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

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1 #37 Notes Unit 5: Thermodynamics/Thermochemistry/Energy Ch. 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 w = -PΔV ΔE = change in internal energy of a system (joules) q = heat, w = work 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 II. Enthalpy (H) -is the heat energy. q = H at constant P Exothermic Heat lost. Endothermic Heat gained. We can measure the difference in heat for the formation of compounds from their elements. See Textbook for ΔH o f values ( o for standard state, f for formation) cr = crystalline solid

3 ΔH reaction = Σ n ΔH products(final) Σ n ΔH reactants(initial) Gr. letter sigma (summation) Ex. 1) Find ΔH reaction for Zn (s) + 2 HCl (aq) ZnCl 2(aq) + H 2(g) reactants products 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) )

4 #38 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?

5 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) Ex. 3) Calculate the final temperature of the mixture when ml of 1.00 M HCl is mixed with ml of M Ca(OH) 2. (T i = 25 o C, final mass of mixture = g, s = 4.18 J /(g oc) 2 HCl (aq) + Ca(OH) 2(aq) CaCl 2(aq) + 2 H 2 O (l) ΔH = -112 kj for the whole reaction, i.e. for 1 mol CaCl 2, for 2 H 2 O etc. Find limiting reagent: HCl Ca(OH) 2 M = mol/l M= mol/l 1.00 M HCl = mol M Ca(OH) 2 = mol L L = mol HCl = mol Ca(OH) mol HCl 1 mol CaCl 2 = mol CaCl 2 limiting reagent 2 mol HCl mol Ca(OH) 2 1 mol CaCl 2 = mol CaCl 2 1 mol Ca(OH) mol CaCl kj = kj heat produced 1 mol CaCl 2 ΔE = q + w, no work, so ΔE = q = smδt **since heat is produced, T f should be larger than T i So change heat to positive q = smδt 16.8 kj 1 X10 3 J = (4.18 J /(g oc)) (650.0 g) (T f 25 o C) 1 kj 6.18 = T f 25 o C 31 o C = T f

6 #39 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.)

7 #40 Notes Ch. Energy I. Entropy (S) -is the disorder in a system. In nature systems get more messed up (gain entropy). Ex. 1) Is entropy increasing or decreasing? a) melting Al Al (s) Al (l) More ordered More messy (lower entropy) (higher entropy) ΔS = (+) increasing b) sublimating I 2 I 2(s) I 2(g) More ordered More messy (lower entropy) (higher entropy) ΔS = (+) increasing c) 2 CO (g) + O 2(g) 2 CO 2(g) all gases, so look at amounts 3mols 2 mols Three objects can get more disorganized than 2 objects. High entropy Low entropy ΔS = (-) decreasing d) Zn (s) + 2 HCl (aq) ZnCl 2(aq) + H 2(g) aqueous to aqueous is no change Solid Gas More messy, higher entropy ΔS = (+) increasing Ex. 2) Which has more entropy? a) C 6 H 12 O 6(s) or H 2 O (s) C 6 H 12 O 6(s) is more complex b) He at 0 o C or He at 200 o C 200 o C is a higher temperature, more motion c) He at 1 atm or He at 2 atm 1 atm has lower pressure, less organized ** The higher the temperature and the lower the pressure, the higher the entropy! STP = 273 K & 1 atm II. 2 nd Law of Thermodynamics In any spontaneous process there is an increase in the Universe s entropy. **spontaneous immediately reacts, but it could be a fast or slow process.

8 ΔS universe = ΔS system + ΔS surroundings (+) ΔS universe is spontaneous, since it gains entropy, ΔS surroundings is usually determined by heat flow to/from the system. For example: Melting Ice For the system: the ice gains heat (ΔH = (+)), the ice goes from solid to liquid (it gains entropy, ΔS = (+)) For the surroundings: it loses heat to the ice (ΔH = (-)), so it loses entropy as well (ΔS = (-)) ΔS universe = ΔS system + ΔS surroundings (+) to be spontaneous. (+) (-) Which is stronger? Will the ice melt? This is determined by temperature. When the surroundings is > 0 o C, the ice melts. (ΔS system is stronger. ΔS universe be (+).) When the surroundings is < 0 o C, the ice doesn t melt. (ΔS surroundings is stronger. ΔS universe be (-).) ΔS surroundings = -ΔH system but ΔS system = +ΔH system T T T is in Kelvin Ex. 1) The melting point of iron is 1535 o C and the enthalpy of fusion is 13.8 kj /mol. What is the entropy of fusion (ΔS system )? What is the ΔS surroundings? ΔS system = 13.8 kj/mol = 7.63 X10-3 kj 1 X10 3 J = 7.63 J/mol K 1808 K mol K 1 kj So, ΔS surroundings = J/mol K

9 #41 Notes III. 3 rd Law of Thermodynamics The entropy of a perfect crystal at 0 Kelvin is ZERO. (Absolute Zero temperature = no vibrations, no random movement, no Kinetic Energy) IV. Gibb s Free Energy (G) -is the chemical potential energy. ΔG (-) = spontaneous ΔG (+) = nonspontaneous ΔG = ΔH TΔS ΔG in kj/mol, ΔH in kj/mol, T in Kelvin, ΔS in j/(mol K) **so change to kj/(mol K) Ex.1) Will a reaction be spontaneous at 25 o C, if ΔH = 542 kj/mol and ΔS = -14 J/(mol K)? -14 J 1 kj = kj/(mol K) mol K 1 X10 3 J ΔG = ΔH TΔS = (542 kj/mol) [(298 K) ( kj/(mol K))] ΔG = 542 kj/mol kj/mol = 546 kj/mol (+) so nonspontaneous ΔG reaction = Σ n ΔG products(final) Σ n ΔG reactants(initial) ΔS reaction = Σ n ΔS products(final) Σ n ΔS reactants(initial) Ex. 2a) Calculate ΔG reaction for: 2 HF (g) H 2(g) + F 2(g) ΔG reaction = Σ n ΔG products(final) Σ n ΔG reactants(initial) ΔG reaction = [(1 mol H 2 ) ( 0 kj/mol) + (1 mol F 2 ) ( 0 kj/mol) ] [( 2 mol HF) ( kj/mol)] ΔG reaction = kj = kj nonspontaneous Ex. 2b) Calculate ΔS reaction ΔS reaction = Σ n ΔS products(final) Σ n ΔS reactants(initial) ΔS reaction = [(1 mol H 2 ) ( J/(mol K)) + (1 mol F 2 ) ( J/(mol K)) ] [( 2 mol HF) ( J/(mol K))] ΔS reaction = J/K J/K J/K = J/K decreasing *End of Notes* (Assignments #42-43 are Review Assignments. There are no notes for these assignments.)

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