15 & 16 - Chemical Equilibria

15 & 16 - Chemical Equilibria - Today Chemical...

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Unformatted text preview: Today Chemical Equilibrium Example of Acid/Base Buffer system Electrochemistry & Equilibrium (today or Thursday) Chemistry in Context Gibbs Free Energy & Chemical Equilibrium This is the equation that we want to get to aA bB cC dD G RT ln K K C c A a D d B b Reaction Coordinate Diagrams transition state H---I---I E Ea = energy of activation (not a state function) Ea reactants H + I2 E products HI + I Reaction Coordinate E = energy difference (usually G) between reactants and products (a state function) G = Go + Gstep1 + Gstep3 G G1 G3 P2 nRT ln P1 TS Pref RT ln PO2 2 RT ln PH 2O Pref Pref RT ln RT ln RT ln PH 2O Pref Pref 2 G1 2H2O (PH2O) G3 2 O2 (Pref) + 2H2 (Pref) G2 2H2O (Pref) 2 At equilibrium G=0 Pref PH 2 O2 (PO2) + 2H2 (PH2) 2 PH 2 PH 2O Gsteps1& 2 G PO2 Pref So Go = - Gsteps1&2 The response of an Equilibrium to a Perturbation (Le Chatelier Principle) A B (exothermic) Low Temperature High Temperature G0 G0 0 0 G0 G0 Q A A B e G0 RT b K eq B a A B Q 0 RT ln K eq G H TS G Depending on the sign & magnitude of S, G can:  may go from + to  valued w/ increasing T  may go from  to + valued w/ increasing T  may decrease in magnitude  may increase in magnitude  may remain unchanged For many systems, G is relatively invariant to at least small T changes (i.e. the entropic contributions to G are often relatively small) Effect of Temperature On Equilibrium A B Your book explains it like this. I don’t believe that this is a particularly intuitive picture. Chemical Equilibrium & Acids & Bases pK a log10 K a Ka AB AB The strengths of some acids and their conjugate bases in aqueous solution. pKa + pKb = 14 (Oxtoby) Example: Strong Acids Prepare solutions with pH=3 from nitric acid For nitric acid (HNO3) pKa = -1.44 101.44 = 27.54 H3O+ Ka AB AB 27.5 H NO3 HNO3 10 3 10 M 10 3 3 Solve for M; M=1.003x10-3 M This is the sign of a strong acid; the concentration of [H+] [HA] added Acids and Bases: Buffers A buffer contains both a weak acid and its conjugate base. Example: for an acetic acid/acetate buffer you would add acetic acid and sodium acetate to water. Buffers resist change in pH. The weak acid and its conjugate base react with either acid or base HA(aq) + OH (aq) A (aq) + H2O(l) HA(aq) A (aq) + H+(aq) Both of these reactions have large equilibrium constants. Example Make a buffer solution that remains at approximately pH=5 from acetic acid & sodium acetate, and show that it remains close to pH=5 upon addition of 20 mM HNO3 weak acid conjugate base Start with [1M] CH3CO2H and, with Na+CH3CO2-, make the buffer Example Make a buffer solution that remains at approximately pH=5 from acetic acid & sodium acetate, and show that it remains close to pH=5 upon addition of 20 mM HNO3 weak acid conjugate base Start with [1M] CH3CO2H and, with Na+CH3CO2-, make the buffer H3O+ Ka AB AB 1.75 10 5 H H 3CO2 H 3CO2 H 10 5 x 10 1 5 Acetic acid is only weakly ionized (at pH=5) and so we assume its concentration stays at [1M] Solving for x yields x=1.75M CH3CO2- Na+ Now add 20 mM HNO3 20 mM HNO3 = [0.02M] Assume that all HNO3 ionizes (it is a strong acid) [0.02M] H+ is taken up by H3CO2-, thereby reducing the concentration of acetate from 1.75 M to 1.73 M Ka AB AB 1.75 10 5 H H 3CO2 H 3CO2 H x 1.73 x 1.02 x Now set up the equilibrium equation, and let the concentration of [H+] (or, actually [H3O+]) vary, and solve for x Example, Part II Make a buffer solution that remains at approximately pH=5 from acetic acid & sodium acetate, and show that it remains close to pH=5 upon addition of 20 mM HNO3 Assume that all HNO3 ionizes (it is a strong acid) [0.02M] H+ is taken up by H3CO2-, thereby reducing the concentration of acetate from 1.75 M to 1.73 M Ka AB AB Solve for x; 1.75 10 5 x 1.73 x 1.02 x [x] = 1.032x10-5; pH = 4.98 The pH has hardly changed, even though we added a strong acid! This is the point of a buffer!! The rest of the lecture will be an extended Chemistry in Context Today Electrochemistry & Equilibrium Prof Reisman takes over class next Monday Electrochemistry Light (with 50% less knowledge than your regular electrochemistry) Oxidation/Reduction Reactions 2 Fe( s ) 3 2 O2 Fe2O3 OGC Unit IV: Chemical Equilibrium: Chapter 17 (only the bits and pieces we will cover today). Redox Reactions Redox reactions are among the simplest chemical transformations of which electron transfer reactions are one type. Oxidation: loss of electrons Reduction: gain of electrons reduction + oxidation = redox reaction Electrochemistry Light (with 50% less knowledge than your regular electrochemistry) An extension of Chemical Equilibria Oxidation/Reduction Reactions 2 Fe( s ) 3 2 O2 Fe2O3 Breaks down into these two ½ reactions 2 Fe( s ) 3 6e 2 O2 2 Fe 3 3O 2 6e Iron is transformed into the +3, or Fe(III) oxidation state Oxygen is transformed into the -2 oxidation state oxidation states: The oxidation state of an element is always 0  F atom just about always has the oxidation state of F-, or an oxidation number of -1, regardless of what molecular or compound it is in (other than F2). Other halogens are typically -1 as well, except when they form molecules with oxygen and with other halogens.  The Oxidation state of hydrogen is +1 in all compounds except when bound to metal as a hydride (e.g. LiH), where it is -1.  The oxidation state of oxygen is -2 in all compounds except peroxides (H2O2) where it is -1, and in compounds with fluorine. Examples of Redox Reactions Example #1: Zn(s) in aqueous acid: Example Zn(s) oxidation state: (0) reducing agent, reductant + 2H+ (+1) oxidizing agent, oxidant Zn2+(aq) (+2) oxidized + H2(g) (0) reduced Example #2: Reaction of phosphine gas with oxygen - no ions: 2PH3(g) + 4O2 (-3)(+1) (0) red. agent ox. agent P2O5(s) (+5)(-2) oxidized + 3H20(g) (+1)(-2) reduced Oxidizing agent is not always oxygen!! (note example 1 above) Back to simple redox processes Examples Examples of Redox Reactions Example #3: Zn strip in a CuSO4(aq) solution; over time blue color of solution disappears, copper plates out  and Zn strip slowly disappears Overall: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) We introduce idea of two half-reactions: oxidation: reduction: Zn(s) Cu2+(aq) + 2 e- Zn2+(aq) + 2 eCu(s) Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Notice that the half reactions explicitly display the number of electrons transferred in the redox process. Example: Balancing reactions Balancing RedoxRedox Equations Decolorization of permanganate solution by oxalate in aqueous acid solution (unbalanced): MnO4 ( aq.) 2 C2O4 ( aq.) Mn 2 CO2 ( g ) Incomplete half reactions: oxidant : MnO4 ( aq.) Mn 2 (aq.) reductant : C2O4 (aq.) CO2 ( g ) Balance first with respect to atom undergoing redox; balance O with H2O: MnO4 ( aq.) Mn 2 (aq.) 4 H 2O (l ) Then balance H with H+ (from acid): 8 H ( aq.) MnO4 (aq.) Mn 2 ( aq.) 4 H 2O (l ) Example: Balancing reactions Balancing RedoxRedox Equations Decolorization of permanganate solution by oxalate in aqueous acid solution (unbalanced): MnO4 ( aq.) 2 C2O4 ( aq.) Mn 2 CO2 ( g ) Incomplete half reactions: oxidant : MnO4 ( aq.) Mn 2 (aq.) reductant : C2O4 (aq.) CO2 ( g ) Balance first with respect to atom undergoing redox; balance O with H2O: MnO4 ( aq.) Mn 2 (aq.) 4 H 2O (l ) Then balance H with H+ (from acid): 8 H ( aq.) MnO4 (aq.) Mn 2 ( aq.) 4 H 2O (l ) Example: Balancing reactions Balancing RedoxRedox Equations Decolorization of permanganate solution by oxalate in aqueous acid solution (unbalanced): MnO4 ( aq.) 2 C2O4 ( aq.) Mn 2 CO2 ( g ) Incomplete half reactions: oxidant : MnO4 ( aq.) Mn 2 (aq.) reductant : C2O4 (aq.) CO2 ( g ) Balance first with respect to atom undergoing redox; balance O with H2O: MnO4 ( aq.) Mn 2 (aq.) 4 H 2O (l ) Then balance H with H+ (from acid): 8 H ( aq.) MnO4 (aq.) Mn 2 ( aq.) 4 H 2O (l ) Then balance charge with electrons: 5e 8H MnO4 ( aq ) Mn 2 ( aq ) 4 H 2O( l ) Continued II): Example (part Balancing Redox Equations Balance charge with electrons: 5e 8H MnO 4 ( aq.) Mn 2 (aq.) 4 H 2O (l ) Similarly, for the half reaction involving oxalate: 2 C2O4 (aq.) 2 CO2 ( g ) 2e Multiply half-reactions as appropriate to give same number of electrons in each (gain = loss), then add: 10 e 16 H ( aq.) 2 MnO 4 ( aq.) 2 5 C2O4 ( aq.) 2 Mn 2 ( aq.) 8 H 2O (l ) 10 CO2 ( g ) 10 e 16H (aq.) 2MnO4 (aq.) 5C2O2 (aq.) 4 2Mn2 (aq.) 8H2O(l) 10CO2 (g) Redox chemistry in biology: The ATP cycle The simple equation below is comprised of many steps but G is a state function - we don't care about the pathway, just the balanced final reaction, as long as all under standard conditions C6H12O6 + O2 + ADP + phosphate CO2 + H2O + ATP G=-423 kcal/mol of glucose What exactly is oxidized and what exactly is reduced in this cycle? + 6O2 + - - 36 ADP + 36 ATP 42 H2O + 6 CO2 + OH OH OH 6O2 + + P2O6H2 glucose C6H12O6 36 ADP (probably ADP3-) CHOPe glu ox ADP phos 6 0 0 0 6 12 0 72 0 84 6 12 216 144 378 + 00 00 72 108 36 36-72 108 144-180 42 H2O + 6 CO2 CO2 H2O ATP 36 Phosphate this is typically HPO42and H2PO4- H3PO4 H2PO4HPO42PO43- CHOPe 36 ATP OH OH (how do we know?) look up pKa's for P3O9 + OH 6 0 0 6 0 12 0 0 84 42 0 0 0 324 108 144 84 378 108 144 6O2 + + P2O6H2 glucose C6H12O6 36 ADP (probably ADP3-) CHOPe glu ox ADP phos 6 0 0 0 6 12 0 72 0 84 6 12 216 144 378 + 00 00 72 108 36 36-72 108 144-180 P3O9 + 42 H2O + 6 CO2 CHOPe 36 ATP CO2 H2O ATP 6 0 0 6 0 12 0 0 84 42 0 0 0 324 108 144 84 378 108 144 OH OH OH 36 Phosphate this is typically HPO42and H2PO4- + oxidant 6O2 + - - oxidized 36 ADP reduced + 36 ATP + OH OH OH reductant 42 H2O + 6 CO2 Note  this is a pretty complicated problem  I had to piece all of this together from many sources and so you aren't likely going to see anything quite this complex This car battery is an electrochemical cell. The chemical reaction occurring inside the battery is an oxidation/reduction process That reaction is exothermic. The energy generated from the reaction can be used for work. The electrochemical cell produces a cell voltage Electrochemical cells typically operate at constant T, P, and V Cell Voltage Cell voltage reflects the driving force of the reaction. How does the cell potential ( ) relate to G? G = H - TS dG = dH - TdS - Sdt dG = dq + dw + PdV + VdP - TdS - SdT At constant T and P: dG = dq + dw + PdV - TdS For a reversible process, dG = TdS + dw + PdV - TdS = dw + PdV Cell Voltage Cell voltage reflects the driving force of the reaction. How does the cell potential ( ) relate to G? G = H - TS dG = dH - TdS – Sdt This appears a bit crazy, but it is a wholly rationale logical flow  look at the derivation carefully dH = dU + d(PV) = dq + dw + PdV + VdP so dG = dq + dw + PdV + VdP - TdS - SdT At constant T and P: dG = dq + dw + PdV - TdS For a reversible process (dqrev/T = dS, so dq = TdS), dG = TdS + dw + PdV - TdS = dw + PdV Negligible volume change Cell Voltage dG = dw + PdV = dw since dV = 0. G = w = welec welec = charge x potential G = w = welec= -nF F is Faraday s constant and is equal to the charge of one mole of electrons (6.02 x 1023)(1.60 x 10-19) = 9.65 x 104 C/mol. G = w = welec= -nF Gº -nF º Cell Voltage and the Nernst Equation Gº = -nF Recall that -nF º G = Gº + RTln(Q) = -nF º + RTln(Q) This gives the Nernst equation: = = º - (RT/nF)ln(Q) º - (0.059/n)log10(Q) at 25 C The Nernst Equation = º - (0.059/n)log10(Q) , in analogy with G, indicates the direction of spontaneous change: > 0 reaction goes left to right = 0 no net reaction (equilibrium) < 0 reaction goes right to left Electrochemical Cells, Go, and equilibrium Cu Cu2+(aq) + Zn Cu(s) + Zn2+ Fig. ElCh.2. A laboratory version of a Daniell Cell Note reading of voltmeter: 1.10 V The salt bridge conducts ions between both chambers to maintain electrical neutrality. Electrochemical Cells Electrochemistry  Electrolytic cells are those in which electrical energy from an external source causes nonspontaneous chemical reactions to occur.  Voltaic cells are those in which spontaneous chemical reactions produce electricity and supply it to an external circuit. Electrodes  Electrodes are surfaces upon which oxidation and reduction half  reactions occur.  The cathode is the electrode at which reduction occurs.  The anode is the electrode at which oxidation occurs. Daniell Galvanic Cell - Laboratory Version Cu2+(aq) + Zn Suppose we attach a power supply that opposes the current flow. What happens as the potential is increased? Cu(s) + Zn2+ Power - Supply + At sufficiently high potential the reaction reverses and the galvanic cell becomes an < 1.10 V, reaction proceeds as before electrolysis cell. = 1.10 V, no current flows (reaction stopped) > 1.10, current and reaction reverse. A galvanic cell based on the combined reduction of copper and oxidation of silver, Cu2+ + 2Ag Cu + 2Ag+ has half cell reactions (both written as reductions) : Cu2+ + 2e Cu = 0.34 V Ag+ + e Ag = 0.80 V G 0 o o is a measure of G of this reaction It is also a measurement of how much work can be extracted from this reaction To oxidize copper and reduce 2 moles of silver: Cu Cu2+ + 2e 2Ag 2Ag+ + 2e Cu + 2Ag+ = -0.34 V = 0.80 V Cu2+ + 2Ag G 0 = -0.34 + 0.80 = 0.46 V nF o -(2 mol e-)(9.65e4 C/mol)(0.46 J/C)(1 kJ/103J) = -88.8 kJ Note: G depends upon the number of charges (requires balancing eqtn) only depends upon the summation of the ½ cell voltages F = Faraday = 9.65e4 Coulombs/mole Conventions for (Half-cell Potentials) 1. Assign standard hydrogen electrode a potential of 0.0 V. 2. When half reactions are written as reductions, reactions that proceed as reductions more readily than H+/H2 are assigned positive voltages, those that proceed less readily are assigned negative voltages. 3. If the direction of the half reaction is reversed, the sign of the standard half cell potential is reversed. If the reaction is multiplied by some factor, the standard half cell potential remains unchanged. 4. The value of the half cell potential is a measure of the tendency of the reaction to proceed from left to right as written. Reduction ½ Cells A Sampling of Half Cell Potentials Half Reaction F2 + 2e o(V) 2F- 2.89 MnO4- + 8H+ + 5e O2 + 4H+ + 4e Ag+ + e Cu2+ + 2e Mn2+ + 4H2O 2H2O Ag Cu 1.51 Oxidizing agents 1.23 0.80 0.34 2H+ + 2e H2 0.00 Fe2+ + 2e Fe -0.44 Zn2+ + 2e Zn -0.76 Mn2+ + 2e Mn -1.18 Na+ + e Na -2.70 Reducing agents Not favorable in this direction The Hydrogen Electrode The hydrogen electrode is the standard reference electrode for aqueous half-cell potentials. The half-cell reaction is the reduction 2 H+(aq) + 2e H2 (g) The standard potential of this half-cell is set equal to zero. Potentials measured for a cell containing the standard hydrogen electrode will be equal to the half-cell potential of the other electrode. A table of standard half-cell potentials is given by Oxtoby, Gillis and Nachtrieb in Appendix E. ...
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