Thermodynamics filled in class notes_Part_87

Thermodynamics filled in class notes_Part_87 - 181 5.5....

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Unformatted text preview: 181 5.5. CHEMICAL KINETICS OF A SINGLE ISOTHERMAL REACTION 3 k(T) (m /kmole/s) 10000 0.0001 1. x 10-12 1. x 10-20 1. x 10-28 1. x 10-36 1000 1500 2000 3000 T (K) 7000 10000 5000 Figure 5.1: k (T ) for Nitrogen dissociation example. Now the general equation for kinetics of a single reaction, Eq. (5.279), reduces for N2 molar concentration to dρN2 dt −E RT = νN2 aT β exp ′ ′ (ρN2 )νN2 (ρN )νN 1− 1 (ρ )νN2 (ρN )νN Kc N2 . (5.302) ′ ′ Realizing that νN2 = 2, νN = 0, νN2 = −1, and νN = 2, one gets dρN2 dt = −E RT − aT β exp ρ2 2 1 − N 1 ρ2 N K c ρN 2 . (5.303) =k ( T ) Examine the primary temperature dependency of the reaction k (T ) = = = aT β exp −E RT 7.0 × 1018 , (5.304) m3 K 1.6 kmole s 7.0 × 1018 exp T 1.6 kJ −941550 kmole kJ 8.314 kmole K T T −1.6 exp −1.1325 × 105 T . , (5.305) (5.306) Figure 5.1 gives a plot of k (T ) which shows its very strong dependency on temperature. For this problem, T = 6000 K , so −1.1325 × 105 7.0 × 1018 exp 60001.6 6000 m3 = 40071.6 . kmole s Now, the equilibrium constant Kc is needed. Recall k (6000) = Kc = Po RT PN i=1 νi exp −∆Go RT . , (5.307) (5.308) (5.309) CC BY-NC-ND. 18 November 2011, J. M. Powers. 182 CHAPTER 5. THERMOCHEMISTRY OF A SINGLE REACTION f ( ρN ) (kmole/m 3 /s) 2 unstable equilibrium 1 0.0005 0.001 0.0015 0.002 ρ -1 N2 unstable equilibrium 0.0025 (kmole/m 3 ) stable, physical equilibrium -2 Figure 5.2: Forcing function, f (ρN2 ), which drives changes of ρN2 as a function of ρN2 in isothermal, isochoric problem. For this system, since = Po RT = Kc Po RT = = N i=1 νi = 1, this reduces to exp −(2go − g o 2 ) N N exp −(2(hN − T so ) − (hN2 − T so 2 )) T,N T,N RT , (5.310) o o RT 100 exp (8.314)(6000) kmole 0.000112112 . m3 , (5.311) −(2(597270 − (6000)216.926) − (205848 − (6000)292.984)) ,(5.312) (8.314)(6000) (5.313) The differential equation for N2 evolution is then given by dρN2 dt = − 40071.6 m3 kmole ρ2 2 N 1− 1 0.000112112 kmole m3 2 2.00465 × 10−3 ρN 2 kmole m3 2 − ρN2 , ≡f (ρN2 ) (5.314) (5.315) = f (ρN2 ). The system is at equilibrium when f (ρN2 ) = 0. This is an algebraic function of ρN2 only, and can be plotted. Figure 5.2 gives a plot of f (ρN2 ) and shows that it has three potential equilibrium points. It is seen there are three roots. Solving for the equilibria requires solving 0 = − 40071.6 m3 kmole ρ2 2 N 1− 1 0.000112112 kmole m3 2 2.00465 × 10−3 ρN 2 kmole m3 − ρN 2 2 . (5.316) CC BY-NC-ND. 18 November 2011, J. M. Powers. ...
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.

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