Spontaneity, Entropy, & Free Energy Chapter 16

1st Law of Thermodynamics The first law of thermodynamics is a statement of the law of conservation of energy: energy can neither be created nor destroyed. The energy of the universe is constant, but the various forms of energy can be interchanged in physical and chemical processes.

Spontaneous Processes and Entropy Thermodynamics lets us predict whether a process will occur but gives no information about the amount of time required for the process. A spontaneous process is one that occurs without outside intervention. This is also considered thermodynamically favored.

Entropy The driving force for a spontaneous process is an increase in the entropy of the universe. Entropy, S, can be viewed as a measure of randomness, or disorder. Nature spontaneously proceeds toward the states that have the highest probabilities of existing.

The expansion of an ideal gas into an evacuated bulb.

Positional Entropy A gas expands into a vacuum because the expanded state has the highest positional probability of states available to the system. Therefore, S solid < S liquid << S gas

Positional Entropy Which of the following has higher positional entropy? a) Solid CO 2 or gaseous CO 2 ? b) N 2 gas at 1 atm or N 2 gas at 1.0 x atm?

Entropy What is the sign of the entropy change for the following? a) Solid sugar is added to water to form a solution?  S is positive b) Iodine vapor condenses on a cold surface to form crystals?  S is negative

The Second Law of Thermodynamics...in any spontaneous(thermodynamically favored) process there is always an increase in the entropy of the universe.  S univ > 0 for a spontaneous process.

 S Universe  S universe is positive -- reaction is spontaneous.  S universe is negative -- reaction is spontaneous in the reverse direction.  S universe = 0 -- reaction is at equilibrium.

 S  =  n p S  (products)   n r S  (reactants)  H  =  n p H  (products)   n r H  (reactants)

 S o reaction Calculate  S  at 25 o C for the reaction 2NiS (s) + 3O 2(g) ---> 2SO 2(g) +2NiO (s) SO 2 (248 J/Kmol) NiO (38 J/Kmol) O 2 (205 J/Kmol) NiS (53 J/Kmol)

 S  =  n p S  (products)   n r S  (reactants)  S  = [(2 mol SO 2 )(248 J/Kmol) + (2 mol NiO)(38 J/Kmol)] - [(2 mol NiS)(53 J/Kmol) + (3 mol O 2 )(205 J/Kmol)]  S  = 496 J/K + 76 J/K J/K J/K  S  = -149 J/K # gaseous molecules decreases!

Effect of  H and  S on Spontaneity

 G -- Free Energy Two tendencies exist in nature: tendency toward higher entropy --  Stendency toward higher entropy --  S tendency toward lower energy --  Htendency toward lower energy --  H If the two processes oppose each other (e.g. melting ice cube), then the direction is decided by the Free Energy,  G, and depends upon the temperature.

Free Energy  G =  H  T  S (from the standpoint of the system) A process (at constant T, P) is spontaneous in the direction in which free energy decreases:  G sys means +  S univ Entropy changes in the surroundings are primarily determined by the heat flow. An exothermic process in the system increases the entropy of the surroundings.

Free Energy  G  G =  H  T  S  G = negative -- spontaneous  G = positive -- spontaneous in opposite direction  G = 0 -- at equilibrium

 G,  H, &  S Spontaneous reactions are indicated by the following signs:  G = negative  H = negative  S = positive

Free Energy Change and Chemical Reactions  G  = standard free energy change that occurs if reactants in their standard state are converted to products in their standard state.  G  =  n p  G f  (products)   n r  G f  (reactants)

Temperature Dependence  H o &  S o are  H o &  S o are not temperature dependent.  G o is temperature dependent.  G =  H  T  S

 G  Calculations Calculate ,  S  G  for the reaction  SO 2(g) + O 2(g) ----> 2 SO 3(g)  =  n p  H f  (products)   n r  H f  (reactants)  = [(2 mol SO 3 )(-396 kJ/mol)]-[(2 mol SO 2 )(-297 kJ/mol) + (0 kJ/mol)]  H  = kJ kJ  H  = -198 kJ

 G  Calculations Continued  S  =  n p S  (products)   n r S  (reactants)  S  = [(2 mol SO 3 )(257 J/Kmol)]-[(2 mol SO 2 )(248 J/Kmol) + (1 mol O 2 )(205 J/Kmol)]  S  = 514 J/K J/K J/K  S  = -187 J/K

 G  Calculations Continued  G o =  H o  T  S o  G o = kJ - (298 K)(-187 J/K)(1kJ/1000J)  G o = kJ kJ  G o = kJ The reaction is spontaneous at 25 o C and 1 atm.

The Third Law of Thermodynamics... The third law of thermodynamics: the entropy of a perfect crystal at 0K is zero. [not a lot of perfect crystals out there so, entropy values are RARELY ever zero—evenelements] So what? This means the absolute entropy of a substance can then be determined at any temp. higher than 0 K. (Handy to know if you ever need to defend why G & H for elements = BUT S does not!).

Hess’s Law &  G o C diamond(s) + O 2(g) ---> CO 2(g) C diamond(s) + O 2(g) ---> CO 2(g)  G o = -397 kJ C graphite(s) + O 2(g) ---> CO 2(g) C graphite(s) + O 2(g) ---> CO 2(g)  G o = -394 kJ Calculate  G o for the reaction C diamond(s) ---> C graphite(s) C diamond(s) + O 2(g) ---> CO 2(g) C diamond(s) + O 2(g) ---> CO 2(g)  G o = -397 kJ CO 2(g) ---> C graphite(s) + O 2(g) CO 2(g) ---> C graphite(s) + O 2(g)  G o = +394 kJ C diamond(s) ---> C graphite(s)  G o = -3 kJ Diamond is kinetically stable, but thermodynamically unstable.

SoSo SoSo S o increases with: solid ---> liquid ---> gassolid ---> liquid ---> gas greater complexity of molecules (have a greater number of rotations and vibrations)greater complexity of molecules (have a greater number of rotations and vibrations) greater temperature (if volume increases)greater temperature (if volume increases) lower pressure (if volume increases)lower pressure (if volume increases)

 G o & Temperature  G o depends upon temperature. If a reaction must be carried out at temperatures higher than 25 o C, then  G o must be recalculated from the  H o &  S o values for the reaction.

Free Energy & Pressure The equilibrium position represents the lowest free energy value available to a particular system (reaction).  G is pressure dependent  S is pressure dependent  is not pressure dependent

Free Energy and Pressure  G =  G  + RT ln(Q) Q = reaction quotient from the law of mass action.

Free Energy Calculations CO (g) + 2H 2(g) ---> CH 3 OH (l). CO (g) + 2H 2(g) ---> CH 3 OH (l) Calculate Calculate  G o for this reaction where CO (g) is 5.0 atm and H 2(g) is 3.0 atm are converted to liquid methanol.  G  =  n p  G f  (products)   n r  G f  (reactants)  G  =  mol CH 3 OH)(- 166 kJ/mol)]-[(1 mol CO)(-137 kJ/mol) + (0 kJ)]  G  =  kJ kJ  G  =  x 10 4 J

Free Energy Calculations Continued = 2.2 x 10 -2

Free Energy Calculations Continued  G =  G  + RT ln(Q)  G = (-2.9 x 10 4 J /mol rxn) + ( J/Kmol)(298 K) ln(2.2 x )  G =  x 10 4 J/mol rxn) - (9.4 x 10 3 J/mol rxn)  G = - 38 kJ/ mol rxn Note:  G is significantly more negative than  G  implying that the reaction is more spontaneous at reactant pressures greater than 1 atm. Why?

A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.

As A is changed into B, the pressure and free energy of A decreases, while the pressure and free energy of B increases until they become equal at equilibrium.

Graph a) represents equilibrium starting from only reactants, while Graph b) starts from products only. Graph c) represents the graph for the total system.

Free Energy and Equilibrium  G  =  RT ln(K) K = equilibrium constant This is so because  G = 0 and Q = K at equilibrium.

 G o & K  G o = 0K = 1  G  1 (favored)  G  not favored)

Equilibrium Calculations 4Fe (s) + 3O 2(g) 2Fe 2 O 3(s) Calculate K for this reaction at 25 o C.  G o = x 10 6 J  H o = x 10 6 J  S o = -543 J/K  G  =  RT ln(K) K = e -  G  R  K = e 601 or K is very large because  G  is very negative.

Temperature Dependence of K y = mx + b y = mx + b (  H  and S   independent of temperature over a small temperature range) If the temperature increases, K decreases for exothermic reactions, but increases for endothermic reactions.

Free Energy & Work The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy: w max =  G

Reversible vs. Irreversible Processes Reversible: The universe is exactly the same as it was before the cyclic process. Irreversible: The universe is different after the cyclic process. All real processes are irreversible -- (some work is changed to heat).  w < All real processes are irreversible -- (some work is changed to heat).  w <  G Work is changed to heat in the surroundings and the entropy of the universe increases.

Laws of Thermodynamics First Law: You can’t win, you can only break even. Second Law: You can’t break even.