Download presentation

Presentation is loading. Please wait.

Published byDulce Dibble Modified over 2 years ago

1
1 Chapter 6 Chemical Equilibrium

2
2 Spontaneous Chemical Reactions The Gibbs Energy Minimum Consider the simple equilibrium reaction: A B The equilibrium concentrations (or pressures) will be at the extent of reaction at which the Gibbs function of the system will be at a minimum. The equilibrium may lie: (1)Close to pure A: The reaction "doesn't go" (2)Close to pure B: The reaction "proceeds to completion" (3)At a point where there are finite concentrations of both A and B (equilbrium) Extent of Reaction ( ): This parameter is a measure of how far the reaction has proceeded from reactants towards products. = 0: Reactants only = 1: Products only

3
3 The change in may be related to the change in the number of moles of 'reactants and products. For the simple reaction: A B dn A = -d and dn B = +d or n A = - and n B = + Let's say the stoichiometry is different; i.e. A 2 B In this case: dn A = -d and dn B = +2d or n A = - and n B = + 2 Back to: A B We showed in Chapter 3 that for a system with two species, the infinitesimal change, dG is given by: For processes at constant T and p (e.g. many reactions), this reduces to:

4
4 For the simple reaction: A B dn A = -d and dn B = +d From this, one gets: Implications: Note: r G is a function of the mixture's composition (relative amounts of A and B)

5
5 If: r G < 0, the forward reaction is spontaneous r G > 0, the reverse reaction is spontaneous r G = 0, the reaction is at equilibrium The reaction is exergonic The reaction is endergonic

6
6 The Description of Equilibrium Let's first reconsider the simple equilibrium, A B, and assume that A and B are both Perfect Gases. We learned in Chapter 5 that for a mixture of Perfect Gases, the chemical potential of each component is given by: p o is the reference state, 1 bar We then have: Therefore: where and

7
7 Therefore: where and Equilibrium At equilibrium, r G = 0. The equilibrium constant is: We have developed the expression, for the simplest of equilibria. However, as we'll see, one gets the same expression for more complex equilibria.

8
8 The General Case Here, we will: (1) Use activities, rather than pressures (2) Consider a more general reaction aA + bB cC + dD A = A o + RTln(a A ) B = B o + RTln(a B ) r G = {c C + d D } - {a A + b B } r G = r G o + RTln(Q) C = C o + RTln(a C ) D = D o + RTln(a D ) r G o = {c C o + d D o } - {a A o + b B o } It can be shown that:

9
9 r G o = {c C o + d D o } - {a A o + b B o } Standard Gibbs Energy Change ( r G o ) r G o is the Gibbs Energy change when reactants and products are in the standard state. aA + bB cC + dD Gas Phase: 1 bar in the standard state Solution: 1 M aA + bB cC + dD r G = r G o + RTln(Q)

10
10 aA + bB cC + dD r G = r G o + RTln(Q) Reaction Quotient (Q) Gas Phase Solution Ref. State: P o = 1 bar Ref. State: c o = 1 M

11
11 aA + bB cC + dD r G = r G o + RTln(Q) Reaction Quotient (Q) Standard State: All a J =1 Q = 1 r G = r G o Reactants Only:Q = 0 r G = - a C = 0 a D = 0 Products Only: Q = + r G = + a A = 0 a B = 0

12
12 Text Notation aA + bB cC + dD r G = r G o + RTln(Q) Reaction: J is a chemical component (rct. or prod.) j is the stoichiometric coefficient j < 0 for reactant j > 0 for product Gibbs Energy: where Equilibrium: where

13
13 An Illustrative Example r G = r G o + RTln(Q) r G o = +4430 J T = 298 K A(g) B(g) + C(g) Consider: Calculation of K

14
14 Standard State: P A =1 bar P B =P C =1 bar Q = 1 r G = r G o = +4430 J Reactant Only: Q = 0 r G = - Products Only: Q = + r G = + P A =2 bar P B =P C =0 Reaction proceeds to Left Reaction proceeds to Right P A = 0 P B =P C =2 bar Reaction proceeds to Left Reactants and Products: P A =1.5 bar P B =P C =0.5 bar Q = 0.167 r G = 0 Reaction at Equilibrium r G = r G o + RTln(Q) r G under various conditions

15
15 Some Equilibrium Calculation Examples 2 NO 2 (g) N 2 O 4 (g) (a)The Gibbs Energies of formation of NO 2 and N 2 O 4 are 51.3 kJ/mol and 97.9 kJ/mol, respectively. Calculate the equilibrium constant, K, at 25 o C G o = -4.7 kJ = -4700 J K = 6.7 (b)If the initial pressures of NO 2 and N 2 O 4 are both 10 4 Pa (= 0.1 bar), calculate the direction of spontaneity under these conditions. G = -4700 J + 5700 J = +1000 J: Spontaneous to Left (c) Calculate the pressures of NO 2 and N 2 O 4 at equilibrium. 26.8x 2 + 3.68x -0.033 = 0 x = 0.00845 p(NO 2 ) = 0.1169 p(N2O4) = 0.09155 Check: K = 6.70

16
16 Some Equilibrium Calculation Examples H 2 O(g) H 2 (g) + (1/2) O 2 (g) (a) G o for the dissociation of water vapor at 2300 K is +118.1 kJ/mol Calculate K for this reaction. (b)The fraction dissociation,, is defined as the fraction of molecules which which have dissociated at equilibrium; i.e. = 1 - n eq /n, where n is the initial amount of reactant prior to dissociation, and n eq is the amount of reactant present at equilibrium. Calculate for H 2 O gas at equilibrium at 2300 k and a total pressure of 1 bar. You may assume that << 1. K = 2.07x10 -3

17
17 Some Equilibrium Calculation Examples H 2 O(g) H 2 (g) + (1/2) O 2 (g) (b)The fraction dissociation,, is defined as the fraction of molecules which have dissociated at equilibrium; i.e. = 1 - n eq /n, where n is the initial amount of reactant prior to dissociation, and n eq is the amount of reactant present at equilibrium. Calculate for H 2 O gas at equilibrium at 2300 k and a total pressure of 1 bar. You may assume that << 1. K = 2.07x10-3 Strategy: 1. Express number of moles of reactants and products in terms of. 2. Determine mole fraction of each component. 3. Use Dalton's law to determine partial pressures of the components. 4. Calculate from the equilibrium expression. = 0.0205 0.021

18
18 Some Equilibrium Calculation Examples A(g) 2 B(g) The equilibrium constant for the gas phase dissociation above is: K = 2.0 If one introduces pure A(g) into a vessel, calculate the fraction dissociation,, and the partial pressures of A(g) and B(g) at a total pressure of 5. bar. Note: You may NOT assume that << 1. Strategy: 1. Express number of moles of reactants and products in terms of. 2. Determine mole fraction of each component. 3. Use Dalton's law to determine partial pressures of the components. 4. Calculate from the equilibrium expression.

19
19 Some Equilibrium Calculation Examples A(g) 2 B(g) The equilibrium constant for the gas phase dissociation above is: K = 2.0 If one introduces pure A(g) into a vessel, calculate the fraction dissociation,, and the partial pressures of A(g) and B(g) at a total pressure of 5. bar. Note: You may NOT assume that << 1.

20
20 Some Equilibrium Calculation Examples A(g) B(g) + C(g) HOMEWORK: The equilibrium constant for the gas phase dissociation above is: K = 2.0 If one introduces pure A(g) into a vessel, calculate the fraction dissociation,, and the partial pressures of A(g), B(g) and C(g) at a total pressure of 5. bar. Note: You may NOT assume that << 1.

21
21 Some Equilibrium Calculation Examples For the above gas phase reaction at 25 o C, it is found that if one mixes 1.0 mol A, 4.0 mol B and 3.0 mol D in a vessel, and the reaction is allowed to come to equilibrium, the mixture contains 0.60 mol C at a total pressure of 2.0 bar. Calculate the following quantities for this equilibrium: (A) the mole fraction of each species n(A) = 0.40 mol x(A) = 0.054 n(B) = 2.20 mol x(B) = 0.297 n(C) = 0.60 mol x(C) = 0.081 n(D) = 4.20 mol x(D) = 0.568 n(tot) = 7.40 mol A(g) + 3 B(g) C(g) + 2 D(g) Example: (Similar to Text Exer. 6.8a). Consider the gas phase equilibrium:

22
22 Some Equilibrium Calculation Examples For the above gas phase reaction at 25 o C, it is found that if one mixes 1.0 mol A, 4.0 mol B and 3.0 mol D in a vessel, and the reaction is allowed to come to equilibrium, the mixture contains 0.60 mol C at a total pressure of 2.0 bar. Calculate the following quantities for this equilibrium: A(g) + 3 B(g) C(g) + 2 D(g) Example: (Similar to Text Exer. 6.8a). Consider the gas phase equilibrium: (B) the equilibrium constant, K x(A) = 0.054 x(B) = 0.297 x(C) = 0.081 x(D) = 0.568 p(A) = x(A) p = 0.108 bar p(B) = x(B) p = 0.594 bar p(C) = x(C) p = 0.162 bar p(D) = x(D) p = 1.136 bar

23
23 Some Equilibrium Calculation Examples For the above gas phase reaction at 25 o C, it is found that if one mixes 1.0 mol A, 4.0 mol B and 3.0 mol D in a vessel, and the reaction is allowed to come to equilibrium, the mixture contains 0.60 mol C at a total pressure of 2.0 bar. Calculate the following quantities for this equilibrium: A(g) + 3 B(g) C(g) + 2 D(g) Example: (Similar to Text Exer. 6.8a). Consider the gas phase equilibrium: (C) r G o K = 9.24

24
24 The Response of Equilibrium to Pressure Condensed Phase Reactions: The external pressure has no effect. Gas Phase Reactions: Upon increase in the external pressure, the equilibrium will shift in the direction of fewer moles of gas. However, the equilibrium constant is unchanged!! In the slides below, we will: 1. Present a quantitative treatment of the effect of pressure on amount of reactants and products. 2. Present some qualitative examples The above trend for gas phase reactions is an example of LeChatelier's Principle.

25
25 The Response of Equilibrium to Pressure Quantitative Treatment A(g) 2 B(g) Among the previous examples, we analyzed the dissociation equilibrium: We found that the equilibrium constant is a function of (a)the fraction dissociation, (b)the total pressure, p This equation was solved for :

26
26 A(g) 2 B(g) Let's apply this to the gas phase equilibrium, N 2 O 4 (g) 2 NO 2 (g) K = 0.146 at 25 o C p 0.1 bar 0.52 0.5 0.26 1.0 0.19 10.0 0.06 Notice that the equilibrium moves to the left (i.e. the fraction dissociation decreases) with an increase in the total pressure. This is consistent with LeChatelier's Principle.

27
27 A(g) 2 B(g) Different K values The graph above shows that the same trend holds for different values of the equilibrium constant, K.

28
28 The Response of Equilibrium to Pressure Qualitative Examples Consider the equilibrium: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) What happens to the equilibrium when: (a) The volume is decreased: (b) NO 2 (g) is added at constant volume: (c) NO 2 (g) is added at constant total pressure: The equlibrium is shifted to the right There is no effect on the equilibrium The equlibrium is shifted to the left Note: In all cases, the equilibrium constant, K, is unchanged

29
29 The Response of Equilibrium to Pressure Qualitative Examples Consider the equilibrium: H 2 (g) + Br 2 (g) 2 HBr(g) What happens to the equilibrium when: (a) The volume is decreased: (b) Cl 2 (g) is added at constant volume: (c) Cl 2 (g) is added at constant total pressure: There is no effect on the equilibrum There is no effect on the equilibrium Note: In all cases, the equilibrium constant, K, is unchanged

30
30 The Response of Equilibrium to Pressure Qualitative Examples Consider the equilibrium: H 2 (g) + I 2 (s) 2 HI(g) What happens to the equilibrium when: (a) The volume is decreased: (b) Cl 2 (g) is added at constant volume: (c) Cl 2 (g) is added at constant total pressure: The equilibrium is shifted to the left There is no effect on the equilibrium The equilibrium is shifted to the right Note: In all cases, the equilibrium constant, K, is unchanged

31
31 The Response of Equilibrium to Temperature Le Chateliers Principle The equilibrium constant, K, and hence the ratio of products to reactants, shifts in the endothermic direction as the as the temperature is increased N 2 O 4 (g) 2 NO 2 (g) r H o = +57.2 kJ With rising temperature, K increases and the ratio of NO 2 to N 2 O 4 increases; i.e. equilibrium shifts towards right. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) r H o = -92.0 kJ With rising temperature, K decreases and the ratio of NH 3 to N 2 /H 2 decreases; i.e. equilibrium shifts towards left. Qualitative Considerations

32
32 The Response of Equilibrium to Temperature Quantitative Treatment: The van't Hoff Equation In Chapter 3, when discussing the effect of temperature on the Gibbs Energy, we derived the formula: We also have: Gibbs Helmholtz Equation van't Hoff Equation Therefore: Which yields:

33
33 Alternate form of the van't Hoff Equation Consider that: Now: Therefore: or: Alternate form

34
34 ln(K) 1/T Slope = - r H o /R The second form of the van't Hoff equation illustrates that if one plots ln(K) vs. 1/T, the tangent at a given point can be used to calculate r H o at that temperature.

35
35 Application: Calculation of K at a second temperature For a given reaction, A B, the equilibrium constant is 0.05 at 25 o C. r H o for the reaction is given by: a = 125 kJ/mol and b = 2x10 4 kJ-K/mol Calculate the value of the equilibrium constant at 75 o C K = 2600 at 75 o C

36
36 HOMEWORK For a given reaction, C D, the equilibrium constant is 200. at 25 o C. r H o for the reaction is given by: a = -80 kJ/mol and b = 0.12 kJ/mol-K Calculate the value of the equilibrium constant at 75 o C K = 18 at 75 o C

37
37 Integrated van't Hoff Equation with constant r H o This integrates to: Identification of the Integration Constant, C In a plot of lnK vs. 1/T, (a) the slope is - r H/R (b) the intercept is + r S/R

38
38 If one measures the equilibrium constant at two temperatures, K 1 at T 1 and K 2 at T 2, the data can be used to determine r H and r S and Subtraction of the second equation from the first equation (to eliminate r S) yields: This equation can be used to determine r H and then either of the first two equations can be used to calculate r S.

39
39 Application: The Dissociation of N 2 O 4 (g) For the equilibrium reaction, N 2 O 4 (g) 2 NO 2 (g), the equilibrium constant is 0.0176 at 0 o C and 15.05 at 100 o C. (a) Calculate r H and r S for this reaction. (b) Calculate the equilibrium constant at 25 o C (c)Calculate the temperature, in o C, at which the equilibrium constant is K = 100. r H = +57.1 kJ/mol r S = + 176 J/mol-K K = 0.146 T = 143 o C

40
40 Equilibrium Electrochemistry Redox Reactions and Half-Reactions Oxidation-Reduction (Redox) reactions are important in many areas of Chemistry. One particularly useful application is to harness spontaneous redox reactions to provide electric energy in an electrochemical cell. Consider the reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq)+ Cu(s) This can be split into: Zn(s) Zn 2+ (aq)+ 2 e - : Oxidation Half-Reaction Cu 2+ (aq) + 2 e - Cu(s): Reduction Half-Reaction A useful pneumonic is: OIL: Oxidation Is Loss (of electrons) RIG: Reduction Is Gain (of electrons)

41
41 e-e- e-e- Oxidation: Zn Zn 2+ + 2 e - Reduction: Cu 2+ + 2 e - Cu K 2 SO 4 "Salt Bridge": Flow of K + and SO 4 2- into the half-cells keeps the two halves electrically neutral. Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) Electrochemical Cell

42
42 Anode Oxidation Negative (-) Charge Anions flow towards anode Electrons leave anode Mneumonic: Both A and O are vowels Cathode Reduction Positive (+) Charge Cations flow towards cathode Electrons enter cathode Mneumonic: Both C and R are consonants Zn Zn 2+ + 2 e - Cu 2+ + 2 e - Cu

43
43 Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu (s) Current flows from anode to cathode. | = phase boundary || = salt bridge Details (e.g. concentration) are listed after each species. anode cell cathode cell Electrochemical Cells: Compact Notation Current Flow Write the half-cell reactions and balanced redox equation for the reaction characterized by: Cu(s)|Cu 2+ (aq)||Fe 3+ (aq)|Fe(s) Anode (Oxid): Cu(s) Cu 2+ (aq) + 2 e - Cathode (Red): Fe 3+ (aq) + 3 e - Fe(s) Overall Rxn: 3 Cu(s) + 2 Fe 3+ (aq) 3 Cu 2+ (aq) + 2 Fe(s)

44
44 Standard Cell Potential The voltage of an electrochemical cell varies with the conditions (i.e. state of reactants and products, concentrations, etc.) The cell potential has a standard voltage (E o or E o cell ) if: Solute concentrations are 1 M Gases have a pressure of 1 bar Solids are pure Sign of the Cell Potential (by convention) If a reaction is Spontaneous, E o cell > 0 If a reaction is Non-Spontaneous, E o cell < 0 Note: Absolute Cell Potentials of a half-cell cannot be measured. They are measured relative to a standard, the Standard Hydrogen Electrode The Cell Potential

45
45 Standard Hydrogen Electrode Hydrogen gas at 1 bar, H 2 (g, 1 bar), is bubbled over a platinum electrode immersed in 1 M aqueous acid solution. The cell potential for this half-reaction, is defined as 0 (for either oxidation or reduction). Reduction: 2 H 3 O + (aq, 1 M) + 2 e - H 2 (g, 1 bar) + 2 H 2 O(l) E o (red) = 0 Oxidation: H 2 (g, 1 bar) + 2 H 2 O(l) 2 H 3 O + (aq, 1 M) + 2 e - E o (oxid) = 0 The reduction potentials of other half-cell reactions can be determined with this convention, by measuring the voltage of a cell containing the standard hydrogen electrode (or indirectly) Using the above convention, extensive tables of reduction potentials for many species have been derived. A partial list is given on the next slide.

46
46 Reduction Half-Cell Potentials (Partial List) Easier to reduce than H 3 O + Harder to reduce than H 3 O + F 2 (g) + 2 e - 2 F - (aq) +2.87 H 2 O 2 (aq) + 2 H 3 O + 2 e - 4 H 2 O( l )+1.77 MnO 4 - (aq)+8 H 3 O + + 5 e - Mn 2+ (aq) + 12 H 2 O( l )+1.51 C l 2 (g) + 2 e - 2 C l - (aq) +1.36 Br 2 (g) + 2 e - 2 Br - (aq) +1.07 Ag + (aq) + e - Ag(s) +0.80 Cu 2+ (aq) + 2 e - Cu(s) +0.34 2 H 3 O + (aq) + 2 e - H 2 (g) + 2 H 2 O( l ) 0.00 Ni 2+ (aq) + 2 e - Ni(s) -0.25 Fe 2+ (aq) + 2 e - Fe(s) -0.44 Zn 2+ (aq) + 2 e - Zn(s)-0.76 A l 3+ (aq) + 3 e - A l (s)-1.66 Li + (aq) + e - Li(s)-3.05

47
47 Consider the reaction: Cu 2+ + Zn Cu + Zn 2+ (I am leaving out the "aq" and "s" for convenience) The two half-cell reactions are: Reduction (Cathode): Cu 2+ + 2 e - Cu E o Red (Cu 2+ )= +0.34 V Oxidation (Anode): Zn Zn 2+ + 2 e - E o Oxid (Zn) = -E o Red (Zn 2+ ) = +0.76 V Therefore, the overall cell potential is: E o cell = E o Red (Cu 2+ ) + E o Oxid (Zn) = +0.34 + 0.76 = +1.10 V Equivalently, one could write: E o cell = E o Red (Cu 2+ ) - E o Red (Zn 2+ ) = +0.34 - (-0.76) = +1.10 V Because E o cell > 0, this reaction is spontaneous. Using Standard Reduction Potentials

48
48 In general, one can calculate E o cell as either: E o cell = E o Red (Cathode) + E o Oxid (Anode) or E o cell = E o Red (Cathode) - E o Red (Anode) The two half-cell reactions are: Reduction (Cathode): Cu 2+ + 2 e - Cu E o Red (Cu 2+ )= +0.34 V Oxidation (Anode): Zn Zn 2+ + 2 e - E o Oxid (Zn) = -E o Red (Zn 2+ ) = +0.76 V Therefore, the overall cell potential is: E o cell = E o Red (Cu 2+ ) + E o Oxid (Zn) = +0.34 + 0.76 = +1.10 V Equivalently, one could write: E o cell = E o Red (Cu 2+ ) - E o Red (Zn 2+ ) = +0.34 - (-0.76) = +1.10 V

49
49 The standard reduction potentials of Sn 2+ and Al 3+ are -0.14 V and -1.66 V, respectively. Write the balanced redox equation and determine the cell potential for the reaction, Al(s)|Al 3+ (aq)||Sn 2+ (aq)|Sn(s) 2 Al(s) + 3 Sn 2+ (aq) 2 Al 3+ (aq) + 3 Sn(s) E o cell = +1.52 V Note that even though the Al half-reaction was multiplied by 2 and the Sn 2+ reaction was multiplied by 3, the cell potentials were NOT multiplied by any factor. Review Example How many electrons are transfered in this reaction? 6 electrons

50
50 The Nernst Equation It can be shown that the Gibbs Energy change, r G, for a reaction is related to the cell potential, E cell, by the equation: r G = -nFE cell F is Faraday's Constant. This is the charge, in Coulombs (C) of one mole of electrons. F = 96,485 C/mol 96,500 C/mol n is the number of electrons transfered in the reaction.* * The text uses the symbol,, to represent the number of transfered electrons.

51
51 We learned earlier in this chapter that the Gibbs Energy change for a reaction depends upon the reactant and product concentrations (or pressures for gases) and can be determined from the equation: r G o is the Standard Gibbs Energy change (concentrations = 1 M), and Q is the reaction quotient. It is straightforward to use the above equation, with the relations r G = -nFE cell and r G o = -nFE o cell to derive the following relationship between cell potential and concentrations: which yields: Nernst Equation

52
52 Nernst Equation Alternate Form In classical Analytical texts, it is common to rewrite the Nernst Equation at the specific temperature of 25 o C (=298.15 K) in terms of base 10 logarithms [ln(x) = 2.303 log(x)]. In this case, the equation is commonly written: I'll use the more general form (which allows for variable temperature).

53
53 Example The standard reduction potentials of Al 3+ and Zn 2+ are -1.66 V and -0.76 V, respectively. Consider the electrochemical cell, Al(s)|Al 3+ (5.0 M)||Zn 2+ (0.02 M)|Zn(s) (A)Write the oxidation and reduction half-reactions + the balanced overall reaction. Oxidation: Al(s) Al 3+ (5.0 M) + 3 e - Reduction: Zn 2+ (0.02 M) + 2 e - Zn(s) Overall: 3 Zn 2+ (0.02 M) + 2 Al(s) 3 Zn(s) + 2 Al 3+ (5.0 M) (B) Determine the Standard Cell Potential, E o cell.

54
54 Example The standard reduction potentials of Al 3+ and Zn 2+ are -1.66 V and -0.76 V, respectively. Consider the electrochemical cell, Al(s)|Al 3+ (5.0 M)||Zn 2+ (0.02 M)|Zn(s) Overall: 3 Zn 2+ (0.02 M) + 2 Al(s) 3 Zn(s) + 2 Al 3+ (5.0 M) (C) Determine the Cell Potential at 25 o C under the given conditions E o cell = 0.90 V R = 8.31 J/mol-K = 8.31 C-V/mol-K T = 298 K F = 96,500 C/mol Note: You get the same answer if you use the alternate form of the Nernst Equation:

55
55 Cell Potential and the Equilibrium Constant One interesting application of electrochemical cell potentials is to calculate the equilibrium constant for a reaction. Remember from earlier in the chapter that, at equlibrium: Because the reaction Gibbs Energy change is related to the cell potential, r G = -nFE cell, one also has that, at equilibrium:

56
56 Example Determine the equilibrium constant, at 25 o C for the reaction: Fe(s) + Cd 2+ (aq) Fe 2+ (aq) + Cd(s) R = 8.31 J/mol-K = 8.31 C-V/mol-K T = 298 K F = 96,500 C/mol The standard reduction potentials of Fe 2+ and Cd 2+ are -0.44 V and -0.40 V, respectively.

57
57 The Determination of Thermodynamic Functions Gibbs Energy Change ( r G o ) One can use standard cell potentials to calculate the standard Gibbs Energy change for a reaction from the formula: r G o = -nFE o cell As we see below, the dependence of cell potential on temperature can be used to determine the Entropy change, r S o, and Enthalpy change, r H o, of the reaction. Entropy Change ( r S o ) Remember that the total differential for dG is: This leads to:

58
58 Entropy Change ( r S o ) This leads to: For a reaction under standard conditions, one can write r G o = G o (Prod) - G o (Rct) and r S o = S o (Prod) - S o (Rct) Thus, for a reaction, the above equation can be rewritten as: Therefore, r S o for a reaction can be determined from the measured temperature dependence of the standard electrochemical cell potential.

59
59 Enthalpy Change ( r H o ) We have already seen that the Gibbs Energy change and the Entropy change for a reaction are related to the electrochemical cell potential by: and We recall from Chapter 3 that the relation between the Enthalpy, Entropy and Gibbs Energy changes for a reaction is: which yields: If we wish, we can plug in the expressions for r G o and r S o to get: Alternately, we can first calculate r G o and r S o and then insert the numerical values into the equation:

60
60 Consider the electrochemical cell reaction: AgBr(s) + ½ H 2 (g) Ag(s) + HBr(aq) The standard cell potential is temperature dependent and is given by: Example (similar to Text Example 6.5) a = -0.0884 V, b = +47.6 V-K, T = temperature in K n = 1 (A) r G o Calculate r G o, r S o and r H o for this reaction at 25 o C = 298 K F = 96,500 C/mol 1 C-V = 1 J

61
61 Consider the electrochemical cell reaction: AgBr(s) + ½ H 2 (g) Ag(s) + HBr(aq) The standard cell potential is temperature dependent and is given by: Example (similar to Text Example 6.5) a = -0.0884 V, b = +47.6 V-K, T = temperature in K n = 1 (B) r S o Calculate r G o, r S o and r H o for this reaction at 25 o C = 298 K F = 96,500 C/mol 1 C-V = 1 J

62
62 Consider the electrochemical cell reaction: AgBr(s) + ½ H 2 (g) Ag(s) + HBr(aq) The standard cell potential is temperature dependent and is given by: Example (similar to Text Example 6.5) a = -0.0884 V, b = +47.6 V-K, T = temperature in K n = 1 (C) r H o Calculate r G o, r S o and r H o for this reaction at 25 o C = 298 K F = 96,500 C/mol 1 C-V = 1 J r G o = -6880 J/mol r S o = -51.7 J/mol-K

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google