C) = (0.856 J/(g oc)) ( 8.77 g) ( 19.3 o C) = 1.45 X10 2 J Ex. 1b) Calculate the molar heat capacity J g CCl 4 = 132 J/(mol oc) g oc 1 mol CCl 4 Ex.1c) Calculate the heat of raising 24.1 mol CCl 4 from 25 o C to 325 o C. q = smδt = (132 J/(mol oc)) ( 24.1 mol) ( 325 o C 25 o C) = (132 J/(mol oc)) ( 24.1 mol) ( 300 o C) = 9.5 X10 5 J Ex. 2) A 25.0 g sample of metal was heated to 90.0 o C and added to 50.0 cm 3 of water at 20.0 o C. What is the specific heat of the metal, if the final temperature was 22.0 o C? Metal Water D = m/v Loses heat Gains heat (1.0 g/cm 3 ) = m -q = + q (50.0 cm 3 ) - smδt = + smδt 50.0 g = m -s ( 25.0 g) ( 22.0 o C 90.0 o C) = (4.184 J/(g oc)) ( 50.0 g) ( 22.0 o C 20.0 o C) -s (25.0 g) (-68 o C) = (4.184 J/(g oc)) ( 50.0 g) ( 2.0 o C) s (1700) = (418.4) s = J/(g oc)

4 #49 Notes II. Enthalpy (H) -is the heat energy. q = H at constant P Exothermic Heat lost. Endothermic Heat gained. Ex. 1) Are the following endothermic or exothermic. a) boiling water: absorbing heat to boil, so endothermic b) burning paper: gives off more heat, than it absorbs to begin burning, so exothermic c) melting ice cube: absorbing heat to melt, so endothermic We can measure the difference in heat for the formation of compounds from their elements. See P. A21-A24 for ΔH o f values ( o for standard state, f for formation) cr = crystalline solid (Appendix Four) ΔH reaction = Σ n ΔH products(final) Σ n ΔH reactants(initial) Gr. letter sigma (summation) Ex. 2) Find ΔH reaction for 2 Ag (s) + Cl 2(g) 2 AgCl (s) reactants products ΔH reaction = Σ n ΔH products(final) Σ n ΔH reactants(initial) ΔH reaction = [(2mol AgCl) ( kj/mol)] [(2 mol Ag) (0 kj/mol) + (1 mol Cl 2 )(0 kj/mol)] ΔH reaction = kj 0 0 = kj exothermic

5 Ex. 3) Find ΔH reaction for Zn (s) + 2 HCl (aq) ZnCl 2(aq) + H 2(g) ΔH reaction = Σ n ΔH products(final) Σ n ΔH reactants(initial) ΔH reaction = [(1 mol ZnCl 2 ) ( kj/mol) + (1 mol H 2 ) ( 0 kj/mol)] - [(1 mol Zn) ( 0 kj/mol) + ( 2 mol HCl) ( kj/mol)] ΔH reaction = kj kj = kj exothermic **Make sure subscripts match ( (cr) = (s) )

6 #50 Notes IV. Hess s Law The enthalpy change (ΔH r o ) for a reaction is the sum of the enthalpy changes for a series of reactions, that add up to the overall reaction. Steps: For each reaction: 1) Check to see, if the compounds are on the correct sides of the reaction. **If not, reverse the entire reaction, and change the sign of ΔH. 2) Check to see, if all of the unwanted compounds will cancel completely. **If not, multiply an entire reaction by a number so that they do cancel completely and multiply ΔH by that same number. Ex.1) Given: 2 HF (g) H 2(g) + F 2(g) ΔH = +537 kj ½ C (s) + F 2(g) ½ CF 4(g) ΔH = kj 2 C (s) + 2 H 2(g) C 2 H 4(g) ΔH = 52 kj Find ΔH r o for: C 2 H 4(g) + 6 F 2(g) 2 CF 4(g) + 4 HF (g) H 2(g) + F 2(g) 2 HF (g) ΔH = -537 kj (F 2 & HF were on wrong side, now -537)) ½ C (s) + F 2(g) ½ CF 4(g) ΔH = kj (F 2 & CF 4 on correct sides already) C 2 H 4(g) 2 C (s) + 2 H 2(g) ΔH = -52 kj (C 2 H 4 was on wrong side, now -52) 2 ( H 2(g) + F 2(g) 2 HF (g) ) ΔH = -537 kj X 2 (need to cancel H 2 ) = kj 4 (½ C (s) + F 2(g) ½ CF 4(g) ) ΔH = kj X 4 (need to cancel C) = kj C 2 H 4(g) 2 C (s) + 2 H 2(g) ΔH = -52 kj C 2 H 4(g) + 6 F 2(g) 2 CF 4(g) + 4 HF (g) ΔH = kj Ex. 2) Given the following data: SO 3(g) S (s) + 3/2 O 2(g) ΔH =+395 kj 2 SO 2(g) + O 2(g) 2 SO 3(g) ΔH = -198 kj Calculate ΔH for the reaction: S (s) + O 2(g) SO 2(g) S (s) + 3/2 O 2(g) SO 3(g) ΔH = -395 kj (S & O 2 on wrong side, now -395) 2 SO 3(g) 2 SO 2(g) + O 2(g) ΔH = +198 kj (SO 2 was on the wrong side, now +198)* *Now for the second reaction O 2 is on the wrong side, but there is other O 2 in the first reaction! 2 (S (s) + 3/2 O 2(g) SO 3(g) ) ΔH = -395 kj X 2 (need to cancel SO 3 ) = -790 kj 2 SO 3(g) 2 SO 2(g) + O 2(g) ΔH = +198 kj 2S (s) + 2O 2(g) * 2SO 2(g) ΔH = -592 kj *(3 O 2 on the left reduces to 2 O 2 by cancelling the 1 O 2 on the right) S (s) + O 2(g) SO 2(g) ΔH = -296 kj (Divided by 2 to get answer.)