(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).

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

AndDGr DGFtRTlnQ _lo 26.62 at equilibrium

8Gr 0 oEtRTlnk CID

Then RThnkEDGF 44526.653

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 0.735^atm

PastPNE10 0.735 PHE6.949atm PNE2.3162 co735K 97

ka p.ua F.pftsp.E _2.316 46.9491312

0.261 perMol^e 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 oftenfromAHfan^dDsf

or fromsom^ethingcalled 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

What if 8HEchanges

i

w

th

T

W

Ocp

F

i

V

Cp

i

Y p Ehifftp

PIET_p DCP_r

Example2673 integrate

aHr^CT DHr Ti t f Cp _r DT 64

w o change into121

lnkaltd lnko.lt AIL ^f t9 Iln ¹²⁵¹

Where A DHr^CT 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

Problem 105formixtures are differentthan 01_s ^for pure components LewisandRandall estimate Is formixturesbased on thetotal pressure fi Oli Pi j di fugacity wet based _on thetotal

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

Coz T0957e T5^r0⁰⁷ 171j aPEtm j failPE2812001j 01256157atm foe eye Haber Process ZHdgltEMHFNH3blfi.ch kpKoPo dVgs.Ka

0.00372lKafound byextrapolati_on 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 system^{s y} ₂₈₁

A pur_esolid P activitypark

pturseolid

RTlnT VmdP 29

it incompressible f exp V^m 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‘sLaw^holds for all components

PEXi P and Kp IT pi X POY GD define KE.ITXii 134 thenKy KpP _o

heChatelier‘s 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 NollBrie^d B2Mr od totNollaldt2D no Ita

Ea Ea

Kp ^Bp fIIP 91 Go

5 4k _Kp 1600^K 1450.255 atm

7 I in

Thereca_nbe ^a_Felatilo^t_v^of_e Br at lowpressures

0 is very corrosive it exists in space

Ex cg t CLG

^Kpap_

Et 50I NidOkt1OI HgcgatmTNN zHztlatmHza0const.VLg KfO336atm1

Cal_c fiPnFaMlpressurePapaand 0.336NHzpreatmse1nt₂ 0.113^DoubleRynatm2

_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

wKx2a_KIsk_P40.33614115eptco^recallnsKp _Kpitantwomoles_moretiofthen^product

NH p

arx Nofx7 z TEtxsokx.IE E YioIYiTNu HIsxn p

solvingforX X ⁰¹¹⁷⁶j 4 50.0331

So it depends ho^wyour 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).