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CHAPTER 3:Review of Chemical Equilibrium — Introduction

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CHAPTER 3:Review of Chemical Equilibrium — Introduction CHAPTER 3:Review of Chemical Equilibrium | Introduction Copyright © 2020 by Nob Hill Publishing, LLC This chapter is a review of the equilibrium state of a system that can undergo chemical reaction Operating reactors are not at chemical equilibrium, so why study this? Find limits of reactor performance Find operations or design changes that allow these restrictions to be changed and reactor performance improved 1 / 75 Thermodynamic system T ; P; nj Variables: temperature, T , pressure, P, the number of moles of each component, nj , j = 1;:::; ns . Specifying the temperature, pressure, and number of moles of each component then completely specifies the equilibrium state of the system. 2 / 75 Gibbs Energy The Gibbs energy of the system, G, is the convenient energy function of these state variables The difference in Gibbs energy between two different states X dG = −SdT + VdP + µj dnj j S is the system entropy, V is the system volume and µj is the chemical potential of component j. 3 / 75 Condition for Reaction Equilibrium Consider a closed system. The nj can change only by the single chemical reaction, X ν1A1 + ν2A2 )−*− ν3A3 + ν4A4 νj Aj = 0 j Reaction extent. dnj = νj d" Gibbs energy. X dG = −SdT + VdP + (νj µj ) d" (3.2) j For the closed system, G is only a function of T ; P and ". 4 / 75 Partial derivatives X dG = −SdT + VdP + (νj µj ) d" j @G S = − (3.3) @T P;" @G V = (3.4) @P T ;" X @G ν µ = (3.5) j j @" j T ;P 5 / 75 G versus the reaction extent G @G X = ν µ = 0 @" j j j " A necessary condition for the Gibbs energy to be a minimum X νj µj = 0 (3.6) j 6 / 75 Other forms: activity, fugacity ◦ µj = Gj + RT ln aj aj is the activity of component j in the mixture referenced to some standard state ◦ Gj is the Gibbs energy of component j in the same standard state. The activity and fugacity of component j are related by ◦ aj = fj =fj fj is the fugacity of component j ◦ fj is the fugacity of component j in the standard state. 7 / 75 The Standard State The standard state is: pure component j at 1.0 atm pressure and the system temperature. ◦ ◦ Gj and fj are therefore not functions of the system pressure or composition ◦ ◦ Gj and fj are strong functions of the system temperature 8 / 75 Gibbs energy change of reaction ◦ µj = Gj + RT ln aj X X ◦ X νj µj = νj Gj + RT νj ln aj (3.8) j j j P ◦ ◦ The term j νj Gj is known as the standard Gibbs energy change for the reaction, ∆G . ◦ Y νj ∆G + RT ln aj = 0 (3.9) j 9 / 75 Equilibrium constant ◦ K = e−∆G =RT (3.10) Another condition for chemical equilibrium Y νj K = aj (3.11) j K also is a function of the system temperature, but not a function of the system pressure or composition. 10 / 75 Ideal gas equilibrium The reaction of isobutane and linear butenes to branched C8 hydrocarbons is used to synthesize high octane fuel additives. isobutane + 1{butene )−*− 2; 2; 3{trimethylpentane I + B )−*− P Determine the equilibrium composition for this system at a pressure of 2.5 atm and temperature of 400K. The standard Gibbs energy change for this reaction at 400K is −3:72 kcal/mol [5]. 11 / 75 Solution The fugacity of a component in an ideal-gas mixture is equal to its partial pressure, fj = Pj = yj P (3.12) ◦ fj = 1:0 atm because the partial pressure of a pure component j at 1.0 atm total pressure is 1.0 atm. The activity of component j is then simply P a = j (3.13) j 1 atm a K = P (3.14) aIaB 12 / 75 Solution (cont.) ◦ K = e−∆G =RT K = 108 P y K = P = P PIPB yIyBP in which P is 2.5 atm. Three unknowns, one equation, X yj = 1 j Three unknowns, two equations. What went wrong? 13 / 75 Ideal-gas equilibrium, revisited Additional information. The gas is contained in a closed vessel that is initially charged with an equimolar mixture of isobutane and butene. Let nj0 represent the unknown initial number of moles nI = nI0 − " nB = nB0 − " nP = nP0 + " (3.16) Summing Equations 3.16 produces nT = nT 0 − " in which nT is the total number of moles in the vessel. 14 / 75 Moles to mole fractions The total number of moles decreases with reaction extent because more moles are consumed than produced by the reaction. Dividing both sides of Equations 3.16 by nT produces equations for the mole fractions in terms of the reaction extent, nI0 − " nB0 − " nP0 + " yI = yB = yP = nT 0 − " nT 0 − " nT 0 − " Dividing top and bottom of the right-hand side of the previous equations by nT 0 yields, y − "0 y − "0 y + "0 y = I0 y = B0 y = P0 I 1 − "0 B 1 − "0 P 1 − "0 0 in which " = "=nT 0 is a dimensionless reaction extent that is scaled by the initial total number of moles. 15 / 75 One equation and one unknown 0 0 (yP0 + " )(1 − " ) K = 0 0 (yB0 − " )(yI0 − " )P 0 0 0 0 (yB0 − " )(yI0 − " )KP − (yP0 + " )(1 − " ) = 0 0 Quadratic in " . Using the initial composition, yP0 = 0; yB0 = yI0 = 1=2 gives "02(1 + KP) − "0(1 + KP) + (1=4)KP = 0 The two solutions are 1 ± p1=(1 + KP) "0 = (3.19) 2 16 / 75 Choosing the solution The correct solution is chosen by considering the physical constraints that mole fractions must be positive. The negative sign is therefore chosen, and the solution is "0 = 0:469. The equilibrium mole fractions are then computed from Equation 3.18 giving −2 yI = 5:73 × 10 −2 yB = 5:73 × 10 yP = 0:885 The equilibrium at 400K favors the product trimethylpentane. 17 / 75 Second derivative of G. Please read the book for this discussion. I will skip over this in lecture. 18 / 75 Evaluation of G. Let's calculate directly G(T ; P;"0) and see what it looks like. X G = µj nj (3.34) j ◦ µj = Gj + RT [ln yj + ln P] X ◦ X G = nj Gj + RT nj [ln yj + ln P] (3.35) j j For this single reaction case, nj = nj0 + νj ", which gives X ◦ X ◦ ◦ nj Gj = nj0Gj + "∆G (3.36) j j 19 / 75 Modified Gibbs energy P ◦ 0 G − j nj0Gj Ge(T ; P;" ) = (3.37) nT 0RT Substituting Equations 3.35 and 3.36 into Equation 3.37 gives ◦ 0 ∆G X nj Ge = " + [ln yj + ln P] (3.38) RT nT 0 j Expressing the mole fractions in terms of reaction extent gives 0 0 X 0 (yj0 + νj " ) Ge = −" ln K + (yj0 + νj " ) ln + ln P 1 +ν" ¯ 0 j 20 / 75 Final expression for the modified Gibbs Energy 0 0 0 X 0 yj0 + νj " Ge = −" ln K + (1 +ν" ¯ ) ln P + (yj0 + νj " ) ln (3.39) 1 +ν" ¯ 0 j T and P are known values, so Ge is simply a shift of the G function up or down by a constant and then rescaling by the positive constant 1=(nT 0RT ). The shape of the function Ge is the same as G The minimum with respect to "0 is at the same value of "0 for the two functions. 21 / 75 Minimum in G for an ideal gas Goal: plot Ge for the example and find the minimum with respect to "0 I + B )−*− P P For this stoichiometry: j νj =ν ¯ = −1. Equimolar starting mixture: yP0 = 0; yI 0 = yB0 = 0:5 0 0 0 Ge(T ; P;" ) = −" ln K(T ) + (1 − " ) ln P+ "0 ln("0) + 2(0:5 − "0) ln(0:5 − "0) − (1 − "0) ln(1 − "0) (1) 22 / 75 A picture is worth 1000 words Recall that the range of physically significant "0 values is 0 ≤ "0 ≤ 0:5 and what do we see... 1 −1:88 −1:89 0:5 −1:9 −1:91 0 −1:92 −1:93 Ge −0:5 −1:94 −1:95 0:45 0:46 0:47 0:48 0:49 0:5 −1 −1:5 −2 0 0:1 0:2 0:3 0:4 0:5 "0 23 / 75 A closer look Good agreement with the calculated value 0:469 The solution is a minimum, and the minimum is unique. 24 / 75 Effect of pressure From Equation 3.39, for an ideal gas, the pressure enters directly in the Gibbs energy with the ln P term. Remake the plot for P = 2:0. Remake the plot for P = 1:5. How does the equilibrium composition change. Does this agree with Le Chatelier's principle? For single liquid-phase or solid-phase systems, the effect of pressure on equilibrium is usually small, because the chemical potential of a component in a liquid-phase or solid-phase solution is usually a weak function of pressure. 25 / 75 Effect of temperature The temperature effect on the Gibbs energy is contained in the ln K(T ) term. This term often gives rise to a large effect of temperature on equilibrium. We turn our attention to the evaluation of this important temperature effect in the next section. 26 / 75 Evaluation of the Gibbs Energy Change of Reaction We usually calculate the standard Gibbs energy change for the reaction, ∆G ◦, by using the Gibbs energy of formation of the species. The standard state for the elements are usually the pure elements in their common form at 25◦C and 1.0 atm.
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