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

Reversible 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

Thermodynamics vs. Kinetics Tro, Chemistry: A Molecular Approach

Diamond → Graphite Tro, Chemistry: A Molecular Approach

4 particles partition in 16 microstates 1 mole of particles partition in 2N (N~1023) 1/16 4/16 6/16 4/16 1/16

“There is a tide in the affairs of men, Which taken at the flood, leads on to fortune. Omitted, all the voyage of their life is bound in shallows and in miseries. On such a full sea are we now afloat. And we must take the current when it serves, or lose our ventures.” William Shakspeare Letters put together form words, in their specific organization, convey timeless meaning! Tro, Chemistry: A Molecular Approach

Increases in Entropy Tro, Chemistry: A Molecular Approach

The 2nd Law of Thermodynamics DSuniverse = DSsystem + DSsurroundings 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

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

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:

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 Tro, Chemistry: A Molecular Approach

Relative Standard Entropies Allotropes Tro, Chemistry: A Molecular Approach

Relative Standard Entropies Molecular Complexity 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 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

Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

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

DGº and K Because DGrxn = 0 at equilibrium, then DGº = −RTln(K). When K < 1, DGº is positive and the reaction is spontaneous in the reverse direction under standard conditions. Nothing will happen if there are no products yet! When K > 1, DGº is negative and the reaction is spontaneous in the forward direction under standard conditions. When K = 1, DGº is 0 and the reaction is at equilibrium under standard conditions.

Tro, Chemistry: A Molecular Approach

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

Why Is the Equilibrium Constant Temperature Dependent? Combining these two equations DG° = DH° − TDS° DG° = −RTln(K) It can be shown that This equation is in the form y = mx + b. The graph of ln(K) versus inverse T is a straight line with slope and y-intercept .