Chapter 17 Free Energy and Thermodynamics Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 17 Free Energy and Thermodynamics Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall

First Law of Thermodynamics you can’t win! First Law of Thermodynamics: Energy cannot be Created or Destroyed the total energy of the universe cannot change though you can transfer it from one place to another DEuniverse = 0 = DEsystem + DEsurroundings Tro, Chemistry: A Molecular Approach

First Law of Thermodynamics Conservation of Energy For an exothermic reaction, “lost” heat from the system goes into the surroundings two ways energy “lost” from a system, converted to heat, q used to do work, w Energy conservation requires that the energy change in the system equal the heat released + work done DE = q + w DE = DH + PDV DE is a state function internal energy change independent of how done Tro, Chemistry: A Molecular Approach

Energy Tax you can’t break even! to recharge a battery with 100 kJ of useful energy will require more than 100 kJ every energy transition results in a “loss” of energy conversion of energy to heat which is “lost” by heating up the surroundings Tro, Chemistry: A Molecular Approach

fewer steps generally results in a lower total heat tax Tro, Chemistry: A Molecular Approach

Thermodynamics and Spontaneity thermodynamics predicts whether a process will proceed under the given conditions spontaneous process nonspontaneous processes require energy input to go spontaneity is determined by comparing the free energy of the system before the reaction with the free energy of the system after reaction. if the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable. spontaneity ≠ fast or slow Tro, Chemistry: A Molecular Approach

Comparing Potential Energy The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end. Tro, Chemistry: A Molecular Approach

Reversibility of Process any spontaneous process is irreversible it will proceed in only one direction a reversible process will proceed back and forth between the two end conditions equilibrium results in no change in free energy if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction Tro, Chemistry: A Molecular Approach

Thermodynamics vs. Kinetics Tro, Chemistry: A Molecular Approach

Diamond → Graphite Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in your lifetime (or through many of your generations). Tro, Chemistry: A Molecular Approach

Factors Affecting Whether a Reaction Is Spontaneous The two factors that determine the thermodynamic favorability are the enthalpy and the entropy. The enthalpy is a comparison of the bond energy of the reactants to the products. bond energy = amount needed to break a bond. DH The entropy factors relates to the randomness/orderliness of a system DS The enthalpy factor is generally more important than the entropy factor Tro, Chemistry: A Molecular Approach

Enthalpy related to the internal energy DH generally kJ/mol stronger bonds = more stable molecules if products more stable than reactants, energy released exothermic DH = negative if reactants more stable than products, energy absorbed endothermic DH = positive The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions. Hess’ Law DH°rxn = S(DH°prod) - S(DH°react) Tro, Chemistry: A Molecular Approach

Entropy entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S S generally J/mol S = k ln W k = Boltzmann Constant = 1.38 x 10-23 J/K W is the number of energetically equivalent ways, unitless Random systems require less energy than ordered systems Tro, Chemistry: A Molecular Approach

W Energetically Equivalent States for the Expansion of a Gas Tro, Chemistry: A Molecular Approach

Macrostates → Microstates These microstates all have the same macrostate So there are 6 different particle arrangements that result in the same macrostate This macrostate can be achieved through several different arrangements of the particles Tro, Chemistry: A Molecular Approach

Macrostates and Probability There is only one possible arrangement that gives State A and one that gives State C There are 6 possible arrangements that give State B Therefore State B has higher entropy than either State A or State B The macrostate with the highest entropy also has the greatest dispersal of energy Tro, Chemistry: A Molecular Approach

Changes in Entropy, DS entropy change is favorable when the result is a more random system. DS is positive Some changes that increase the entropy are: reactions whose products are in a more disordered state. (solid > liquid > gas) reactions which have larger numbers of product molecules than reactant molecules. increase in temperature solids dissociating into ions upon dissolving Tro, Chemistry: A Molecular Approach

Increases in Entropy Tro, Chemistry: A Molecular Approach

The 2nd Law of Thermodynamics the total entropy change of the universe must be positive for a process to be spontaneous for reversible process DSuniv = 0, for irreversible (spontaneous) process DSuniv > 0 DSuniverse = DSsystem + DSsurroundings if the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount when DSsystem is negative, DSsurroundings is positive the increase in DSsurroundings often comes from the heat released in an exothermic reaction Tro, Chemistry: A Molecular Approach

Entropy Change in State Change when materials change state, the number of macrostates it can have changes as well for entropy: solid < liquid < gas because the degrees of freedom of motion increases solid → liquid → gas Tro, Chemistry: A Molecular Approach

Entropy Change and State Change Tro, Chemistry: A Molecular Approach

Heat Flow, Entropy, and the 2nd Law Heat must flow from water to ice in order for the entropy of the universe to increase Tro, Chemistry: A Molecular Approach

Temperature Dependence of DSsurroundings when a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings when a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings the amount the entropy of the surroundings changes depends on the temperature it is at originally the higher the original temperature, the less effect addition or removal of heat has Tro, Chemistry: A Molecular Approach

DGreaction = S nDGprod – S nDGreact Gibbs Free Energy, DG maximum amount of energy from the system available to do work on the surroundings G = H – T∙S DGsys = DHsys – TDSsys DGsys = – TDSuniverse DGreaction = S nDGprod – S nDGreact when DG < 0, there is a decrease in free energy of the system that is released into the surroundings; therefore a process will be spontaneous when DG is negative Tro, Chemistry: A Molecular Approach

Ex. 17.2a – The reaction C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) has DHrxn = -2044 kJ at 25°C. Calculate the entropy change of the surroundings. Given: Find: DHsystem = -2044 kJ, T = 298 K DSsurroundings, J/K Concept Plan: Relationships: DS T, DH Solution: Check: combustion is largely exothermic, so the entropy of the surrounding should increase significantly

Free Energy Change and Spontaneity Tro, Chemistry: A Molecular Approach

Gibbs Free Energy, DG process will be spontaneous when DG is negative DG will be negative when DH is negative and DS is positive exothermic and more random DH is negative and large and DS is negative but small DH is positive but small and DS is positive and large or high temperature DG will be positive when DH is + and DS is − never spontaneous at any temperature when DG = 0 the reaction is at equilibrium Tro, Chemistry: A Molecular Approach

DG, DH, and DS Tro, Chemistry: A Molecular Approach

Ex. 17.3a – The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has DH = +95.7 kJ and DS = +142.2 J/K at 25°C. Calculate DG and determine if it is spontaneous. Given: Find: DH = +95.7 kJ, DS = 142.2 J/K, T = 298 K DG, kJ Concept Plan: Relationships: DG T, DH, DS Solution: Since DG is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature. Answer:

Ex. 17.3a – The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has DH = +95.7 kJ and DS = +142.2 J/K. Calculate the minimum temperature it will be spontaneous. Given: Find: DH = +95.7 kJ, DS = 142.2 J/K, DG < 0 T, K Concept Plan: Relationships: T DG, DH, DS Solution: The temperature must be higher than 673K for the reaction to be spontaneous Answer:

The 3rd Law of Thermodynamics Absolute Entropy the absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles the 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy therefore, the absolute entropy of substances is always + Tro, Chemistry: A Molecular Approach

Standard Entropies S° extensive entropies for 1 mole at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution Tro, Chemistry: A Molecular Approach

Relative Standard Entropies States the gas state has a larger entropy than the liquid state at a particular temperature the liquid state has a larger entropy than the solid state at a particular temperature Substance S°, (J/mol∙K) H2O (g) 70.0 H2O (l) 188.8 Tro, Chemistry: A Molecular Approach

Relative Standard Entropies Molar Mass the larger the molar mass, the larger the entropy available energy states more closely spaced, allowing more dispersal of energy through the states Tro, Chemistry: A Molecular Approach

Relative Standard Entropies Allotropes the less constrained the structure of an allotrope is, the larger its entropy Tro, Chemistry: A Molecular Approach

Relative Standard Entropies Molecular Complexity larger, more complex molecules generally have larger entropy more available energy states, allowing more dispersal of energy through the states Substance Molar Mass S°, (J/mol∙K) Ar (g) 39.948 154.8 NO (g) 30.006 210.8 Tro, Chemistry: A Molecular Approach

Relative Standard Entropies Dissolution dissolved solids generally have larger entropy distributing particles throughout the mixture Substance S°, (J/mol∙K) KClO3(s) 143.1 KClO3(aq) 265.7 Tro, Chemistry: A Molecular Approach

Substance S, J/molK NH3(g) 192.8 O2(g) 205.2 NO(g) 210.8 H2O(g) 188.8 Ex. 17.4 –Calculate DS for the reaction 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l) Given: Find: standard entropies from Appendix IIB DS, J/K Concept Plan: Relationships: DS SNH3, SO2, SNO, SH2O, Solution: Check: DS is +, as you would expect for a reaction with more gas product molecules than reactant molecules

DGreaction = DHreaction – TDSreaction Calculating DG at 25C: DGoreaction = SnGof(products) - SnGof(reactants) at temperatures other than 25C: assuming the change in DHoreaction and DSoreaction is negligible DGreaction = DHreaction – TDSreaction Tro, Chemistry: A Molecular Approach

Substance DGf, kJ/mol CH4(g) -50.5 O2(g) 0.0 CO2(g) -394.4 H2O(g) -228.6 O3(g) 163.2 Ex. 17.7 –Calculate DG at 25C for the reaction CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4 O3(g) Given: Find: standard free energies of formation from Appendix IIB DG, kJ Concept Plan: Relationships: DG DGf of prod & react Solution:

Ex. 17. 6 – The reaction SO2(g) + ½ O2(g)  SO3(g) has DH = -98 Ex. 17.6 – The reaction SO2(g) + ½ O2(g)  SO3(g) has DH = -98.9 kJ and DS = -94.0 J/K at 25°C. Calculate DG at 125C and determine if it is spontaneous. Given: Find: DH = -98.9 kJ, DS = -94.0 J/K, T = 398 K DG, kJ Concept Plan: Relationships: DG T, DH, DS Solution: Since DG is -, the reaction is spontaneous at this temperature, though less so than at 25C Answer:

DG Relationships if a reaction can be expressed as a series of reactions, the sum of the DG values of the individual reaction is the DG of the total reaction DG is a state function if a reaction is reversed, the sign of its DG value reverses if the amounts of materials is multiplied by a factor, the value of the DG is multiplied by the same factor the value of DG of a reaction is extensive Tro, Chemistry: A Molecular Approach

Free Energy and Reversible Reactions the change in free energy is a theoretical limit as to the amount of work that can be done if the reaction achieves its theoretical limit, it is a reversible reaction Tro, Chemistry: A Molecular Approach

Real Reactions in a real reaction, some of the free energy is “lost” as heat if not most therefore, real reactions are irreversible Tro, Chemistry: A Molecular Approach

DG under Nonstandard Conditions DG = DG only when the reactants and products are in their standard states there normal state at that temperature partial pressure of gas = 1 atm concentration = 1 M under nonstandard conditions, DG = DG + RTlnQ Q is the reaction quotient at equilibrium DG = 0 DG = ─RTlnK Tro, Chemistry: A Molecular Approach

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DG = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7) Example - DG Calculate DG at 427°C for the reaction below if the PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0 atm N2(g) + 3 H2(g) ® 2 NH3(g) Q = PNH32 PN21 x PH23 (2.0 atm)2 (33.0 atm)1 (99.0)3 = = 1.2 x 10-7 DH° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J DS° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K DG° = -92380 J - (700 K)(-198.2 J/K) DG° = +46400 J DG = DG° + RTlnQ DG = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7) DG = -46300 J = -46 kJ

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Example - K Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C N2(g) + 3 H2(g) Û 2 NH3(g) DH° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J DS° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K DG° = -92380 J - (700 K)(-198.2 J/K) DG° = +46400 J DG° = -RT lnK +46400 J = -(8.314 J/K)(700 K) lnK lnK = -7.97 K = e-7.97 = 3.45 x 10-4 since K is << 1, the position of equilibrium favors reactants

Temperature Dependence of K for an exothermic reaction, increasing the temperature decreases the value of the equilibrium constant for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant Tro, Chemistry: A Molecular Approach