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Chapter 18: Thermodynamics

Vanessa N. Prasad-Permaul Valencia College CHM 1046. Chapter 18: Thermodynamics. Thermodynamics. Thermodynamics: The study of interconversion of heat and other forms of energy.

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Chapter 18: Thermodynamics

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  1. Vanessa N. Prasad-Permaul Valencia College CHM 1046 Chapter 18: Thermodynamics

  2. Thermodynamics Thermodynamics: The study of interconversion of heat and other forms of energy. Internal energy (U): the sum of the kinetic and potential energies of the particles making up a system. State Function: a property of a system that depends only on its present state which is determined by variables such as temperature and pressure.

  3. Thermodynamics Heat (q): energy that moves into or out of the system because of a temperature difference between the system and it’s surroundings Work (w): energy exchange that results when a force (F)moves an object through distance (d) w = F x d 1st Law of Thermodynamics: DU = q + w The change in internal energy of a system is equal to heat plus work. if heat evolves from system (-) if heat is absorbed by the system (+) work done by the system (-) work done on a system (+)

  4. Thermodynamics Spontaneous process: Process that proceeds on its own without external influence Non-spontaneous: Needs continuous external influence to take place

  5. Thermodynamics Entropy (S): a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy. 2nd Law of Thermodynamics: DS = q T For a spontaneous process at a given temperature, the change in entropy of a system is greater than the heat divided by the absolute temperature. The total entropy of a system and its surroundings always increases for a spontaneous process

  6. Entropy and temperature 3rd Law of Thermodynamics a) The entropy of a perfectly ordered crystalline substance at 0 K is zero b) As the temperature increases, the KE increases, Molecular motion increases, entropy increases

  7. Thermodynamics Entropy (S): Disorder, molecular randomness S = Sfinal - Sinitial When disorder increases +S When disorder decreases -S Enthalpy (H): Heat flow In to the system +H Out of the system -H Enthalpy change: H = TS

  8. Thermodynamics The Entropy of a system usually increases in the following situations: • A reaction in which a molecule is broken down into 2 or more smaller molecules • A reaction in which there is an increase in moles of gas • A process on which a solid changes to liquid or gas or a liquid changes to a gas

  9. Example 1: Predict the sign of S in the system for each of the following a)      H2O(g) H2O(l) b)      I2(g) 2I-(g) c)      CaCO3(s) CaO(s) + CO2(g) d) Ag+(aq) + Br-(aq) AgBr(s)

  10. Example 2: Which of the following reactions has an increase in entropy? • H2O(g) H2O(l) • H2O(l) H2O(g) • H2O(g) H2O(s)

  11. Standard Molar Entropies and Standard Entropies of Reaction Standard Molar Entropy, S The entropy of one mole of the pure substance at 1 atm pressure and a specific temperature usually 25C Standard Entropy of Reaction, S Entropy change for a chemical reaction S = Sproducts - Sreactants Based on 1 mole of substance so you have to multiply S by the number of moles present

  12. Standard Entropy of Reaction, S aA + bB  cC + dD S = [c S(C) + d S(D)] – [a S(A) + b S(B)] Units: coefficients are moles S = J/K mol S = J/K

  13. Example 3: Calculate the standard entropy of reaction at 25C for the decomposition of calcium carbonate CaCO3(s) CaO(s) + CO2(g) Substance S (J/K mol) CaCO3(s) 92.9 CaO(s) 39.7 CO2(g) 213.6

  14. Entropy and the Second Law of Thermodynamics 1st Law of Thermodynamics In any process, spontaneous or nonspontaneous, the total energy of a system and its surroundings is constant 2nd Law of Thermodynamics In any spontaneous process, the total entropy of a system and its surroundings always increases

  15. Entropy and the Second Law of Thermodynamics Stotal = Ssystem + Ssurroundings if S > 0 spontaneous if S< 0 non spontaneous if S = 0 equilibrium Ssurr = -H / T

  16. Entropy and the Second Law of Thermodynamics a)  Exothermic reaction: H<0, because the surroundings gain heat (entropy increases), heat is lost from the system b) Endothermic reaction: H>0, surroundings lose heat (entropy decreases), and system gains the heat

  17. Example 4: Which of the following reactions is endothermic? • N2O4(g) 2 NO2(g)H = +57.1 kJ • 2 NO2(g) N2O4(g)H = -57.1 kJ

  18. Free-Energy Free energy (G): A thermodynamic quantity defined by the equation: G = H – TS G = H - TS if G < 0 spontaneous if G > 0 nonspontaneous if G = 0 equilibrium

  19. Example 5: Consider the decomposition of gaseous N2O4 N2O4(g) 2 NO2(g)H = +57.1 kJ S = +175.8 J/K • Is this reaction spontaneous under standard-state conditions at 25C? • Estimate the temperature at which the reaction becomes spontaneous

  20. Standard Free-Energy Changes for Reactions • Standard State Conditions: Solids, liquids, and gases in pure form at 1 atm pressure, Solutes at 1M concentration, specified temperature, usually 25 celsius • Standard Free Energy Change, G: The change in free energy that occurs when reactants in their Std. States are converted to products in their Std. States. • G = H - TS G = H - TS

  21. Example 6: Consider the thermal decomposition of calcium carbonate CaCO3(s)CaO(s) + CO2(g)H = 178.3 kJ S = 160.4 J/K a) Calculate the standard free energy change for this reaction at 25C b)    Will a mixture of solid CaCO3, CaO, and gaseous CO2 at 1 atm pressure react spontaneously at 25C? c) Assuming that H and S are independent of temperature, estimate the temperature at which the reaction becomes spontaneous

  22. Standard Free Energies of Formation 1. Standard Free Energy of Formation, Gf The free energy change for formation of one mole of the substance in its standard state from the most stable form of its constituent elements in their standard states • Gf measures the substances thermodynamic stability with respect to its constituent elements • -Gf are stable and do not decompose to their constituent elements under standard state conditions

  23. Standard Free Energies of Formation 4. +Gf are thermodynamically unstable with respect to their constituents elements a) There is no point in trying to synthesize a substance that has a +Gf because it would degrade into it’s constituents b) You would need to synthesize it at different temperatures and or pressures or start with different starting materials that has a reaction with a -Gf • G = Gf(products) - Gf(reactants) 6. General reaction: aA +bB  cC + dD G = [cGf(C) + dGf(D)] – [aGf(A) + bGf(B)]

  24. Standard Free Energies of Formation To judge the spontaneity of a reaction: • When DGo is a large negative number (more than -10kJ), the reaction is spontaneous as written and reactants transform almost entirely to products when equilibrium is reached. • When DGo is a large positive number (more than 10kJ), the reaction is nonspontaneous as written and reactants do not give significant amounts of products at equilibrium. • When DGo has a small negative or positive value, the reaction gives an equilibrium mixture with significant amounts of both reactants and products.

  25. Example 7: Calculate the standard free energy change for the reaction of calcium carbide with water. Might this reaction be used for the synthesis of acetylene (C2H2)? CaC2(s) + 2 H2O(l) C2H2(g) + Ca(OH)2(s) Gf (CaC2) = -64.8 kJ/mol Gf (H2O(l)) = -237.2 kJ/mol Gf (C2H2) = 209.2 kJ/mol Gf (Ca(OH)2) = -898.6 kJ/mol

  26. Free Energy Changes and Composition of the Reaction Mixture Standard state conditions are unrealistic, the reaction itself will change the temperature and pressure, so what can we use to calculate the free energy change under non-standard state conditions? G = G + RT ln Q R = gas constant T = temperature in Kelvins Q = reaction quotient (Qc or Qp)

  27. Free Energy Changes and Composition of the Reaction Mixture Thermodynamic Equilibrium Constant (K): the equilibrium constant in which gases are expressed in partial pressures (atm) whereas the concentrations of solutes in liquid solutions are expressed in molarities.

  28. Example 8: Calculate the Free energy change for the formation of ethylene (C2H4) from carbon and hydrogen at 25C when the partial pressures are 100 atm H2 and 0.10 atm C2H4 2 C(s) + 2 H2(g) C2H4(g)G = 68.1 kJ/mol Is the reaction spontaneous in the forward or reverse direction?

  29. Free Energy and Chemical Equilibrium 1. When the Reaction Mixture is mostly reactants Q<<1 RT lnQ <<0 G<0 2. When the Reaction Mixture is mostly products Q>>1 RT lnQ >>0 G>0 3. G= -RT ln K K = equilibrium constant Kc or Kp

  30. Example 9: Calculate the Kp at 25C for the reaction CaCO3(s) CaO(s) + CO2(g) G= 130.5 kJ/mol

  31. Example 10: Which of the following can always predict the spontaneity of a reaction? • H • S • G

  32. Spontaneity and Temperature Change Effect of Temperature on the Spontaneity of Reactions

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