Thermodynamics Chapter 18

The First Law of Thermodynamics q = heat w = work E = q + w = heat + work = Energy Everything that is not part of the system (sys) is the surroundings (surr), and vice versa: Euniv = Esys + Esurr The total energy of the universe is constant and, therefore, energy cannot be created or destroyed.

Thermodynamics One of the main objectives of thermodynamics is to predict whether or not a reaction will occur when reactants are brought together under specific conditions. A reaction that does occur under the specific set of conditions is called a spontaneous reaction. The reverse of a spontaneous reaction cannot occur under the same set of conditions. What kind of rule can we come up with to characterize spontaneous processes? Perhaps spontaneous processes decrease energy.

Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0C and ice melts above 0 0C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous 18.2

Sign of H Cannot Predict Spontaneous Change All combustion reactions are spontaneous and exothermic: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O(g) Horxn = -802 kJ Iron rusts spontaneously and exothermically: 2 Fe(s) + O2 (g) Fe2O3 (s) 3 2 Horxn = -826 kJ Ionic compounds form spontaneously from their elements with a large release of heat: 1 2 Horxn = -411 kJ Na(s) + Cl2 (g) NaCl(s) At ordinary pressure, water freezes below 0°C but melts above 0°C. Both processes are spontaneous, but the first is exothermic and the second endothermic. H2O(l) H2O(s) Horxn = -6.02 kJ (exothermic; spontaneous at T < 0oC) H2O(s) H2O(l) Horxn = +6.02 kJ (endothermic; spontaneous at T > 0oC)

A Spontaneous, Endothermic Chemical Reaction

EH Assignment Due Unit 21 Minimum score Sec 1 Enthalpy Change 2 points extra credit Sec 2 Entropy Change 60 Sec 3 Free Energy Change Sec 4 Spontaneity of Rxns 90 Sec 5 Free Energy and Conc.

Does a decrease in enthalpy mean a reaction proceeds spontaneously? Spontaneous reactions CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = -890.4 kJ H+ (aq) + OH- (aq) H2O (l) DH0 = -56.2 kJ H2O (s) H2O (l) DH0= 6.01 kJ NH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJ H2O 18.2

No energy change! Probability factor. spontaneous nonspontaneous No energy change! Probability factor. 18.2

Entropy and the Second Law of Thermodynamics Entropy (S)–Entropy is a measure of the randomness or disorder of a system A system with relatively few equivalent ways to arrange its components, such as a crystalline solid or a deck of cards in a specific sequence, has relatively small disorder and low entropy. A system with many equivalent ways to arrange its components, such as a gas or a shuffled deck of cards, has relatively large disorder and high entropy. Sdisorder > Sorder Ssys = Sfinal - Sinitial

Entropy (S) is a measure of the randomness or disorder of a system. DS = Sf - Si If the change from initial to final results in an increase in randomness Sf > Si DS > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state Ssolid < Sliquid << Sgas H2O (s) H2O (l) DS > 0 18.2

Processes that lead to an increase in entropy (DS > 0) 18.2

Predicting Relative Entropy Values Problem: Choose the member with the higher entropy in each of the following pairs, and justify your choice. (a) 1 mol NaCl(s) or 1 mol NaCl(aq) (b) 1 mol O2 and 2 mol H2 or 1 mol H2O (c) 1 mol H2O(s) or 1 mol H2O(g) (d) Lipton’s noodle soup at 24oC or at 95oC Is the entropy change greater or less than zero for: (a) freezing ethanol (b) evaporating a beaker of bromine (c) dissolving urea in water (d) cooling N2 gas

(a) Condensing water vapor How does the entropy of a system change for each of the following processes? (a) Condensing water vapor Randomness decreases Entropy decreases (DS < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (DS < 0) (c) Heating hydrogen gas from 600C to 800C Randomness increases Entropy increases (DS > 0) (d) Subliming dry ice Randomness increases Entropy increases (DS > 0) 18.2

The Second Law of Thermodynamics - the Entropy and the Second Law of Thermodynamics The Second Law of Thermodynamics - the entropy of the universe (system + surroundings) always increases in a spontaneous process and remains unchanged in an equilibrium process. Any process in which the entropy of the universe would decrease is forbidden. DSuniv = DSsys + DSsurr > 0 spontaneous DSuniv = DSsys + DSsurr < 0 forbidden

Entropy Changes in the System (DSsys) The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aA + bB cC + dD DS0 rxn dS0(D) cS0(C) = [ + ] - bS0(B) aS0(A) DS0 rxn nS0(products) = S mS0(reactants) - What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g) S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol DS0 rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] DS0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol 18.3

Entropy Changes in the System (DSsys) When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes, DS0 > 0. If the total number of gas molecules diminishes, DS0 < 0. If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s) The total number of gas molecules goes down, DS is negative. 18.3

Entropy Changes in the Surroundings (DSsurr) Exothermic Process DSsurr > 0 Endothermic Process DSsurr < 0 18.3

Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature. 18.3

For a constant-temperature process: Gibbs Free Energy Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 For a constant-temperature process: Gibbs free energy (G) DG = DHsys -TDSsys DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium. 18.4

Entropy and Free Energy Gibbs free energy–This is a function that combines the systems enthalpy and entropy: G = H - TS The change in free energy of a system at constant temperature and pressure is given by the Gibbs equation: Gsys = Hsys - T Ssys The Second Law can be stated in terms of G for the system. Suniv > 0 for a spontaneous process G < 0 for a spontaneous process Suniv < 0 for a nonspontaneous process G > 0 for a nonspontaneous process Suniv = 0 for a process at equilibrium G = 0 for a process at equilibrium

Reaction Spontaneity and the Signs of Ho, So, and Go Ho So -T So Go Description - + - - Spontaneous at all T + - + + Nonspontaneous at all T + + - + or - Spontaneous at higher T; Nonspontaneous at lower T - - + + or - Spontaneous at lower T; Nonspontaneous at higher T An endothermic reaction can be spontaneous only if there is an entropy increase (an increase in disorder).

Effect of Temperature on Reaction Spontaneity The temperature at which a reaction occurs influences the magnitude of the T S term. By scrutinizing the signs of H and S, we can predict the effect of temperature on the sign of G and thus on the spontaneity of a process at any temperature. Temperature-independent cases (opposite signs) 1. Reaction is spontaneous at all temperatures: Ho < 0, So > 0 2. Reaction is nonspontaneous at all temperatures: Ho > 0, So < 0 2 H2O2 (l) 2 H2O(l) + O2 (g) Ho = -196 kJ and So = 125 J/K 3 O2 (g) 2 O3 (g) Ho = 286 kJ and So = - 137 J/K Temperature-dependent cases (same signs) 3. Reaction is spontaneous at higher temperature: Ho > 0 and So > 0 4. Reaction is spontaneous at lower temperature: Ho < 0 and So < 0 2 N2O(g) + O2 (g) 4 NO(g) Ho = 197.1 kJ and So = 198.2 J/K 2 Na(s) + Cl2 (g) 2 NaCl(s) Ho = - 822.2 kJ and So = - 181.7 J/K

DG0 of any element in its stable form is zero. The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD DG0 rxn dDG0 (D) f cDG0 (C) = [ + ] - bDG0 (B) aDG0 (A) DG0 rxn nDG0 (products) f = S mDG0 (reactants) - Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f DG0 of any element in its stable form is zero. f 18.4

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) What is the standard free-energy change for the following reaction at 25 0C? 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DG0 rxn nDG0 (products) f = S mDG0 (reactants) - DG0 rxn 6DG0 (H2O) f 12DG0 (CO2) = [ + ] - 2DG0 (C6H6) DG0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0C? DG0 = -6405 kJ < 0 spontaneous 18.4

There is an important relationship between the Gibbs Free Energy change and the equilibrium constant. Go = -RT ln K Homework - pages 598 - 601 #1, 2, 5, 6, 7, 10, 14, 15, 16, 19, 22, 44, 66

The Relationship Between Go and K at 25oC Go (kJ) K Significance 200 9 x 10 -36 Essentially no forward reaction; 100 3 x 10 -18 reverse reaction goes to 50 2 x 10 -9 completion. 10 2 x 10 -2 1 7 x 10 -1 0 1 Forward and reverse reactions -1 1.5 proceed to same extent. -10 5 x 101 -50 6 x 108 -100 3 x 1017 Forward reaction goes to -200 1 x 1035 completion; essentially no reverse reaction.

DG = DH - TDS 18.4

Temperature and Spontaneity of Chemical Reactions CaCO3 (s) CaO (s) + CO2 (g) DH0 = 177.8 kJ DS0 = 160.5 J/K DG0 = DH0 – TDS0 At 25 0C, DG0 = 130.0 kJ DG0 = 0 at 835 0C 18.4

DG0 = - RT lnK 18.4