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Sustainable Resource Engineering Course Note 1-1 (Reaction Principles)

This course note explores the dynamics of transformation processes in sustainable resource engineering, including stoichiometry, thermodynamics, and kinetics.

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Sustainable Resource Engineering Course Note 1-1 (Reaction Principles)

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  1. Sustainable Resource Engineering Course Note 1-1 (Reaction Principles) Joonhong Park Yonsei CEE Department 2015. 9. 14. CEE3330 Y2013 WEEK3

  2. Contents: 1. Dynamics of Transformation Processes - stoichiometry - thermodynamics - kinetics CEE3330 Y2013 WEEK3

  3. Intro: Governing Concepts of Transformation Processes - Stoichiometry: how much can you get? • Thermodynamics: can the reaction occur? • Kinetics: how fast? aA +bB  cC + dD Ex. CH4 + 2O2 = 1CO2 + 2H2O CEE3330 Y2013 WEEK3

  4. Stoichiometry The application of the principle of material balance to a chemical transformation. bR1 + cR2 => mP1 + nP2 here R1 and R2: reactants P1 and P2: products b, c, m, and n: stoichiometric coefficients Meaning: b molecules of R1 combine with c molecules of R2 to form m molecules of P1 and n molecules of P2 RULE: A chemical reaction must conserve (i) the number of atoms for each element involved in the reaction and (ii) the electrical charge associated with ions. CEE3330 Y2013 WEEK3

  5. Stoichiometry Example: A general photosynthesis reaction my be written in the form shown below. Use the principle of stoichiometry to determine the coefficients a-f. aCO2 + bNO3- + cHPO42- + dH+ + eH2O => C106H263O110N16P1+ fO2 C: a =106 O: 2a + 3b + 4c + e = 110 + 2f N: b=16 H: c + d + 2e = 263 P: c =1 +/-: -b -2c + d = 0 a=106 b=16 c=1 d= b + 2c = 16 + 2 = 18 e = 0.5*(263-c–d) =122 f=0.5*(110-2a-3b-4c-e)=138 CEE3330 Y2013 WEEK3

  6. Stoichiometry Example: Benzene (C6H6) can be oxidized in the presence of oxygen. The complete oxidization of benzene produces CO2 and H2O. Calculate how many moles of oxygen molecules have to be needed for one mole of benzene to be completely oxidized. C6H6 + aO2 => mCO2 + nH2O Since # of C should be 6 in the left-hand term, m =6. (MB for C atoms) Since # of H should be 6 in the left-hand term, n=3. (MB for H atoms) Therefore, total # of O in the right-hand term is 6*2 + 3*1 = 15. And a should be 15/2 (MB for O atoms) CEE3330 Y2013 WEEK3

  7. Chemical Energy (Gibbs Free Energy, G) Enthalpy, H = U + W (M L2 T-2) U: internal energy W: external energy (W = PV; 압력 X 부피) endothermic (흡열반은, dH > 0) exothermic (발열반응, dH < 0) H = G +TS (M L2 T-2) G: Gibbs Free Chemical Energy TS: Entropy Gain Function (T = absolute temp.; S = entropy) CEE3330 Y2013 WEEK3

  8. Thermodynamics/Gibbs Free Energy Free energy of species participating in a reaction (G) Activation energy (forward rxn) Reactants ΔG Change in Gibbs free energy Products Progress of reaction ΔG = Σ Gf,i (products) -Σ Gf,i (reactants) ΔG < 0: thermodynamically favorable. CEE3330 Y2013 WEEK3

  9. Thermodynamics/Gibbs Free Energy ΔG = Σ Gf, i (products) - Σ Gf, i (reactants) Gf,i = ni [Gof,i + RT ln {i}] n: # of moles of species i. {i}: activity of species i. (when diluted, {i} = [i]) Gof,i : Gibbs free energy for species i at reference conditions (T=298 K, P =1atm) CEE3330 Joonhong Park Copy Right

  10. Thermodynamics/Gibbs Free Energy Example: Using free energy to determine the direction of a reaction. Water that is in contact with solid lime (Ca(OH)2) at 25oC contains concentrations of calcium and hydroxide ions of 10-5 M and 10-4 M, respectively. Given: Gof for Ca(OH)2 = -898.4kJ/mol; Gof for Ca2+ = -553.1 kJ/mol; Gof for OH- = -157.3 kJ/mol) Ca(OH)2 (s)  Ca2+ + 2 OH- Solution) Ca(OH)2: Gf,= 1*[-898.4 kJ/mol + RT ln[Ca(OH)2] = -898.4 kJ/mol Ca 2+: Gf = 1*[-553.1kJ/mol + RT ln[Ca2+]] = -582.1 kJ/mol OH-: Gf = 2 * [-157.3 kJ/mol + RT ln[OH-]] = -360.2 kJ/mol ΔG = Σ G f, i (products) - Σ G f, i (reactants) = (-582.1-360.2) – (- 898.4) = - 43.9 kJ/mol < 0 Forward direction! CEE3330 Y2013 WEEK3

  11. Thermodynamics/Gibbs Free Energy aA + bB  cC + dD ( at a given temperature, T) ΔG = Σ Gf, i (products) - Σ Gf, i (reactants) Gf,A = a[Gof,A + RT ln {A}] Gf,B = b[Gof,B + RT ln {B}] Gf,C = c[Gof,C + RT ln {C}] Gf,D = d[Gof,D + RT ln {D}] ΔG = -a*Gof,A - b*Gof,B + c*Gof,C + d*Gof,D+ RT ln[{A}-a{B}-b{C}+c{D}+d] = ΔGo + RT lnQ Environmental factors (온도, 농도) Chemical properties CEE3330 Joonhong Park Copy Right

  12. Chemical Equilibrium/Thermodynamics When a system is at “chemical equilibrium”, the distribution of elements among chemical species and the distribution of species among physical states (phases) satisfies thermodynamic relationships, regardless of the history of the system. A system at chemical equilibrium must satisfy: • It does not vary with time. • It is internally uniform (at least at the level of discrete subsystems). • There are not net flows of mass, heat, or species within the system or between the systems and its surroundings. • The net rate of all chemical reactions is zero. The net rate of a reaction is the difference between the rates of the forward and reverse reactions. CEE3330 Y2013 WEEK3

  13. Chemical Equilibrium/Thermodynamics aA + bB  cC + dD ( at a given temperature, T) ΔG = 0 = Σ Gf, i (products) - Σ Gf, i (reactants) Gf,A = a[Gof,A + RT ln {A}] Gf,B = b[Gof,B + RT ln {B}] Gf,C = c[Gof,C + RT ln {C}] Gf,D = d[Gof,D + RT ln {D}] 0 = -a*Gof,A - b*Gof,B + c*Gof,C + d*Gof,D+ RT ln[{A}-a{B}-b{C}+c{D}+d] = ΔGo + RT lnK here, Equilibrium Constant, K = Exp [-ΔGo /RT] CEE3330 Joonhong Park Copy Right

  14. Thermodynamics/Gibbs Free Energy Example: CNO- + 2H3O+ NH4+ + H2CO3 (1) Calculate K value Given: Gof for CNO- = -98.7kJ/mol; Gof for H3O+ = -237.2 kJ/mol; Gof for NH4+ = -79.5 kJ/mol; Gof for H2CO3 = -623.4 kJ/mol) (R=8.314 J/mol/K; RT ln K = 5.7 log10 K when unit is kJ/mol) (2) Calculate pH when [CNO-]=10-8 M and [NH4+]=[H2CO3]= 1 M CEE3330 Joonhong Park Copy Right

  15. Kinetics aA + bB => cC + dD ( at a given temperature, T) (intrinsic) Reaction Rate Rf = - (1/a) (d[A]/dt) = -(1/b)(d[B]/dt) = (1/c)(d[C]/dt) = (1/d)(d[D]/dt) In general, Rf∝ - ΔG or f (ΔG) Rate Laws Rf = kf [A]α[B]β here kf = rate constant α = reaction order with respect with [A] β = reaction order with respect with [B] CEE3330 Y2013 WEEK3

  16. Elementary versus Non-elementary Reactions Elementary Reaction Reaction orders correspond to the stoichiometric coefficients Non-elementary Reaction CEE3330 Joonhong Park Copy Right

  17. Reaction Order • The order of reaction refers the powers to which the concentrations are raised in the kinetic rate law. • The reaction is α order with respect to reactant A, and β order with respect to reactant B. • The overall order is n (= α + β ). CEE3330 Y2013 WEEK3

  18. Reaction Order (II): Unit of Rate Constant Oth order {k} = mol/L/sec 1st order {k} = 1/sec 2nd order {k} = L/mol/sec CEE3330 Joonhong Park Copy Right

  19. Empirical Determination of Reaction Order and Rate Constant (I) Oth order [A] = Ao -kt [A] = Ao -kt 1st order [A] = Ao Exp[-kt] Ln[A] =LnAo -kt 2nd order [A] = Ao /(1 + ktAo) 1/[A] =LnAo -kt Integrated Linearlized Differential CEE3330 Joonhong Park Copy Right

  20. Kinetics Example: solving a simple kinetic problem Radon-222 => polonium-218 + alpha particle (elementary reaction) The rate constant for this rxn is k = 2.1 x 10-6 s-1, independent of temperature. At t=0, a batch reactor is filled with air containing radon at concentration Co. How does the radon concentration in the reactor change over time? Solution) R = - dCRn-222/dt = kCRn-222 dCRn-222/CRn-222 = - k dt lnCRn-222 - lnCo = -k (t – 0) => CRn-222 = Co Exp(-kt) CEE3330 Joonhong Park Copy Right

  21. Rate Constants and Equilibrium Constant (I) Example: Reversible element rxn. CEE3330 Joonhong Park Copy Right

  22. Rate Constants and Equilibrium Constant (II) Where Kf = equilibrium constant on a basis of the forward rxn. CEE3330 Joonhong Park Copy Right

  23. Summary I: Dynamics of Transformation Processes Stoichiometry: aA +bB = cC + dD => Predicting the equilibrium concentrations of reactants and products of a reaction Thermodynamics:ΔG <0 ? or Q vs. K • Predicting the direction of a reaction • Comparing the feasibility of candidate reactions Kinetics:Rf = kf [A]α[B]β => Predicting the reaction rate of a reactor, and being useful in calculating non-equilibrium concentrations in a system CEE3330 Y2013 WEEK3

  24. Summary II: System and Process Dynamics • System and Process Identification aA + bB  cC + dD ( at T) • System Dynamics: Material/Mass Balance Accumulationrate = Input rate – Output rate + Reaction rate • Transformation Process Dynamics • Chemical species - Reaction and Stoichiometry • Thermodynamics - Kinetics (Reaction rate) CEE3330 Y2013 WEEK3

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