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Transcript
1/21/14 OCN 623: Thermodynamics, Chemical Equilibrium, and Gibbs Free Energy or how to predict chemical reactions without doing experiments Introduction • We want to answer these questions: • Will this reaction go? • If so, how far can it proceed? We will do this by using thermodynamics. • This lecture will be restricted to a small subset of thermodynamics... 1 1/21/14 Outline • Definitions • Thermodynamics basics • Chemical Equilibrium • Free Energy • Aquatic redox reactions in the environment Definitions • Extensive properties -depend on the amount of material – e.g. # of moles, mass or volume of material – examples in chemical thermodynamics: – G -- Gibbs free energy – H -- enthalpy • Intensive properties - do not depend on quantity or mass e.g. – temperature – pressure – density – refractive index 2 1/21/14 Reversible and irreversible processes • Reversible process occurs under equilibrium conditions e.g. gas expanding against a piston Net result - system & surroundings back to initial state P p p = P + ∂p reversible p = P + ∆p irreversible Reversible Process • No friction or other energy dissipation • System can return to its original state • Very few processes are really reversible in practice – e.g. Daniell Cell Zn + CuSO4 = ZnS04 + Cu with balancing external emf 3 1/21/14 Daniell Cell Zn + CuSO4 = ZnS04 + Cu with balancing external emf -cell voltage is a measure of: ability of oxidant to gain electrons, ability of reductant to lose electrons. Zn oxidation, Cu reduction Zn(s) = Zn2+(aq) + 2e- Cu2+(aq) +2e- = Cu(s) Compression of gases Vaporization of liquids Reversibility is a concept used for comparison Will get more of this next lecture, covering redox potential Spontaneous processes • Occur without external assistance – e.g. - expansion of a gas from region of high pressure to low pressure - diffusion of a solute in a solvent Are irreversible by definition -- they occur at a finite rate 4 1/21/14 Thermodynamics: The Basics • Nature of energy: W = F x d capacity to do work or transfer heat • Goals of (re) learning thermodynamics – To compute the equilibrium of a system – To determine how far a given system is from equilibrium – To evaluate the energy available from a given reaction 1st law of thermodynamics • Energy cannot be created or destroyed – when one kind of energy disappears it is transformed into an equivalent quantity of another kind ∆U = q-w dU = dq - dw dU = change in internal energy of the system dq = heat absorbed by the system + dw = work done by the system + Energy is conserved 5 1/21/14 • U is a thermodynamic function dU depends only on the initial and final states of the system-not the path taken q and w are NOT thermodynamic functions Internal energy of system is increased by gaining heat (q) Internal energy of system is decreased when work is done by the system dU = dq - dw wrev. > wirrev. • Work done by system during irreversible process is less than could be obtained if process was reversible e.g. if piston moved at finite rate, friction dissipates energy 6 1/21/14 Chemical thermodynamics Heat • Heat (q) is measured in terms of: amount of substance x molar heat capacity x temp rise • For phase change q = moles of substance x molar latent heat • q is the heat of reaction • Molar heat capacity = Heat energy per mole required to raise temperature of 6.022x1023 molecules 1oC WORK (1) Work of expansion = P ∆V where P = pressure and ∆V is change in volume at constant volume ∆V = 0, therefore w =0 no work is done 7 1/21/14 (2) Electrical work – done on charged species • Occurs in electrical cells e.g. Zn + Cu2+ = Zn2+ + Cu • Electrical energy = I * E * t E is the emf (voltage), t = time, I = current flowing It = zF for 1 mol z = # of electrons transferred, F (Faraday)= 96,490 Coulombs/mol Electrical energy = zEF • System at constant volume (all non-gas reactions are at constant volume) P∆V = w= 0 ∆U = qv qv= heat at constant volume • System at constant pressure (all reactions open to atmosphere) ∆U = qp - P∆V Ideal gas law P∆V = ∆nRT, (n = amount of substance) R is the gas constant = 8.314 Joules mol-1 K-1 therefore ∆U = qp - ∆nRT rearranging qp = ∆U + ∆nRT 8 1/21/14 • qp is called the enthalpy H change in enthalpy ∆H = ∆U + P∆V in absolute terms H = U + PV • H is a thermodynamic property, is defined in terms of thermodynamic functions: U, P and V For an infinitesimal change at constant pressure: dH = dU + PdV • ∆Hf o298 is the heat of formation of 1 mole of a compound from its elements at 298oK • ∆H is neg = exothermic reaction • ∆H is pos = endothermic reaction (seen from system perspective) • ∆H is proportional to amount of material is an extensive property H2(g) + 1/2O2(g) --> H2O(g) This reaction releases 241.8 kJ of heat, therefore heat of formation is: -241.8 kJ 9 1/21/14 • ∆H is equal in magnitude but opposite in sign for the reverse reaction, because it is a thermodynamic quantity ∆Hproducts - ∆Hreactants ∆H for reaction is same regardless of number of steps between reactants and products ∆H = ∆U + ∆nRT • Can use component reactions to calculate enthalpy change for unknown reaction Reaction Enthalpy H2 + 0.5 O2 = H2O ∆H = -285.8 kJ mol-1 (a) C + O2 = CO2 ∆H = -393.3 kJ mol-1 (b) C2H6 + 3.5 O2 = 2 CO2 + 3H2O ∆H = -1559.8 kJ mol-1 (c) can use above to calculate ∆H for ethane formation 2*b + 3*a - c 2C + 2O2 + 3H2 + 1.5 O2 - C2H6 - 3.5 O2 = 2CO2 + 3H2O - 2 CO2 - 3H2O canceling yields 2C + 3 H2 = C2H6 ∆H = 2* -393.3 + 3* -285.8 - (-1559.8) = -84.2 kJ mol-1 10 1/21/14 • But enthalpy change alone is insufficient to allow prediction of likelihood of reaction • Entropy change is also needed 2nd law of thermodynamics dS = dqrev T change in entropy = amount of reversible heat absorbed by system divided by temperature Entropy of an isolated system which is not in equilibrium will tend to increase over time (a) Entropy is the degree of disorder of the system chance of finding something in a fixed volume- higher pressure, greater chance, liquid lower entropy than gas, solid lower than liquid (b) Change in entropy measures capacity for spontaneous change diffusion of a solute from high concentration to low concentration, or a gas from high pressure to low pressure • A system undergoing spontaneous change is moving to a greater degree of disorder 11 1/21/14 • Entropy is an extensive property • Units heat/temperature = Joules K-1 mol-1 for solids dq = moles x molar heat capacity x temperature rise dq = CmdT Cm = molar heat capacity and dT = temperature change molar entropy change: ∆Sm = from T1 to T2 ∫ CmdT T ∆Sm = Cm ln T2 T1 • For phase change ∆S = L T where L is the latent heat of fusion e.g. for water L = 6 kJ mol-1 ∆S = 6000 = 22.0 J K-1mol-1 for melting ice 273.1 • reversible reaction the net change in entropy is zero • irreversible reaction there is a net increase in entropy 12 1/21/14 • Since nearly all reactions are irreversible, entropy is increasing -- called times arrow • always increasing --cannot run time backwards Third law of thermodynamics The entropy of a perfectly crystalline material at the absolute zero (-273.1˚ C) is zero • entropy allows prediction of the feasibility of reactions Why study chemical equilibrium? • Is a geochemical system at chemical equilibrium? • If not, what reactions are most likely to occur? • Solubility - • Redox - • Complexation - • Carbonate system - 13 1/21/14 Why study chemical equilibrium? • Is a geochemical system at chemical equilibrium? • If not, what reactions are most likely to occur? • Solubility - diatoms in surface seawater • Redox - organic matter oxidation • Complexation - iron speciation & plankton growth • Carbonate system - CaCO3 in marine sediments Equilibrium vs. kinetics • Equilibrium calculations give the energy available from a reaction & the direction it will proceed - Usually simpler & require less info than those for kinetics • Often a good approximation for many systems • In some cases, equilibrium calculations can predict kinetic rate constants • HOWEVER, most natural water reactions are not at equilibrium and knowledge of kinetics is often required - Steady state, states of dynamic equilibrium, reaction rates, are topics outside of the scope of this course, but will be covered in 643 in Fall. 14 1/21/14 Chemical Equilibrium - 1 Consider a reversible reaction taking place at constant temperature: aA + bB cC + dD The reactants A and B combine to form products C and D. The concentrations of A and B decrease until they reach values that do not change with time: Chemical Equilibrium – 2 • The time-invariant concentrations of reactants and products are called equilibrium concentrations. • The ratio of these concentrations (or activities – active concentrations) is characteristic for each reaction, and is called the equilibrium constant, K: {C}c {D}d K≡ { A}a {B}b • Note that at equilibrium, the forward and reverse reactions proceed at the same, stable rate. 15 1/21/14 Free Energy – 1 • A criterion for equilibrium is that the total free energy (Gibbs free energy, Gr) of the reaction is at a minimum: • If we add more reactant or more product, the reaction will proceed spontaneously (without external help) as long as the value for Gr decreases. • Thus, a reaction in the direction of decreasing Gr is spontaneous. A reaction in the direction of increasing Gr is not spontaneous, and will not occur in a closed system. Free Energy - 2 As any reaction proceeds an incremental amount, the change in Gr can be calculated as: & # & # ΔGr = $ ∑υi G fi ! − $ ∑υi G fi ! % i " products % i " reactants where νi is the stoichiometric coefficient (a,b,c,d) and Gfi is the free energy of formation per mole. 1. If ΔG < 0, (i.e., ΔG is negative and thus Gr decreases as the reaction proceeds), then the reaction proceeds spontaneously as written. 2. If ΔG > 0, (i.e., ΔG is positive and thus Gr increases as the reaction proceeds), then the reaction proceeds spontaneously in the opposite direction as written. 3. If ΔG= 0, (i.e., ΔG is at a minimum), then the reaction is at equilibirium and will not proceeds spontaneously in either direction. 16 1/21/14 Free Energy - 3 Values for ΔG for a reaction give us a powerful tool to predict if a reaction is possible. We calculate in-situ ΔGr using this equation: ΔG r = ΔG r ° + RT ln In-situ Std. state {C}c {D}d { A}a {B}b where & # & # ΔGr ° = $ ∑υi G° fi ! − $ ∑υi G° fi ! % i " products % i " reactants (The the superscript zero (°) indicates standard state: 25°C (298°K),1 atm pressure, and activity = 1.) Gfi° is the standard-state free energy of formation per mole of species i. { } = activity (active concentration) R = the ideal gas constant = 1.987 cal °K-1 mol-1 = 8.31 J °K-1 mol-1 T = °K Free Energy - 4 Standard free energy of formation (ΔG°fi): • G°fi = 0 at standard state for all pure elements (solid reference). • G°fi = 0 for H+ at a concentration of 1 mole/liter at standard state (solution reference). • Allows the measure of the energy change involved in forming compounds at standard state from their component elements at standard state. • Measured values are listed in tables. Units are: • kJ/mol (SI units) • kcal/mol 17 1/21/14 Table is included in reading (handout) See complete table in handout. 18 1/21/14 Sample Calculation #1 Consider the dissolution of CaCO3 (calcite) in aqueous solution: CaCO3 → Ca2+ + CO32Does the reaction proceed spontaneously as written? Calculate the free energy of reaction at standard state (products and reactants at activity = 1, P = 1 atm, T = 25°C): & # & # ΔG°r = $ ∑υi G° fi ! − $ ∑υi G° fi ! % i " products % i " reactants ΔG°r = (1(-132) + 1(-126) – (1(-270)) kcal/mol = = (-258 + 270) kcal/mol +12 kcal/mol Therefore, at standard state, the reaction spontaneously proceeds in the opposite direction to what is written. 19 1/21/14 Activities – How to Calculate Activity of water = 1 Equilibrium at In-Situ Conditions - 1 We have already seem that: ΔG = ΔG ° + RT ln In-situ Std. state {C}c {D}d { A}a {B}b We can define a reaction quotient, Q, using in-situ conditions: Q≡ Thus: {C}c {D}d { A}a {B}b ΔG = ΔG° + RT ln Q In the case of equilibrium, Q = K and ΔG = 0: Thus: 0 = ΔG° + RT ln K ΔG° = -RT ln K Substituting (for the general case): ΔG = -RT ln K + RT ln Q = RT ln Q/K 20 1/21/14 Equilibrium at In-Situ Conditions - 2 ΔG = RT ln Q/K This allows us to develop a set of criteria to determine in which direction a reaction will proceed under non-standard conditions. This is because Q/K will determine the sign of ΔG: 1. If Q/K < 1, then ΔG is negative, and the reaction is spontaneous as written. 2. If Q/K = 1, then ΔG = 0 and the system is at equilibrium. 3. If Q/K > 1, then ΔG is positive, and the reaction is spontaneous in the opposite direction as written. Sample Calculation #2 Again consider the dissolution of CaCO3 (calcite): CaCO3 → Ca2+ + CO32In the surface ocean, does the reaction proceed spontaneously as written? Use ΔG° = -RT ln K From Sample Calculation #1, we know: ΔG° = +12 kcal/mole Thus: cal & 1 kcal # 12 kcal mol = − 1.9987 ° K mol $ cal !(298° K )ln K % 1000 " ( ) (all units cancel out) ln K = -20.27 K = 1.58 x 10-9 (“equilibrium constant”) 21 1/21/14 Sample Calculation #2 (cont.) Surface seawater: {CaCO3} ≡ 1 [Ca2+] = 0.01 mol/L; γ = 0.28; {Ca2+} = 0.0028 [CO32-] = 45 µmol/L; γ = 0.21; {CO32-} = 9.45 x 10-6 {Ca }{CO }= (2.8x10 )(9.45x10 )= 2.65x10 2− 2+ Q= −3 −6 3 {CaCO3 } −8 1 Q 2.65x10 −8 = = 17.6 K 1.58x10 −9 Therefore: • The reaction goes in the opposite direction as written (Q/K > 1) • CaCO3 precipitates in the surface ocean • Surface seawater is supersaturated with respect to calcite Note: No information is given on the kinetics of the reaction! Temperature and Pressure Effects - 1 The amount of heat that is released or taken up by a reaction is called the enthalpy change (ΔH°). Similarly to ΔG°, we can calculate ΔH° using tabulated data: & # & # ΔH r ° = $ ∑υi H ° fi ! − $ ∑υi H ° fi ! % i " products % i " reactants where νi is the stoichiometric coefficient (a,b,c,d), and H°fi is the enthalpy of formation of species i (kcal/mol or kJ/mol) at standard state. The van’t Hoff equation gives the temperature dependence of K: ΔH r ° ' ∂ ln K $ % " = 2 ∂ T & # P RT For a small change in T, ΔH° does not change much, so we can integrate directly: & K # ΔH r ° & 1 1 # ln$$ T * !! = − $ ! R % 298 T * " % K 298 " where T* is the temperature of interest. 22 1/21/14 Temperature and Pressure Effects - 2 The change in the partial molar volume (ΔV°) during a reaction at standard state is also calculated using tabulated data: & # & # ΔVr ° = $ ∑υiV ° fi ! − $ ∑υiV ° fi ! % i " products % i " reactants where νi is the stoichiometric coefficient (a,b,c,d), and V°fi is the partial molar volume of formation of species i (cm3/mol) at standard state. The pressure dependence of K is also known: ΔVr ° ' ∂ ln K $ % " = RT & ∂P #T Again integrating directly: ( K % ΔVr ° (P * −1atm ) ln&& P* ## = ' K1atm $ RT where P* is the pressure of interest (in atm). Temperature and Pressure Effects - 3 See the following example in the handout for an example of temperature-correcting the equilibrium constant: 23 1/21/14 • For exothermic reactions lnK increases as 1/T increases, (T decreases), favouring product formation • For endothermic reactions lnK increases as 1/T decreases (T increases) favouring product formation Exothermic lnK Endothermic 1/T …putting it all together in terms of redox… 24 1/21/14 Redox Reactions in Nature • Example: net reaction for aerobic oxidation of organic matter: CH2O + O2 → CO2 + H2O • In this case, oxygen is the electron acceptor – the half-reaction is: O2 + 4H+ + 4e- → 2H2O • Different organisms use different electron acceptors, depending on availability due to local redox potential • The more oxidizing the environment, the higher the energy yield of the OM oxidation (the more negative is ΔG, the Gibbs free energy) • The higher the energy yield, the greater the benefit to organisms that harvest the energy • In general: – There is a temporal and spatial sequence of energy harvest during organic matter oxidation – Sequence is from the use of high-yield electron acceptors to the use of low-yield electron acceptors 25 1/21/14 Why is organic matter an electron donor? Z-scheme for photosynthetic electron transport Falkowski and Raven (1997) ADP→ATP Energy Scale Ferredoxin ADP→ATP to Krebs cycle and carbohydrate formation photooxidation of water energy from sun converted to C-C, energy rich, chemical bonds 26 1/21/14 Environmentally Important Organic Matter Oxidation Reactions Eh (V) ΔG +0.812 -29.9 +0.747 -28.4 +0.526 -23.3 -0.047 -10.1 -0.221 -5.9 -0.244 -5.6 Reduction of O2 + - O2 + 4H +4e --> 2H2O Reduction of NO3- + - 2NO3 + 6H + 6e --> N2 + 3H2O Reduction of Mn (IV) MnO2 + 4H+ + 2e- --> Mn2+ +2H2O Reduction of Fe (III) + 2+ Fe(OH)3 + 3H + e --> Fe +3H2O Reduction of 2- SO42+ - SO4 + 10H + 8e --> H2S + 4H2O Reduction of CO2 CO2 + 8H+ + 8e- --> CH4 + 2H2O DECREASING ENERGY YIELD Reducing Half-reaction The Next Lecture: Thursday January 23 “Redox & pE-pH” 27 1/21/14 Thermodynamics and kinetics Lecture 2. Energy and life Catalysis Catalysis = Speeding reactions without getting used in the process. K = equilibrium constant k = rate constant Energy Exergonic direction Endergonic direction kf kb Energy Catalysis (Cat) = the catalyzed reactions proceed faster toward equilibrium (K). Catalyst Energy Energy Kinetic inhibition may have been used by life to create an energy capacitor George Cody citing Everet Shock, OLEB, 1990. 28