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Thermodynamics - The study of heat changes that accompany chemical reactions and phase changes. - It determines three factors about the reaction: -the direction -the degree (extent of reaction) -its spontaneity - It does not determine the speed of the reaction. -Speed is determined by kinetics. Thermodynamic Law - First Law: Energy can be converted to one form or another but it can not be created or destroyed. Enthalpy -Enthalpy (ΔH): The heat change of a system at constant pressure. -It determines the ability of a reaction to produce heat. -It is a state function. -A state function’s value is: 1.Dependent on temperature & pressure 2.Independent on the path. -Letter has a °as a superscript -There are two types of heat (q) involved in a reaction: 1.Exothermic: heat is released 2.Endothermic: heat is absorbed -Not a good indication of spontaneity since both can be spontaneous. -Standard Conditions Temperature:25oC or 298K Pressure: 1.00 atm Element in its stable state Standard Molar Enthalpy of Formation -The heat change for the formation of one mole substance from its elements in their standard states (ΔHf). -Units are kJ/K mol Enthalpy -Units are kJ/K mol - Equation: - Equation: ΔHrxn = ∑nΔH°f products - ∑nΔH°f reactants n = number of moles -Elements inEnthalpy their standard state have a ΔH°f of zero. where n is the number of moles Example: Find the ΔHf for the reaction below C (graphite) + O2 (g) CO2 (g) ΔHf for CO2= -393.5 kJ/K mol ΔHof for the rxn = Enthalpy -393.5 kJ/K mole Find the ΔH°f for the reaction below: 2 CH3OH (l) + 3 O2 (g) 2 CO2 (g) + 4 H2O (g) Enthalpy Compound CH3OH (l) O2 (g) CO2 (g) H2O (g) Total ΔH°f nΔH°f (kJ/K mol) (kJ/K mol) -238.7 0 -477.4 0 -393.5 -241.8 -787 -967.2 n is the number of moles Enthalpy ΔHrxn = ∑nΔH°f products - ∑nΔH°f reactants (-787 + -967.2) - (-477.4) -1754.2 - (-477.4) -1276.8 kJ/K mol Entropy - A measure of randomness or disorder of a system. -For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than the gas state. (ΔS is positive) Ssolid SliquidSgas Predict which compound has the highest S°. A. H2O (s) B. Na (s) C. Br2 (l) D. H2O (g) Answer: C3H8 most complex molecule E. C3H8 (g) -Behavior in gases If a reaction -Produces more gas molecules than it consumes, ΔS°> 0. Predict the sign of ΔS° 2SO2 (g) + O2(g) 2SO3(g) 2H2 (l) + O2 (g) 2H2O (s) CS2 (l) CS2 (g) + CaCO3 (s) CaO (s) + CO2 (g) + Thermodynamic Law -Second Law: The entropy of the universe always increases for a spontaneous process and remains unchanged in an equilibrium process. Spontaneous Processes -Occur naturally -Tendency to want to go to a lower energy state -Increase in entropy -Spontaneity tells the direction of the energy flow. -It does not tell anything about the speed of the reaction. For a spontaneous process SUNIVERSE = SSYS + SSURR > 0 For an equilibrium process SUNIVERSE = SSYS + SSURR = 0 - If ΔSuniverse > 0 spontaneous ΔSuniverse < 0 not spontaneous -The reverse reaction is ΔSuniverse = 0 at equilibrium Entropy -The standard entropy of reaction – Neither the forward nor the reverse reaction is favored (ΔS) is the entropy change for a Entropy reaction carried out at 1 atm and -The25°C. standard entropy of reaction (ΔS) is the entropy change for a reaction carried out at 1 atm and 25°C. -Equation: -Equation: ΔSrxn entropy = ∑nΔS°f products - ∑nΔS°f reactants -Unit: J/K mol Example: Find the ΔS° for the reaction below: 2 H2 (g) + O2 (g) 2 H2O (l) Unit: J/K mol entropy Total nΔS° (J/K mol) Compound ΔS° (J/K mol) H2 (g) O2 (g) H2O(l) 131 205 262 205 69.9 139.8 Enthalpy n is the number of moles ΔSrxn = ∑nΔS°products - ∑nΔS°reactants 139.8 - (262 + 205) 139.8 - 467 -327.2 J/K mol Entropy and Enthalpy -The entropy of the surroundings is related to the heat gained or lost by the system. Entropy Equation: Equation: Used ONLY for phase changes Used ONLY for phase changes ΔSsurr = ΔHsys T ΔS is in J/K mol is *All in J/K phase mol changes are at equilibrium ΔS ΔH is in kJ/K mol T is in Kelvin *All phase changes are at equilibrium ΔH is in kJ/K mol T is in Kelvin Example: Entropy Estimate the normal boiling point of H2O (l) H2O (g) H2O (l) H2O (g) ΔH°f -285.8 -241.8 S° 69.9 188.7 kJ/K mol J/K mol Entropy Units ΔHsys =∑nΔH°f products-∑nΔH°f reactants = -241.8 – (-285.8) = 44 kJ/K mol ΔSsurr =∑nΔS°products-∑nΔS°reactants = 188.7 – (69.9) Entropy = 118.8 J/K mol Covert kJ to J 44 kJ = 44000 J ΔS = ΔH T 118.8 J/K mol = 44000 J/K mol T T = 370.4 K Thermodynamic Law -Third Law: The entropy of a perfect crystal is zero at 0 K. -Defines absolute zero. Gibbs Free Energy (G) -Combines ΔH & ΔS -Represents the energy that is free (released from the system) to do work. -Direct indicator of spontaneity. Equation: (Gibbs Helmholtz Equation) ΔG = ΔH –free T ΔS energy Units: J/K mol Gibbs Can use under any conditions What is the G for the reaction at 25°C ? N2 (g) + 3 H2 (g) 2 NH3 (g) Compound ΔH°f (kJ/K mol) ΔS° (J/K mol) 0 N2 (g) 0 H2 (g) Gibbs energy NH3 (g) free-46.3 191.5 131 193 ΔH°f =∑nΔH°f products-∑nΔH°f reactants = (2 x -46.3)-(0) = -92.6 kJ/K mol ΔS°rxn =∑nΔS°products-∑nΔS°reactants = (2x193) – (191.5)+(3x131) Gibbs free energy = -198.5 J/K mol ΔH°= -96600 J/K mol ΔS°= -198.5 J/K mol ΔG° = ΔH°-TΔS° = (-96600) – (298 x -198.5) = -37447 J/K mol Relationship with spontaneity -If G<0, the reaction is spontaneous (forward dir.) -If G>0, the reaction is not spontaneous (forward dir.) -If G=0,the reaction is at equilibrium -A reaction is spontaneous in the forward direction only if ΔG is negative. -Spontaneity is controlled by enthalpy and entropy. If ΔH (-) & ΔS (+),then what is the sign of ΔG? Spontaneous? Answer: ΔG is negative & spontaneous Standard Free Energy (ΔG°): The free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. -Occurs under standard state conditions. Gibbs free energy - Equation: Equation: ΔG°rxn = ∑nΔG°f products - ∑nΔG°f reactants where n is the number of moles -Units J/K mol -Units are molstate have a ΔG°f of zero. -Elements in theirJ/K standard Gibbs Free Energy & Equilibrium Equation: ΔG° = -2.303RT log Keq ΔG°= Free Energy R = 8.31 J/K mol (Gas Constant) T = Temperature in Kelvin Keq = Equilibrium Constant Gibbs free energy Reaction is.. IF ΔG°>0 & Keq < 1 Not Spontaneous ΔG°<0 & Keq > 1 Spontaneous Gibbs free energy ΔG°=0 & K = 1 At Equilibrium eq Find Keq at 298 K for CaCO3 (s) CaO (s) + CO2 (g) Gibbs free energy Solve for ΔG° first ΔG° = 130900 J ΔG° = -2.303 RT log Keq 130900 J=(-2.303)(8.31)(298)log Keq log Keq = -22.95 = 1.12 x 10 -23