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

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

CHEMICAL THERMODYNAMICS  H > 0

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

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

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

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 N 2 and H 2 N 2 (g) + 3H 2 (g)  2NH 3 (g)  S  = [2S  (NH 3 )]- [ S  (N 2 ) + S  (H 2 )]  S  = [2(192 J/mol-K)]- [ 191.5 J/mol-K+ 3(130.58 J/mol-K)]  S  = -198.2 J/K

4 Ice melting 4 water condensing 4 mixing salt water 4 diffusion 4 butane burning 4 CH 4g + 2O 2g → CO 2g + 2H 2 O g

CHEMICAL THERMODYNAMICS Gibbs Free Energy Whether a reaction occurs spontaneously is determined by the changes in enthalpy and entropy for that reaction such that: 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 I get it!!

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

CHEMICAL THERMODYNAMICS

Sample exercise : Using the data from Appendix C, calculate the  G  using  H  and  S  for the following reaction: BaO(s) + CO 2 (g)  BaCO 3 (g)  H  = [-1216.3 kJ/mol] - [ -553.5 kJ/mol - 393.5 kJ/mol] = -269.3 kJ  G  = [-1137.6 kJ/mol] - [ -525.1 kJ/mol - 394.4 kJ/mol] = -218.1 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

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

Calculate  G at 298 K for the following reaction if the reaction mixture consists of 1.0 atm N 2, 3.0 atm H 2,and 1.0 atm NH 3 N 2 (g) + 3H 2  2NH 3 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

CHEMICAL THERMODYNAMICS E k =  mv 2  E = E final - E initial  E = q + w if  V = 0 then,  E = q v if P is constant,  H =  E + P  V or,  H = q p  H   E  H= H products - H reactants   H f  H rxn =  n n (products) -   H f  m m (reactants) q = n C  T where C is J/mol-  C q = m S  T where is C J/g-  C Chapter 5 Hess’s Law 1st Law of Thermodynamics

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

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