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Chapter 19 Chemical Thermodynamics. Introduction 1 st Law of Thermodynamics: Energy can be neither created nor destroyed. Energy of the Universe is constant.

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Presentation on theme: "Chapter 19 Chemical Thermodynamics. Introduction 1 st Law of Thermodynamics: Energy can be neither created nor destroyed. Energy of the Universe is constant."— Presentation transcript:

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)] = J/K.mol S[CO 2 (g)] = J/K.mol S[NH 3 (g)] = 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° = 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?


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