chm 112 summer 2007 m. prushan chapter 17 thermodynamics: entropy, free energy, and equilibrium

CHM 112 Summer 2007 M. Prushan

Chapter 17

Thermodynamics: Entropy, Free Energy, and Equilibrium

Thermodynamics: Entropy, Free Energy, and Equilibrium


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 400-500ºC results in

an equilibrium mixture

• Extent of particular reaction described by K– But what determines value of K? → thermodynamics


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.



• A spontaneous reaction will always move a reaction mixture toward equilibrium.

• Reaction spontaneity is independent of reaction rate.



• 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


• Standard Entropy is the absolute entropy of a substance at 1 atm and 25°C.

• Change in entropy is given by ∆S = Sf – Si



• The decomposition of N2O4 (O2N–NO2) is also accompanied by an increase in randomness.

• Whenever molecules break apart, randomness increases.




• When NaCl dissolves in water, the crystal breaks up, and the ions are surrounded by water molecules.


Example (a)The rusting of cars is spontaneous, but it

occurs slowly

(b)Spontaneous reaction occurs slowly if it has high Ea


Example

• The combustion of natural gas

(mainly CH4) in air is a

spontaneous, exothermic

reaction


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.


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.


Entropy and TemperatureEntropy and Temperature


Entropy and Temperature


• 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)

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)



• 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)


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.



• 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)



• Entropy Changes to the Surroundings:

• Exothermic increase ∆Ssurr

Endothermic decrease

∆Ssurr

Ssurr Hsys

T


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


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.


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.


Gibbs Free Energy


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?


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)?


• 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.

Gibbs Free Energy


Gibbs Free Energy


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.



• 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.


• 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 chapter 17 thermodynamics: entropy, free energy, and equilibrium

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