Rate Law for Second Order & Free Energy and Equilibrium Constants Questions

 (10 pts) Rate law for second order. (Basically M&S P28.24) As in class, start 22 eq. 25 and show how to get to equation 27. 2. (10 pts) The decomposition dimethyl ether results in methane, carbon monoxide and hydrogen gases. This gas phase reaction was studied by studying the time it took for the total pressure to double based on the initial pressure of the ether (presumably in a constant volume container). From the following data, i) determine the order of the reaction (rate law)* and ii) its rate constant. Initial P (torr) 28 58 150 171 261 321 394 422 509 586t (s) 1980 1500 900 824 670 625 590 508 465 4843. (10 pts) M&S P29.15 (Reaction of ethyl acetate) 4. (10 pts) Arrhenius Problem. Calculate the energy of activation for the following data for the decomposition of N2O5-> N2O4 + ½ O2, T (K) 270 370 470 570 670k x 109 (cm6 mol-2 s-1) 9.12 4.67 3.28 2.75 2.495. (10 pts) M&S P28.17 – Parallel First Order Reactions* What I did (and there are other approaches) was to first figure out what extent of decomposition (I called it a) gave 2X the original pressure. Then I used the rate laws for -(dA/dt) = kAn, where n was likely to be half an integer, i.e., 0, ½, 1, 3/2, 2, etc. and A is the pressure of the system. I plotted the data for each order and eventually, found that one of these exponents fit the data very well over the entire concentration range. (Note: From Excel, you can plot things from rows or columns, if you like columns, you can paste special -> transpose).

word image 414 MoreChemical Equilibria Chapt 26 Dependsonedition office3 5WedHRs

I FreeEnergy and Equilibrium constants

A FreeEnergy

1 Extentof reaction 5 lowercase Xi for A B

dnaEdEgjdnB dEgE26D

const.T P

DG Madha 1 Msdn 4143 Mad5 nN264

IFpp473 Ma DGr Gibbs reactionenergy 2

word image 415

word image 416

AndDGr DGFtRTlnQ lo 26.62 at equilibrium

8Gr 0 oEtRTlnk CID

Then RThnkEDGF 44526.653

word image 417 SurfaceTension D H theory B Equilibriumconstants ah tbB

cctaa.at dbifequilDQibriumQ KafI g

Ka is the thermodynamic equilibriumconstant For ideal substances

Ailsolidfl jail idealgas a ip

4

before

One could define Kp as

kp Pp Pp kKa E ougschangein

the ofmolesof

gasin bat eg

Ex ZHztINeNHzj F 620K F Datin

J j

Found 7.3ideal5 NHbehbyavmiooler look ex26.1 assume gas

PNHz 0.0735110atm 0.735atm

word image 418 PastPNE10 0.735 PHE6.949atm PNE2.3162 co735K 97

ka p.ua F.pftsp.E 2.316 46.9491312

0.261 perMole NHS’s S S 1bar

Kp Kalpa_0,0264 atm

at 50atm canestimate Kalforidealeaseland amount

of NH’s produced

Butwhen measured 5OatmKa

O.O272snloi9nhitfYealEx.NHy

Secs NHzeg t Asse Cg

F 23.1 torr 30.1 C What is off

Pam.jpase tCIIotioiifn l 5 iEatmCidealsas

ACH2Se H3 PNHzPhase 4

Ka 9 4Hse TE 2 37,10

IL 0.98712

from R DGf 8.3141303 2 In 2.3710

21.04k

C Estimation of OG

DG RT luke Vi DG i IIImung

oGor r DHr TITDsr oftenfromAHfandDsf

word image 419 or fromsomethingcalled a free energyfunction skip from Stat Mech

II TemperatureandPressureeffects

A Temperature

f H 118 526.28 p

cofficeandishoassuersx3am5 for DE wiGE

pgq.gg p jviyEfI ViHi 0HF

9 26.281

or since dffIEtta

dhgk NII or dfn GEET EEI.my

LechateliersPrinciple for an exothermic reaction theequilibriumis shifted to the left with an

increase in temperature and viceversa

In KaCI hrKACI tfhfttfdTGDkb.ua

word image 420

What if 8HEchanges

i

w

th

T

W

Ocp

F

i

V

Cp

i

Y p Ehifftp

PIETp DCPr

Example2673 integrate

aHrCT DHr Ti t f Cp r DT 64

w o change into121

lnkaltd lnko.lt AIL f t9 Iln 1251

Where A DHrCT T Depr assfamffggartdoesnot

otherwise integrate

B the urgecometRTentpo forsassfpEIS.ca

4aAtbBFcCtf didPipoDCHEKf

HE KpKp Pe Kf

Where K daa Cfpb Ca

Where Cfs arethefugacity coefficients inthe equilibriummixture bookuses 8

word image 421 Problem 105formixtures are differentthan 01s for pure components LewisandRandall estimate Is formixturesbased on thetotal pressure fi Oli Pi j di fugacity wet based on thetotal

word image 422

40 60

Cot CO2 400K 5000atm

Example What are the fi’s

CO TE134,17 3.0 12 35 17 14.3 0re2a1lly.281.55

word image 423 Coz T0957e T5r007 171j aPEtm j failPE2812001j 01256157atm foe eye Haber Process ZHdgltEMHFNH3blfi.ch kpKoPo dVgs.Ka

0.00372lKafound byextrapolation to Fo at773K

What is Kp at 773K and 600ohm

NHz TE Tr j PE PE j 0 0.89 Nz TE TE R PE j 01 1.31

Asi 01 1.26

kp ka µ po j f g 6.85403bar

measured Kp 651.153bar

Heterogeneous systems y 281

A puresolid P activitypark

pturseolid

RTlnT VmdP 29

it incompressible f exp Vm 301

Thisis notgoodwhen gases aredissolvedin liquids

Ex F expEjP weaziioEI.a10MPa mfYIg5m 103

T i P5OMIa 3.62

III Extent of Reaction

A Molefactionconst

If Dalton‘sLawholds for all components

word image 424 PEXi P and Kp IT pi X POY GD define KE.ITXii 134 thenKy KpP o

heChateliers Principle equilibriumshifts to less gaseous

species w pressure

Ex B Rxn

Csl t 029152 COG Kp Peftp.ECY P l4 llatm kpI CxEoIlatm7Xcol2xEh3kkx xcdP Pee

1100I 0 0.9OO367631820.93867631.2

  1. Moleand concentration constants

04 134

Ni T j kn IT ni j kn KpRf 135

Eat Kcft Ei j ki Kp Rt ons

E Tig

  1. Moreexamples

Brig 2 Br

given no omolesofBra howmuchBr present

F degreeofdissociation

N NollBried B2Mr od totNollaldt2D no Ita

Ea Ea

Kp Bp fIIP 91 Go

5 4k Kp 1600K 1450.255 atm

7 I in

Therecanbe aFelatilotvofe Br at lowpressures

0 is very corrosive it exists in space

Ex cg t CLG

Kpap_

word image 425 Et 50I NidOkt1OI HgcgatmTNN zHztlatmHza0const.VLg KfO336atm1

Calc fiPnFaMlpressurePapaand 0.336NHzpreatmse1nt2 0.113DoubleRynatm2

K’p

p

Ne SHI ZHAI total initP lo l O 11

finalP 10 Y 134 24 1124

Kp’s10 4 13473 0.113

fly 0 442Kpho4 I 3713

1124 10.652atm 4 0.174atm

PHE0.478atm

9oNH5toopiffbff.IE 3.27

P

OneMore Suppose inthe aboveexample

word image 426 wKx2aKIskP40.33614115eptcorecallnsKp Kpitantwomolesmoretiofthenproduct

NH p

arx Nofx7 z TEtxsokx.IE E YioIYiTNu HIsxn p

solvingforX X 01176j 4 50.0331

So it depends howyour operationworks fixedPorv which one You use

Chem 3433-63342, Physical Chemistry 1

Problem Set #10

When possible, use the values for constants from our text. If you get them from other sources (if they are not in the book) cite where you got them from.

Do, but do not hand in P26.1, P26.2 to get you started.

1. (10 pts) Rate law for second order. (Basically M&S P28.24)

As in class, start 22 eq. 25 and show how to get to equation 27.

2. (10 pts) The decomposition dimethyl ether results in methane, carbon monoxide and hydrogen gases. This gas phase reaction was studied by studying the time it took for the total pressure to double based on the initial pressure of the ether (presumably in a constant volume container). From the following data, i) determine the order of the reaction (rate law)* and ii) its rate constant.

Initial P (torr) 28 58 150 171 261 321 394 422 509 586

t (s) 1980 1500 900 824 670 625 590 508 465 484

3. (10 pts) M&S P29.15 (Reaction of ethyl acetate)

4. (10 pts) Arrhenius Problem. Calculate the energy of activation for the following data for the decomposition of N2O5 -> N2O4 + ½ O2,

T (K) 270 370 470 570 670

k x 109 (cm6 mol-2 s-1) 9.12 4.67 3.28 2.75 2.49

5. (10 pts) M&S P28.17 – Parallel First Order Reactions

* What I did (and there are other approaches) was to first figure out what extent of decomposition (I called it a) gave 2X the original pressure. Then I used the rate laws for -(dA/dt) = kAn, where n was likely to be half an integer, i.e., 0, ½, 1, 3/2, 2, etc. and A is the pressure of the system. I plotted the data for each order and eventually, found that one of these exponents fit the data very well over the entire concentration range.

(Note: From Excel, you can plot things from rows or columns, if you like columns, you can paste special -> transpose).

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