Unit 5 - Chpt 17 - Thermochemistry Part II
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Transcript of Unit 5 - Chpt 17 - Thermochemistry Part II
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Unit 5 - Chpt 17 - Thermochemistry Part II
• Thermo - Entropy and Free Energy• HW set1: Chpt 17 - pg. 807-815 #24, 25, 27-30,
32, 34, 36 Due Tues. Jan 29• HW set2: Chpt 17 pg. 807-815 # 40 - 42, 50, 54
Due Fri Feb 1 • HW set3: Chpt 17 pg. 807-815 # 60, 64, 66, 71,
72, 109 Due Tues Feb 5
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Spontaneous Process
• Reminder... 1st law of thermodynamics - the energy of the universe is a constant.
• Spontaneous reactions occur without outside intervention. Can be fast of slow. Thermodynamics can tell us direction of reaction, but say nothing about speed.
• Driving forces of reactions... 1. energy (exothermic) and 2. increase in entropy (chaos)... ice melting is endothermic
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Entropy (chaos, disorder)
Statistically which distribution is more likely?
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Entropy of phases of matter
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Concept check
Predict the sign of ΔS for each of the following, and explain:a)The evaporation of alcohol
b)The freezing of water
c)Compressing an ideal gas at constant temperature
d)Heating an ideal gas at constant pressure
e)Dissolving NaCl in water
+
–
–
+
+
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2nd Law of Themodynamics
• In any spontaneous reaction there is always an increase in the entropy of the universe. THE ENTROPY OF THE UNIVERSE IS INCREASING.
• System vs. Surroundings?
ΔSuniverse = ΔSsystem + ΔSsurroundings
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Ice Melting example
• Exo or Endo ?• Increasing or decreasing Entropy• Will ice melt spontaneously?
– depends on temperature!!
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ΔSsurroundings
• Heat flow (constant P) = change in enthalpy = ΔH
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Spontaneous Reaction? (Entropy driving force)
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Thermochem Relationships
• Entropy
ΔSuniv = ΔSsys + ΔSsurr
ΔSsurr = - ΔH / T
• Free Energy, G
G = H - TS at constant T
ΔG = ΔH - TΔS
ΔSuniv = - ΔG / T at constant T & P
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Freezing & Melting conditions
• Melting point, boiling point are equilibrium between states ΔSuniv = 0 and thus ΔG = 0
• So ΔGo = ΔHo - TΔSo = 0o means each substance in its standard state.
• Given enthalpy and entropy can calculate the boiling point or freezing pt.
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Exercise 1
The value of ΔHvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C.
Calculate ΔS, ΔSsurr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm.
ΔS = 132 J/K·mol
ΔSsurr = -132 J/K·molΔG = 0 kJ/mol
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Third Law of Thermodynamics
• The entropy of a pure perfect crystal (every atom aligned - one possible lowest energy configuration) is zero at absolute zero temperature.
• Entropy increases with temperature• Absolute zero cannot be attained. • 2nd Law prohibits heat can never spontaneously move from a
colder body to a hotter body. So, as a system approaches absolute zero, it will eventually have to draw energy from whatever systems are nearby. It would take an infinite number of steps and infinite energy to attain.
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Spontaneous Reaction, G
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State Functions
• Entropy is a state function.• Standard entropy is defined at 298K and 1
atm. Recall at 0 K S = zero
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
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Exercise 2
Calculate ΔS° for the following reaction:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Given the following information: S° (J/K·mol)
Na(s) 51
H2O(l) 70 NaOH(aq) 50
H2(g) 131ΔS°= –11 J/K
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Concept Check
Consider the following system at equilibrium at 25°C.
PCl3(g) + Cl2(g) PCl5(g) ΔG° = −92.50 kJ
What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCl3 if the temperature is raised? Explain.
The ratio will decrease.
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State Function - Free Energy
• The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.
ΔG° = ΔH° – TΔS°
ΔG°reaction = ΣnpG°products – ΣnrG°reactants
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Dependence of G on pressure
G = G° + RT ln(P)G is free energy of a particular gas at current
pressure and G° is at 1 atm.
Which can be expanded for a total reaction as…
ΔG = ΔG° + RT ln(Q)Q is the equilibrium reaction quotient, we still
need ΔG° from the standard free energies
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Exercise 3 (example 17.13)
Calculate ΔG at 25oC for the reaction with CO at 5.0atm and H2 at 3.0atm ΔG = ΔG° + RT ln(Q) use thermo data in Appendix 4
CO(g) + 2H2(g) --> CH3OH(l)
ΔGf (CH3OH) = -166kJ ; ΔGf (H2) = 0 ; ΔGf (CO) = -137kJ ΔG° = -29kJ now plug into above equation… T in Kelvin
ln(Q) = 1 / [CO]x[H2]2 = 1/45 = 2.2x10-2
ΔG = -38kJ/mol of reaction
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G and Equilibrium
• The equilibrium point occurs at the lowest value of free energy available to the reaction system.
• At equilibrium ΔG = 0 and Q becomes K
ΔG° = - RT ln(K)
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G illustrationA system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.
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Change in free energy to reach equilibrium
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Temp dependence of K
ΔG° = - RT ln(K) combining with
ΔG° = ΔH° – T ΔS°
ln(K) = - ΔH° ( 1 ) + ΔS°
R ( T ) R
Plotting ln(K) vs 1/T gives slope and intercept of enthalpy and entropy
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Qualitatitive : ΔG° and K
• Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction
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Free Energy and Work
• Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy.
wmax = ΔG
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Reversible car battery charge
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G and Work ramifications
• Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy.
• All real processes are irreversible.
• First law: You can’t win, you can only break even.
• Second law: You can’t break even.