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Causes of Change

Causes of Change. Ch.11. (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. Calorimeter.

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Causes of Change

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  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

  4. Specific Heat Capacity • (Cp): 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 = CpmΔT • m = mass (g) • ΔT = Tfinal - Tinitial = temp. change (ºC) • q = heat (J)

  5. 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 Cp of Al is 0.902 J/g• °C ? • List the eq. q = CpmΔT 2. Substitute & solve q = (0.902 J/g• °C)(300 g)(70°C - 20°C) = 13,530 J

  6. 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 H2O(measured w/ calorimeter)

  7. Heat Exchange Practice Find the Cp of 100 g of an unknown metal when it’s removed from H2O at 100°C & placed into 200 g of H2O at 20°C. The final T of the mixture is 23.5°C. • Write eq. Heat lost (by metal) = heat gained (by water) q(metal) = q(water) CpmΔT = CpmΔT

  8. Heat Exchange Practice • Substitute & solve Cp(100 g)(100°C – 23.5°C) = (4.18 J/g• °C)(200 g)(23.5°C - 20°C) Cp = 0.382 J/g• °C Note: this side must remain +, Therefore ΔT = Ti - Tf (only in these types of problems)

  9. 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

  10. Molar Heat Capacity Practice If C of H2O is 76 J/K·mol, calculate the amt of heat E needed to raise the T of 90.0 g of H2O from 35°C to 45°C. 1. List eq. q = nCΔT

  11. 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

  12. (11-2) Thermodynamics • Study of E flow • Thermo = “heat” • Dynamics = “motion”

  13. 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.

  14. 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

  15. 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? • List eq. ΔH = q = CΔT n Can’t use w/out q

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

  17. 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

  18. (11-3) Change of State • S & H change dramatically during a state change • Heat of fusion (ΔHfus): heat absorbed when 1 mol of a subst. melts • Molar enthalpy of fusion • Heat of vaporization(ΔHvap): heat absorbed when 1 mol of a liquid vaporizes • Molar enthalpy of vaporization

  19. (g) ΔHvap (l) (s) ΔHfus

  20. 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)

  21. Gibb’s Practice If ΔH° is 41.2 kJ/mol & ΔS° is 0.0418 kJ/K is the following rxn spontaneous at 25°C? H2 + CO2 H2O + CO • List the eq. ΔG = ΔH – TΔS • Subst. & solve ΔG = 41.2 kJ/mol – (298 K)(0.0418 kJ/K) = 28.7 kJ, nonspontaneous

  22. (11-4) Hess’s Law • Overall enthalpy change in a rxn is = to the sum of the individual steps CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) • Actually occurs in 2 steps: • CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802 kJ • 2H2O(g)  2H2O(l) DH = -88 kJ • ΔH = -802 kJ + -88 kJ = -890 kJ

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

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

  25. Equations • ΔHº = ∑ΔHºf(products) - ∑ΔHºf(reactants) • ΔSº = ∑ΔSºf(products) - ∑ΔSºf(reactants) • ΔGº = ∑ΔGºf(products) - ∑ΔGºf(reactants)

  26. Enthalpy Practice Calculate ΔHº for the following rxn. Is the rxn exo. or endothermic? H2(g) + CO2(g)  H2O(g) + CO(g) • List the eq. ΔHº = ∑ΔHºf(products) - ∑ΔHºf(reactants)

  27. Enthalpy Practice • 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

  28. Entropy Practice Calculate ΔSº for the following rxn. Does the rxn proceed toward a more ordered or disordered state? H2(g) + CO2(g)  H2O(g) + CO(g) • List the eq. ΔSº = ∑ΔSºf(products) - ∑ΔSºf(reactants)

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

  30. Gibb’s Practice Calculate ΔGº for the following rxn. Is the rxn spontaneous at 25°C? H2(g) + CO2(g)  H2O(g) + CO(g) • List the eq. ΔGº = ∑ΔGºf(products) - ∑ΔGºf(reactants)

  31. Gibb’s Practice • 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

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