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Chapter 18: Thermodynamics. Renee Y. Becker Valencia Community College. Thermodynamics. Thermodynamics: The study of interconversion of heat and other forms of energy Spontaneous process: Process that proceeds on its own without external influence (- G)

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Chapter 18: Thermodynamics


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    1. Chapter 18: Thermodynamics Renee Y. Becker Valencia Community College

    2. Thermodynamics Thermodynamics: The study of interconversion of heat and other forms of energy Spontaneous process: Process that proceeds on its own without external influence (-G) Non-spontaneous: Needs continuous external influence to take place (+G)

    3. 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 Gibbs Free-Energy: Measure of spontaneity G = H-TS

    4. 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)

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

    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. 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

    8. 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

    9. 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

    10. 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

    11. 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

    12. 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

    13. 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

    14. Free-Energy Free energy, G G = H – TS G = H - TS if G < 0 spontaneous Only direct relationship thatif G > 0 nonspontaneousalways holds up! if G = 0 equilibrium

    15. 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

    16. 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 SS are converted to products in their SS • G = H - TS G = H - TS

    17. 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

    18. 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

    19. 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)]

    20. 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

    21. 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)

    22. 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?

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

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

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