Chemical Reactions II

Transcription

1 Mass Transfer Chemical Reactions II Lecture, 6..7, r. K. Wegner

2 . Homogeneous chemical reactions Chemical reactions can alter mass fluxes by orders of magnitude. Recall NH 3 absorption into water: log(c How much is mass transfer altered by chemical reaction? ph + H l ) mol

3 3. Mass transfer with st order homogen. reactions First order reactions are typical - even though two reactants may participate as usually one of them is in excess. In the latter case we have pseudo first-order reactions. CH 4 + O CO + H O where [ κ c ] (reactionrate) O [ κ c O ] c CH 4 is the pseudo-first order rate constant. As mass transfer and reaction are intimately coupled when homogeneous reactions are involved, detailed calculations of k are extremely hard! Instead, the mass transfer correlations will be corrected to account for reaction. This is rather easy for st order reactions.

4 Irreversible Reactions A liquid is in contact with well mixed gas: Mass balance in the film (dilute concentrations) without chemical reaction (steady state): dn dz dj dz Boundary conditions: d c dz z : c c i and z l: c c c i( z l ) j (c l i ) k l 4

5 5 Now including chemical reaction (conversion of species in the film): c dz c d r dz c d κ + Solving this equation and inserting the B.C.s gives ] / sinh[ )] ( / sinh[ l z l c c i κ κ This profile results in the curvature in our schematic instead of the standard straight line. Now when the reaction becomes rather slow: You recover the diffusion result! ( ) l z l z l l z l c c : i + κ + κ κ

6 6 Now back to the full problem: The flux in the presence of reaction is: j dc dz κ c i cosh[ κ sinh[ / ( l z)] κ / ] l j κ coth( κ / l) c ( ) At the interface z: i where ( κ coth( κ )) / l is k with chemical reaction. Again as κ then k k l

7 7 The problem now is to determine l in order to obtain k. However, we don t have to, we only need the correction to k. For this reason we only need to calculate k/k : k κ coth κ k k k ( ) ( ) When reaction is slow When the reaction is fast Note that now f ( k ) k κ k 3 (k ) κ and k κ κ so coth and k but only of κ

8 Is the simple film model adequate? Predicted corrections for chemical reaction are almost the same, independent of the model chosen! 8

9 9 Coupling between mass transfer and reaction greatly affects the temperature-dependence of k. If κ doubles every o C, then k doubles every o C In the absence of reaction k doubles every 5 o C! Example: Variation of mass transfer with fluid flow A spinning disk of reagent is immersed in a dilute solution of reagent. We measure reagent lost from the spinning disk as a function of rotation speed of the disk. How can we distinguish between the two possible reaction mechanisms: a) reagent dissolution and irreversible homogeneous reaction b) irreversible heterogeneous reaction at the solid surface

10 Remember from Chapter 4: «Generalized Mass Balances» j j / d ω υ.6 d υ / / 3.6 Re Sc d k (mass transfer coefficient) More in the upcoming chapters / 3 c c (sat) (sat) ϴ k d.6 d ω ν ν 3

11 Key dependence (no chemical reactions): k f(fluid flow) or Sh f(re) Spinning disk: k d.6 dv ν / ν / 3 a dv ν / or k a v d ν a) For st order irreversible HOMOGENEOUS reaction (see above): k κ κ coth / k ν a(v / d) ( k ) /

12 k κ coth κνd v a For a homogeneous reaction, there is a clear dependence of k with v. At low v, k const., while at high v, k~v /. b) For a st order irreversible HETEROGENEOUS reaction: + k κ k (d ν ) + κ av / /

13 3 Here, when v is small, the second RHS term becomes large, so k~v /. When now v is large, the second RHS term is insignificant and k κ. This is nicely shown in the diagram

14 4. Mass transfer with nd order homogen. reactions How do second-order chemical reactions enhance mass transfer? Mass balances: d c Species : κ c c dz Species : d c κ c c dz B.C. s: z : c c dc /dz z l: c c c l

15 5 These are difficult to solve in the general case. Of course, if one of the reactants is in excess then it reduces to the first order reaction problem. Another special case is when the reaction is FAST and irreversible: The two reactants disappear forming a reaction FRONT between TWO films.

16 (species ) + ν (species ) products n (c ) i zc n ( c ) l l zc where z c is the location of the reaction front ν n + From a mass balance: n The unknown z c can be eliminated to give: n c l ci l + ν c i As k l then: k k + c l ν c i 6

17 7 Example: SO absorption in a packed tower Known from absorption in water (no chemical reaction): Gas-side mass transfer coefficient: k P a.7s a in m /m 3 Liquid-side mass transfer coefficient: k L a s p SO mmhg in equilibrium with solution containing. wt-% SO

18 8 If we change the absorbing medium from water to a dilute NaOH, by how much will the mass transfer be improved as a function of c NaOH,? SO+ NaOH NaSO3 + HO Species : SO Species : NaOH First of all recognize that you have an acid-base second order reaction that is FAST as all acid-base reactions. At steady state: j k ( p ) P p i k k L L ( c ) i + c l ν c i c i From gas-liquid equilibrium at the interface: p i Hc i

19 9 The interfacial concentration can be calculated: kk pp pp HH cc ii kk LL cc ii + cc ll νν cc ii kk pp HH + kk LL kk pp pp + kk LL cc ll νν cc ii kk pp pp kk cc ll LL νν kk pp HH + kk LL However, the interfacial concentration is usually not of interest. We would like to know the enhancement of the overall mass transfer. Following what we learned about the overall mass transfer coefficient: jj KK LL cc KK LL pp HH

20 Now as we have a packed tower KK LL KK LL aa with the specific surface area (surface/volume) aa In absence of reaction: In the presence of reaction: Then KK LL aa KK LL aa + cc,ll HH νν pp H a k a k a K p L L + H a k a k p H c a K p L,L L + ν +

21 Calculate H from our equilibrium data pp HH cc atm.g mol SO mmhg H 3 76 mmhg cm 64 g H 84 cm 3 atm / mol With diffusion coefficients in water:.9 SO 5 NaOH. cm 5 cm s s K K L L a a + (. 5 (.9 cm 5 cm / s) (84 cm / s) 3 atm / mol) mmhg 76 mmhg /atm c + 35 liter / mol c (mol / liter)

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23 3. iffusion-controlled homogeneous reactions Examples: Acid base reactions, combustion (e.g. H or CH 4 ) These reactions are so fast that they are always diffusion-controlled. Goal: To determine the overall reaction rate As the reaction is fast, it is determined by the rate at which molecules collide with each other. κ ( species ) + ( species ) ( species 3)

24 4 Consider now a single molecule with radius a of perfect spherical shape in a control volume / (c N Av ), where c is the concentration of species and N AV is the Avogadro number. Molecules of radius a are in Brownian motion and diffuse to the surface of. Molecules are consumed by reaction after collision. A mass balance gives: r MMMMMMMMMMMMMMMMMM cccccccccccccccc bbbb rrrrrrrrrrrrrrrr MMMMMMMMMMMMMMMMMM tttttttttttttttttttttt bbbb dddddddddddddddddd

25 5 r 4π rj cn AV We would like to calculate the concentration profile c away from the surface of a sphere of radius σ so we can calculate the flux of molecules to the surface of molecules. This will give the rate of collisions of molecules and per unit area. Now we know the flux of molecules using the earlier result (e.g. dissolution of sphere, Chapter ): j + (a a ) c r Combining the equations gives: r 4π (a + a )NAVcc

26 6 Now consider that the molecule of radius a is in Brownian motion. Then we introduce the diffusion coefficient describing the relative motion of the two molecules: Einstein equation (lecture on iffusivity): x t x x t + x t ( ) x x t + Thus the reaction rate can be written as r 4π ( + )(a + a )NAVc c

27 7 κ cc As the reaction is: r κ 4π( + ) σ NAV The κ becomes Where σ is the intermolecular distance that is tabulated in the Cussler book and in various handbooks as the molecular collision diameter: σ ( σ + σ ) κ gives a good approximation of very fast rates (within a factor of ).