We have already referred to a certain reaction, which takes place at high temperature after rich combustion.

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1 ME 41 Day 13 Topcs Chemcal Equlbrum - Theory Chemcal Equlbrum Example #1 Equlbrum Costats Chemcal Equlbrum Example #2 Chemcal Equlbrum of Hot Bured Gas 1. Chemcal Equlbrum We have already referred to a certa reacto, whch takes place at hgh temperature after rch combusto. CO2 + H2 CO + H2O What s meat by equlbrum? The products are turg to reactats as well as the reactats turg to products. The reacto goes both ways at the same rate. The cosequece s that all speces have reached costat cocetratos the mxture. A more geeral case. M s a speces volved a reacto ad s ts molar coeffcet. 1M 1 + 2M2 +! mmm + M +! Aga, the reacto goes both ways. Speces reach costat cocetratos. Ths s the kd of stuato we have at the ed of combusto. The reacto rates are faster tha the ege s kematcs, so we assume, at least to beg wth that the bured gases are equlbrum of ths sort at the begg of the power stroke.

2 Some theory... Closed system of reactg gases. Oly PV work may be exchaged wth surroudgs. We assume the gases equlbrum are at temperature T. We also assume that the pressure, P, s costat. Image a cremetal process whch reactats go to products. The frst law of thermodyamcs smplfes to Q H The secod law of thermodyamcs smplfes to Q T S So the combato reveals that H T S Recall from last tme that the property G H - T S s called the Gbbs Free Eergy. The reacto takg place mples that G That s the system lowers ts G by reactg. Cosequetly, we have foud a hady measure for establshg equlbrum. That s whe the chage G s zero. G At the temperature ad pressure specfed, the Gbbs free eergy of the products wll match that of the reactats.

3 2. Example Problem Ths s a varato of Example 3.4 (p ) A stochometrc mx of carbo mooxde ad oxyge was bured. 1 CO + O 2 CO 2 2 The combusto product CO 2 wll the dsassocate back to the orgal speces. So, a state of chemcal equlbrum, all three speces are preset. The challege s to fd the mole fractos of each. Assume that we have T eq 25 K ad p eq 5.76 atm. The soluto wll be doe EES. We assume that the product s α CO + + 2O2 3CO2 The varable α s used to facltate comparso wth the text. We also wrte two elemet balace equatos 1 α α The total moles of dsassocated mxture s T α So we ca set up the followg molar fractos y 1 α T y 2 2 T y 3 3 T I EES we work o a per mole ut bass. Note that whe we wat to get the etropy of a speces we have to supply a pressure as well as a temperature.

4 Due to the way etropy s calculated, ths must be the partal pressure of the speces. So for CO we supply the pressure p y 1 P eq Remember that the partal pressure of a gas s obtaed by multplyg the total pressure by the mole fracto. The EES soluto ca ow be demostrated. Please dowload t from the webste. 3. Use of Equlbrum Costats Here s a more complete explaato of chemcal equlbrum. Thk of the reacto volvg 1 mole of some speces. Not mportat whch. We dvde out the molar coeffcet of that speces. We also subtract the reactat sde from the product sde. I.e., thk of wrtg the equato descrbg equlbrum as follows: M + + 1N + 1 +! 1M 1 2M2! So f a molar amout, δ, of our speces reacted (.e. chaged), the umber of moles of each of the other speces that chaged would be proportoal to t, δ δ The total chage Gbbs free eergy f we let these molar composto chages take place would be G g δ + g δ + +! g δ g δ!

5 G g δ g δ The g s are specfc Gbbs Free Eerges, the G per mole of each speces at the temperature ad pressure where equlbrum s occurrg. The text refers to these as the chemcal potetals. Symbol µ. I ll keep t g these otes. It ca be show, (the text suggests how) that for deal gases we ca wrte the followg expresso for g g g p + RTl p (Remember we are gvg these quattes o a molar bass.) The quatty g s evaluated at the temperature of terest ad stadard pressure of 1 atm. It s the Gbbs Free Eergy of formato of the substace. G g p + RTl p δ Now for arbtrary. G to be zero the sum must be zero. Remeber δ s R s the uversal gas costat, p s stadard 1 atm.

6 p l p g RT Let G be the Gbbs free eergy of the reacto as t takes place uder stadard codtos (1 atm ad the desred temperature). It ca be regarded as a costat, actually. G g See how t s based o the specfc Gbbs Free Eergy of Formato for each speces. l p p p l p G RT Let ( ) l K p G RT where we wll call Kp the equlbrum costat at costat pressure ad temperature T. We the fd that the sum s coverted to a product. K p p p " Equlbrum costats lke ths are tabulated JANAF tables. We wll eed them to uderstad our ext challege:

7 Equlbrum of Bured Gas Just AFTER Combusto Ths later. Rght ow aother example. Example 3.5 (p. 89-9) I fuel-rch combusto product mxtures, equlbrum betwee the speces CO2, H 2 O, CO ad H 2 s ofte assumed to determe the bured gas composto. For φ 1.2, for C 8 H 18 combusto products determe the mole fractos of the product speces at 17 K. I cte here the author s combusto equato usg hs ow otato. It s based o a mole of C 8 H C N 2 ( O N2 ) aco2 + bh2o + cco 2 8H18 + dh Balace C: a c 8 + H: 2b 2d 18 + O: /1.2 2a b c N: already doe. Ths descrbes the stuato completely. What s gog o the products s CO2 + H2 CO + H2O

8 Ths reacto s deemed to be chemcal equlbrum. We wll employ EES to calculate the Gbbs free eerges of products ad reactats. Please see the EES documet o the Webste.