1. CHEMICAL THERMODYNAMICS CHEM171 – Lecture Series Three : 2012/01  Spontaneous processes  Enthalpy (H)  Entropy (S)  Gibbs Free Energy (G)  Relationship between G and K Brown et al., Chapter 4.9-4.15

2. A spontaneous chemical reaction is one that, given sufficient time, will achieve chemical equilibrium, with an equilibrium constant greater than 1, by reacting from left to right. SPONTANEOUS REACTIONS HOW CAN WE TELL IF A REACTION WILL PROCEED OR NOT? Some chemical and physical changes take place by themselves, given enough time. CHEM171 – Lecture Series Three : 2012/02

3. Cu(s) + Cl 2 (g)  CuCl 2 (s)  spontaneous  2H 2 (g) + O 2 (g)  2H 2 O(g) spontaneous reaction but occurs only if you ignite the mixture O 3 (g)  O(g) + O 2 (g)  nonspontaneous  This does not mean that it does not occur at all. CHEM171 – Lecture Series Three : 2012/03

4. It means that, when equilibrium is achieved, not many O 3 molecules have broken down into products. That is Spontaneous does not mean instantaneous and has nothing to do with the rate. If a change is spontaneous in one direction, it is not spontaneous in the other. CHEM171 – Lecture Series Three : 2012/04

5. System – portion of the universe we wish to study. Surroundings – everything else. Universe = System + Surroundings Example: In the chemistry lab, system is usually a flask or a beaker and the surrounding is the rest of the laboratory. CHEM171 – Lecture Series Three : 2012/05

6. ENTHALPY The enthalpy (H) is a measure of the total energy of a system. Can we use the enthalpy to predict the direction of chemical change? Some scientists thought that the sign of  H determined spontaneity. CHEM171 – Lecture Series Three : 2012/06

7. Exothermic Reactions Physical CaCl 2 (s)  Ca 2+ (aq) + 2Cl - (aq) Chemical 8Al(s) + 3Fe 3 O 4 (s)  4Al 2 O 3 (s) + 9Fe(l) reactants products The enthalpy (energy) of the chemical system is lowered. H CHEM171 – Lecture Series Three : 2012/07

8. Endothermic Reactions Physical H 2 O(s)  H 2 O(l)  H 2 O(g) NH 4 Cl(s)  NH 4 + (aq) + Cl - (aq) Chemical Ba(OH) 2  8H 2 O(s) + 2NH 4 NO 3 (s)  Ba(NO 3 ) 2 (aq) + 2NH 3 (g) + 10H 2 O(l) reactants products Some spontaneous chemical reactions are endothermic (enthalpy increases) H CHEM171 – Lecture Series Three : 2012/08

9.  H, alone cannot be used to predict if a reaction or process will go. The reason is that it represents the total energy of the system. We need to examine the available energy, the free energy (G) available to do useful work. The unavailable energy per degree kelvin is known as the entropy (S) of a system. CHEM171 – Lecture Series Three : 2012/09

10. ENTROPY Entropy (S) is used to quantify the extent of disorder resulting from the dispersal of energy and matter. The greater the disorder in a system, the greater the entropy and the larger the value of S. When comparing the same or similar substances, entropies of gases are much larger than those for liquids and entropies for liquids are larger than those for solids. CHEM171 – Lecture Series Three : 2012/10

11. Larger molecules have a large entropy than smaller molecules and molecules with more complex structures have larger entropies than simpler molecules. EXAMPLE Which substance has the higher entropy? Explain your reasoning. (a) NO 2 (g) or N 2 O 4 (g) (b) I 2 (g) or I 2 (s) CHEM171 – Lecture Series Three : 2012/11

12. SOLUTION (a) Both NO 2 and N 2 O 4 are gases, since N 2 O 4 is the larger molecule, it is expected to have the higher standard entropy. S o value for N 2 O 4 = 304.38 J K -1 mol -1 whereas for NO 2 it is 240.04 J K -1 mol -1 (b) Gases have higher entropies than solids S o value for I 2 (g) = 260.69 J K -1 mol -1 whereas for I 2 (s) it is 116.135 J K -1 mol -1 CHEM171 – Lecture Series Three : 2012/12

13. The entropy change (  S o ) for chemical and physical changes under standard conditions can be calculated from values of S o.  S o rxn =  S o (products) -  S o (reactants) EXAMPLE Calculate  S o rxn for the oxidation of NO with O 2 2NO(g) + O 2 (g)  2NO 2 (g) CHEM171 – Lecture Series Three : 2012/13 SOLUTION  S o rxn = [2(240.0 J K -1 mol -1 )] – [2(210.8 J K -1 mol -1 ) + 205.1 J K -1 mol -1 ] = -146.7 J K -1 or -73.35 J K -1 for 1 mol of NO 2 formed

14. G = H – TS where G = available energy, Gibbs energy H = total energy, enthalpy S = unavailable energy per Kelvin, entropy T = temperature The direction of chemical change is the direction which lowers the Gibbs energy. GIBBS FREE ENERGY CHEM171 – Lecture Series Three : 2012/14

15. Δ G = Δ H – T Δ S A process is spontaneous in the direction in which the Gibbs energy decreases: i.e. Δ G is negative. At constant temperature, Δ G < 0 for a spontaneous process Δ G > 0 for a nonspontaneous process Δ G = 0 for a process at equilibrium CHEM171 – Lecture Series Three : 2012/15

16. EXAMPLE Using values of  H o f and S o to find  H o rxn and S o rxn respectively, calculate the free energy change,  G o, for the formation of 2 mol NH 3 (g) from the elements at standard conditions (and 25°C): N 2 (g) + 3H 2 (g)  2NH 3 (g) SOLUTION  H o rxn = 2  H o f [NH 3 (g)] – {  H o f [N 2 (g)] + 3  H o f [H 2 (g)] = 2(-45.9 kJ mol -1 ) - (0 + 0) = -91.8 kJ CHEM171 – Lecture Series Three : 2012/16

17.  S o rxn = [2(192.77 J K -1 mol -1 ) – [191.56 J K -1 mol -1 + 3(130.7 J K -1 mol -1 )] = -192.12 J K -1 = -0.19212 kJ K -1  G o rxn =  H o rxn - T  S o rxn = -91.8 kJ – (298 K)( -0.19212 kJ K -1 ) = -34.5 kJ CHEM171 – Lecture Series Three : 2012/17

18. The free energy is related to the equilibrium constant, K, by the equation: Δ G o rxn = -RT ln K EXAMPLE From the standard free energy change, Δ G o rxn, which is - 16.37 kJ mol -1 for the formation of 1.00 mol of ammonia from nitrogen and hydrogen, calculate the equilibrium constant for this reaction at 25°C. CHEM171 – Lecture Series Three : 2012/18

19. SOLUTION The balanced equation for the chemical reaction under investigation is: The free energy change for this reaction is -16.37 kJ mol -1 From Δ G o rxn = -RT ln K, ln K = Δ G o rxn /-RT ln K = (-16 370 J mol -1 )/-(8.3145 J K -1 mol -1 )(298.15 K) = 6.604  K = 7.38 × 10 2 ½N 2 + 1½H 2  NH 3 CHEM171 – Lecture Series Three : 2012/19