1. Causes of Change Ch.11

2. (11-1) Governing Principles Nature favors rxns that proceed toward lower E & greater disorder Heat: total KE of particles –Quantity of thermal E –Joules Temp.: avg. KE of particles –Intensity of thermal E –Kelvin

3. Calorimeter Measures heat absorbed or released in a rxn Exo = E released Endo = E absorbed

6. Specific Heat Capacity (C p ): amt of heat needed to raise the T of 1 g of a substance by 1ºC –Units = J/g· ºC –Water = 4.18 J/g· ºC q = C p mΔT –m = mass (g) –ΔT = T final - T initial = temp. change (ºC) –q = heat (J)

7. Specific Heat Practice How many joules are needed to raise the T of 300 g of Al from 20 °C to 70°C if the C p of Al is 0.902 J/g °C ? 1.List the eq. q = C p mΔT 2. Substitute & solve q = (0.902 J/g °C)(300 g)(70°C - 20°C) = 13,530 J

8. Law of Heat Exchange Heat flows from hot to cold The law: –Heat lost = heat gained –Heat lost by a metal will be gained by the surrounding H 2 O (measured w/ calorimeter)

9. Heat Exchange Practice Find the C p of 100 g of an unknown metal when it’s removed from H 2 O at 100°C & placed into 200 g of H 2 O at 20°C. The final T of the mixture is 23.5°C. 1.Write eq. Heat lost (by metal) = heat gained (by water) q (metal) = q (water) C p mΔT = C p mΔT

10. Heat Exchange Practice 2.Substitute & solve C p (100 g)(100°C – 23.5°C) = (4.18 J/g °C)(200 g)(23.5°C - 20°C) C p = 0.382 J/g °C Note: this side must remain +, Therefore ΔT = T i - T f (only in these types of problems)

11. Molar Heat Capacity (C): heat required to inc. the T of 1 mol of a substance by 1 K –Units = J/K·mol –Table 11-1, p.389 q = nCΔT –n = moles

12. Molar Heat Capacity Practice If C of H 2 O is 76 J/K·mol, calculate the amt of heat E needed to raise the T of 90.0 g of H 2 O from 35°C to 45°C. 1. List eq. q = nCΔT

13. Molar Heat Capacity Practice 2. Substitute (make sure to convert g to mol) & solve q = (90 g x 1 mol ) )(76 J/K·mol)(45°C - 35°C) 18.02 g = 3,792 J

14. (11-2) Thermodynamics Study of E flow Thermo = “heat” Dynamics = “motion”

15. Entropy Total disorder in a substance or system Molar entropy (S): quantity of entropy in 1 mol of a subst. –Units = J/K·mol ΔS > O (+), disorder inc. ΔS < O (-), disorder dec.

16. Enthalpy E “inside” an atom or molecule Molar enthalpy (H): total E content of a system –Units = kJ/mol or J/mol ΔH > O (+), endo ΔH < O (-), exo ΔH = q = nCΔT = CΔT n n

17. Enthalpy Practice How much does the molar enthalpy change when a 92.3 g block of ice is cooled from –0.2°C to –5.4°C? 1.List eq. ΔH = q = CΔT n Can’t use w/out q

18. Enthalpy Practice 2.Find C for ice in Table 11-1 37.4 J/K·mol 3.Convert °C to K -0.2°C + 273 = 272.8 K -5.4°C + 273 = 267.6 K 4.Subst. & solve ΔH = (37.4 J/K·mol)(267.6 K – 272.8 K) = -194 J/mol Most ΔH are in kJ/mol

19. Properties of Matter Extensive property: depends on amt. of subst. –S, H, m, V, C Intensive property: does not depend on amt. of material –D, P, T

20. (11-3) Change of State S & H change dramatically during a state change Heat of fusion (ΔH fus ): heat absorbed when 1 mol of a subst. melts –Molar enthalpy of fusion Heat of vaporization (ΔH vap ): heat absorbed when 1 mol of a liquid vaporizes –Molar enthalpy of vaporization

21. ΔH fus ΔH vap (s)(s) (l)(l) (g)(g)

22. Gibbs Energy Molar Gibbs E (G): “free E”; determines spontaneity of a rxn –Units = kJ Spontaneous rxns occur w/out outside assistance –ΔG < O (-), spont. rxn –ΔG > O (+), nonspont. rxn ΔG = ΔH - TΔS, (T in K)

23. Gibb’s Practice If ΔH° is 41.2 kJ/mol & ΔS° is 0.0418 kJ/K is the following rxn spontaneous at 25°C? H 2 + CO 2  H 2 O + CO 1.List the eq. ΔG = ΔH – TΔS 2.Subst. & solve ΔG = 41.2 kJ/mol – (298 K)(0.0418 kJ/K) = 28.7 kJ, nonspontaneous

24. (11-4) Hess’s Law Overall enthalpy change in a rxn is = to the sum of the individual steps CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l) Actually occurs in 2 steps: –CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ –2H 2 O(g)  2H 2 O(l)  H = -88 kJ ΔH = -802 kJ + -88 kJ = -890 kJ

25. Heat of Rxn E absorbed or released during a chemical rxn Std. heat of formation (ΔHº f ): change in enthalpy when 1 mol of a cmpd is produced from free elements –Table A-13, p.802 (check state of matter)

26. Standard Conditions (°) are generally: –Temp: 298 K or 25°C –Pressure: 1 atm or 760 mmHg

27. Equations ΔHº = ∑ΔHº f(products) - ∑ΔHº f(reactants) ΔSº = ∑ΔSº f(products) - ∑ΔSº f(reactants) ΔGº = ∑ΔGº f(products) - ∑ΔGº f(reactants)

28. Enthalpy Practice Calculate ΔHº for the following rxn. Is the rxn exo. or endothermic? H 2 (g) + CO 2 (g)  H 2 O(g) + CO(g) 1.List the eq. ΔHº = ∑ΔHº f(products) - ∑ΔHº f(reactants)

29. Enthalpy Practice 2.Using Table A-13, subst. & solve (account for # of mols (coef.) of each cmpd) ΔHº = [(1 mol)(-241.8 kJ/mol) + (1 mol)(-110.5 kJ/mol)] - [ (1 mol)(0 kJ/mol) + (1 mol)(-393.5 kJ/mol)] = 41.2 kJ, endothermic

30. Entropy Practice Calculate ΔSº for the following rxn. Does the rxn proceed toward a more ordered or disordered state? H 2 (g) + CO 2 (g)  H 2 O(g) + CO(g) 1.List the eq. ΔSº = ∑ΔSº f(products) - ∑ΔSº f(reactants)

31. Entropy Practice 2.Using Table A-13, subst. & solve (account for # of mols (coef.) of each cmpd) ΔSº = [(1 mol)(188.7 J/Kmol) + (1 mol)(197.6 J/Kmol)] - [ (1 mol)(130.7 J/Kmol) + (1 mol)(213.8 J/Kmol)] = 41.8 J/K, disorder

32. Gibb’s Practice Calculate ΔGº for the following rxn. Is the rxn spontaneous at 25°C? H 2 (g) + CO 2 (g)  H 2 O(g) + CO(g) 1.List the eq. ΔGº = ∑ΔGº f(products) - ∑ΔGº f(reactants)

33. Gibb’s Practice 2.Using Table A-13, subst. & solve (account for # of mols (coef.) of each cmpd) ΔGº = [(1 mol)(-228.6 kJ/mol) + (1 mol)(-137.2 kJ/mol)] - [ (1 mol)(0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] = 28.6 kJ, nonspontaneous