Ch. 19: Spontaneity (“Thermodynamically Favored”), Entropy and Free Energy

Spontaneous Processes –“Thermodynamically Favored” or “Product Favored” Spontaneous processes may be fast or slow Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0 C and ice melts above 0 0 C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous

© 2009, Prentice-Hall, Inc. Spontaneous Processes Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. –Above 0  C it is spontaneous for ice to melt. –Below 0  C the reverse process is spontaneous.

Entropy (S)  A measure of the randomness or disorder or “ S preadioutiness”  Entropy is a thermodynamic function describing the number of arrangements that are available to a system  Nature proceeds toward the states that have the highest probabilities of existing

Second Law of Thermodynamics  "In any spontaneous process there is always an increase in the entropy of the universe"  "The entropy of the universe is increasing"  For a given change to be spontaneous,  S universe must be positive  S univ =  S sys +  S surr

Increase in Entropy  S = S final - S initial  S is positive when:  Phase changes S solid < S liquid << S gas

Solutions  S is positive when: –a solid is dissolved in a solvent

Entropy Changes  S is positive when: –The number of gas molecules increases; –The number of moles increases.

Standard Entropies Larger and more complex molecules have greater entropies.

Processes that lead to an increase in entropy (  S > 0) 18.2

The Dependence of Entropy on Pressure and Volume S large volume > S small volume S low pressure > S high pressure

Standard Entropies Increased Temperature: Increased Entropy (molecules move faster)

Third Law of Thermodynamics  “Entropy of a pure crystalline substance at absolute zero (0 K) is zero."  Entropy decreases as temperature decreases.

Entropy Changes Entropy changes for a reaction:  S  =  n  S  (products) —  m  S  (reactants) where n and m are the coefficients in the balanced chemical equation.

Entropy Changes in the System (  S sys ) aA + bB cC + dD S0S0 rxn dS 0 (D) cS 0 (C) = [+] - bS 0 (B) aS 0 (A) [+] S0S0 rxn nS 0 (products) =  mS 0 (reactants)  - The standard entropy of reaction (  S 0 ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. rxn What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = J/K mol S 0 (O 2 ) = J/K mol S 0 (CO 2 ) = J/K mol S0S0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S0S0 rxn = – [ ] = J/K mol

Entropy Practice Identify the following reactions or processes as having a positive or negative  S. 1.Melting Ice 2. Freezing 3.A + B  C 4.A  C + B 5.Which has the largest entropy C 2 H 4 or C 6 H 6 ?

Gibbs Free Energy The maximum amount of energy that COULD be converted to work Used to determine if a reaction or process is product favored When  S universe is positive,  G is negative. Therefore, when  G is negative, a process is spontaneous.

Gibbs Free Energy  G < 0 The reaction is spontaneous in the forward direction.  G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.  G = 0 The reaction is at equilibrium

Calculating Free Energy Using standard free energy of formation (  G f 0 ):  G f 0 of an element in its standard state is zero

For reactions at constant temperature:  G 0 =  H 0 - T  S 0 Calculating Free Energy Temperature affects  G only with the entropy term

Free Energy and Temperature HH SS Product-Favored… ++ at higher temperatures -- at lower temperatures -+ at all temperatures +- never (reactant-favored at all temps)  G 0 =  H 0 - T  S 0 Units:  H is usually in kJ/mol;  S is usually in J/mol·K

Gibbs Free Energy & the Equilibrium Constant  At equilibrium,  G = 0 and Q = K G0G0 K  G 0 = 0K = 1  G 0 < 0K > 1  G 0 > 0K < 1  G° =  RT ln (K)

Gibbs Free Energy & the Equilibrium Constant Calculate the equilibrium constant at 25  C if  G = kJ/mol (R = J/mol K)  G° =  RT ln (K)

Calculations with  G  G° =  RT ln (K)  G 0 =  H 0 - T  S 0