Three laws of thermodynamics, Lecture notes of Thermodynamics

Download Three laws of thermodynamics and more Lecture notes Thermodynamics in PDF only on Docsity! Three laws of thermodynamics First law of thermodynamics: law of conservation of energy; energy can neither be created nor destroyed; total energy of universe is constant; energy is conserved. (Chapter 6) ΔE(gained by system) + ΔE(lost by surroundings) = 0 ΔE(gained by surroundings) + ΔE(lost by system) = 0 ΔEuniverse = 0 ΔEsystem + ΔEsurroundings = 0 ΔEsystem = -ΔEsurroundings Third law of thermodynamics: a perfect crystal of any substance at 0 K has zero entropy. S = 0 at 0 K Chemistry 103 Spring 2011 2 Second law of thermodynamics: the total entropy of the universe is increasing. The energy of the universe constantly becomes more dispersed and spread out over time. ΔSuniverse > 0 ΔSsystem + ΔSsurroundings > 0 We could decrease the entropy of a system, e.g., organize a messy living room. (ΔSsystem < 0) But it could only happen if the entropy of the surroundings increased (ΔSsurroundings > 0) and increased by more than the magnitude of the decrease in the system: ΔSsystem + ΔSsurroundings > 0 … ΔSsystem < 0, ΔSsurroundings > -ΔSsystem so -ΔSsystem > 0 In other words, the decrease only happens by dispersing more energy in the surroundings (we spend energy to clean up the room), so the total entropy of the universe increases. Chemistry 103 Spring 2011 5 Practice: For the combustion of propane (C3H8), calculate the standard enthalpy change and standard entropy change for the reaction. What is the total entropy change for the universe? Chemistry 103 Spring 2011 6 Clearly for ΔSsystem > 0, the process favors products, because it increases entropy. Ex: producing more moles of gas. Also for ΔHsystem < 0, the process favors products, because it is exothermic, which increases entropy (disperses energy). ΔHsystem = -ΔHsurroundings so ΔHsurroundings > 0 ΔSsurroundings = (ΔHsurroundings/T) > 0. Ex: releasing energy to the surroundings by burning fuel (exothermic). These principles are summarized in one concept, Gibbs free energy, given the symbol G. Essentially, free energy is the amount of potential energy available within a chemical system to do useful work at constant pressure and constant temperature. Chemistry 103 Spring 2011 7 As the entropy of universe increases (ΔSuniverse > 0), energy is being dispersed, so free (useful) energy decreases (ΔG < 0). This relationship defines Gibbs free energy for any process: ΔGsystem = -TΔSuniverse Therefore, a process that decreases free energy (ΔGsystem < 0) is favored because it increases the entropy of universe (ΔS > 0). Doing some algebra … ΔSuniverse = ΔSsurroundings + ΔSsystem = -ΔHsystem/T + ΔSsystem ΔGsystem = -TΔSuniverse = -T (-ΔHsystem/T + ΔSsystem) = ΔHsystem – TΔSsystem ΔGsystem = ΔHsystem – TΔSsystem Chemistry 103 Spring 2011 10 Practice: For the combustion of propane (C3H8), calculate the standard free energy change using the standard enthalpy and standard entropy change you calculated earlier. Chemistry 103 Spring 2011 11 With the standard enthalpies of formation (ΔHfº) and standard entropies (Sº) tabulated, then the standard Gibbs free energy of formation (ΔGfº) for each substance can be tabulated: Formation reactions: reactants in their standard states make products in their standard states. C(s) + O2(g)  CO2(g) 2H2(g) + O2(g)  2H2O(l) ΔGsystem = ΔHsystem – TΔSsystem ΔGfº = ΔHfº – TSº (ΔSº = Sº for a pure substance) With ΔGfº values tabulated, we can also do a Hess’s Law-type of calculation to calculate ΔGºsystem for a process (i.e., calculate ΔGºrxn). ΔG°rxn = ΣnΔG°f(products) - ΣnΔG°f(reactants) Chemistry 103 Spring 2011 12 Practice: For the combustion of propane (C3H8), calculate the standard free energy change using Gibbs free energy of formation values, and compare to your answer you calculated before.