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Thermodynamics and Heat Exchange Principles in Chemistry

Understand the governing principles of nature affecting chemical reactions, such as heat transfer, specific heat capacity, and calorimetry. Learn about molar heat capacity, entropy, enthalpy, Gibbs energy, and Hess's Law in thermodynamics. Explore calculations for heat exchange and heat of reaction.

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Thermodynamics and Heat Exchange Principles in Chemistry

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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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