chm 112 summer 2007 m. prushan chapter 17 thermodynamics: entropy, free energy, and equilibrium
TRANSCRIPT
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CHM 112 Summer 2007 M. Prushan
Chapter 17
Thermodynamics: Entropy, Free Energy, and Equilibrium
Thermodynamics: Entropy, Free Energy, and Equilibrium
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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 400-500ºC results in
an equilibrium mixture
• Extent of particular reaction described by K– But what determines value of K? → thermodynamics
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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.
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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.
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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
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CHM 112 Summer 2007 M. Prushan
• Standard Entropy is the absolute entropy of a substance at 1 atm and 25°C.
• Change in entropy is given by ∆S = Sf – Si
Spontaneous Processes & Entropy
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CHM 112 Summer 2007 M. Prushan
• The decomposition of N2O4 (O2N–NO2) is also accompanied by an increase in randomness.
• Whenever molecules break apart, randomness increases.
Spontaneous Processes & Entropy
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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.
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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
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CHM 112 Summer 2007 M. Prushan
Example
• The combustion of natural gas
(mainly CH4) in air is a
spontaneous, exothermic
reaction
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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.
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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.
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CHM 112 Summer 2007 M. Prushan
Entropy and TemperatureEntropy and Temperature
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CHM 112 Summer 2007 M. Prushan
Entropy and Temperature
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CHM 112 Summer 2007 M. Prushan
• 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
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CHM 112 Summer 2007 M. Prushan
Standard Molar Entropies
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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)
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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)
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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.
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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)
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CHM 112 Summer 2007 M. Prushan
2nd Law of Thermodynamics
• Entropy Changes to the Surroundings:
• Exothermic increase ∆Ssurr
Endothermic decrease
∆Ssurr
Ssurr Hsys
T
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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
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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.
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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.
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CHM 112 Summer 2007 M. Prushan
Gibbs Free Energy
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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?
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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)?
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CHM 112 Summer 2007 M. Prushan
• 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
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CHM 112 Summer 2007 M. Prushan
Gibbs Free Energy
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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.
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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.
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CHM 112 Summer 2007 M. Prushan
Free Energy and Chemical Equilibrium
Free Energy and Chemical Equilibrium
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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.