Chapter 17
Thermodynamics: Entropy, Free
Energy, and Equilibrium
CHM 112 Summer 2007
M. Prushan
Questions
• What factors determine the direction and extent of a
chemical reaction?
– Combustion of hydrocarbon fuels goes almost to
completion
– Combination of gold and oxygen occurs hardly at all
– Industrial synthesis of ammonia from N2 and H2 at 400500ºC results in an equilibrium mixture
• Extent of particular reaction described by K
– But what determines value of K? →
thermodynamics
CHM 112 Summer 2007
M. Prushan
Spontaneous Processes &
Entropy
• The 2nd Law explains why chemical reactions
tend to favor a particular direction.
• It is important to predict whether a reaction will
occur.
• A reaction that does occur under specific
conditions is called a spontaneous reaction.
• A reaction that does not occur under specific
conditions is called a nonspontaneous reaction.
CHM 112 Summer 2007
M. Prushan
Spontaneous Processes &
Entropy
• A spontaneous reaction will always move
a reaction mixture toward equilibrium.
• Reaction spontaneity is independent of
reaction rate.
CHM 112 Summer 2007
M. Prushan
Spontaneous Processes &
Entropy
• To predict spontaneity we need to know the energy
change and the entropy.
• Entropy (S) is a “measure” of the randomness or
disorder of a system.
• The greater the disorder, the greater the entropy.
• Nature tends to the greatest entropy.
Ssolid < Sliquid < Sgas
CHM 112 Summer 2007
M. Prushan
Spontaneous Processes &
Entropy
• Standard Entropy is the absolute entropy of a
substance at 1 atm and 25°C.
• Change in entropy is given by ∆S = Sf – Si
CHM 112 Summer 2007
M. Prushan
Spontaneous Processes &
Entropy
• The decomposition of N2O4 (O2N–NO2) is also
accompanied by an increase in randomness.
• Whenever molecules break apart, randomness
increases.
CHM 112 Summer 2007
M. Prushan
Spontaneous Processes &
Entropy
• When NaCl dissolves in water, the crystal
breaks up, and the ions are surrounded by water
molecules.
CHM 112 Summer 2007
M. Prushan
Example
(a)The rusting of cars is spontaneous, but it
occurs slowly
(b)Spontaneous reaction occurs slowly if it has
high Ea
CHM 112 Summer 2007
M. Prushan
Example
• The combustion of natural gas
(mainly CH4) in air is a
spontaneous, exothermic
reaction
CHM 112 Summer 2007
M. Prushan
Question:
• Consider the gas phase reaction of A2
molecules (red) with B atoms (blue).
– (a) Write a balanced equation for the reaction.
– (b) Predict the sign of ∆S for the reaction.
CHM 112 Summer 2007
M. Prushan
Entropy and Temperature
• Entropy is associated with molecular motion.
• As temperature increases, entropy increases.
• Third Law of Thermodynamics: Entropy of a
perfectly ordered crystalline substance at 0 K is
zero.
• At the melting and boiling point there is a
discontinuous jump in entropy.
CHM 112 Summer 2007
M. Prushan
Entropy and Temperature
CHM 112 Summer 2007
M. Prushan
Entropy and Temperature
CHM 112 Summer 2007
M. Prushan
Standard Molar Entropies
• Standard Molar Entropy, S°: The entropy of 1 mol of
the pure substance at 1 atm pressure and a specified
temperature, usually 25°C.
• Standard molar entropies are absolute entropies
measured against an absolute reference point.
• Standard entropy of reaction:
– ∆S° = ∑nS°(products) – ∑nS°(reactants)
CHM 112 Summer 2007
M. Prushan
Standard Molar Entropies
CHM 112 Summer 2007
M. Prushan
Standard Molar Entropies
• Calculate the standard entropy of reaction at 25°C for the
synthesis of ammonia:
– N2(g) + 3 H2(g)  2 NH3(g)
• Calculate the standard entropy of reaction at 25°C for the
decomposition of calcium carbonate:
– CaCO3(s)  CaO(s) + CO2(g)
CHM 112 Summer 2007
M. Prushan
Standard Molar Entropies
• Predict the entropy change, and then calculate the standard
entropy change for the following reactions at 25°C.
– a. 2 CO(g) + O2(g)  2 CO2(g)
– b. 3 O2(g)  2 O3(g)
– c. 2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)
CHM 112 Summer 2007
M. Prushan
2nd Law of Thermodynamics
• The total entropy increases in a spontaneous process and remains
unchanged in an equilibrium process.
– Spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0
– Equilibrium: ∆Stotal = ∆Ssys + ∆Ssur = 0
• The system is what you observe; surroundings are everything else.
CHM 112 Summer 2007
M. Prushan
2nd Law of Thermodynamics
• For ∆Stotal, we need to know ∆Ssys and ∆Ssurr.
• ∆Ssys is often determined from the standard entropy of
reaction, ∆S°rxn:
∆S°rxn = ∑nS°(Products) – ∑nS°(Reactants)
CHM 112 Summer 2007
M. Prushan
2nd Law of Thermodynamics
• Entropy Changes to the Surroundings:
•
DSsurr =
-D H sys
T
Exothermic  increase ∆Ssurr
Endothermic  decrease ∆Ssurr
CHM 112 Summer 2007
M. Prushan
Gibbs Free Energy
• The 2nd law tells us a process will be spontaneous if
∆Stotal > 0 which requires a knowledge of ∆Ssurr.
• Since we only concern ourselves with the system we
derive an expression using only ∆Ssys.
–T∆Stotal = ∆Hsys – T∆Ssys < 0
CHM 112 Summer 2007
M. Prushan
Gibbs Free Energy
• The expression –T∆Stotal is equated as Gibbs free
energy change (∆G), or simply free energy change:
•
∆G = ∆H – T∆Ssys
∆G < 0 Reaction is spontaneous in forward direction.
∆G = 0 Reaction is at equilibrium.
∆G > 0 Reaction is spontaneous in reverse direction.
CHM 112 Summer 2007
M. Prushan
Gibbs Free Energy
• Using ∆G = ∆H – T∆S, we can predict the sign of ∆G
from the sign of ∆H and ∆S.
• If both ∆H and ∆S are positive, ∆G will be negative
only when the temperature value is large.
• If ∆H is positive and ∆S is negative, ∆G will always be
positive.
• If ∆H is negative and ∆S is positive, ∆G will always be
negative.
• If both ∆H and ∆S are negative, ∆G will be negative
only when the temperature value is small.
CHM 112 Summer 2007
M. Prushan
Gibbs Free Energy
CHM 112 Summer 2007
M. Prushan
Gibbs Free Energy
•
Iron metal can be produced by reducing iron(III) oxide
with hydrogen:
Fe2O3(s) + 3 H2(g)  2 Fe(s) + 3 H2O(g)
∆H° = +98.8 kJ; ∆S° = +141.5 J/K
1. Is this reaction spontaneous at 25°C?
2. At what temperature will the reaction become
spontaneous?
CHM 112 Summer 2007
M. Prushan
Gibbs Free Energy
• What are the signs (+, –, or 0) of ∆H, ∆S,
and ∆G for the following spontaneous
reaction of A atoms (red) and B atoms
(blue)?
CHM 112 Summer 2007
M. Prushan
Gibbs Free Energy
• Standard Free Energy (∆G˚rxn) is the free energy for a
reaction occurring under standard state conditions.
Reactants in their standard states are converted to
products in their standard states.
∆G˚rxn = ∑ n ∆G˚ƒ (products) – ∑ n ∆G˚ƒ (reactants)
• ∆G˚ƒ is the standard free energy of formation of a
compound from its elements in their standard states.
CHM 112 Summer 2007
M. Prushan
Gibbs Free Energy
CHM 112 Summer 2007
M. Prushan
Free Energy and Chemical
Equilibrium
• The sign of ∆G° tells the direction of spontaneous
reaction when both reactants and products are present
at standard state conditions.
• Under nonstandard conditions, ∆G˚ becomes ∆G.
∆G = ∆G˚ + RT lnQ
• The reaction quotient is obtained in the same way as an
equilibrium expression.
CHM 112 Summer 2007
M. Prushan
Free Energy and Chemical
Equilibrium
• If ∆G˚ is a large negative value, RT lnQ will not be
positive enough to match ∆G˚ until a significant amount
of product is formed.
• If ∆G˚ is a large positive value, RT lnQ will only be more
negative when very little product has been formed.
CHM 112 Summer 2007
M. Prushan
Free Energy and Chemical
Equilibrium
CHM 112 Summer 2007
M. Prushan
• Using the solubility product of silver chloride at 25°C (1.6 x
10–10), calculate ∆G˚ for the process:
AgCl(s)
Ag+(aq) + Cl–(aq)
• The ∆G˚ for the reaction H2(g) + I2(g) 2 HI(g) is 2.60 kJ at
25°C. In one experiment, the initial pressures are PH2 = 4.26
atm, PI2 = 0.024 atm, and PHI = 0.23 atm. Calculate ∆G for
the reaction and predict the reaction direction.
CHM 112 Summer 2007
M. Prushan