chapter 07 equilibrium

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Equilibrium

181

Equilibrium7.1 Dynamic equilibrium

7.2 The position of equilibrium

17.1 Liquid-vapour equilibrium (AHL)

17.2 The equilibrium law (AHL)

The Vapour pressure of mixtures (EXT)

Distillation and fractional distillation (EXT)

Raoults law and deviations from ideality (EXT)

Colligative properties of solutions (EXT)

Other equilibrium constants (EXT)

7

Many cemca reacons go o compeon because eproducs are muc more energecay avourabean e reacans and e acvaon energy s ow enoug

o aow or a rapd reacon a e amben emperaure,

or exampe, e neurasaon o aqueous sodum

ydroxde by ydrocorc acd. Oer poena reacons

do no occur eer because, oug energecay easbe,

e acvaon energy barrer s oo grea or sgncan

reacon a e amben emperaure (suc as e

combuson o sucrose a room emperaure), or because as

we as e acvaon energy beng oo g, e reacansare muc more energecay sabe an e producs (as

n e decomposon o waer o ydrogen and oxygen).

W some cemca sysems owever e energes o e

reacans and producs are o a smar order o magnude

so a e reacon s reversbe, a s can occur n

eer drecon. An exampe s e reacon o ammona

and ydrogen corde o orm ammonum corde. I

ammona gas and ydrogen corde gas are mxed ey

reac o orm a we smoke o sod ammonum corde.

Conversey ammonum corde s eaed, en some o e

sod dsappears, because as been convered no ammona

and ydrogen corde gas. We ndcae suc a reversible

reaction by means o a doube arrow as sown beow:

NH3(g) + HC (g) NH4C (s)

I suc a sysem s esabsed n a cosed vesse (so a

no gases can escape) and a a consan emperaure, en a

chemical equilibrium s esabsed.

Chemical equilibrium s e sae o dynamc equbrum

a occurs n a cosed sysem wen e orward and

reverse reacons o a reversbe reacon occur a e same

rae. I we consder mxng ogeer wo reacans A and B

n e reversible reaction:

A + B C + D

en nay e orward reacon occurs rapdy, bu as e

concenraons o e reacans a s rae decreases. he

reverse reacon nay canno occur a a, bu as soon as

C and D sar o orm, s rae ncreases. Evenuay e rae

o e wo reacons becomes equa, e concenraons

reac consan vaues and equbrum s esabsed. hs

s sown n Fgure 701.

7.1 Dynamic Equilibrium7.1.1 Outline the characteristics of chemical

and physical systems in a state of

equilibrium.

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CHAPTER 7

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Time

Concen

tration

Products

Reactants

(a)

Equilibrium

Time

Forward

Reverse

(b)

Equilibrium

Rate

Figure 701 (a), (b) The change of concentration (a) and

rate of reaction (b) with time in establishing a chemicalequilibrium

In an equbrum a o e speces nvoved, bo reacans

and producs, are presen a a consan concenraon. As a

consequence, macroscopc properes o e sysem (a s

ose a can be observed or measured, suc as s coour,

densy, pH) are consan, even oug on a moecuar

scae ere s connua nerconverson o reacans and

producs. he concenraons o e speces a equbrum

w relec ow ready ey reac on coson. I wo

speces reac on every coson, en e concenraon

requred o produce a gven rae o reacon w be muc

ess an ey ony ave a 10% cance o reacng. InFgure 701 above, e reacans (A & B) reac ogeer ar

more easy an e producs (C & D) because a smaer

concenraon s requred o gve e same rae o reacon.

A smar equbrum coud obvousy be esabsed by

mxng ogeer C and D.

A specc exampe o suc a sysem woud be e

nroducon o one moe o qud dnrogen eroxde

(N2O

4) no an evacuaed, seaed one dm3 lask a ~80 C.

he coouress dnrogen eroxde w nay vapourse

and en sar o decompose no brown nrogen doxde

(NO2). he rae o s decomposon w a as e

concenraon o dnrogen eroxde decreases. Inay

ere s no nrogen doxde presen o dmerse, bu as

more s produced e rae o e reverse reacon o orm

dnrogen eroxde w ncrease. Evenuay e wo raes

w become equa and cemca equbrum s esabsedas ndcaed by e ac a e brown coour o e gas

does no cange any urer and e pressure n e lask

remans consan. Under ese condons equbrum

woud occur wen abou 60% o e dnrogen

eroxde as been convered o nrogen doxde, so

a e concenraon o dnrogen eroxde w a o

0.4 mo dm3 and e concenraon o nrogen doxde

w ncrease o 1.2 mo dm3, because eac dnrogen

eroxde moecue decomposes o gve wo nrogen

doxde moecues. hs s sown n e equaon beow

and n Fgure 702.

N2O4 (g) 2 NO2 (g)

Ina concenraon 1 mo dm3 0 mo dm3

Equbrum concenraon 0.4 mo dm3 1.2 mo dm3

Time

Concentration/moldm3

(a)

0

0.5

1.0

1.5NO

2

N2O

4

2.0

Time

Ra

te

N2O

42 NO

2

2 NO2

N2O

4

(b)

Figure 702 (a), (b) The equilibrium established by

heating dinitrogen tetroxide

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Equilibrium

183

I 2 moes o nrogen doxde were cooed o 80 C rom a

muc ger emperaure, a wc ere was no dnrogen

eroxde presen, en e brown coour woud ade o a

consan vaue, bu evenuay exacy e same poson o

equbrum woud be reaced, as sown n Fgure 703.

Time

Concentration(moldm3

) (a)

0

0.5

1.0

1.5NO

2

N2O

4

2.0

Time

Rate

N2O

42 NO

2

(b)

2 NO2

N2O

4

Figure 703 (a), (b) The equilibrium established by

cooling nitrogen dioxide

Exercise 7.1

1. In a sysem a equbrum, wc o e oowng s

no aways rue?

A here are bo reacans and producspresen.

B he orward and reverse reacons occur

a e same rae.

C he concenraons o reacans and

producs are equa.

D he concenraons o reacans and

producs reman consan.

2. Wen sod posporus(V) corde s eaed

decomposes o sod posporus(III) corde and

corne gas. Conversey wen posporus(III)

corde s saken n an amospere o corne, orms posporus(V) corde.

a) Wre a baanced equaon or s reversbe

reacon, w posporus(V) corde on e

et and sde.

b) Consder warmng some posporus(V)

corde n an empy, seaed lask:

Wa w appen o e posporus(V)

corde?

As me passes wa w appen o e

rae a wc s occurs? Wy?

Inay, wa s e rae o reacon

beween posporus (III) corde andcorne? Wy?

v As me passes wa w appen o e

rae a wc s occurs? Wy?

v Evenuay wa w appen o e raes

o ese wo processes?

v Wa name s gven o s sae?

v A s pon wa speces w be

presen n e lask? W er

concenraons a be equa?

c) Woud ere be a dference e

posporus(V) corde was eaed n an

open beaker? I so expan wy and predc

wa woud n ac appen.

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he rae a wc a reacon occurs depends upon e

concenraon o e speces nvoved. Le us make e

assumpon a bo e orward and reverse reacon n

e equbrum

A + B C + D

o be rs order n eac speces (e na resu can aso beproved or a more genera case, bu e proo s s muc

more compcaed!) en e rae expressons or e

orward and reverse reacons are:

Forward rae = kf.[A][B]

Reverse rae = kr.[C][D]

A equbrum ese raes are equa, so

kf.[A][B] = k

r.[C][D] wc rearranges o

k

f

__

k

r= [C][D]______[A][B]

As kf

and krare consans, a a gven emperaure, er rao

mus aso be a consan. hs s known as e equbrum

consan, Kc.

More generay, e equilibrium constant s gven by

e concenraon o e producs rased o e power o

er socomerc coeicens (a s e numbers a

appear beore em n e baanced equaon) dvded by

e concenraons o e reacans aso rased o ese

powers. For a genera reacon:

a A + b B + c C + ... p P + q Q + r R + ...

he equbrum consan s gven by

Kc=

[P ]p[Q ]q[R ]r ...____________[A ]a[B ]b[C ]c ...

For exampe n e equbrum beween ammona gas

and oxygen gas o gve nrogen monoxde gas and waer

vapour:

4 NH3(g) + 5 O

2(g) 4 NO (g) + 6 H

2O (g)


Kc=

[NO]4[H2O ]6___________[NH3]4[O2]5 mo dm3he equbrum consan does no ave xed uns and

ey mus be cacuaed n eac case rom e equaon

or Kc, usng e ac a concenraons ave uns o

mo dm3.

I e concenraons o e speces nvoved were a

1 mo dm3, en Kc

woud ave a vaue o one. I Kc

s

greaer an one en e concenraons o producs are

greaer an ose o e reacans and e equbrum s

sad o e on e rg and sde. I Kc s very arge ereacon can be regarded as gong o compeon. I K

cs

ess an one e oppose s rue and e equbrum s

sad o e on e et and sde. IKc

s very sma, en e

reacon may be consdered no o occur.

he concenraons o ceran subsances reman consan,

so ese are omed rom e equbrum consan

expresson. A sods ave a xed densy and ence a

consan concenraon, so ese are omed. For exampe

n e equbrum:

NH4C (s) NH

3(g) + HC (g)

e equbrum consan s smpy gven by

Kc

= [NH3][HC] mo2 dm6

because e consan concenraon o e sod ammonum

corde s omed. he concenraon o any pure

qud s aso consan as oo as a xed densy. hs

s parcuary mporan or e concenraon o waer

7.2 ThE pOsiTiOn Of Equilibrium

7.2.1 Deduce the equilibrium constant

expression (Kc) from the equation for a

homogeneous reaction.

7.2.2 Deduce the extent of a reaction from the

magnitude of the equilibrium constant.

7.2.3 Apply Le Chateliers principle to predict

the qualitative effects of changes of

temperature, pressure and concentration

on the position of equilibrium and on the

value of the equilibrium constant.

7.2.4 State and explain the effect of a catalyst

on an equilibrium reaction.

7.2.5 Apply the concepts of kinetics and

equilibrium to industrial processes.

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Equilibrium

185

wc s aken as consan and omed or equbra n

due aqueous souons.

he equbrum consan or e ormaon o e

eracorocobaae(II) on n due souon s ereore

wren as sown beow:

Co(H2O)62+(aq)+ 4 C(aq) CoC42(aq)+ 6 H2O()

Kc=

[CoC42]_______________

[Co(H2O )62+][C]4

I owever e concenraon o waer can vary because

e reacon s no n aqueous souon, or exampe n

e equbrum beow, were e pure quds woud be

mxed:

CH3COOH () + C

2H

5OH ()

CH3COOC

2H

5() + H

2O ()

Kc =

[CH3COOC2H5][H2O ]____________________[CH3COOH ][C2H5OH ]

Waer mus aso be ncuded s n e gas pase, as

or exampe n e reducon o carbon doxde sown n

equaon (1) beow, en e waer mus be ncuded, or

wc e vaue or e equbrum consan a 1000 K s

gven:

(1) H2(g) + CO

2(g) H

2O (g) + CO (g)

Kc=

[CO ][H2O ]__________[H2][CO2]

= 0.955

(Noe - ere Kcs uness as e uns cance)

hs equbrum coud easy be wren e oer way

round, as sown n equaon (2) beow, and n s case e

equbrum consan s e recprocao e vaue gven

above:

(2) H2O (g) + CO (g) H

2(g) + CO

2(g)

Kc=

[H2][CO2]__________[CO ][H2O ]

= 1_____0.955

= 1.05 (agan e uns cance)

Anoer neresng suaon o consder s wen e

reacan n one equbrum s e produc o a prevous

equbrum. For exampe e carbon monoxde n (2)

coud be e produc o e reacon o carbon w seam

sown n (3) beow (e vaue o e equbrum consan

agan beng a a 1000 K):

(3) H2O (g) + C (s) H

2(g) + CO (g)

Kc=

[H2][CO ]________[H2O ]

= 4.48 104 mo dm3

hese wo equbra can en be combned, as sown n

(4) and n s case e equbrum consan s e produco ose or e wo separae equbra:

(4) 2 H2O (g) + C (s) 2 H

2(g) + CO

2(g)

Kc=

[H2]2[CO2]__________

[H2O ]2

=[H2][CO2]__________[CO ][H2O ]

[H2][CO ]________[H2O ]

= 1.05 4.48 104

= 4.70 10

4

mo dm

3

ThEEffEcTOfcOnDiTiOnsOnThE

pOsiTiOnOfEquilibrium

I e condons (suc as emperaure, pressure, or e

concenraons o e speces nvoved) under wc e

equbrum s esabsed are canged, en e raes

o e orward and reverse reacons w no onger be

equa. As a resu e equbrum s dsurbed and e

concenraons o e speces w cange un e raes

once agan become equa and equbrum s once moreesabsed.

Le Chateliers principle s a way o predcng e drecon

(orward or reverse) n wc e poson o equbrum

w cange e condons are aered. I saes

If a change is made to the conditions of achemical equilibrium, then the position ofequilibrium will readjust so as to minimisethe change made.

hs means a ncreasng a concenraon o a speces

w resu n a cange a w cause a concenraon odecrease agan; ncreasng pressure w resu n a cange

a w cause e pressure o decrease agan; ncreasng

e emperaure w resu n a cange a w cause e

emperaure o decrease agan. he efecs o canges n

e condons o equbrum are summarsed n Fgure

704.

Noe a ws canges n concenraon and pressure

afec e poson o equbrum and e amouns o e

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CHAPTER 7

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varous speces presen, ey ave no efec on e vaue

o e equbrum consan, Kc, because e vaues o e

orward and reverse rae consans kf

and kr

respecvey

do no cange. A cange n emperaure does oweverafec e rae consans, so a e vaue oKc

canges as

we as e poson o equbrum.

I an unreacve gas s added o a xed voume o e

equbrum, so a concenraons reman consan, en

ere s no efec. I owever e gas s added a a consan

pressure, so a e oa voume as o ncrease, en e

concenraons o a speces w decrease, so e efec s

smar o a o reducng e oa pressure.

he presence o a caays reduces e acvaon energy

o bo e orward and reverse reacons by e same

amoun. hs means a bo e orward and reversereacons are speeded up by e same acor, so even oug

e equbrum s esabsed more rapdy, neer e

poson o equbrum nor e vaue o e equbrum

consan are afeced.

coettoI e concenraon o a speces s ncreased, en e

equbrum moves owards e oer sde causng e

concenraon o a o a vaue beween e orgna

concenraon and e ncreased vaue. Conversey e

concenraon o a speces s reduced e equbrum sts

owards e sde o e equbrum on wc occurscausng s concenraon o ncrease o a vaue beween

e orgna concenraon and e reduced vaue.

Consder e equbrum:

Fe(H2O)

63+(aq) + SCN(aq)

Yeow-Brown Coouress

[Fe(H2O)

5SCN]2+(aq) + H

2O ()

Boodred

I aqueous ocyanae ons are added o an aqueous

souon o an ron(III) sa, en a boodred coouraon

s observed owng o e ormaon o e compex on

sown. I e concenraon o eer e ocyanae

on or e ron(III) on s ncreased, en e nensy

o e coouraon ncreases. hs s n keepng w Le

Caeers prncpe because e st o e equbrum

o e rg causes e concenraon o e added reacan

o a agan. I aso sows a e reacon as no gone o

compeon because addon o eer reacan causes an

ncrease n e amoun o produc. I e concenraon oron(III) ons s decreased by addng luorde ons (wc

orm e very sabe FeF6

3 compex on) en e nensy

o e coouraon decreases. hs s n keepng w Le

Caeers prncpe because e st o e equbrum

o e et produces more aqueous ron(III) ons o

counerac e reducon caused by e luorde ons. I

aso sows a e reacon s reversbe. Noe a even

oug e poson o equbrum s aered, e cange

n concenraons s suc a e vaue o Kc remans

uncanged.

peeI e oa pressure o a sysem (P) s ncreased en e

equbrum sts o e sde w eas moes o gas, so

causng e pressure o a o a vaue beween e orgna

pressure and e ncreased vaue. Conversey e oa

pressure o e sysem s reduced e equbrum sts

owards e sde w e mos moes o gas, causng

e pressure o ncrease o a vaue beween e orgna

pressure and e reduced vaue. Consder e exampes

beow:

Change Efect on Equilibrium Does Kc

change?

Increase concenraon Sts o e oppose sde No

Decrease concenraon Sts o a sde No

Increase pressure Sts o sde w eas moes o gas No

Decrease pressure Sts o sde w mos moes o gas NoIncrease emperaure Sts n endoermc drecon Yes

Decrease emperaure Sts n exoermc drecon Yes

Add caays No efec No

Figure 704 The effect of changes in conditions on the position of an equilibrium

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Equilibrium

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2 SO2

(g) + O2

(g) 2 SO3(g)

3 moes o gas s convered o o 2 moes o gas

Increased P, equbrum;

Decreased P, equbrum

C (s) + H2O (g) CO(g) + H2 (g)

1 moe o gas s convered o o 2 moes o gas(noe: carbon s a sod)

Increased P, equbrum;

Decreased P, equbrum

H2(g) + I

2(g) 2 HI (g)

2 moes o gas s convered o 2 moes o gasCangng P as no efec

Noe a even oug e poson o equbrum s

aered, e canges n e concenraons a resu

rom e canges n pressure are suc a e vaue oKc

remans uncanged.

TeeteI e emperaure o a sysem s ncreased en e

equbrum sts n e drecon o e endothermic

change, so absorbng ea and causng e emperaure

o a o a vaue beween e orgna emperaure and

e ncreased vaue. Conversey e emperaure o e

sysem s reduced e equbrum sts n e drecon

o e exothermic change, so reeasng ea and causng

e emperaure o ncrease o a vaue beween e orgna

emperaure and e reduced vaue. Consder e exampes

beow:

N2(g) + O2(g) 2 NO (g)H= +180 kJ mo1(.e. orward reacon endoermc)

Increased T, Kc

ncreases, equbrum;

Decreased T, Kc

decreases, equbrum

2 SO2(g) + O

2(g) 2 SO

3(g)

H= 197 kJ mo1(.e. orward reacon exoermc)Increased T, K

cdecreases, equbrum;

Decreased T, Kc

ncreases, equbrum

Noe a canges n emperaure afec e rae consans

o e orward and reverse reacons o dferen exens,

so e acua vaue oKc

canges.

EXplanaTiOnsLe Caeers prncpe s jus a memory ad a eps us

o predc e efec a a cange n condons w ave

on e poson o an equbrum. I s not an expanaon

owhyese canges occur. he expanaon come rom a

consderaon o e efec o e cange n condons on

e raes o e orward and reverse reacons.

Consder e efecs o e canges n condons on e

equbrum beween sod posporus penacorde,

qud posporus rcorde and gaseous corne:

PC5(s) PC

3() + C

2(g) H= +88 kJ mo1

I e concenraon o corne s ncreased, en e

rae o e reverse reacon w ncrease, bu e orward

reacon w be unafeced, so a e reacon raes are

no onger equa. In order or equbrum o be resored,

e amoun o e penacorde mus ncrease and eamoun o e rcorde decrease so a e poson

o e equbrum w st o e et, ence e amoun

o corne decreases o beow s new ger eve, bu

s above e orgna eve (see Fgure 705). hs s n

agreemen w e predcons o Le Caeers prncpe

a e poson o equbrum w st o e oppose

sde o e speces wose concenraon as been ncreased.

Obvousy a decrease n e concenraon o corne as

e oppose efec, oug neer afecs e vaue oKc.

I e oa pressure s ncreased en e rae o e reverse

reacon w agan ncrease, because s s e ony

one a nvoves a gas and an ncrease n pressure onyncreases e concenraon o gases. he more moes o

gas nvoved on e sde o e equbrum, e more a

cange n pressure w afec e rae. he resu o s

ncrease n e rae o e reverse reacon s e same as

ose expaned above and e poson o equbrum

w agan st o e et. hs resus n a reducon n

e amoun o corne and ence e oa pressure as

beow e new ger vaue (see Fgure 706). A decrease n

oa pressure w ave e oppose efec, bu once agan

e vaue oKc

s uncanged.

I e emperaure s ncreased, en e raes o bo

e orward and reverse reacons w ncrease, buey w no do so by e same amoun. he ger e

acvaon energy, e greaer e efec o emperaure on

reacon rae, so a an ncrease n emperaure w speed

up e reacon n e endoermc drecon (wc

mus ave e greaer acvaon energy) more an e

exoermc reacon. In s exampe e orward reacon

s endoermc (H posve), so e reverse reacon s

exoermc. he efec o an ncrease n emperaure

s ereore o ncrease e rae consan o e orward

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reacon more an a o e reverse reacon. hereore,

e vaue oKcncreases and e reacon sts o e rg,

producng more corne and posporus rcorde

un e reacon raes agan become equa (see Fgure

707). hs endoermc cange absorbs ea energy and

causes e emperaure o e sysem o a o beow s

new ger vaue. A decrease n emperaure w ave e

oppose efec and e vaue oKc w decrease.

Fgures 705, 706 and 707 sow e efec o canges n

condons on e equbrum esabsed by eang

posporus penacorde.

Concentration

Rate

Time Time

(a) (b)

PCl5

Cl2

PCl3

Reverse

Forward

Figure 705 Increased concentration of chlorine

Con

centration

Rate

Time Time

(a) (b)

PCl5

Cl2

PCl3

Reverse

Forward

Figure 706 Increased total pressure

C

oncentration

Rate

Time Time

(a) (b)

PCl5

PCl3

& Cl2

Forward

Figure 707 Increased temperature

(N.B. the flat sections on the rate graph have been exaggerated to distinguish

between the effects of temperatures and concentration changes)

inDusTrialprOcEssEs

Many ndusra processes nvove equbra. he am o

e process s o produce e desred produc as eiceny

as possbe, a s rapdy, bu w e mnmum amoun

o wase and e mnmum npu o energy. hs requres a

sudy o bo knecs (ow as e produc s made rome reacans) and equbrum (ow muc o e desred

produc s presen n e mxure produced) consderaons.

Two processes o wc ese consderaons ave been

apped are e Haber Process, or e producon o

ammona, and e Conac Process, or e producon o

suurc acd. hese are consdered separaey n Fgure 707.

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Equilibrium

189

he poehe Haber process nvoves e drec combnaon o

nrogen and ydrogen o produce ammona. A mxure

o nrogen and ydrogen n a 1:3 rao by voume s

compressed and passed over a eaed ron caays were

e oowng equbrum s esabsed.

N2(g) + 3 H2(g) 2 NH3(g)H= 92 kJ mo-1

he coce o condons or s equbrum s crca. I

can be seen a n e reacon 4 moes o gas are convered

o 2 moes o gas, ence a g pressure w avour e

ormaon o e produc, as Fgure 708 conrms. he

provson o a g pressure s owever expensve, bo

n erms o e capa cos o provdng a pan a w

ress g pressures and n erms o e operang coss

o compressng gases o g pressures. he na coce

w ereore be a compromse pressure a akes no

accoun ese acors.

I can be seen rom e equaon above a e orward

reacon s exoermc (H s negave), ence a ow

emperaure woud avour e producs, as can be seen

rom Fgure 708. Unorunaey ow emperaures resu n

ow raes o reacon so a even oug ere may be a

g proporon o ammona n e produc may ake

a ong me or e converson o occur. A compromse

emperaure s ereore cosen so as o produce e

maxmum mass o ammona per our. he use o a ney

dvded caays conanng ron aso ncreases e reacon

rae.

%

conversiontoammonia

Pressure / atm

20

40

60

80

100

20 40 60 80 100

500 K600 K

700 K

800 K

900 K

Figure 708 The effect of conditions on the proportion of

ammonia at equilibrium

Typca condons cosen or e Haber process are

pressures n e range 200 1000 am (20 100 MPa) and

emperaures 700 K. he reacon s owever no et

or suicen me o reac equbrum (remember e

reacon rae w decrease as equbrum s approaced

see Fgure 701) and ypcay n e converer ony

abou 20% o e nrogen and ydrogen s convered o

ammona. I woud be very uneconomca o wase e

uncanged reacans, so e mxure o gases s cooed

causng e ammona o condense ( can ydrogen bond,

unke e reacans) so a can be separaed and e

gaseous nrogen and ydrogen recyced.

Nrogen s an eemen va or pan grow, so e

major use o ammona s e manuacure o erzers,

suc as ammonum sas and urea. I s aso used n e

manuacure o nrogen conanng poymers suc as

nyon. Ammona can aso be oxdsed o produce nrc

acd, nvovng e na oxdaon o ammona over a

panum caays (see e equaons beow), wc s used

n e producon o exposves suc as TNT, dyname ec.

and n e dye ndusry.

4 NH3(g) + 5 O2(g) 4 NO (g) + 6 H2O (g)

2 NO (g) + O2(g) 2 NO

2(g)

4 NO2(g) + 2 H

2O () + O

2(g) 4 HNO

3(aq)

cott poehe Contact process s e producon o suurc acd by

e oxdaon o suur. Frsy pure suur s burn n ar o

orm suur doxde:

S (s) + O2(g) SO

2(g)

Suur doxde s mxed w ar and passed over a

vanadum(V) oxde caays o produce suur roxde:

2 SO2(g) + O

2(g) 2 SO

3(g)

H= 196 kJ mo1

As w e Haber Process a g pressure woud avour

e ormaon o e produc (3 moes o gas gong o 2),

bu n s case exceen converson s aceved wou e

expense o a g pressure process. Hence e reacans are

ony compressed o e pressure needed (abou 2 am) o

aceve e desred low rae n e reacor. Smary usng

pure oxygen raer an ar woud drve e equbrumo e rg, bu agan woud be an unnecessary

expense. Anoer smary s a e orward reacon

s exoermc, so a ow emperaure avours e producs.

As w e Haber Process e emperaure canno be oo

ow oerwse e process becomes uneconomcay sow.

he resu s e coce o a compromse emperaure

(700 800 K) and e use o a caays (ney dvded

V2O

5) o enance e reacon rae. Aso e oxdaon o

suur roxde s usuay done by a number o converers a

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cOrE

successvey ower emperaures, so as o make use o g

emperaure o gve a as na rae o reacon as we as a

ow emperaure o gve a g na equbrum yed. he

resu s we over 90% converson o e roxde. Ater

absorpon o e suur roxde, e gases are oten passed

roug one more converer o ensure a e wase gases

conan so e suur doxde a ey can be reeased

drecy no e ar.

he suur roxde mus now be reaced w waer o

produce suurc acd:

SO3(g) + H

2O () H

2SO

4()

Suurc acd as numerous uses n e cemca

ndusry ndeed e onnage o used annuay gves

a good ndcaon o e exen o a counrys cemca

ndusry. hese uses ncude manuacure o erzers

(especay converng nsoube pospae rock o soube

superpospae), poymers, deergens, pans andpgmens. I s aso wdey used n e perocemcas

ndusry and n e ndusra processng o meas. One

o s mnor, oug possby mos amar uses, s as e

eecroye n auomobe baeres.

Exercise 7.2

1. he equbrum consan or a reacon a occurs

oay n e gas pase s gven beow. Wa s e

cemca equaon or s equbrum?

Kc=

[CO2][CF4]__________[COF2]2A CO

2(g) + CF

4(g) COF

2(g)

B CO2(g) + CF

4(g) 2 COF

2(g)

C 2 COF2(g) CO

2(g) + CF

4(g)

D COF2(g) CO

2(g) + CF

4(g)

2. Wc one o e oowng w ncrease e rae

a wc a sae o equbrum s aaned wou

afecng e poson o equbrum?

A Increasng e emperaure.

B Increasng e pressure.C Decreasng e concenraon o e

producs.

D Addng a caays.

3. In e manuacure o meano, ydrogen s reaced

w carbon monoxde over a caays o znc and

cromum oxdes and e oowng equbrum s

esabsed:

2 H2(g) + CO (g) CH3OH (g)H= 128.4 kJ mo1

Wc one o e oowng canges woud ncrease

e percenage o carbon monoxde convered o

meano a equbrum?

A Decreasng e oa pressure.

B Increasng e emperaure.

C Increasng e proporon o ydrogen

n e mxure o gases.

D Increasng e surace area o e

caays.

4. Wen eaed n a seaed vesse, ammonum cordes n equbrum w ammona and ydrogen

corde accordng o e equbrum:

NH4C (s) NH

3(g) + HC (g)

Increasng e emperaure ncreases e proporon

o e ammonum corde a s dssocaed. he

bes expanaon o s observaon s a:

A s ncreases e rae o bo reacons,

bu e orward reacon s afeced

more an e reverse reacon.

B s ncreases e rae o bo reacons,bu e reverse reacon s afeced more

an e orward reacon.

C s ncreases e rae o e orward

reacon, bu decreases e rae o e

reverse reacon.

D s decreases e rae o bo reacons,

bu e reverse reacon s afeced more

an e orward reacon.

5. Wen meane and seam are passed over a eaed

caays e equbrum beow s esabsed.

CH4(g) + H2O (g) CO (g) + 3 H2(g)

Wc one o e oowng w resu n a cange n

e vaue o e equbrum consan (Kc)?

A Increasng e pressure.

B Addng more meane (CH4).

C Decreasng e concenraon o seam.

D Increasng e emperaure.

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Equilibrium

191

6. Wen 0.1 mo dm3 aqueous souons o sver

nrae and ron(II) nrae are mxed, e oowng

equbrum s esabsed:

Ag+(aq) + Fe2+(aq) Fe3+(aq) + Ag (s)

Wc o e oowng canges woud produce more

sver?

A Addng some ron(III) nrae souon.

B Addng more ron(II) nrae souon.

C Removng some o e Ag+ ons by

ormng nsoube sver corde.

D Increasng e oa pressure.

7. In e converson o nrogen o ammona usng e

Haber process, e man reason wy e emperaure

s med o abou 450 C s because

A a ger emperaure woud cause ecaays o break down.

B a ger emperaure woud cause e

reacon o occur oo sowy.

C a ger emperaure woud decrease

e amoun o ammona presen a

equbrum.

D a ger emperaure woud cos oo

muc money o manan.

8. For eac o e oowng equbrum reacons,

baance e equaon w woe number coeicens

and deduce s Kc

expresson:

a) N2(g) + H2(g) NH3(g)(Haber process)

b) SO2(g) + O

2(g) SO

3(g)(Conac process)

c) NH3(aq) + H

2O () NH

4+(aq) + OH(aq)

d) H2O () H+(aq) + OH(aq)

e) NO (g) + C2(g) NOC (g)

) NH3(g) + O

2(g) H

2O (g) + NO (g)

g) CH3NH2(aq) + H2O ()CH3NH3

+(aq) + OH(aq)

) CH3OH () + CH3COOH ()CH

3COOCH

3() + H

2O ()

9. he cenra reacon n e Haber process s e

equbrum beween nrogen, ydrogen and

ammona.

a) Wre a baanced equaon or s

equbrum.

b) he enapy cange n s reacon s

92.6 kJ mo1 and e acvaon energy s

+335 kJ mo1. Draw an energy eve dagram

or s equbrum.

c) he reacon usuay akes pace n e

presence o an ron caays. On e dagramrom b), mark e reacon paway or e

caaysed reacon w a doed ne.

d) Woud you expec e ron o be presen

as sod umps or n a ney dvded sae?

Expan wy.

e) he reacon s usuay carred ou a a

pressure we above amosperc pressure.

How woud you expec s o afec e rae

o reacon? Expan wy as s efec.

10. Nrogen monoxde and oxygen reac ogeer n a

reversbe reacon o orm nrogen doxde.

a) Descrbe, n erms o e raes o e

reacons and e concenraons o e

speces presen, e way n wc equbrum

s esabsed nrogen monoxde and

oxygen are suddeny mxed n an empy lask.

b) Wre a baanced equaon or s

equbrum.

c) Nrogen doxde s brown, wereas nrogen

monoxde s coouress. I e oxygen s

repaced w an equa voume o ar, e

mxure o gases becomes ger cooured.

Expan wy s occurs.

11. For eac o e oowng equbra, sae:

I weer cange () woud st e poson o

equbrum o e rg or e et.

and

II ow you coud cange e second acor ()

so as o st e poson o equbrum n

e oppose drecon.

a) C (s) + H2O (g) CO (g) + H

2(g)

Forward reacon endoermc

() ncreasng e oa pressure.

() cangng e emperaure.

b) Br2(aq) + H

2O ()

HOBr (aq) + H+(aq) + Br(aq)

() addng poassum bromde.

() cangng e pH.

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192

cOrE

c) N2O

4(g) 2 NO

2(g)

Forward reacon endoermc

() decreasng e emperaure.

() cangng e oa pressure.

d) CO (g) + C2(g) COC2(g)

Forward reacon exoermc

() addng more corne.

() cangng e emperaure.

e) NH4HS (s) NH

3(g) + H

2S (g)

() reducng e pressure.

() cangng e concenraon o

ammona.

12. A gaseous mxure o ydrogen, odne and ydrogen

odde are n equbrum accordng o e equaon:

H2(g) + I

2(g) 2 HI (g) H= +56 kJ mo1

Sae ow e poson o equbrum w be

afeced by e oowng canges and expan your

reasonng:

a) Decreasng e emperaure.

b) Addng more ydrogen a consan pressure.

c) Increasng e oa pressure.

13. A lask conans odne monocorde, a brown

qud, odne rcorde, a yeow sod, and cornegas n equbrum accordng o e equaon:

IC () + C2(g) IC

3(s)

a) I e voume o e lask was reduced so as o

ncrease e oa pressure, expan wa you

woud expec o appen o e amouns o

brown qud and yeow sod?

b) Wen e lask s cooed n ced waer e

amoun o yeow sod ncreases and ere

s ess brown qud. Expan wa s sows

abou e equbrum?

14. Suurc acd s manuacured by e reacon beween

suur roxde and waer. he suur roxde s

ormed by e reacon o suur doxde and oxygen

rom ar n e presence o a caays. hs s known

as e Conac process and esabses an equbrum

n wc e orward reacon s exoermc.

a) Wre a baanced equaon or e

equbrum.

b) Wa efec does e caays ave upon:

) e rae o e orward reacon?

) e rae o e reverse reacon?

) e proporon o suur doxde

convered o suur roxde?

c) I a g pressure was used, wa efec woud

s ave on e reave proporons o suur

doxde and suur roxde? Expan.d) Muc greaer reacon raes coud be

aceved e emperaure was ncreased.

Expan wy s s no done.

hiGhEr lEVEl

Consder an evacuaed conaner w a ayer o a voae

qud n e boom o . Moecues o e qud w

escape rom e surace and ener e vapour pase. hese

moecues n e vapour pase w code w e was o

e conaner and exer a pressure. Some o e moecues

w aso srke e surace o e qud and condense

back no e qud pase. Inay s rae o reurn w

be ow, bu as more and more moecues escape no e

vapour pase and e pressure ncreases, e rae o reurnaso ncreases un becomes equa o e rae a wc

e parces vapourse rom e surace so a

Rae o vapoursaon = Rae o condensaon

A s pon e sysem s n a sae o dynamc equbrum,

smar o a cemca equbrum, and e pressure

exered by e parces n e vapour pase s known as

e vapour pressure o e qud. Aerng e surace

17.1 liquiD-VapOurEquilibrium (ahl)

17.1.1 Describe the equilibrium established

between a liquid and its own vapour

and how it is affected by temperature

changes.

17.1.2 Sketch graphs showing the relationship

between vapour pressure and

temperature and explain them in terms of

the kinetic theory.

17.1.3 State and explain the relationship

between enthalpy of vapourization,

boiling point and intermolecular forces.

IBO 2007

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Equilibrium

193

Compound

Enthalpy ovapourisation

/ kJ mol1

Boilingpoint / K

Intermolecular orces

Meane 9.0 109 van der Waas ony

Meoxymeane 27.2 248van der Waas and dpoe-

dpoe

Eano 38.6 352van der Waas, dpoe-dpoe,

and ydrogen bonds

Figure 710 Enthalpy of vapourisation and boiling point data of some compounds

area o e qud afecs bo o e raes equay, so a

as no overa efec on e vapour pressure, oug w

afec e me aken o reac equbrum - e greaer e

surace area, e more rapdy equbrum s aceved.

Moecues on e surace need a ceran mnmum amoun

o knec energy beore ey can escape rom e aracve

orces o e oer surace moecues. hs w depend one sreng o e nermoecuar orces and s smar o

e concep o acvaon energy or a cemca reacon.

Vapoursaon s an endoermc process, as requres e

overcomng o e aracve orces beween e parces.

he amoun o energy requred or s pase cange s

known as e enapy o vapoursaon.

More precsey e enthalpy o vapourisation s e

amoun o energy requred o conver one moe o e

subsance rom e qud o e gaseous sae. Usng waer

as an exampe, s e enapy cange assocaed w

e ranson:

H2O () H

2O (g)

H= +40.7 kJ mo1 (a 373 K and 101.3 kPa)

hs energy s many requred o overcome nermoecuar

orces, oug some s requred o do work agans e

amospere. Wen a subsance bos, s emperaure

does no ncrease (ere s no ncrease n knec energy),

so a e energy absorbed s nvoved n ncreasng

poena energy by overcomng aracve orces beween

e parces.

he sronger e orces beween e parces, e greaere enapy o vapoursaon, e ower e vapour

pressure a a gven emperaure, and e ger e bong

pon. hs s usraed by e daa n Fgure 710.

A a ger emperaure, more moecues w ave e

requred knec energy o escape no e vapour pase

and e rae o vapoursaon w ncrease (see Fgure

711). hs means a more moecues are requred n e

gas pase or e rae o condensaon o equa e rae

o vapoursaon, ence an ncrease n emperaure resus

n an ncrease n vapour pressure, as sown n Fgure

711. A qud w bo wen s vapour pressure s equao e pressure on e surace o e qud, because s

aows bubbes o vapour o orm n e body o e qud.

he norma boiling point s e emperaure a wc

e vapour pressure s equa o sandard amosperc

pressure (101.3 kPa), as sown n Fgure 711. A qud

w, owever, bo a a ower emperaure e exerna

pressure s reduced, as woud be e case on op o a

mounan. Smary e exerna pressure s ncreased,

as n a pressure cooker, e bong pon o e qud

ncreases.

Temperature

Vapourpressure

Normalb.p.

Atmospheric pressure

Reduced pressure

Increased pressure

Increasedb.p.

Reducedb.p.

Figure 711 The relationship between temperature and

vapour pressure showing the boiling point

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CHAPTER 7

194

ahl

A homogeneous equilibrium s one n wc a e

reacans and producs are n e same pase. I ere are

wo or more pases en s a heterogeneous equilibrium.

he concenraons, wen subsued no e equbrum

consan ormua, w ony equa e equbrum

consan e sysem s n ac a equbrum. For oer

suaons, wen equbrum s no presen, e vaue

produced by reang e concenraons n e same way

as e equbrum consan (somemes reerred o as e

reacon quoen, Qc) w ndcae wc way e reacon

needs o st n order o aan equbrum. I Qc

> Kc,

en e vaue oQc mus a, ence some producs musbe convered o reacans and e sysem mus st o e

et (n e reverse drecon) un Qc= K

cand equbrum

s esabsed. Conversey Qc

< Kc, en e sysem mus

st o e rg (n e orward drecon) and some

reacans mus be convered o producs un Qc

= Kc

and

equbrum s esabsed. Consder e reacon:

CO2(g) + H

2(g) CO (g) + H

2O (g)

Kc

= 0.955 a 1000 K

I equa amouns o e gases are mxed a 1000 K en

Qc = 1, wc s greaer an Kc a s emperaure, soe sysem mus st o e et (n e reverse drecon)

o aceve equbrum. In oer words, some o e waer

and carbon monoxde w reac o orm carbon doxde

and ydrogen, aerng e concenraons so a Qc

= Kc.

As w any maemaca expresson, a o e erms

excep one n e equbrum expresson are known, e

unknown erm may be cacuaed by subsuon. hs s

bes usraed by means o exampes:

Solution

2 SO2(g) + O2(g) 2 SO3(g)

Assumng, or smpcy, a voume o 1 dm3, en as 80%

o e suur doxde urns no e roxde:

Equbrum [SO3] = 0.0200 0.800

= 0.0160 mo dm3

Eac suur roxde moecue s ormed rom one suur

doxde moecue, so:

Equbrum [SO2] = 0.0200 0.0160

= 0.0040 mo dm3

Eac suur roxde moecue requres ony a an oxygen

moecue, so:

Equbrum [O2] = 0.0200 ( 0.0160)

= 0.0120 mo dm3

Subsung no e equbrum consan expresson, e

vaue o e equbrum consan may be cacuaed:

Kc=

[SO3]2_________[SO2]2[O2]= 0.0160

2______________0.00402 0.0120

= 1333

= 1330 mo1 dm3(3 s..)

Smary e equbrum consan s known, en

gven approprae normaon e equbrum or sarng

concenraon o one o e reacans may be ound, e bes

ecnque usuay beng o subsue x or e unknownquany. In many cases oug e resung equaon may

conan powers o x and ence requre speca ecnques

or er souon. hs s no owever e case w e

exampe gven beow:

17.2 ThE Equilibrium law (ahl)

Example

17.2.1 Solve homogeneous equilibrium

problems using the expression for Kc.

IBO 2007

Wen a mxure nay conanng 0.0200 mo dm3

suur doxde and an equa concenraon o oxygen

s aowed o reac equbrum n a conaner o xed

voume a 1000 K, s ound a 80.0% o e suur

doxde s convered o suur roxde. Cacuae e vaue

o e equbrum consan a a emperaure.

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ahl

4. Wen ammonum ydrogensude s eaed

dssocaes accordng o e equbrum beow.

he vaue o Kc

or s equbrum a a parcuar

emperaure s 0.00001 mo2 dm6.

NH4HS (s) NH3(g) + H2S (g)

a) Expan e uns o e equbrum consan.b) Cacuae e concenraon o ammona a

equbrum.

c) I some ammona gas was njeced a consan

pressure and emperaure, ow woud s

afec

e mass o sod presen?

e concenraon o ydrogen sude?

e vaue oKc?

5. Wen nrogen and ydrogen reac ogeer n

e presence o a caays, ey produce ammona.

he oowng abe gves e percen o ammonan e mxure wen a 3:1 H2:N

2mxure reaces

equbrum under varous condons:

Pressure / MPa10 20

Temperaure / oC

400 25% 36%

500 10% 17%

a) Wa do ese daa sow abou e

equbrum?b) A 10 MPa and 400C, e equbrum

concenraons o e speces presen are:

[N2] = 0.335 mo dm3,

[H2] = 1.005 mo dm3,

[NH3] = 0.450 mo dm3.

Cacuae a vaue or e equbrum consan

and gve approprae uns.

6. For e gaseous equbrum:

CO (g) + C2(g) COC

2(g)

a) Wre e expresson or e equbrum

consan Kc.

b) he concenraons o e varous speces

a a parcuar emperaure are gven beow.Cacuae e vaue o e equbrum

consan Kc, gvng approprae uns.

[CO] = 0.800 mo dm3;

[C2] = 0.600 mo dm3;

[COC2] = 0.200 mo dm3

c) I e pressure o e sysem s suddeny

ncreased so a e voume aves, cacuae

e new concenraons.

d) I ese are subsued no e equbrum

consan expresson, wa s e numerca

resu.e) Is e sysem s a equbrum? I no n

wc drecon w e reacon proceed?

How dd you deduce s?

) Is s conssen w Le Caeers

prncpe? Expan.

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Equilibrium

197

I e sysem conans wo componens a do no mx

(or exampe, waer and rcoromeane), ey w eacexer er own vapour pressure, provded s saken

so bo come no conac w e vapour pase. In

oer words e oa vapour pressure s e sum o e

ndvdua vapour pressures. he suaon s jus e same

as one sde o e conaner conaned one componen

and e oer sde e oer componen, because parces

o a parcuar subsance can ony eave rom, and reurn

o er own surace.

I e sysem conans wo subsances a do mx, en

bo o em w ave vapour pressures n e mxure

a are ower an n e pure quds. hs s becauseere w be ess moecues o eac componen near e

surace, reducng e rae o escape, bu a o e surace

s avaabe or parces n e gas pase o reurn o, ence

e rae o reurn s unafeced by beng a mxure. I bo

o e quds are voae, en e oa vapour pressure

s e sum o ese reduced vapour pressures and s

provdes e bass or racona dsaon (see beow).

I owever one o e componens s non-voae, as s

usuay e case or a sod dssoved n a qud, en e

oa vapour pressure o e sysem w be reduced, as

sown n Fgure 712, and s s e bass or e eevaon

o bong pon and depresson o reezng pon.

Vapourpressure

ofsolvent

Temperature

Pure Solvent

Solution

Figure 712 Illustrating the effect of a dissolved solid on

the vapour pressure of a solvent

ThE VapOur prEssurEOf miXTurEsA voae qud can be separaed rom a nonvoae

soue by simple distillation. he apparaus or s s

sown n Fgure 713. he vapour rom e eaed laskpasses over no e condenser, were s cooed by e

crcuang cod waer and urns back no a qud, wc

s coeced n e recever. I necessary, o reduce erma

decomposon, e process can be carred ou a a ower

emperaure by reducng e pressure n e apparaus so

as o reduce e bong pon o e qud (reer o Fgure

711).

A mxure o wo mscbe, voae quds, w bo wen

e sum o e vapour pressures o e wo componens

equas e exerna pressure. he vapour w aways conan

a greaer proporon o e more voae componen ane qud pase does. I e qud conans an equa number

o moes o wo quds, a e bong pon o e mxure,

e more voae componen w be conrbung more

an 50% o e vapour pressure. For exampe e vapour

above an equmoar mxure o benzene (b.p. = 353 K)

and meybenzene (b.p. = 384 K) w conan more an

50% benzene, bu s no possbe o oban pure benzene

by smpe dsaon as ere woud s be sgncan

amouns o meybenzene vapour. Suc a mxure coud

owever be more compeey separaed by successve

dsaons e greaer e dference n bong pons

e easer e separaon.

Reurnng o e equmoar mxure o benzene and

meybenzene, e vapour pressures o e wo

componens over e mxure a varous emperaures

are sown n Fgure 714. A a parcuar emperaure

(~ 370 K) e oa vapour pressure over e mxure

equas amosperc pressure and e mxure bos. he

vapour evoved w conan e wo componens n e

rao o er vapour pressures a a emperaure (~ 70%

benzene and ~ 30% meybenzene).

VapourPress

ure

Temperature

Boilingpoint

Atmospheric pressureBenzene

Methylbenzene

Total vapour pressure

30

70

100

Figure 714 The vapour pressures in an equimolar

benzenemethylbenzene mixture

EXTEnsiOns DisTillaTiOn anD fracTiOnalDisTillaTiOn

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EXTEns

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Thermometer

Mixture

Anti-bumpinggranules

Water out

Water in

Condenser

Adaptor

Receiver

Distillate

Figure 713 Typical simple distillation apparatus

I s 70-30 mxure s now dsed, w bo a a

ower emperaure (~ 365 K), as s rcer n e more

voae componen, and e vapour a dss of w

be even rcer n benzene (~ 85% benzene and ~ 15%

meybenzene). hs can agan be dsed gvng a

produc s rcer n benzene and s process can be

connued un e requred degree o pury s obaned.

hs s bes usraed n e orm o a bong pon

composon grap or e sysem, sown n Fgure 715,

wc ndcaes e bong pon o mxures o dferng

composons (qud curve) and e composon o e

vapour a w ds rom s (vapour curve). he resuo successve dsaons can be seen by drawng nes

(somemes caed e nes) parae o e axes, as sown

n Fgure 715:

Temperature

of vapour

of liquid

0 50 100

boil

condense

Composition

Composition

Figure 715 The boiling pointcomposition graph for

benzene-methylbenzene mixtures

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Equilibrium

199

Thermometer

Water in

Water out

Condenser

Adaptor

ReceiverDistillate

Mixture

Anti-bumpinggranules

Fractionatingcolumn

Figure 716 Typical fractional distillation apparatus

In ractional distillation, e vapour rom e bong

qud rses up e raconang coumn, coos andcondenses. I en runs down e coumn and mees

o vapours rsng up, causng o bo agan. hus as

rses up e coumn e qud undergoes a number

o vapoursaoncondensaonvapoursaon cyces,

equvaen o avng been dsed a number o mes

e onger e coumn e more dsaons. I a suabe

coumn s used, e qud dsng over w be a pure

sampe o e more voae componen o e mxure andevenuay, n eory, e dsaon lask w conan e

ess voae componen. Fracona dsaon s carred

ou usng apparaus smar o a sown n Fgure 716.

Fracona dsaon s used n many ndusra processes

suc as e separaon o qud ar and o peroeum (reer

o Caper 14).

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EXTEns

iOn

he vapour pressure above a mxure o wo mscbe

quds s predced byRaoults law. hs assumes a e

orces beween parces o e wo componens n e

mxure (AB orces) are denca o ose presen n e

pure componens (AA and BB orces), ence e vapourpressure o componen A over e mxure (P

A) s equa o

e vapour pressure o e pure componen PA0 muped

by s moe racon:

PA

= PA0 moes o A_________

oa moes

In many cases wen e moecues ony ave weak van der

Waas orces beween em, suc as e cases o benzene

and meybenzene above, e suaon approxmaes we

o s dea beavour and e oa vapour pressure vares

n a near manner w composon (see a n Fgure 717).

In oer cases ony mnor devaons rom deay occurand e vapour pressure/composon grap s a smoo

curve (see b n Fgure 717). In some cases owever, were

srong ner-parce orces occur, as a resu o ydrogen

bondng or dssocaon o one o e speces, ere are

major devaons rom deay.

I e ner-parce orces n e mxure are weaker

an ose n e pure quds, en s easer or e

parces o escape rom e mxure, ncreasng e

vapour pressure above e vaue predced by Raous

aw (a posve devaon rom Raous Law). In exreme

cases o a positive deviation from Raoults Law, e

vapour pressure/composon grap w pass roug amaxmum, and as a resu e bong pon/composon

dagram w pass roug a mnmum (see c n Fgure

717). In suc cases separaon o bo componens by

racona dsaon s no possbe because e qud

a dss over s no a pure componen, bu e mxure

w e mnmum bong pon. he mxure w s

composon s known as an azeotrope. An exampe o s

ype o beavour s a mxure o eano and waer, were

e azeoropc mxure (96% eano, 4% waer) bos a a

emperaure o 78.2 C, wereas e bong pon o pure

eano s 78.5 C.

I e ner-parce orces n e mxure are srongeran ose n e pure quds, en s more dicu

or e parces o escape rom e mxure, decreasng

e vapour pressure beow e vaue predced by Raous

aw (a negave devaon). In exreme cases o a negative

deviation from Raoults Law e vapour pressure/

composon grap w pass roug a mnmum, and as

a resu e bong pon/composon dagram w pass

roug a maxmum (see d n Fgure 717). Agan s no

possbe o separae e mxure no e wo componens

because e qud a remans n e lask s e mxure

w e maxmum bong pon. hs mxure s asoknown as an azeorope. An exampe o s ype o

beavour s a mxure o nrc acd and waer, were e

azeoropc mxure (68% nrc acd, 32% waer) bos a a

emperaure o 121 C, wereas e bong pon o pure

nrc acd s 83 C.

raOulTs lawanD DEViaTiOnsfrOm iDEaliTy

Exercise

1. As e emperaure ncreases, e vapour pressure oa qud ncreases because

A e nermoecuar orces become

weaker.

B expanson causes e surace area o e

qud o ncrease.

C a greaer proporon o e moecues

ave e knec energy requred o

escape rom e surace.

D e number o moecues n e gas

pase s consan, bu ey are movng a

a greaer veocy.

2. A 50 C, wc one o e oowng quds woud

you expec o ave e greaes vapour pressure?

A Eano

B Lubrcang o

C Mercury

D Waer

3. Wc one o e oowng does no afec e bong

pon o a qud?

A he sreng o e nermoecuar

orces.B Is surace area.

C he exerna pressure.

D he presence o dssoved mpures.

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Equilibrium

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a) An ideal mixture

b) A slight positive deviation from Raoults law

c) A major positive deviation from Raoults law (e.g. ethanol-water)

d) A major negative deviation from Raoults law (e.g. nitric acid-water)

Mole fraction

Vapourpressure

Boilingpo

int

Mole fraction

Mole fraction

Vapourpressure

Boilingp

oint

Mole fraction

Mole fraction

Vapourp

ressure

Boiling

point

Mole fraction

Mole fraction

Vapourpressure

Boilingp

oint

Mole fraction

Vapour

Liquid

Vapour

Liquid

Azeotrope

Vapour

Liquid

Azeotrope

Vapour

Liquid

Figure 717 a, b, c, d Vapour pressure/composition and boiling point/composition graphs for ideal and non-ideal solutions

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4. A mxure conanng equa numbers o moes o

exane (b.p. 69 C) and cycoexane (b.p. 81 C) s

eaed.

a) Wa condons deermne e emperaure

a wc e qud bos?

b) Wa do e bong pons sow abou e

reave srengs o e nermoecuar orcesn e wo componens?

c) Woud you expec s mxure o bo beow

69 C, beween 69 C and 81 C, or above

81 C? Expan wy.

d) How woud you expec e composon

o e vapour o compare w a o e

qud? Expan wy s s so.

e) I e vapour s condensed and en urer

dsed a number o mes, wa w appen

o e proporons o e wo componens n

e dsae?

) Wa separaon ecnque depends on sprncpe?

g) Gve one ndusra appcaon o s

separaon meod.

As was dscussed prevousy (Secon 7.4), e addon

o a nonvoae soue o a soven owers s vapour

pressure and s w afec bo s reezing point and s

boiling point. As a qud bos wen s vapour pressures equa o e exerna pressure, by owerng e vapour

pressure, a soue w ncrease e emperaure requred

or s o occur, a s eevaes e bong pon, as

sown n Fgure 718.

A e reezng pon o a subsance e vapour pressure o

bo e qud and e sod saes mus be equa. I s

were no e case en, bo e qud and sod were

paced n an evacuaed conaner a e reezng pon,

equbrum woud no exs. Normay wen a souon

reezes s e pure soven a separaes rom e

souon, eavng e soue n e qud pase. hs means

e vapour pressure o e sod sae s unafeced, soowerng e vapour pressure o e qud sae means a

s no equa o a o e sod un a ower emperaure,

a s depresses (owers) e reezng pon, as sown n

Fgure 718.

Normalboilingpoint

Normalfreezingpoint

Lowe

redfreezingpoint

Temperature

Vapourpressure

Pure liquid

Solution

Solid

Atmosphericpressure

Raised

boilingpoint

Figure 718 The effect of a nonvolatile solute on

freezing point and boiling point

An aernave expanaon depends on e ac a wen

one pase s n equbrum w anoer G = 0, a s,

H = TS. Addng e nonvoae soue o e qud

pase ncreases e enropy o e qud pase wou

afecng a o eer e sod or vapour. hs decreases

S or e pase cange no e vapour (as e vapouras a greaer enropy an e qud S = Svap

Sliq

ges

smaer), ence a greaer emperaure s requred or

TS o equa H. I w owever ncrease S or e

pase cange rom sod o qud (n s case e sod

as a ower enropy an e qud so S = Sliq

Ssol

ges

greaer), so a requres a ower emperaure or TS o

equa H.

he decrease n e vapour pressure o a parcuar soven

s proporona o e concenraon o soue parces

because ey occupy spaces on e surace and ence

reduce e surace avaabe or soven moecues o escape

rom. I s e concenraon o suc parces raer ane naure o e parces a s mporan, as a resu n

aqueous souon one moe o sodum corde (wc

dssocaes no Na+ and C) w ave wce e efec

o one moe o sucrose (wc does no dssocae) and

aumnum suae w ave ve mes e efec (2 A3+

and 3 SO4

2). he magnude o e efec vares rom

soven o soven, dependen many on s moar mass,

bu consans (cryoscopic constants or reezng and

ebullioscopic constants or bong) ave been measured

or mos common sovens. hese usuay gve e cange

n bong/reezng pon wen one moe o parces s

dssoved n 1 kg o e soven, and s s e case, e

cange n bong/reezng pon can be cacuaed usnge ormua:

T = K n m

st

______M

st

1000____msv

Were Ks e reevan consan (uns K mo1 kg), n e

number o parces e soue dssocaes no, mst

e mass

o e soue, msv

e mass o e soven (bo masses n g)

andMst

e moar mass o e soue.

cOlliGaTiVEprOpErTiEsOfsOluTiOns

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Equilibrium

203

Example

he reezng pon depresson consan or waer s

1.86 K mo1 kg. By ow muc w dssovng 5.00 g o

sodum corde n 100 g o waer ower e reezng

pon?

Solution

T = K n m

st

______M

st

1000____msv

= 1.86 2 5.00_______58.5

1000____100

= 3.18 K

Because er magnude s proporona o e number

o moes o soue parces and s no dependen on e

naure o ese parces, reducon n vapour pressure,

eevaon o bong pon and depresson o reezng pon

are a known as colligative properties o souons. One

oer propery a aso vares n s way, and ence s aso

couned as a cogave propery, s e osmotic pressure

o a souon, dscussed beow. In a cases, e drec

proporonay o e concenraon o soue parces s

ony an approxmaon, e accuracy o wc decreases ase concenraon ncreases.

I a soven and a souon are separaed by a seecvey

permeabe membrane (one a aows soven moecues o

pass roug, bu no soue parces) en, e pressure

on bo sdes o e membrane s equa, e rae a wc

soven moecues pass no e souon w be greaer an

e rae a wc ey pass ou o . hs s because on e

souon sde, some o e cosons beween parces and

e membrane nvove soue parces a canno pass

roug (makng e pressure o e soven ess an e

oa pressure), wereas on e soven sde e pressure s

oay generaed by e coson o soven moecues. heresu s e ne ranser o soven moecues rom e pure

soven o e souon a process known as osmosis. he

sysem may be broug no equbrum and e ranser

o soven sopped by appyng addona pressure o e

souon sde. he addona pressure requred s known

as e osmoc pressure. hs s usraed n Fgure 719:

Puresolvent

Solution

Selectivelypermeablemembrane

Additionalpressure

(= osmoticpressure, )

required forsolventtransfer ratesto be equal

Figure 719 Osmotic pressure

he osmoc pressure o a souon may be cacuaed rom

e expresson:

V = nRT

Were s e osmoc pressure n kPa, Ve voume osoven n dm3, n e number o moes o soue parces,

R e gas consan (8.314 J K1 mo1) and Te absoue

emperaure.

Example

Wa s e osmoc pressure o a souon o 2.64 g o

ammonum suae n 250 cm3 o waer a 300 K?

Solution

Amoun o ammonum suae = m__M

= 2.64____132

= 0.0200 moes

Amoun o soue parces = 3 0.0200= 0.0600 moes

(as (NH4)

2SO

4dssocaes no 2 NH

4+ and 1 SO

42)

= nRT____V

= 0.0600 8.314 300_________________0.250

= 599 kPa ((neary sx mes amosperc pressure!)

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Exercise

Hsorcay a o e cogave properes were mporan

ways o deermnng e moar mass o a subsance.

Example

Wen 1.00 g o a naura o s dssoved n 50.0 g o

eracoromeane (bong pon eevaon consan

5.02 K mo1 kg), e bong pon s ncreased by 0.500C.

Wa s e moar mass o e o?

Solution

T = K n m

st

______M

st

1000____msv

= 5.02 1 1.00_______M

st

1000____50

= 0.500

Mst

= 5.02 1 1.00_______0.500 1000____50= 200.8 g mo1

1. Wc one o e oowng s no a cogave

propery o a souon?

A Vapour pressure

B Eevaon o bong ponC Depresson o reezng pon

D Osmoc pressure

2. Wc o e oowng w ave approxmaey e

same efec on e reezng pon o waer as 0.1 moe

o sodum corde?

A 0.3 moe o sodum suae

B 0.1 moe o gucose

C 0.1 moe o copper(II) nrae

D 0.05 moe o aumnum corde

3. Wen 1.50 g o napaene s dssoved n 50.0 go cycoexane (reezng pon depresson consan

20.1 K mo1 kg), e reezng pon s decreased by

4.70 C. Wa s e moar mass o napaene?

4. Wen 0.0135 moes o a non-voae soue s added

o 20.0 g o a soven n wc does no dssocae,

e bong pon o e soven ncreases by 1.20 C.

Cacuae e ebuoscopc consan o e soven,

gvng approprae uns.

i te o t ee Kp

he concenraon o a gas s proporona o s partial

pressure and s someme more convenen o wre

e equbrum consan or a gas pase equbrum n

erms o s. hs equbrum consan s dferenaed

rom a n erms o concenraon by usng e subscrp

p raer an e subscrp c. hereore or a genera

equbrum

a A + b B + c C + ... s S + T + u U + ...

n wc a e componens are n e gas pase, e

equbrum consan n erms o para pressures s gven

by

Kp

=p(S )sp(T )tp(U )u..._______________p(A )ap(B )bp(C )c...

Werep(A) represens e para pressure o A, wc can

be cacuaed rom e oa pressure (Ptot

) and e amoun

o A n e mxure usng e expresson:

p(A) = Po

moes o A_________oa moes

For exampe n e equbrum beween ammona and

oxygen o gve nrogen monoxde and seam:

4 NH3

(g) + 5 O2

(g) 4 NO (g) + 6 H2O (g)


Kp

=p(NO)4p(H2O )6_____________p(NH3)

4p(O2)5

Kpa

As w Kc, K

pdoes no ave xed uns and ey mus be

cacuaed n eac case.

he concenraon o a gas s nked o s para pressureusng e dea gas equaon:

[A ] = na__v =p(A )____RThs reaonsp can be used o nerconver K

pand K

c

vaues, by subsung n e reevan equaon. hs eads

o Kp

= Kc(RT)n were n s e cange n e number

o moes o gas.

OThErEquilibriumcOnsTanTs

5. Wen 5.00 g o a nonvoae soue o moar mass

150 g mo1 s dssoved n 500 cm3 o waer a

25.0 C, e osmoc pressure s ound o be 330 kPa.

Wa can you deduce abou e subsance rom s

normaon?

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Equilibrium

205

I w ave been noed by many sudens a e

equbrum consan and e ree energy cange or a

reacon (G) are bo measures o e exen o wc

e reacans are convered o e producs n a cemca

reacon a equbrum. I s no surprsng ereore a

e wo quanes are nked. he exac reaonsp s:

G = RTnKp

Te ot odt Ksp

[EXTENSION, but note that though not required in the Core

or AHL sections of the IB syllabus, the solubility product is

referred to in Section E12 of the Environmental Chemistry

option, see Chapter 16.]

he solubility product s e name gven o e equbum

consan (Kc) or an onc sod n equbrum w s

aqueous ons (rememberng a e concenraons osods are omed rom suc expressons). For exampe or

a sauraed souon o ead(II) corde, e equbrum

and e souby produc s:

PbC2

(s) Pb2+ (aq) + 2 C (aq)

Ksp

= [Pb2+][C]2 mo3 dm9

I s reay ony a useu concep or sparngy soube

eecroyes as concenraed onc souons exb

sgncan devaons rom dea beavour. he souby

produc can be cacuaed rom e souby o a subsance

and vce versa.

Example

he souby o ead(II) corde a 298 K s 3.90 104

mo dm3, wa s e souby produc or ead(II)

corde a s emperaure?

Solution

From e equaon above, eac ormua un o ead(II)corde orms one ead on and wo corde ons,

ereore:

[Pb2+] = 3.90 104 mo dm3 and[C] = 2 3.90 104

= 7.80 104 mo dm3;subsung:

Ksp

= [Pb2+] [C]2= 3.90 104 (7.80 104)2= 2.37 1010 mo3 dm9

Ions beave ndependeny n souon and ence, n e

above exampe, e corde ons need no necessarycome rom e ead(II) corde, ey coud aso come

rom some oer soue, or exampe ydrocorc acd.

hs means a an onc sod s sgncany ess soube

n a souon a aready conans one o s componen

ons, an s n pure waer. hs s known as e common

ion efect.

Example

Cacuae e souby o ead(II) corde, n g dm3,

n 0.100 mo dm3 ydrocorc acd a 298 K, gven s

souby produc deermned above.

Solution

[C] = 0.100 mo dm3, (assumng any onsrom e ead(II) corde are neggbe),ereore

Ksp

= [Pb2+] [C]2= [Pb2+] (0.1)2

(= 2.37 10= 2.37 1010 mo3 dm9)

[Pb2+] = 2.37 1010__________

0.0100

= 2.37 108 mo dm3

= [PbC2] (noe s s muc ess an n waer)

Mr(PbC

2) = 278

m = n Mr

= 2.37 108 278= 6.60 106 g dm3

he souby produc may be used o predc weer a

sparngy soube sa w be precpaed under parcuar

crcumsances. he concenraons o e ons a

woud be presen s cacuaed and en subsued no

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CHAPTER 7

EXTEns

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e souby produc expresson. I e resu o s s

greaer an e souby produc, en e sod w be

precpaed.

Example

he souby produc or sver suae s 1.60 105 mo3dm9. Woud sver suae be precpaed wen 20.0 cm3

o 0.0100 mo dm3 aqueous sver nrae s mxed w

30.0 cm3 o 2.00 mo dm3 suurc acd?

Solution

[Ag+] = 0.01 20.0____50.0

= 0.004 mo dm3;

[SO42

] = 2.00 30.0____50.0

= 1.20 mo dm3

(Noe e aowance made or e duon efec o mxng

souons);

subsung:

Ksp

= [Ag+]2 [SO4

2]

= (0.00400)2 1.20

= 1.92 105

mo3

dm9

hs s jus greaer an e souby produc (1.60 105

mo3 dm9), so a sma quany o sod sver suae woud

be precpaed.

Exercise

1. For wc one o e oowng woud Kp and Kc avee same numerca vaue a e same emperaure?

A 2 SO2(g) + O

2(g) 2 SO

3(g)

B C (s) + H2O (g) CO (g) + H

2(g)

C N2(g) + 3 H

2(g) 2 NH

3(g)

D H2(g) + I

2(g) 2 HI (g)

2. In wc o e oowng souons w sver corde

be eas soube?

A 0.1 mo dm3 sodum corde

B 0.1 mo dm3 gucose

C 0.1 mo dm3 copper(II) nrae

D 0.1 mo dm3 aumnum corde

3. Wen ar (assume 20% oxygen, 80% nrogen) s eaed

o 2000 K a a pressure o 100 kPa, 3.0% o e oxygen s

convered o nrogen monoxde n e equbrum:

N2(g) + O

2(g) 2 NO (g)

Cacuae e para pressures o O2, N

2and NO a

equbrum.

4. he souby o cacum suae s 6.34 g dm3.

a) Cacuae s souby produc, sang e uns.b) I equa voumes o 0.1 mo dm3 souons o

cacum corde and suurc acd are mxed

woud you expec a precpae o cacum

suae o orm. Expan your reasonng.

5. he souby produc o magnesum ydroxde s

2.0 1011 mo3 dm9.

a) Cacuae s souby n g dm3.

b) Wa s e concenraon o ydroxde ons

n a sauraed souon?

c) How many grams o e sod woud dssove

n 50 cm3

o waer?d) How many grams woud dssove n 50 cm3 o

0.010 mo dm3 aqueous sodum ydroxde?

e) Expan wy e wo vaues dfer.

6. he souby producs o znc carbonae and znc

ydroxde are:

1.4 1011 mo2 dm6 and2.0 1017 mo3 dm9 respecvey.

a) Wre cemca equaons, ncudng sae

symbos, or e wo equaons nvoved.

b) Wre souby produc expressons or ese.c) Sauraed aqueous souons are made o

ese compounds. Wc as e ger

concenraon o znc ons?

d) A souon conanng znc ons s added

o a souon a s 0.10 mo dm3 n bo

ydroxde and carbonae ons. Wc sod w

precpae ou rs and wa concenraon o

znc ons woud be requred or s o occur?

(A b )

chapter 07 equilibrium

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