Lecture27A - Microscopic Connection to the Equilibrium...

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Unformatted text preview: Microscopic Connection to the Equilibrium Constant(26.8-10) Once again let's return to our generic reaction v A A ( g ) + vB B ( g ) vY Y ( g ) + vz Z ( g ) B A We showed earlier: dA = - SdT - PdV + dn for a single component system 1 T, n V A A at constant T,V 2 dA = dn and 1 T, n V n V T, ln Q ln Q Since A = - k BT ln Q = - k BT = - RT 1n V N V T, T, For a multi-component system: dA = A dnA + B dnB + Y dnY + Z dnZ so when and dni =G vi d + for products - for reactants A We have reached equilibrium and there is no = 0 mass flow between reactants and products. = vY Y + vZ Z - v A A - vB B = 0 Lecture 27 1 The partition function for the system, Q, is Q ( N A , N B , NY , N Z , T , V ) = Q ( N A , V , T ) Q ( N B , V , T ) Q ( NY , V , T ) Q ( N Z , V , T ) = qA ( V , T ) NA ! NA qB ( V , T ) NB ! NB qY ( V , T ) NY ! NY qZ ( V , T ) NZ ! = - RT NZ for ideal gases ln Q 1 For one component: A = - RT 1N A ,T , N V j A NA ( N A ln qA - ln N A !) Recall Stirling's Approx: ln N ! = N ln N - N A = - RT NA ( N A ln q A - N A ln N A + N A ) T ,V qA 1 = - RT q A - ln N A - N A ln + 1 - RT ln = NA NA vY vZ vA vB Z A B qY q q q 1A + ln - ln - ln so = 0 = - RT ln Y 1 N N N N Z A B Lecture 27 2 Separate the q's and the N's and drop the -RT Y vY qz vZ Y vY N z vZ q N qY vY qz vZ NY vY N z vZ ln vA vB ln vA vB 0 - = = vA vB q A qB A N B N q A qB N A v A N B vB This is a fairly interesting result which at equilibrium relates the ratio of populated levels (qi's) to the ratio of the populations (Ni) of molecules populating those levels. Recall for an ideal gas: q ( V , T ) = f ( T ) V so q ( V , T ) / V = f ( T ) is only a function of Temp N N q q Y Z Y Z V V V V = vA vB vA vB N N q q A B A B V V V V Note that N i / V = i = number density, which is proportional to concentration Lecture 27 3 vY vZ vY vZ q q Y z vY + vZ - v A - vB o V V and K T c RT Kc ( T ) = = KP ( T ) ) o c( vA vB q q A B P V V So we can calculate K(T) from "first principles", i.e. from molecular (i.e. microscopic properties) independent of thermodynamics. The example from the text is a good one: H 2 ( g ) + I 2 ( g ) 2 HI ( g ) We need to remember a few key points: 3/ 2 qdiatomic mk BT 2 T e -vib / 2T = 2 -vib / T e e De / kBT = hv , = hB g 1st vib rot 1 24 V h 44 2 4 rot 1 - e 43 4 q 3 kB kB 14 2 1 4 { 3 elec qtrans / V qrot qvib vY vZ 2nd For this reaction: 3rd Note e prod reac vib De - hv De - + + 2T k B T 2 k BT k BT v v = i j so K P ( T ) = K C ( T ) = K ( T ) De - hv / 2 k BT =e =e =e D0 k BT 4th H 2 = I 2 = 2; HI = 1 Lecture 27 4 M K (T) = H MI M 2 2 2 HI 4 rot , H 2 rot , I 2 2 rot , HI 3/ 2 1- e ( -vib , H 2 / T ( 1- e )( 1- e -vib , I 2 / T 2 -vib , HI / T ) ) ( e H I HI 2 D0 - D0 2 - D0 2 / k BT ) 1) Note the deviations at high temperature. Why? 2) Slope B r H o = -12.9kJ / mol vs. experimental = -13.4kJ / mol Fig 26.5 Due to H.O.-R.R approx. (HI is probably the major source of error.) But all-in-all, rather good agreement. For polyatomics, it is slightly more complicated since qrot and qvib are more complex. 1 Note in the text example: H 2 ( g ) + O2 ( g ) 1 H 2O ( g ) 2 RT c vi j so K c ( T ) Lecture 27 = K P ( T ) v react o prod P o 1/ 2 5 Our concentration units are number density, m -3 , co = 1m -3 . So co /N A would have the "standard" mol m -3 units. o RT c Thus K c ( T ) o = K P ( T ) N A P 1/ 2 T/K 1000 1500 2000 ln KP (calc) 23.5 13.1 8.52 ln KP (exp) 23.3 13.2 8.15 Table 26.3 Again, note the deviations from theory at higher temperatures. Why? For section 26.9 on the JANAF Tables: This has a lot of details and information, but is way too much depth for our (or at least my) interest. So, moving on, we often gloss over non-ideal behavior in gas-phase systems. However for many industrial processes done at high pressure, ignoring non-ideality is at our peril. As pressure increases, non-ideal behavior sets in and we must use fugacities instead. Lecture 27 6 From before: ( T , P ) = o ( T ) + RT ln f fo f P as P 0. f o = fugacity standard state = 1bar (we will drop this now, but it is understood as our standard state). for our gas mixture: i ( T , P ) = io ( T ) + RT ln fi for v A A ( g ) + vB B ( g ) vY Y ( g + vz Z ( g ) we obtain: ) fYvY f ZvZ r G = r G o + RT ln vA vB f A fB o o o o with r G o = vY Y + vz Z - v A A - vB B at eq. r G = 0 and r G o = - RT ln K f YvY 1f ZvZ f with K f = vA vB f A 1f B eq Thermodynamic equilibrium constant! 7 Lecture 27 At low pressures Kf(T) KP(T) as f P in the ideal gas limit. While we know how to determine fugacities for pure gases (straight forward procedure), this is much more complicated for gas mixtures. Why? It takes a number of measurements on mixtures of different composition and at various pressures and temperatures. 1 3 N2 ( g ) + H 2 ( g ) NH 3 ( g ) ( making NH 3 is a major industrial process ) 2 2 Total P (bar) KP/10-3 Kf/10-3 Wait, what is going on? 10 6.59 6.55 Wasn't KP(T) supposed to 30 6.76 6.59 be only a function of T? 50 6.90 6.50 Why is KP(T) varying so 100 7.25 6.36 rapidly with pressure?! 300 600 8.84 12.94 6.08 6.42 Table 26.5 all values for ammonia synthesis equilibrium at 450C. Lecture 27 8 KP(T) independence of pressure is a result of ideal gas behavior (just like Z=PV/nRT=1 for an ideal gas under any conditions). Kf(T), our thermodynamic equilibrium constant has a much smaller pressure dependence. It is almost constant over a range of 60x in pressure! Lecture 27 9 ...
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