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**Unit 5 - Chpt 17 - Thermochemistry Part II**

Thermo - Entropy and Free Energy HW set1: Chpt 17 - pg #24, 25, 27-30, 32, 34, 36 Due Fri. Feb 3 HW set2: Chpt 17 pg # , 50, 54 Due Mon Feb 6 HW set3: Chpt 17 pg # 60, 64, 66, 71, 72, Due Fri Feb 10

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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 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: The evaporation of alcohol The freezing of water Compressing an ideal gas at constant temperature Heating an ideal gas at constant pressure 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 = 0 o 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·mol ΔG = 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 ΔS°reaction = ΣnpS°products – ΣnrS°reactants**

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) NaOH(aq) H2(g) ΔS°= –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) CO(g) + 2H2(g) --> CH3OH(l)**

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 illustration A 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.

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