1. Second law of Thermodyna mics - 2

2. If an irreversible process occurs in a closed system, the entropy S of the system always increase; it never decreases. There are two equivalent ways to define the change in entropy of a system: (1) In terms of the system’s temperature and the energy it gains or loses as heat (2) By counting the ways in which the atoms or molecules that make up the system can be arranged.

3. Substituting n=1.00 mol and V f /V i =2

4. Step 1. Step 2.

5. To predict spontaneity we need to know the energy change and the entropy. Entropy (S) is a measure of the randomness or disorder of a system. The greater the disorder, the greater the entropy. Nature tends to the greatest entropy. S solid < S liquid < S gas

6. Entropy & Chemical Reactions The decomposition of N 2 O 4 (O 2 N–NO 2 ) is also accompanied by an increase in randomness. Whenever molecules break apart, randomness increases. Consider the gas phase reaction: 3A 2 + 6B = 6AB (a) Write a balanced equation for the reaction. (b) Predict the sign of S for the reaction.

7. Standard Molar Entropies Standard Molar Entropy, S°: The entropy of 1 mol of the pure substance at 1 atm pressure and a specified temperature, usually 25°C. Standard molar entropies are absolute entropies measured against an absolute reference point. Standard entropy of reaction: S°= ΣnS°(products) – ΣnS°(reactants)

8. Standard Molar Entropies S° ( J/K.mol )CpdS° ( J/K.mol )Cpd 92.9CaCO 3 200.8C2H2C2H2 39.7CaO192.3NH 3 160CH 3 COOH197.6CO 69.9H2OH2O213.6CO 2 2.4C diamond 219.5C2H4C2H4 3.6C graphite 130.6H2H2 27.3Fe186.2CH 4

9. QUIZ !!!!!! Calculate the standard entropy of reaction at 25°C for the synthesis of ammonia: N 2 (g) + 3 H 2 (g) = 2 NH 3 (g) Calculate the standard entropy of reaction at 25°C for the decomposition of calcium carbonate: CaCO 3 (s) = CaO(s) + CO 2 (g)

10. Gibbs Free Energy, G Multiply through by -T -T∆S univ = ∆H sys - T∆S sys -T∆S univ = change in Gibbs free energy for the system = ∆G system Under standard conditions — ∆G o sys = ∆H o sys - T∆S o sys ∆S univ = ∆S surr + ∆S sys

11. ∆G o = ∆H o - T∆S o Gibbs free energy change = total energy change for system - energy lost in disordering the system If reaction is -exothermic (negative ∆ H o ) (energy dispersed) and entropy increases (positive ∆S o ) matter dispersed) -then ∆G o must be NEGATIVE reaction is spontaneous (and product-favored).

12. ∆G o = ∆H o - T∆S o Gibbs free energy change = total energy change for system energy lost in disordering the system energy lost in disordering the system If reaction is : endothermic (positive ∆H o ) and entropy decreases (negative ∆S o ) endothermic (positive ∆H o ) and entropy decreases (negative ∆S o ) then ∆G o must be POSITIVE reaction is not spontaneous (and is reactant- favored). then ∆G o must be POSITIVE reaction is not spontaneous (and is reactant- favored).

13. Gibbs Free Energy, G ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o ∆H o ∆S o ∆G o Reaction exo(–)increase(+)–Prod-favored endo(+)ecrease(-)+React-favored exo(–)decrease(-)?T dependent endo(+)increase(+)?T dependent

14. Calculating ∆G o rxn Is the dissolution of ammonium nitrate product-favored? If so, is it enthalpy- or entropy-driven? NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

15. Practice Problem Calculate ∆H° and ∆S° for the following reaction: NH 4 NO 3 (s) + H 2 O(l) = NH 4 + (aq) + NO 3 - (aq) Use the results of this calculation to determine the value of ∆G o for this reaction at 25 o C, and explain why NH 4 NO 3 spontaneously dissolves in water at room temperature. Solution Using a standard-state enthalpy of formation and absolute entropy data table, we find the following information: Compound Hfo(kJ/mol) S°(J/mol-K) NH 4 NO 3 (s) -365.56 151.08 NH 4 + (aq) -132.51 113.4 NO 3 - (aq) -205.0 146.4 This reaction is endothermic, and the enthalpy of reaction is therefore unfavorable: ∆H f o (products) - ∆H f o (reactants) = [1 mol NH 4 + x 132.51 kJ/mol + 1 mol NO 3 - x -205.0 kJ/mol] - [1 mol NH 4 NO 3 x -365.56 kJ/mol] = 28.05 kJ The reaction leads to a significant increase in the disorder of the system, however, and is therefore favored by the entropy of reaction: ∆S o = S o (products) – S o (reactants) = [1 mol NH 4 + x 113.4 J/mol-K + 1 mol NO 3 - x 146.4 J/mol-K] - [1 mol NH 4 NO 3 x 151.08 J/mol-K] = 108.7 J/K

16. To decide whether NH 4 NO 3 should dissolve in water at 25 o C we have to compare the ∆H o and T∆S o to see which is larger. Before we can do this, we have to convert the temperature from o C to kelvin: TK = 25 o C + 273.15 = 298.15 K We also have to recognize that the units of ∆H o for this reaction are kilojoules and the units of So are joules per kelvin. At some point in this calculation, we therefore have to convert these quantites to a consistent set of untis. Perhaps the easiest way of doing this is to convert Ho to joules. We then multiply the entropy term by the absolute temperature and subtract this quantity from the enthalpy term: ∆G o = ∆H o - T∆S o = 28,050 J - (298.15 K x 108.7 J/K) = 28,050 J - 32,410 J = -4360 J At 25 o C, the standard-state free energy for this reaction is negative because the entropy term at this temperature is larger that the enthalpy term: ∆G o = -4.4 kJ The reaction is therefore spontaneous at room temp.

23. Helmholtz Free Energy

24. Gibbs Free Energy

25. The increase in entropy principle can be summarized as follows: