CHAPTER 6 - Chemical Equilibrium. b. composition when equilibrium is reached.

CHAPTER 6 - Chemical Equilibrium I. Thermodynamics of Chemical Reactions. A. Spontaneity of Chemical Reactions: 1. Thermodynamics can tell us: a. direction of spontaneous change. b. composition when equilibrium is reached. 2. In a mixture of potentially reactive chemical species, direction of spontaneous change is toward decreasing total G. (if T, p fixed) 3. Simple example for illustration: A! B Example: cis-butene trans-butene define ξ = extent of reaction (progress variable) ξ = 1 means n A = 0, n B = 1 mol ξ = 0 means n A = 1, n B = 0. dg = µ A dn A + µ B dn B = -µ A dξ + µ B dξ -dξ +dξ dg = (µ B - µ A )dξ or ( G/ ξ) p,t = µ B - µ A if µ B > µ A rx goes spontaneously B! A µ B < µ A rx goes spontaneously A! B 1

Equilibrium is reached when: or µ B = µ A ( G/ ξ) p,t = 0 4. The reaction Gibbs function, ΔG r : ΔG r = ( G/ ξ) p,t at some specified composition = slope of G at point along ξ curve = depends on composition (concentrations, gas pressures) = µ B - µ A in our previous A!B case = 0 at equilibrium = change in G when 1 mol A forms 1 mol of B at some specific fixed composition of the reaction mixture. However, if a vessel originally having n A = 1 and n B = 0 at standard state undergoes rxn until n B = 1 and n A = 0, the ΔG of that process is not equated to ΔG r. That ΔG is ΔG θ (standard Gibbs energy change). ΔG θ = µ B θ - µ A θ Extent of reaction, ξ 2

5. Exergonic vs endergonic reactions: if a mixture of A and B at a specified composition has: ΔG r < 0, will proceed spont. in A --> B (said to be exergonic) ΔG r > 0, will proceed spont. in A <-- B (said to be endergonic) = 0, equilibrium Note: an endergonic process can be made to happen spontaneously, but only by coupling it to another exergonic process. e.g. muscle contraction (endergonic) is coupled to ATP decomposition (exergonic). B. Predicting compositions at equilibrium: 1. Example: again consider A! B rxn, perfect gases. ΔG r = µ B - µ A Rearrange: ΔG r = (µ B θ + RT ln pb /p θ ) - (µ A θ + RT ln pa /p θ ) chem pot of mol pure B at 1 bar p θ = 1 bar θ θ ΔG r = (µ B - µa ) + RT ln pb /p A ΔG θ partial pressure quotient Q p ΔG r = ΔG θ + RT ln Q p Now, as rxn proceeds to equilibrium at fixed T, p: ΔG r 0 ΔG θ = - RT ln K p but ΔG θ is a constant Qp Kp = (p B /p A ) eq or equilibrium value of the quotient 3

2. For more general stoichiometries: A A + B B C C + D D ( G/ ξ) p,t = C µ C + D µ D - A µ A - B µ B partial pressure quotient becomes (where p s are in bar) C p D D Q p = p C p A A p B Note: to keep notation simple B have written p α /p α θ as p for each species Just remember, Q and K are actually dimensionless. Equilibrium constant is then # K p = p C C p D & D % p A $ A p B ( B ' equilibrium 3. Considering real gases, replace the partial pressure by activities: Q p = a C C a D D a A A a B B Remember activity a = f/p θ f = fugacity 4. Now, let s do a problem: Calculate K p, equilibrium constant, for gas phase rxn at 298.15K. H 2 (g) + Cl 2 (g)! 2 HCl(g) ΔG θ = - RT ln K p Find this! Use Table 2.8 ΔG f θ ΔG θ θ θ θ = 2 ΔG f (HCl(g)) - ΔGf (H2 (g) - ΔG f (Cl2 (g) -95.30kJ/mol 0 by def 0 by def 2 x (-95.30) kj/mol = -R x 298.15 ln K p K p = e ΔGθ / RT = e ( 2 95.30kJ / 8.314 10 3 298.15) K p = e +76.9 = 2.47 x 10 33 (very large) Reaction will go to completion. 4

5. Practical expressions for equilibrium constants: In general, Q and K = quotient of activities Factor out activity coefficients: Solution reactions, use a α = γ α m α /m θ " Q = m C m D %" $ C D ' γ C C γ D % $ D ' $ m A A m B ' $ # B & γ A A γ B ' # B & Q = Q m Q γ and therefore: " K = m C m D % $ C D ' $ m A A m B ' # B & K = K m K γ Often Kγ 1 so K K m equilib _ mix " γ C C γ D % $ D ' $ γ A A γ B ' # B & molality 1 molal equilib _ mix II. Response of Equilibria to Conditions. (response to changes in p or T) A. Responses to pressure changes. 1. How K p changes with pressure. It doesn t! ΔG θ = - RT ln K p general expression stnd K p independent Gibbs energy of pressure 2. How equilibrium compositions change with pressure. They do! Consider A(g)!B(g) in equilibrium Increase pressure (e.g., by compressing the system). No change. Special case in which number of moles of gas on both sides are equal. Δn g = 0. Example: H 2 (g) + Cl 2 (g)! 2 HCl(g) No effect of pressure. Δn g = 0. However: H 2 (g) + (1/2)O 2 (g)! H 2 O(g) Δn g = 1-3/2 = -(1/2) 5

So here, an increase applied pressure will favor forward rxn, since this will reduce moles of gas. ~Le Chatelier s principle When a system at equilibrium is disturbed, it responds in a way that tends to minimize the effect of the disturbance. 3. Let s prove this quantitatively for this simple case. A(g)! 2 B(g) K p = (p B /p θ ) 2 /(p A /p θ ) any equilib mixture Let p B = x B p where p = total pressure K p = x 2 B /x A * (p/p θ ) 2-1 Initially had (in moles): n of A; 0 of B Establish equilibrium: A is n(1 - α); B is n2α x A = n(1 - α)/[n(1 - α) + n2α] = (1 - α)/(1 + α) x B = n2α/[n(1 - α) + n2α] = 2α/(1+ α) K p = (p/p θ ) 4α 2 /(1 - α 2 ) Solve for α,you ll get: 1 / 2 # & 1 α = % ( % $ 1 + 4( p / p ) o / K ( p ' α is a function of pressure, so concentrations will change as system is compressed or expanded and new equilibrium is established. 6

4. But now try as different case: A(g)! B(g) initial n 0 equilib n(1 - α) nα K p = (p B /p o )/(p A /p o ) = x B p/x A p = x B /x A x A = n(1 - α)/n = 1 - α x B = nα/n = α K p = α/(1 - α) α = K p /(1+K p ) - does not depend on pressure, as expected! 5. K p in terms of mole fractions: " K p = p C p D % " $ C D ' $ p A A p B ' = x C x D %" $ C D ' (p / p o ) C (p / po ) D % $ ' $ # B & x A A x B ' $ # B & (p / p o ) A (p / po ) B ' # & = K χ (p / p o ) C + D A B K p = K χ (p / p o ) Δn g Δn g = change in gaseous moles Now if Δn g = (+) and we compress system, then pressure term increases So K x must (products reverting to reactants) so that K p constant. 6. Problem: Reaction 2NO 2 (g)! N 2 O 4 (g) K p = 4 at T = 298.15K Initially have n moles of NO 2 in volume V. What will equilib partial pressures be? At equilib: (p dimer /p θ )/(p mono /p θ ) 2 = K p = 4 7

Could also write: K p = (x dimer /x mono 2 )(p/p θ ) -1 = 4 2NO 2! N 2 O 4 n 0 init moles n(1-2α) nα equil moles x dimer = α/(1-2α + α) = α/(1 - α) x mono = (1-2α)/(1 - α) p = p dimer + p mono = (n dimer + n mono )RT/V p = (α + 1-2α)n RT/V = (1 - α)n RT/V = (1 - α)p init Use p θ = 1 bar; p init nrt/v K p = [α/(1-2α) 2 ] p init -1 Rearrange and solve for α = number moles N 2 O 4 formed when equilib. Will get a quadratic eqn. in α. α = (K p p init )(1-4α + 4α 2 ) 0 = K p p init - (4 K p p init + 1)α + 4 K p p init α 2 α = ( 4K p p init + 1) ± ( 4K p + 1 p init ) 2 ( 4K p p init ) 2 8K p p init Only (-) root is physically possible: α = 4K p p init + 1 ( ) 8K p p init + 1 8K p p init 8

B. The response of equilibrium to temperature. 1. Basic idea: If A! B + heat (exothermic) Then as T, K, equilibrium will shift back to reactants T K Therefore, K is a sensitive function of T. Exothermic reactions, increased T favors reactants Endothermic reactions, increased T favors products 2. The van t Hoff equation proves this: Start with ΔG θ = - RT ln K p lnk = ΔG o / RT dlnk dt = 1 R d(δg o / T) dt Now the derivative on the right was given by Gibbs-Helmholtz eqn. in Chapter 3 as -ΔH /T 2 so dlnk dt = ΔHo RT 2 Predicts that ln K varies with T according to the value of ΔH. Could also rewrite as: dlnk d(1 / T) = ΔHo R 3. Implications: K varies with T based on ΔH. K as T if ΔH > 0. (endothermic) K as T if ΔH < 0. (exothermic) When a reaction is endothermic and still spontaneous, it is entropy driven. 9

Entropy becomes a larger factor at higher T. Below: Shift in equilibrium population with T according to Boltzmann: Endothermic reaction Exothermic reaction At right, an entropy driven Endothermic reaction 10

4. Estimating K at T 2 from K at T 1 : 1 / T 2 $ Integrate: dlnk = (ΔH / R)d& 1 ' ) % T( 1 / T 1 lnk( T 2 ) lnk( T 1 ) ΔH R ΔH R 1 / T 2 % d' 1 ( * 1 / T 1 & T) % 1 1 ( ' * & T 2 T 1 ) (assumed here ΔH is T-independent) 5. Problem: Estimate the equilibrium constant for the total oxidation of CO to CO 2 at 1000K. Reaction is: CO(g) + (1/2) O 2 (g)! CO 2 (g) ΔG = - RT ln K p If we knew ΔG at 1000K, we can get K p. We don t. But using Table 2C.5 at T = 298.15K: ΔG 298 = ΔG f (CO 2 (g)) - ΔG f (CO(g)) - (1/2)ΔG f (O 2 (g)) = -394.36 - (-137.17) - (1/2)(0) = -257.2 kj/mol So: ln K p (298) = 257.2/(R x 298.15) = 103.76 Then: ln K p (1000) = ln K p (298) - ΔH $ 1 R 1000 1 ' & ) % 298( assume ΔH T-indep, use Hess s Law and Table 2C.5 11

ΔH = -393.51 - (-110.53) - (1/2)(0) = -282.98 kj/mol exothermic ( ) = lnk p ( 298) + 282.98 # ( 8.314 10 ) 3 1 % lnk p 1000 ln K p (1000) = 23.6 1000 1 & ( $ 298' = 103.76 80.12 K p (1000) = e 23.6 = 1.84 x 10 10 Whereas K p (298.15) = e 103.76 = 1.15 x 10 45 As expected, for exothermic reactions, K p as T. 6. For the gas phase dimerization reaction 2 NO 2 N 2 O 4 use Table 2C.5 to calculate K p at 298 and 600 K. ΔG rx ( 298) = 97.89 2 51.31 = 4.73 lnk p = ΔG RT = 1.908 K p = e " 4.73 % $ ' # RT & ' = 6.74 lnk p ( 600) = lnk 298 p ( ) ΔH R # 1 600 1 & % ( $ 298' ΔH = 9.16 2 33.18 = 57.2 kj mol lnk p ( 600) = 4.88 K p = 0.00759 12

III. Application to Equilibrium Electrochemistry. A. Electrochemical Cells: 1. General terms: cathode = reduction takes place anode = oxidation takes place galvanic cell = produces e - current, since rxn occurs spontaneously: ΔG r < 0, cell can do work electrolytic cell = driven by external source of e - current to drive rxn that is not spontaneous: ΔG r > 0, work done on cell. redox rxn = rxn in which there is transfer of e -. Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) -is gaining 2e- -is losing 2e- -is being reduced -is being oxidized -is the oxidizing -is the reducing agent (is causing agent (is causing the oxid of Zn) the redn of Cu 2+ ) 2. Half reactions: Split up overall rxn into 2 half-rxns: Cu 2+ (aq) + 2e - Cu(s) + Zn(s) Zn 2+ (aq) + 2e - 13

Or could write as difference of two reductions: Cu 2+ + 2e - Cu - [Zn 2+ + 2e - Zn] The cell for this redox couple is the Daniell cell: Notation: anode Zn(s) ZnSO 4 (aq) CuSO 4 (aq) Cu(s) cathode phase salt phase boundary bridge boundary 3. Other examples: AgCl(s)! Ag + (aq) + Cl - (aq) (governed by K sp ) Can be written as two half-reactions: AgCl(s) + e-! Ag(s) + Cl - (aq) Ag + (aq) + e-! Ag(s) (but in reverse) Implication: equilibrium constant K sp for a reaction that is overall not even a redox reaction can be determined by measuring cell potential in which these are the two half-cells. 4. Types of electrodes: 14

a. H 2 gas electrode - based on rxn, here written as reduction: 2H + (aq) + 2e - H 2 (g) SHE P H2 = 1 bar [H + ] = 1 molal Notation - if H + being reduced (cathode): H + (aq) H 2 (g) Pt If H 2 being oxidized (anode): Pt H 2 (g) H + (aq) b. Insoluble-salt electrode = metal M covered by pourous layer of salt MX. Example: AgCl half rx AgCl(s) + e - Ag(s) + Cl - (aq) Example: lead storage battery PbSO 4 (s) + 2e - Pb(s) + SO 4 2- (aq) c. Redox electrode = where both species exist in solution. 5. Types of cells: a. Simplest - both electrodes immersed in the same electrolyte solution. (no liquid junction) 15

Example: Hydrogen electrode & AgCl electrode both immersed in HCl(aq). Pt H 2 (g) HCl(aq) AgCl(s) Ag b. Daniell cell - electrode compartments separated and in different electrolytes. liquid junction potential virtually eliminated by salt bridge: Zn ZnSO 4 CuSO 4 Cu double lines indicate liquid junction potential eliminated by salt bridge 16

c. Electrolyte concentration cell - makes use of free energy difference between 2 solutions at different concentrations. Not a cell. ΔG r = maximum non-expansion work 6. Cell potentials = E (in volts) = electromotive force, emf = driving force If positive, cell is galvanic and can do work. The large E, the more work done per e - that passes. Measured when cell is operating infinitesimally slowly (i.e., balanced by external circuit such as the volt meter) - zero current emf 7. E ΔG r. E depends directly on rxn Gibbs energy ΔG r. 17

ΔG r = - FE negative means ΔG r negative (spont) given E positive = moles of e - in balanced rxn; = 2 for Cu, Zn rxn F = Faraday constant (96485 Coul/mol); E = emf Remember: ΔG r = ΔG + RT ln Q (quotient of activities) = 0 at equilibrium Q = K at equilibrium > 0 if reverse rx is spont. Q > K < 0 if forward rx is spont. Q < K 8. Nernst Equation - variation of emf with concentrations (or activities, to be precise). Start with: ΔG r = ΔG + RT ln Q replace: -FE = -FE + RT ln Q activity quotient at existing conc. emf under emf in hypothet state existing conc when activities of conditions reactants & products are all 1 Rearrange: E = E - (RT/F) ln Q RT/F = 25.7 mv (millivolts) at 298.15K (25 C) E = standard cell potential What happens at equilibrium: Q K So: RTln Q RTln K = -ΔG = +FE E 0 Could write, using equilibrium conditions: 0 = E - RT/F ln K ln K = FE /RT Thus can obtain K from cell potentials. 18

9. Concentration cell. Can now see use of concentration gradients to perform emf work: E = E - RT/F ln Q E = 0 Pt H 2 (g) H + (aq, low conc) H + (aq, high conc) H 2 Pt Write out: H 2 2H + (low) + 2e - 2H + (high) + 2e - H 2 Net rxn: 2H + (high) 2H + (low) Q = a(h + (low conc)) 2 /a(h + (high conc)) 2 = 2 Q < 1 ln Q = neg -RT/F ln Q = positive E = E + positive E > 0 = 0 Problem: assume [H + ] are sufficiently low that can replace activities by concentrations. Suppose [H + ] in high conc part of cell is twice that in low conc part. Calculate emf. E = 0 (RT/F) ln (1/2) 2 = 2; RT/F = 25.7mV = -(25.7/2) * 2 ln(1/2) E = 17.8mV 10. Problem - calculate E of cell from Nernst eqn., use DH theory to get ion activities, use ΔG f s to get E. Cell notation: Zn(s) ZnSO 4 (aq, 0.002molal) CuSO 4 (aq, 0.004 molal) Cu(s) Write out: Zn(s) ZnSO4(aq) + 2e - CuSO 4 (aq) + 2e - Cu(s) + SO 4 2- (aq) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) E = E - RT/F ln Q desired quantity 19

= 2 RT/F = 25.7mV Q = a(zn 2+ )/a(cu 2+ ) E = -ΔG /F ΔG = ΔG f (Zn 2+ (aq)) - ΔG f (Cu 2+ (aq)) = -212.7 kj/mol -147.1 +65.6 ΔG f of Cu(s) and Zn(s) = 0 E = -(-212.7)/(2 * 96.485 kcmol -1 ) = 1.102V (Joul=Coul-Volt) a(solute) = γm/m θ a(zn 2+, 0.002 molal) = γ± (ZnSO 4 )* 0.002m/1m a(cu 2+, 0.004 molal) = γ± (CuSO 4 )* 0.004m/1m DH theory: log γ ± = - Z+Z- A I = - (+2)(-2).509 I I = (1/2)Σ Z i 2 m i I in ZnSO 4 (0.002 molal) = (1/2){(+2) 2 0.002 + (-2) 2 0.002} Zn 2+ SO 4 2- = 0.008 I in CuSO 4 (0.004 molal) = (1/2){(+2) 2 0.004 + (-2) 2 0.004} = 0.016 ZnSO 4 CuSO 4 log γ ± = - (+2)(-2).509.008 = -0.182 γ ± = 10 -.182 = 0.657 log γ ± = - (+2)(-2).509.016 γ ± = 0.552 Q = a(zn 2+ )/a(cu 2+ ) = (0.657 * 0.002)/(0.552 * 0.004) Q = 0.595 E = 1.102 (0.0257V/2) ln 0.595 = 1.102 +.007 = 1.109 only 7 mv different than E 11. Tabulated Reduction Potentials. Consider reduction half-rxn: Ag + + 1e - Ag(s) Impossible to measure E of this rxn by itself. Need a rx couple. To tabulate E for series of species then, let s arbitrarily define: E = 0 at all T for the reaction: 2H + (aq) + 2e - H 2 (g) 20

i.e. electrode Pt H 2 (g) H + (aq) is called the Standard Hydrogen Electrode (SHE) H 2 (g) pressure = 1 bar Now if cell Pt H 2 (g) H + (aq) Ag + (aq) Ag(s) has E = +0.80V, then Ag + + e - Ag redn pot is E = +.80V since E (H 2 /H + ) = 0 In general, overall E of cell is: E (cell) = E (redn half-rxn) - E (ox half-rxn) 12. Now, let s examine Table 6D.1 in Appendix of Atkins - explain how to use. 13. Show how to predict spontaneous direction for a given redox couple. E rxn Cu 2+ + 2e - Cu 0.337 (-) (Zn 2+ + 2e - Zn) -(-0.763) Cu 2+ + Zn Cu + Zn 2+ E rx = 1.100 volts Pb 2+ + 2e - Pb -.126 (-) (Zn 2+ + 2e - Zn) -(-0.763) Pb 2+ + Zn Pb + Zn 2+ E = -0.126 -(-0.763) = 0.637 volts 14. Predicting K eq : Find E by combining half-rxns. Then ΔG = -FE number of e - transferred as the overall rxn is written Then ln K eq = -ΔG /RT = FE /RT 21

B. Application of Reduction Potentials. 1. Electrochemical Series: In list of Standard Reduction Potentials, the higher E is, the more easily reduced. (top of chart) Things lower on the table are more easily oxidized. E total = E reduced species - E oxidized species higher one on lower one on the chart the chart (both E are the listed reduction potentials) Arranging in order of reducability = electrochem series. (Atkins Table 6.2) 2. Solubility Products: salt MX(s)! M + (aq) + X - (aq) solub K sp = a(m + )a(x - ) product (a = 1 for the pure solid) If s = solubility = molality of M + or X - in saturated soln. And if γ± 1 (very dilute, low solub) K sp = (s/m θ ) 2 1/2 s = K sp m θ 1 mol/kg Determine from cell experiment: Problem: Find s of AgCl from cell potential data at 25 C. Strategy - find 2 half-reactions that, when combined, produce the reaction: E AgCl(s) + e - Ag(s) + Cl - (aq) +0.22V (-) (Ag + (aq) + e - Ag(s)) -(+.80V) AgCl(s) Ag + (aq) + Cl - (aq) E total = +0.22-0.80 = -.58V Now ΔG = -RT ln K sp of solub solub equil constant 22

-FE = -RT ln K sp ln K sp = FE /RT = [1 x 96,485 Coul/mol x (-0.58V)]/8.314J/mol K x 298.15 ln K sp = -22.6 K sp = e -22.6 = 1.6 x 10-10 s = K s m θ = 1.25 x 10-5 mol/kg 3. ph electrodes: Measure E of hydrogen electrode in the solution, with other electrode being the saturated calomel reference electrode. 4. Determination of thermodynamic functions from cell potentials (noncalorimetric thermochemistry) Most basic equations: ΔG = -FE ΔG = -FE ln K eq = FE /RT temp coeff of cell = ( E/ T) p but since ( G/ T) p = -S ( E/ T) p = ΔS/F = -( / T)(ΔG/F) or ΔS = F( E/ T) p Since: ΔG = ΔH - TΔS from E can from solve ( E/ T) for this 23

Notes: 24