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Thermodynamics filled in class notes_Part_119

# Thermodynamics filled in class notes_Part_119 - 245 7.1...

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Unformatted text preview: 245 7.1. ISOTHERMAL, ISOCHORIC KINETICS Thus, we once again ﬁnd ρO + 2ρO2 = ρO + 2ρO2 = constant. (7.117) As before, we can say ρO ρO2 = =ρ ρO 1 + −1 ρO2 2 =D b =ρ ξO . (7.118) =ξ This gives the dependent variables in terms of a smaller number of transformed dependent variables in a way which satisﬁes the linear constraints. In vector form, the equation becomes ρ = ρ + D · ξ. (7.119) Once again φ · D = 0. 7.1.1.2.1.2 Thermodynamics Equations (7.103-7.104) are supplemented by an expression for the thermodynamics-based equilibrium constant Kc,13 which is: Kc,13 = Po exp RT −∆Go 13 RT . (7.120) Here Po = 1.01326 × 106 dyne/cm2 = 1 atm is the reference pressure. The net change of Gibbs free energy at the reference pressure for reaction 13, ∆Go is deﬁned as 13 ∆Go = 2g o − g o 2 . O O 13 (7.121) We further recall that the Gibbs free energy for species i at the reference pressure is deﬁned in terms of the enthalpy and entropy as o g o = hi − T so . i i (7.122) o It is common to ﬁnd hi and so in thermodynamic tables tabulated as functions of T . i We further note that both Eqs. (7.103) and (7.104) are in equilibrium when ρe 2 ρe = OM 1 eee ρρρ . Kc,13 O O M (7.123) We rearrange Eq. (7.123) to ﬁnd the familiar Kc,13 = ρe ρe OO = ρe 2 O [products] . [reactants] (7.124) If Kc,13 > 1, the products are preferred. If Kc,13 < 1, the reactants are preferred. CC BY-NC-ND. 18 November 2011, J. M. Powers. 246 CHAPTER 7. KINETICS IN SOME MORE DETAIL Now, Kc,13 is a function of T only, so it is known. But Eq. (7.124) once again is one equation in two unknowns. We can use the element conservation constraint, Eq. (7.117) to reduce to one equation and one unknown, valid at equilibrium: Kc,13 = ρe ρe OO 1 ρO2 + 2 (ρO − ρe ) O (7.125) Using the element constraint, Eq. (7.117), we can recast the dynamics of our system by modifying Eq. (7.103) into one equation in one unknown: dρO = 2a13 T β13 exp dt −E 13 RT 1 1 1 1 × (ρO2 + (ρO − ρO )) (ρO2 + (ρO − ρO ) + ρO ) − ρO ρO (ρO2 + (ρO − ρO ) + ρO ) , 2 2 Kc,13 2 =ρO2 =ρM =ρM (7.126) 7.1.1.2.2 Example calculation Let us consider the same example as the previous section with T = 5000 K . We need numbers for all of the parameters of Eq. (7.126). For O , we ﬁnd at T = 5000 K that o erg , mole erg = 2.20458 × 109 . mole K hO = 3.48382 × 1012 (7.127) so O (7.128) So erg erg − (5000 K ) 2.20458 × 109 , mole mole K erg = −7.53908 × 1012 . mole go = O 3.48382 × 1012 (7.129) For O2 , we ﬁnd at T = 5000 K that o erg , mole erg = 3.05406 × 109 . mole K hO2 = 1.80749 × 1012 (7.130) so 2 O (7.131) So erg erg − (5000 K ) 3.05406 × 109 mole mole K 13 erg = −1.34628 × 10 . mole go 2 = O 1.80749 × 1012 CC BY-NC-ND. 18 November 2011, J. M. Powers. (7.132) ...
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