Presentation on theme: "Unit 5 - Chpt 17 - Thermochemistry Part II"— Presentation transcript:
1 Unit 5 - Chpt 17 - Thermochemistry Part II Thermo - Entropy and Free EnergyHW set1: Chpt 17 - pg #24, 25, 27-30, 32, 34, 36 Due Fri. Feb 3HW set2: Chpt 17 pg # , 50, 54 Due Mon Feb 6HW set3: Chpt 17 pg # 60, 64, 66, 71, 72, Due Fri Feb 10
2 Spontaneous ProcessReminder... 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
3 Entropy (chaos, disorder) Statistically which distribution is more likely?
5 Concept checkPredict the sign of ΔS for each of the following, and explain:The evaporation of alcoholThe freezing of waterCompressing an ideal gas at constant temperatureHeating an ideal gas at constant pressureDissolving NaCl in water+–
6 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
7 Ice Melting example Exo or Endo ? Increasing or decreasing Entropy Will ice melt spontaneously?depends on temperature!!
10 Thermochem Relationships EntropyΔSuniv = ΔSsys + ΔSsurrΔSsurr = - ΔH / TFree Energy, GG = H - TS at constant TΔG = ΔH - TΔSΔSuniv = - ΔG / T at constant T & P
11 Freezing & Melting conditions Melting point, boiling point are equilibrium between states ΔSuniv = 0 and thus ΔG = 0So ΔGo = ΔHo - TΔSo = 0o means each substance in its standard state.Given enthalpy and entropy can calculate the boiling point or freezing pt.
12 Exercise 1The 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·molΔSsurr = -132 J/K·molΔG = 0 kJ/mol
13 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 temperatureAbsolute 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.
15 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 = zeroΔS°reaction = ΣnpS°products – ΣnrS°reactants
16 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) 51H2O(l)NaOH(aq)H2(g)ΔS°= –11 J/K
17 Concept Check Consider the following system at equilibrium at 25°C. PCl3(g) + Cl2(g) PCl5(g) ΔG° = −92.50 kJWhat 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.
18 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
19 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
20 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 4CO(g) + 2H2(g) --> CH3OH(l)ΔGf (CH3OH) = -166kJ ; ΔGf (H2) = 0 ; ΔGf (CO) = -137kJ ΔG° = -29kJ now plug into above equation… T in Kelvinln(Q) = 1 / [CO]x[H2]2 = 1/45 = 2.2x10-2ΔG = -38kJ/mol of reaction
21 G and EquilibriumThe equilibrium point occurs at the lowest value of free energy available to the reaction system.At equilibrium ΔG = 0 and Q becomes KΔG° = - RT ln(K)
22 G illustrationA system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.
28 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.