1 CHEMICAL THERMODYNAMICS Spontaneous reactions Nonspontaneous reactions Enthalpy change (ΔH) Exothermic and endothermic reactions Entropy change (ΔS) Gibbs free energy change, (ΔG) Free energy of formation
2 CHEMICAL THERMODYNAMICS Fundamental Law of Nature Any chemical system will tend to undergo an irreversible change from some initial, nonequilibrium state to a final, equilibrium state. aa + bb cc + dd At equilibrium, the total energy of the system is minimum.
3 What Makes a Reaction Spontaneous? Chemical change and energy are always related. When the reactant material contains large amounts of energy relative to the products, the chemical reaction occurs spontaneously. aa + bb cc + dd (A and B have more energy than C and D. Therefore, reaction can occur spontaneously.) Thus, a spontaneous chemical reaction is a process in which matter moves toward a more stable state.
5 Enthalpy change (ΔH) A chemical reaction that releases heat (that is, for which ΔH is negative) is an exothermic reaction. A chemical reaction that takes up heat (that is, for which ΔH is positive) is an endothermic reaction. When a chemical reaction releases large amounts of heat (has a large negative enthalpy change), it is said to be strongly exothermic. When a chemical reaction absorbs large amounts of heat (has a large positive enthalpy change), it is said to be strongly endothermic.
6 Exothermic Endothermic
7 ΔH rxn =ƩΔH products - ƩΔH reactants ΔH rxn <0 for exothermic reactions ΔH rxn >0 for endothermic reactions Strongly exothermic reactions occur spontaneously. Strongly endothermic reactions do not occur spontaneously.
8 However; Some weakly endothermic reactions occur spontaneously. Some weakly exothermic reactions do not occur spontaneously. This implies that enthalpy change alone cannot be used to judge the spontaneity of a chemical reaction.
9 What else then? Answer: A change in entropy
10 Entropy change (ΔS) Change in entropy is the change in degree of randomness or disorder. A definite amount of entropy is associated with each chemical species. When a given chemical species is in the gaseous state, the molecules are disordered with respect to each other. The entropy of this species in the gaseous state is much greater than when it is in the solid state, where the molecules are highly ordered with respect to each other.
11 So once again, entropy is the amount of disorder in a system. The entropy of the universe increases in a spontaneous process. Therefore, it would not be wrong to assume that the disorderliness of the universe is always increasing.
12 A system will tend towards maximum entropy. The entropy drive of a chemical reaction will be towards the side with the most molecules in the most random phase: gases (g) >> solutions (aq) > liquids (l) >> solids (s)
13 If both the enthalpy drive and the entropy drive favor the products, the reaction will be spontaneous If both the enthalpy drive and the entropy drive favor the reactants, the reaction will be non-spontaneous If the enthalpy and entropy drive are in opposite directions, the reaction will be reversible and is said to be in chemical equilibrium.
14 Summary Favorable Unfavorable Conditions Conditions ΔH < 0 ΔH > 0 ΔS > 0 ΔS < 0
15 What Makes a Reaction Spontaneous? So, based on the above discussion we can conclude that the actual driving force for a chemical reaction is a combination of enthalpy change and entropy change. This driving force is called the GIBBS FREE ENERGY CHANGE, (ΔG) of the reaction.
16 Gibbs Free Energy (ΔG) change in free energy change in enthalpy temperature change in entropy ΔG ΔH TΔS Experimental observations have shown that a decrease in free energy is associated with reactions which occur spontaneously. When ΔG<0 When ΔG>0 supplied. spontaneous, tendency to occur reaction will occur only if energy is When ΔG=0 reaction is at equilibrium.
17 G n G m G rxn products reactants Chemical thermodynamics tells us whether a particular reaction can be performed at all, and if it is possible, is necessary to help us find the best conditions for carrying out the reaction.
18 Free Energy of Formation To provide a uniform basis for the calculation of the free energy change associated with a particular chemical reaction, standard state conditions for chemical systems are assigned. 1 atm pressure & 298 o K (25 o C) The concentrations of all aqueous solutions are 1 M. A free energy change under standard state conditions is denoted by ΔG 0.
19 Free Energy of Formation It is impossible to measure absolute values of free energy. We can only measure changes in free energy. To establish a baseline, a convention is necessary. Therefore, Every element is assigned a free energy of formation of zero at standard state. H + at a concentration of 1 mol/l in an ideal solution has a free energy of formation of zero at standard state.
20 Free Energy of Formation The free energy change for chemical reaction is defined as the difference between free energies of the final and the initial states: G G G final Under standard conditions initial G n G m G rxn f, prod. f, react.
21 Note Free Energy of Formation G n G m G rxn f, prod. f, react. ΔG 0 rxn is expressed in terms of kcal. ΔG 0 f is expressed in terms of kcal/mol. Therefore, ΔG 0 f values are multiplied by the molar coefficients in the balanced chemical equation to get ΔG 0 rxn.
22 C(graphite) + O 2 (g) CO 2 (g) G 0 = G 0 rxn f, CO 2 S(rhombic) + O 2 (g) SO 2 (g) G 0 = G 0 rxn f, SO 2
23 Ex. Fe 2+ is present in groundwater under anaerobic conditions. When exposed to air, Fe 2+ may slowly oxidize to Fe 3+, which forms an insoluble hydroxide precipitate. Is the following reaction spontaneous? 2 Fe 2+ (aq)+ HOCl (aq) + 5 H 2 O (l) 2 Fe(OH) 3(s) + 2 Cl - (aq)+ 5 H + (aq) G m G n G rxn f, prod. f, react.
24 2 Fe 2+ (aq)+ HOCl (aq) + 5 H 2 O (l) 2 Fe(OH) 3(s) + 2 Cl - (aq)+ 5 H + (aq) 0 rxn ΔG 0 form (kcal/mol) Fe HOCl (aq) H 2 O (l) Fe(OH) 3(s) Cl - (aq) G ( 19.1) ΔG 0 rxn = kcal ΔG 0 rxn <0, so the reaction is spontaneous.
25 Our interest in an equilibrium system is often at non-standard conditions. So knowing Δ G is usually not very useful. It applies to the situation at which all reactants and products are at standard state, i.e. 1 M concentration or gases at 1 atm pressure. G m G n G rxn f, prod. f, react.
26 Water and steam H 2 O (l, 1 atm) H 2 O (g, 1 atm) G K = 0 kj/mol The system is at equilibrium at 1 atm (standard conditions) and at the boiling point temperature! 26
27 Water and steam H 2 O (l, 1 atm) H 2 O (g, 1 atm) G K = kj/mol The system is not at equilibrium at 1 atm (standard conditions) and at the room temperature! 27
28 Water and steam H 2 O (l, 1 atm) H 2 O (g, 1 atm) G K = kj/mol The forward process is non-spontaneous ( G > 0) so the reverse process is spontaneous and condensation occurs. 28
29 Water and steam H 2 O (l, atm) G K = 0 kj H 2 O (g, atm) The system is at equilibrium at atm (non-standard conditions) and at the room temperature! 29
30 Water and steam H 2 O (l, atm) H 2 O (g, atm) G K = 0 kj/mol Water CAN evaporate at room temperature, just not to give an equilibrium pressure of 1 atm! 30
31 G G RT ln[i] nonstandard state 0 i f,i
32 Now, let us use this information to determine the free energy change of the reaction in which calcite dissolves in acid to form calcium and bicarbonate ions at non-standard state. CaCO 3 + H + Ca 2+ + HCO 3 - Substance ΔG 0 f (kcal/mol) CaCO 3 Ca 2+ HCO kcal/mol kcal/mol kcal/mol
33 G [ G G G G ] rxn 2 f,hco3 f,ca f,caco3 f,h ( 2 ) 3 G G RT ln[hco ] G RT ln[ca ] rxn f,hco 3 ( ) 0 0 G RT ln[caco ] G RT ln[h ] f,ca f,caco3 3 f,h ( 0 0 ) ( 0 0 ) G G G G G rxn 2 f,hco3 f,ca f,caco3 f,h RT ln[hco ] RT ln[ca ] RT ln[caco ] RT ln[h ] 2 3 3
34 2 0 [HCO 3 ][Ca ] Grxn G rxn RT ln [CaCO 3][H ] rxn 0 rxn G G RT lnq where Q is the reaction quotient.
35 When the chemical system is at equilibrium, G rxn = 0 and Q = K eq 0 G RT lnk 0 rxn 0 rxn G RT lnk eq eq *Note that this equation applies to 25 o C since G 0 rxn is given for 25 o C.
36 This equation shows the relationship between the standard free energy change and the thermodynamic equilibrium constant. It is one of the most important in chemistry because it relates the equilibrium composition of a chemical reaction system to measurable physical properties of the reactants and products.
37 If you know the entropies and the enthalpies of formation of a set of substances, you can predict the equilibrium constant of any reaction involving these substances without the need to know anything about the mechanism of the reaction.
38 Table 3.1 Standard Free Energies and Enthalpies of Formation at 25 C Reference : Sawyer, McCarty, Parkin, Chemistry for Environmental Engineering and Science, 5 th Edition, McGraw-Hill, page 56 Substance State* G f, kj/mol H f, kj/mol Ca 2+ aq CaCO 3 s CaF 2 s Ca(OH) 2 s CaSO 4 s CH 4 g CH 3 CH 3 g CH 3 COOH aq CH 3 COO - aq CH 3 CH 2 OH l C 6 H 12 O 6 aq *aq=aqueous, s=solid, l=liquid Note: 1 kcal = kj
39 Substance State* G f, kj/mol H f, kj/mol Cl 2 g 0 0 Cl 2 aq Cl - aq CO 2 g CO 2-3 aq F - aq Fe 2+ aq Fe 3+ aq HCO - 3 aq H 2 CO 3 aq *aq=aqueous, s=solid, l=liquid Note: 1 kcal = kj
40 Substance State* G f, kj/mol H f, kj/mol H 2 O l H 2 O g HS - aq H 2 S g H 2 S aq H 2 SO 4 l Mg 2+ aq Mg(OH) 2 s Na + aq *aq=aqueous, s=solid, l=liquid Note: 1 kcal = kj
41 Substance State* G f, kj/mol H f, kj/mol NH 3 g NH 3 aq NH + 4 aq NO - 2 aq NO - 3 aq O 2 g 0 0 O 2 aq OH - aq S 2- aq SO - 4 aq Zn 2+ aq ZnS s *aq=aqueous, s=solid, l=liquid Note: 1 kcal = kj
42 Ex. Using the given data, calculate the equilibrium constant for the reaction H 2 O (l) H + (aq) + OH (aq) H + (aq) OH - (aq) H 2 O (l) ΔH 0 f, kj/mol S 0, J/K. mol
43 Solution: From the above data, we can evaluate the following quantities: products reactants H m H n H kj rxn f f products reactants S m S n S J / K rxn f f The value of ΔG 0 at 298K is ΔG 0 = ΔH 0 TΔS 0 =(55800) (298)( 80.8) = J/mol K exp / ( ) e eq ( )
44 Ex. The tap water in a town has the following characteristics. [HCO 3- ] =10-3 M [Ca 2+ ] = 10-3 M ph = 8.0 Will this water dissolve or precipitate CaCO 3? Which way the reaction will proceed at 25 0 C? CaCO 3(s) + H + Ca 2+ + HCO 3 -
45 Substance ΔG 0 f (kcal/mol) CaCO kcal/mol Ca 2+ HCO kcal/mol kcal/mol R= kcal/mol.k 0 G rxn kcal kcal = ( kcal/mol.k) (298 K) (lnk eq ) lnk eq = 4.59 K eq = 98.5 at 298 K or 25 o C
46 Q 2 [HCO 3 ][Ca ] [CaCO 3(s) ][H ] Compare Q with K eq At ph = 8.0 Q= 100 Q > K eq only slightly 100 > 98.5 slightly precipitation At ph = 7.0 Q= 10 Q < K eq 10 < 98.5 no precipitation
47 Temperature Dependence of Equilibrium Constant In most cases the reactions of interest in process chemistry are conducted at temperatures other than 25 o C. Therefore, it may be necessary to adjust equilibrium constants for nonstandard state temperatures. The standard-state free energy of reaction is a measure of how far the standard-state is from equilibrium.
48 0 G RT lnk eq But the magnitude of depends on the temperature of the reaction. ΔG 0 ΔH 0 TΔS 0 As a result, the equilibrium constant must depend on the temperature of the reaction.
49 Combining Eq n.s 1 and 2 we obtain RT lnk ΔH eq TΔS 0 0 and lnk ΔH RT ΔS R 0 0 Remember that the enthalpy and entropy changes are regarded as constants with respect to temperature, so the equilibrium constant is exponentially dependent on temperature.
50 We often want to know how a change in temperature will affect the value of an equilibrium constant whose value is known at some fixed temperature. Suppose that the equilibrium constant has the value K 1 at temperature T 1 and we wish to estimate K 2 at temperature T 2. H S lnk 1 RT R and H S lnk 2 RT R 2 0 0
51 Subtracting ln K 2 from ln K 1 we obtain 0 0 H H lnk 1 lnk 2 RT1 RT2 which is most conveniently expressed as K H 1 1 ln 1 K R T T Above equation is called the van't Hoff equation. 0
52 Ex. Plot the ln K eq versus 1/T graph of H 2 O (l) H + (aq) + OH (aq) ΔH 0 rxn = kj/mol t ( o C) T ( o K) 1/T ln K eq = K eq = , ,0 4,6E , ,2 1,0E , ,5 2,1E , ,8 4,1E , ,2 7,8E-14
54 Note that the reaction is endothermic (ΔH 0 rxn = kj/mol ), and K eq increases with temperature. For exothermic reactions however, K eq decreases with temperature. Change Increase temperature Decrease temperature Exothermic Reaction K eq decreases K eq increases Endothermic Reaction K eq increases K eq decreases
56 Ex. A municipal water supply enters a residence at 15 o C and is heated to 60 o C in the home water heater. If the water is just saturated with respect to CaCO 3(s) at 25 o C, what will be the condition of water (i.e. oversaturated or undersaturated) with respect to CaCO 3(s) i. as it enters the residence and ii. as it leaves the water heater? CaCO 3(s) + H + Ca 2+ + HCO 3 -
57 Substance CaCO 3(s) Ca 2+ HCO 3 - ΔH 0 f (kcal/mol) kcal/mol kcal/mol kcal/mol Using Eq n 2, we have already calculated that K eq is 98.8 at 25 o C. Now, using the van t Hoff equation we can calculate the values of K eq at 15 o C and 60 o C. Then, we can compare K 15 and K 60 with the reaction quotient Q and decide whether the water is undersaturated or oversaturated.
58 If Q > K eq water is oversaturated. If Q < K eq water is undersaturated. 0 K1 H 1 1 ln K2 R T1 T2 products reactants H H H rxn f, f, H KCal / mol 0 rxn
59 K 15 = K 60 = K 15 H 1 1 ln K25 R K ln 3 0 K 60 H 1 1 ln K25 R K ln 3
60 15 o C 25 o C 60 o C K eq What is the value of Q? Q 2 [HCO 3 ][Ca ] [H ]
61 Note that water is saturated with respect to CaCO 3(s) at 25 o C. Therefore, Q is [HCO 3 ][Ca ] [H ] At 15 o C Q < K eq (98.8 < ) At 60 o C Q > K eq (98.8 >30.2 ) water is undersaturated. water is oversaturated.