A.P. Chemistry

Spontaneity, Entropy, and Free Energy

16.1 Spontaneous Processes and EnthalpyFirst Law: Energy cannot be created nor destroyed. (The energy of the

universe is constant.)Spontaneous Processes- processes that occur without outside interventionMay be fast or slow (kinetics not important). Example: conversion of graphite to diamond is spontaneous, but very 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.Ssolid < Sliquid << Sgas

Example: Predict the sign of /\S sys for each of the following reactions:

H2O2(l) H2O(l) + ½ O2(g)

H+(aq) + OH-(aq) H2O(l)

CaO(s) + CO2(g) CaCO3(s)

16.2 Entropy and the Second Law of ThermodynamicsIn 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, /\Suniverse must be positive.

/\Suniverse = /\Ssystem + /\S surroundings > 0

(for a spontaneous process)Example: The solubility of silver chloride is so low that it

precipitates spontaneously from many solutions. The entropy change of the system is negative for this process.Ag+(aq) + Cl-(aq) AgCl(s)/\Ho = -65 kJ/mol

Since /\S decreases in this spontaneous reaction, shouldn’t the reaction be nonspontaneous?

16.3 The Effect of Temperature on SpontaneityDirection of Heat FlowEntropy changes in the surroundings are primarily

determined by heat flow. Exothermic reactions in a system at constant temperature increase the entropy of the surroundings.

Endothermic reactions in a system at constant temperature decrease the entropy of the surroundings.

The impact of the transfer of a given quantity of energy as heat to or from the surroundings will be greater at lower temperatures.

16.4 Free Energy (G), also called Gibbs Free Energy

Calculating Free Energy Change (constant temperature and Pressure)/\G = /\H - T/\SH is enthalpy T is temperature, in Kelvin

Free Energy and SpontaneityReactions proceed in the direction that lowers

their free energy (-/\G)

16.5 Entropy Changes in Chemical ReactionsConstant temperature and pressureReactions involving gaseous moleculesThe change in positional entropy is dominated

by the relative numbers of molecules of gaseous reactants and products.Example:

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)

9 molecules 10 molecules /\S increases

Third Law of Thermodynamics: The entropy of a perfect crystal at 0 K is zero. (NO disorder, since everything is in a perfect position.)

Calculating Entropy changes in a reaction/\So

reaction = ΣnpSoproducts - Σnr So

reactants where n = # of moles (coefficient)

Example: Use absolute entropies to calculate the standard entropy change of this reaction:H2(g) + ½ O2(g) H2O(l)

(NOTE the unit; use Table in Appendix)/\So

reaction = 1 mol(69.9 J/mol-K) -[1 mol(130.1 J/mol-K) + ½ mol(205.0 J/mol-K) = -163.6 J/mol-K

Entropy is an extensive property (a function of the number of

moles)Generally, the more complex the molecule, the higher the

standard entropy value.

16.6 Free Energy and Chemical ReactionsStandard Free Energy ChangeWhat are the conditions of standard state for this quantity?

(G) /\Go if the change in free energy that will occur if the reactants

in their standard states are converted to the products in their standard states.

/\Go cannot be measured directly.The more negative the value for /\Go , 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.

Calculating Free Energy ChangeMethod #1 For reactions at constant temperature/\Go = /\Ho - T/\So

(“Going Home To Supper” equation)Example: Hydrated lime Ca(OH)2 can be reformed into quicklime

CaO by heating:Ca(OH)2(s) CaO(s) + H2O(g)

At what temperatures is this reaction spontaneous under standard conditions (that is, where H2O is formed at 1 atm pressure)? (/\Ho

f for Ca(OH)2 = -986.2 kJ/mol; So for Ca(OH)2 = 83.4 J/mol-K)

Method #2 an adaptation of Hess’s Law (see example in textbook)

Example:

Method #3 Using standard free energy of formation (/\Gof)

Standard Free Energy of Formation is the change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states.

/\Goformation of an element in the standard state is zero.

/\Go = Σnp/\Gof(products) - Σn r/\Go

f(reactants)

Example: Calculate /\Gorxn at 25oC for the following reaction: (Use

data from Table in Appendix)2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s)

16.7 The Dependence of Free Energy on PressureEnthalpy (H) is not pressure dependent.Entropy depends on volume, so, therefore, it depends on

pressure.Slarge volume > S small volume

G = Go + RT ln(KP)

Go is the free energy of the gas at a pressure of 1 atmG is the free energy of the gas at a pressure of P atmR is universal gas constant, T is temperature, in Kelvin

Example:

/\G = /\Go + RT ln (Q)

Q is the reaction quotient (from the law of mass action, chapter 13)R is the gas constant (8.3145 J/K mol)/\Go is the free energy change for the reaction with all reactants and products at a pressure of 1 atm/\G is the free energy change for the reaction for the specified pressures of reactants and products.

Example:

16.8 Free Energy and EquilibriumThermodynamic View of Equilibrium/\Suniv = /\Ssys + /\Ssurr = 0

Equilibrium point occurs at the lowest value of free energy available to the reaction system.

At equilibrium, /\G = 0 and Q = K/\Go = -RT ln (K)

Temperature Dependence of K/\Go = -RT ln(K) = /\Ho - T/\So

ln (K) = -/\Ho + /\So = /\Ho 1 + /\S o

RT R R T R Therefore, ln (K) = 1/T

16.9 Free Energy and WorkRelationship to WorkThe maximum possible useful work obtainable from a process

at constant temperature and pressure is equal to the change in free energy. (“Free Energy” is energy “free” to do work.)

The amount of work obtained is always less than the maximum.

Henry Bent’s First Two Laws of ThermodynamicsYou can’t win; you can only break even.You can’t break even.

(and, posted in the classroom: Third Law- You can’t even get out of the game.)

The amount of available free energy of our system (Earth) is decreasing.