1. Chapter 19 Chemical Thermodynamics

2. Introduction 1 st Law of Thermodynamics: Energy can be neither created nor destroyed. Energy of the Universe is constant. Why does a reaction occur in a given direction? How do we know? Amount of energy involved in a chemical change? Does Energy flow into or out of the system? What form does the energy finally assume? Why processes occur in particular directions – main focus of this chapter

3. Spontaneous processes Occurs without outside intervention Spontaneous  Fast Thermodynamic vs. Kinetics Thermodynamics – allows us to predict if a process is spontaneous or not, based on properties of products and reactants Doesn’t tell us anything about speed of reaction or any other species that are formed (kinetics)

4. Spontaneous processes / changes Conversion diamond to graphite: spontaneous –Can we observe this? Ball rolling down a hill –Rolling back up the hill? Rusting of steel –Conversion of rust into iron and O 2 ? Burning of wood to form H 2 O and CO 2 –Heating together H 2 O and CO 2 to form wood A book falling on its side –Book standing back up Freezing of ice below 0 °C

5. Nonspontaneous processes / changes One that requires continuous help and can only occur if accompanied by and linked to a spontaneous process. A brick wall building itself from a pile of bricks? Likely to happen?

6. Spontaneous changes and energy Exothermic processes (-  H); tend to proceed spontaneously Endothermic processes (+  H); tend to proceed nonspontaneously Generalize this for all processes? No. Melting of ice is spontaneous (above 0 °C), but it’s an endothermic process  H helps determine if a process (chemical reaction) is spontaneous, but there is one other thermodynamic factor too….

7. Entropy (S) Property in common with all spontaneous processes A measure of randomness, or disorder. Naturally, all systems tend towards a state of disorder (without outside interference); From order to disorder +  S = increase in entropy (disorder); spontaneous -  S = decrease in disorder; nonspontaenous Throw a deck of cards into the air. Likelihood of picking the cards up in exactly the same order? Possible, but highly unlikely. S associated with probability. Use your kitchen for a week without cleaning up, what happens?

8. Entropy (S)

9. Leave a petri dish of water in a room for 2 days, what happens to the water? Water evaporates (gaseous water molecules vs. liquid water molecules) Leave dish 2 more days, will water have returned? Gaseous water more randomly distributed than liquid water; liquid  gas = increase in disorder (+  S) S gas >> S liquid > S solid Units of S: J/K or J/K.mol

10. Predicting sign of  S for change 2 nd Law of Thermodynamics: ‘In any spontaneous process, there is always an increase in the entropy of the universe – Entropy of the universe is increasing’ +  S if increase in disorder observed Formation / loss of gases particularly important (more so than for liquids/solids) Increasing volume of a gas increases entropy (more random distribution if gas particles) Increasing T increases entropy Solids  Liquids  Gases with increasing T

11. Predicting relative entropy values For each of the following pairs, choose the substance with the higher entropy (per mole) at a given temperature: Solid CO 2 and gaseous CO 2 N 2 gas at 1 atm and N 2 gas at 1.0x10 -2 atm

12.  S in chemical reactions Sign of  S usually easy to predict for reactions producing / consuming gases. Gas produced, entropy increases (from reactants to products), thus +  S Gas consumed, entropy decreases, thus -  S Overall, look for  n gas +  n gas = +  S -  n gas = -  S

13. Molecular complexity Changes in the degree of complexity of molecules affects  S Decrease in degree of molecular complexity, and an increase in # of particles, gives +  S Increase in degree of molecular complexity, and a decrease in # of particles, gives -  S  S can also be found using  n gas idea

14.  S examples Sublimation of a solid Condensation of steam into liquid water Boiling of water

15. Calculating  S Standard entropy change (  S° - measured under standard conditions of 298 K and 1 atm) can be calculated for a given reaction, knowing S° for each component of reaction.  S° =  (S° products) -  (S° reactants) If 1 mole of a compound is formed from its elements, then  S° for that reaction is std. entropy of formation (  S° f ); (same as for  H° f )

16.  S calculation examples S° values available in Appendix  S° =  (S° products) -  (S° reactants)  S° = Expected to be spontaneous? Same answer from  n gas

17.  S calculation examples Calculate  S° for the reaction of 1 mole of urea with water, to produce gaseous CO 2 and NH 3.  S° =  (S° products) -  (S° reactants) S[CO(NH 2 ) 2 (aq)] = 173 J/K.mol S[H 2 O(l)] = 69.96 J/K.mol S[CO 2 (g)] = 213.6 J/K.mol S[NH 3 (g)] = 192.5 J/K.mol Expected to be spontaneous? Compare with  n gas answer.

18. Gibbs Free Energy and spontaneity Balance exists between  H and  S (and T) for spontaneity G = H-TS (G = Gibbs Free Energy)  G =  H-T  S (for changes that occur at constant T and p)  G = G final – G initial Change is spontaneous only if accompanied by a decrease in free energy of the system (-  G) Spontaneous: -  G Nonspontaneous: +  G And, G ultimately determined by  H and  S.

19.  G sign prediction  G =  H-T  S

20. Standard Free Energy Changes  G° =  H°-T  S°  G° for reaction between urea and H 2 O? Need to calculate  H° and  S°  H° =  (  H f ° products) -  (  H f ° reactants) Recall from chapter 6 Thus,  G° = 119.2 kJ – [298K.(0.355 kJ/K) = 13.4 kJ. Spontaneous process? Large +  H° and small +  S°, giving +  G°

21. Standard Free Energy Changes  G° =  (  G f ° products) -  (  G f ° reactants) Methanol: High octane fuel used in racing engines.  G° for reaction? 2CH 3 OH(g) + 3O 2 (g)  2CO 2 (g) + 4H 2 O(g)  G f °[CH 3 OH(g)] = -163 kJ/mol  G f °[O 2 (g)] = 0  G f °[CO 2 (g)] = -394 kJ/mol  G f °[H 2 O(g)] = -229 kJ/mol  G°= Spontaneous?