Create Presentation
Download Presentation

Download Presentation

Unit 5 - Chpt 17 - Thermochemistry Part II

Download Presentation
## Unit 5 - Chpt 17 - Thermochemistry Part II

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**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**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**Entropy (chaos, disorder)**Statistically which distribution is more likely?**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 + – – + +**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**Ice Melting example**• Exo or Endo ? • Increasing or decreasing Entropy • Will ice melt spontaneously? • depends on temperature!!**ΔSsurroundings**• Heat flow (constant P) = change in enthalpy = ΔH**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**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.**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**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.**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**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) 131 ΔS°= –11 J/K**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.**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**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 thestandard free energies**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**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 becomesK ΔG° = - RT ln(K)**G illustration**A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.**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**Qualitatitive : ΔG° and K**• Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction**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**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.