1. First Law of Thermodynamics – Basically the law of conservation of energy energy can be neither created nor destroyed i.e., the energy of the universe is constant – the total energy is constant – energy can be interchanged » e.g. potential energy (stored in chemical bonds) can be converted to thermal energy in a chemical reaction » CH 4 + O 2 --> CO 2 + H 2 O + energy – Doesn’t tell us why a reaction proceeds in a particular direction Signs (+/-) will tell you if energy is entering or leaving a system + indicates energy enters a system - indicates energy leaves a system

2. Spontaneity, Entropy & Free Energy Thermodynamics – lets us predict whether a process will occur – tells us the direction a reaction will go – only considers the initial and final states – does not require knowledge of the pathway taken for a reaction

3. 3 Spontaneous Processes and Entropy One of the main objectives in studying thermodynamics, as far as chemists are concerned, is to be able to predict whether or not a reaction will occur when reactants are brought together under a special set of conditions (for example, at a certain temperature, pressure, and concentration).

4. 4 A reaction that does occur under the given set of conditions is called a spontaneous reaction. If a reaction does not occur under specified conditions, it is said to be nonspontaneous.

5. Spontaneous Physical and Chemical Processes We observe spontaneous physical and chemical processes every day, including many of the following examples: A waterfall runs downhill A lump of sugar dissolves in a cup of coffee 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

6. spontaneous nonspontaneous 18.2

7. Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

8. Spontaneous Reactions In general, spontaneous reactions are exothermic. Fe 2 O 3 (s) + 2 Al(s) ---> 2 Fe(s) + Al 2 O 3 (s) ∆H = - 848 kJ

9. Spontaneous Reactions But many spontaneous reactions or processes are endothermic or even have ∆H = 0. NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

10. 10 Another example that contradicts our assumption is the dissolution of ammonium nitrate in water: NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq)  H = 25 kJ H2OH2O This process is spontaneous, and yet it is also endothermic.

11. 11 From the study of the examples mentioned and many more cases, we cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis of energy changes in the system. To make this kind of prediction we need another thermodynamic quantity, which turns out to be entropy.

12. 12 Entropy In order to predict the spontaneity of a process, we need to know two things about the system. One is the change in enthalpy, (  H). The other is entropy (S), which is a measure of the randomness or disorder of a system. The greater the disorder of a system, the greater its entropy. Conversely, the more ordered a system, the smaller its entropy.

14. Entropy Change ΔS = q rev T

15. Entropy (S) is a measure of the randomness or disorder of a system. orderS disorder S  S = S f - S i If the change from initial to final results in an increase in randomness S f > S i  S > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state S solid < S liquid << S gas H 2 O (s) H 2 O (l)  S > 0 18.3

16. 16 Standard Entropies, S o Every substance at a given temperature and in a specific phase has a well-defined Entropy At 298 o the entropy of a substance is called S o - with UNITS of J.K -1.mol -1 The larger the value of S o, the greater the degree of disorder or randomness e.g. S o (in J.K -1 mol -1 ) : Br 2 (liq) = 152.2 Br 2 (gas) = 245.5 For any process:  S o =  S o (final) -  S o (initial)  S o (vap., Br 2 ) = (245.5-152.2) = 93.3 J.K -1 mol -1

17. 17 S (gases) > S (liquids) > S (solids) S o (J/Kmol) H 2 O(gas)188.8 H 2 O(liq) 69.9 H 2 O (s) 47.9 Ice Water Vapour Entropy, S

18. Entropy of a substance increases with temperature. Molecular motions of heptane, C 7 H 16 Molecular motions of heptane at different temps. Entropy, S

19. Processes that lead to an increase in entropy (  S > 0) 18.2

20. Increase in molecular complexity generally leads to increase in S. Entropy, S

21. 21 Consider 2 H 2 (g) + O 2 (g)  2 H 2 O(liq)  S o = 2 S o (H 2 O) - [2 S o (H 2 ) + S o (O 2 )]  S o = 2 mol (69.9 J/Kmol) - [2 mol (130.7 J/Kmol) + 1 mol (205.3 J/Kmol)]  S o = -326.9 J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid. Calculating  S for a Reaction  S o =  S o (products) -  S o (reactants) If S DECREASES, why is this a SPONTANEOUS REACTION??

22. Standard Entropies These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass.

23. 23 2nd Law of Thermodynamics  S universe =  S system +  S surroundings  S universe > 0 for product-favored process First calc. entropy created by matter dispersal (  S system ) Next, calc. entropy created by energy dispersal (  S surround ) A reaction is spontaneous (product-favored) if  S for the universe is positive.

24. 24 Calculating  S(universe) 2 H 2 (g) + O 2 (g)  2 H 2 O(liq)  S o system = -326.9 J/K  S o surroundings = +1917 J/K Can calculate that  H o rxn =  H o system = -571.7 kJ

25. 25 Calculating  S(universe) (2) 2 H 2 (g) + O 2 (g)  2 H 2 O(liq)  S o system = -326.9 J/K (less matter dispersal)  S o surroundings = +1917 J/K (more energy dispersal) The entropy of the universe increases so the reaction is spontaneous.  S o universe = +1590 J/K

26. Second Law of Thermodynamics In other words: For reversible processes:  S univ =  S system +  S surroundings = 0 For irreversible processes:  S univ =  S system +  S surroundings > 0

27. Entropy Change in the Universe This becomes:  S universe =  S system + Multiplying both sides by  T,  T  S universe =  H system  T  S system  H system T

28. Gibbs Free Energy  T  S universe is defined as the Gibbs free energy,  G. When  S universe is positive,  G is negative. Therefore, when  G is negative, a process is spontaneous.

29. Gibbs Free Energy 1.If  G is negative, the forward reaction is spontaneous. 2.If  G is 0, the system is at equilibrium. 3.If  G is positive, the reaction is spontaneous in the reverse direction.

30. Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation,  G . f  G  =  n  G  (products)   m  G  (reactants) ff where n and m are the stoichiometric coefficients.

31. Free Energy Changes At temperatures other than 25°C,  G° =  H   T  S  How does  G  change with temperature?

32. Free Energy and Temperature There are two parts to the free energy equation: –  H  — the enthalpy term – T  S  — the entropy term The temperature dependence of free energy, then comes from the entropy term.

33. Free Energy and Temperature

34. Spontaneous or Not? Remember that –∆H˚ sys is proportional to ∆S˚ surr An exothermic process has ∆S˚ surr > 0.

35. 1- Choose the substance with the higher positional entropy: a) CO 2 (s) or CO 2 (g)? b) Ch 4 (g) or C 6 H 6 ? 2- Predict the sign of the entropy change a) solid sugar is added to water b) iodine vapor condenses into a cold surface forming crystals

36. Indicate whether each of the following processes produces an increase or decrease in the entropy of the system: Answer: (a) increase, (b) decrease, (c) decrease, (d) decrease Practice Exercise