Spontaneity, Entropy and Free Energy

Spontaneous Processes and Entropy  First Law “Energy can neither be created nor destroyed" The energy of the universe is constant  Spontaneous Processes Processes that occur without outside intervention Spontaneous processes may be fast or slow –Many forms of combustion are fast –Conversion of diamond to graphite is slow

Entropy (S)  A measure of the randomness or disorder  The driving force for a spontaneous process is an increase in the entropy of the universe  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

Positional Entropy  The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved S solid < S liquid << S gas

Example Which has higher positional S? –Solid CO 2 or gaseous CO 2 ? –N 2 gas at 1 atm or N 2 gas at 0.01 atm? Predict the sign of DS –Solid sugar is added to water + because larger volume –Iodine vapor condenses on cold surface to form crystals - because g  s, less volume

What about Chemical Changes? We have been working with only physical changes so far… Compare using # of independent states N 2 (g) + 3H 2 (g)  2NH 3 (g) Entropy decreases (-∆S) Compare using # molecules in higher entropy states 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Entropy increases (+∆S)

Example Predict the sign of ∆S for the following reactions: CaCO 3 (s)  CaO(s) + CO 2 (g) –Increasing S, +∆S 2SO 2 (g) + O 2 (g)  2SO 3 (g) –Decreasing S, -∆S

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

 H,  S,  G and Spontaneity Value of  H Value of T  S Value of  G Spontaneity Negative PositiveNegativeSpontaneous Positive NegativePositiveNonspontaneous Negative ???Spontaneous if the absolute value of  H is greater than the absolute value of T  S (low temperature) Positive ???Spontaneous if the absolute value of T  S is greater than the absolute value of  H (high temperature)  G =  H - T  S H is enthalpy, T is Kelvin temperature

Calculating Entropy Change in a Reaction  Entropy is an extensive property (a function of the number of moles)  Generally, the more complex the molecule, the higher the standard entropy value

Example Find the ∆S° at 25°C for: 2NiS(s) + 3O 2 (g)  2SO 2 (g) + 2NiO(s)

Standard Free Energy Change   G 0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states   G 0 cannot be measured directly  The more negative the value for  G 0, the farther to the right the reaction will proceed in order to achieve equilibrium  Equilibrium is the lowest possible free energy position for a reaction

For reactions at constant temperature:  G 0 =  H 0 - T  S 0 H 2 O(s)  H 2 O(l) where ∆H°=6.03x103 J/mol, ∆S°=22.1 J/mol.K Since reaction’s values are in opposition, the T determines spontaneity. At -10°C, is it spontaneous? Calculating Free Energy Method #1

Example NO! the reverse is spontaneous Is it spontaneous at 0°C? NO! it is at equilibrium so no shift

Example Is it spontaneous at +10°C? YES! When H and S are in opposition, the spontaneity depends on T.

Calculating Free Energy: Method #2 An adaptation of Hess's Law: C diamond (s) + O 2 (g)  CO 2 (g)  G 0 = -397 kJ C graphite (s) + O 2 (g)  CO 2 (g)  G 0 = -394 kJ CO 2 (g)  C graphite (s) + O 2 (g)  G 0 = +394 kJ C diamond (s)  C graphite (s)  G 0 = C diamond (s) + O 2 (g)  CO 2 (g)  G 0 = -397 kJ -3 kJ

Calculating Free Energy Method #3 Using standard free energy of formation (  G f 0 ):  G f 0 of an element in its standard state is zero

The Dependence of Free Energy on Pressure  Enthalpy, H, is not pressure dependent  Entropy, S  entropy depends on volume, so it also depends on pressure S large volume > S small volume S low pressure > S high pressure

Free Energy and Equilibrium  Equilibrium point occurs at the lowest value of free energy available to the reaction system  At equilibrium,  G = 0 and Q = K G0G0 K  G 0 = 0K = 1  G 0 < 0K > 1  G 0 > 0K < 1

Temperature Dependence of K So, ln(K)  1/T

Free Energy and Work  The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy  The amount of work obtained is always less than the maximum  Henry Bent's First Two Laws of Thermodynamics  First law: You can't win, you can only break even  Second law: You can't break even