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CHEMICAL THERMODYNAMICS

CHEMICAL THERMODYNAMICS. The Second Law of Thermodynamics: The is an inherent direction in which any system not at equilibrium moves. Processes that are spontaneous in one direction are not spontaneous in the reverse direction. CHEMICAL THERMODYNAMICS.

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CHEMICAL THERMODYNAMICS

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  1. CHEMICAL THERMODYNAMICS The Second Law of Thermodynamics: The is an inherent direction in which any system not at equilibrium moves Processes that are spontaneous in one direction are not spontaneous in the reverse direction

  2. CHEMICAL THERMODYNAMICS Spontaneity, Enthalpy, and Entropy • Reactions that are exothermic are generally spontaneous, H < 0 • Reactions that are not exothermic may also be spontaneous H = 0

  3. CHEMICAL THERMODYNAMICS H > 0

  4. CHEMICAL THERMODYNAMICS Spontaneity, Enthalpy, and Entropy Spontaneity must, therefore, be a function of the degree of randomness in a system. Entropy (S) is a “state function” that describes the randomness in a system such that, S = S final - S initial

  5. CHEMICAL THERMODYNAMICS The Second Law of Thermodynamics tells us that In any spontaneous process, there is always an increase in the entropy of the universe: S universe = S system + S surroundings > 0

  6. CHEMICAL THERMODYNAMICS A Molecular Interpretation of Entropy translational vibrational rotational The Third Law of Thermodynamics: S(0 K) = 0 The entropy of the lattice increases with temperature because the number of possible energy states in which the molecules or atoms are distributed is larger

  7. CHEMICAL THERMODYNAMICS Calculation of Entropy Changes The standard entropy (S) is expressed in units of J/mol-K For any given reaction: aA + bB + … cC + dD + ... S = [cS(p) + dS (Q) + …] - [ aS(A) + bS(B) + …] Sample exercise: Calculate the S for the synthesis of ammonia from N2 and H2 N2(g) + 3H2(g) 2NH3(g) S = [2S(NH3)]- [ S(N2) + S(H2)] S = [2(192 J/mol-K)]- [ 191.5 J/mol-K+ 3(130.58 J/mol-K)] S = -198.2 J/K

  8. Ice melting • water condensing • mixing salt water • diffusion • butane burning • CH4g + 2O2g→ CO2g + 2H2Og

  9. CHEMICAL THERMODYNAMICS Gibbs Free Energy Whether a reaction occurs spontaneously is determined by the changes in enthalpy and entropy for that reaction such that: I get it!! G = H -TS The change in free energy is therefore: G = H - TS • IfG is negative, the reaction is spontaneous in the forward direction • IfG is positive, the reaction is not spontaneous in the forward direction • Work must be supplied from the surroundings to make it occur. • IfG is zero, the reaction is at equilibrium, there is no driving force

  10. CHEMICAL THERMODYNAMICS Calculation of G using standard free energy change of formation The standard free energy change ( G) for any reaction: aA + bB + … cC + dD + ... G = [cGf (p) + dGf (Q) + …] - [aGf (A) + bGf (B) + …] Hip Deep In Alligators ?

  11. CHEMICAL THERMODYNAMICS

  12. Sample exercise: Using the data from Appendix C, calculate the G using H and S for the following reaction: BaO(s) + CO2(g)  BaCO3(g) H = [-1216.3 kJ/mol] - [ -553.5 kJ/mol - 393.5 kJ/mol] = -269.3 kJ S = [112.1 J/mol-K] - [ 70.42 J/mol-K - 213.6 J/mol-K] = -171.9 J/mol-K G = H - TS G = -269.3 kJ - 298 K (-171.9 J/mol-K)(1 kJ/ 1000 J ) G = -218.07 kJ/mol G = [-1137.6 kJ/mol] - [ -525.1 kJ/mol - 394.4 kJ/mol] = -218.1 kJ CHEMICAL THERMODYNAMICS

  13. Free Energy and the Equilibrium Constant What happens when you can’t describe G under standard conditions ? G = G ° + RT ln Q (R 8.314 J/K-mol) When G = 0 then G ° =-RT ln K G ° is negative: K > 1 G ° is zero: K = 1 G ° is positive: K < 1

  14. Calculate G at 298 K for the following reaction if the reaction mixture consists of 1.0 atm N2, 3.0 atm H2 ,and 1.0 atm NH3 N2(g) + 3H2  2NH3 Q = (1.0)2/ (1.0)(3.0)3 = 3.7 x 10-2 G = G ° + RT ln Q where R 8.314 J/K-mol G = -33.32 kJ + (8.314x 10-3 kJ/K-mol)(298 K)ln(3.7 x 10-2) G = -41.49 kJ

  15. (products) - (reactants)  m H rxn =  n or, H = q p if P is constant, H = E+ P  V   H f H f CHEMICAL THERMODYNAMICS 1st Law of Thermodynamics Chapter 5 Hess’s Law Ek=  mv2 E = E final - E initial E = q + w if  V = 0 then, E = q v H E H= H products - H reactants q = n CT where C is J/mol-C q = m ST where is C J/g-C

  16. CHEMICAL THERMODYNAMICS 2nd Law of Thermodynamics Chapter 19 3rd Law of Thermodynamics S = S final - S initial S universe = S system + S surroundings > 0 For any given reaction: aA + bB + … cC + dD + ... S = [cS(p) + dS (Q) + …] - [ aS(A) + bS(B) + …] where S= J/mol-K • If G = spontaneous G = H - TS • If G =not spontaneous • If G = zero, K = 0 for any reaction: aA + bB + … cC + dD + ... G = [cGf (p) + dGf (Q) + …] - [aGf (A) + bGf (B) + …] G = G ° + RT ln Q where R 8.314 J/K-mol G ° is negative: K > 1 G ° is zero: K = 0 if G ° = 0 then G ° = RT ln K G ° is positive: K > 1

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