chapter 07 equilibrium
TRANSCRIPT
-
8/2/2019 Chapter 07 Equilibrium
1/26
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.
IBO 2007
070803 Chem Chap 7-2.indd 181 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
2/26
CHAPTER 7
182
cOrE
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
070803 Chem Chap 7-2.indd 182 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
3/26
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.
070803 Chem Chap 7-2.indd 183 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
4/26
CHAPTER 7
184
cOrE
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)
he equbrum consan s gven by
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.
IBO 2007
070803 Chem Chap 7-2.indd 184 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
5/26
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
070803 Chem Chap 7-2.indd 185 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
6/26
CHAPTER 7
186
cOrE
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
070803 Chem Chap 7-2.indd 186 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
7/26
Equilibrium
187
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
070803 Chem Chap 7-2.indd 187 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
8/26
CHAPTER 7
188
cOrE
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.
070803 Chem Chap 7-2.indd 188 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
9/26
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
070803 Chem Chap 7-2.indd 189 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
10/26
CHAPTER 7
190
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.
070803 Chem Chap 7-2.indd 190 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
11/26
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.
070803 Chem Chap 7-2.indd 191 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
12/26
CHAPTER 7
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
070803 Chem Chap 7-2.indd 192 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
13/26
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
070803 Chem Chap 7-2.indd 193 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
14/26
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.
070803 Chem Chap 7-2.indd 194 7/12/2007 8:25:
-
8/2/2019 Chapter 07 Equilibrium
15/26
-
8/2/2019 Chapter 07 Equilibrium
16/26
CHAPTER 7
196
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.
070803 Chem Chap 7-2.indd 196 7/12/2007 8:25:2
-
8/2/2019 Chapter 07 Equilibrium
17/26
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
070803 Chem Chap 7-2.indd 197 7/12/2007 8:25:2
-
8/2/2019 Chapter 07 Equilibrium
18/26
CHAPTER 7
198
EXTEns
iOn
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
070803 Chem Chap 7-2.indd 198 7/12/2007 8:25:2
-
8/2/2019 Chapter 07 Equilibrium
19/26
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).
070803 Chem Chap 7-2.indd 199 7/12/2007 8:25:2
-
8/2/2019 Chapter 07 Equilibrium
20/26
CHAPTER 7
200
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.
070803 Chem Chap 7-2.indd 200 7/12/2007 8:25:2
-
8/2/2019 Chapter 07 Equilibrium
21/26
Equilibrium
201
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
070803 Chem Chap 7-2.indd 201 7/12/2007 8:25:2
-
8/2/2019 Chapter 07 Equilibrium
22/26
CHAPTER 7
202
EXTEns
iOn
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
070803 Chem Chap 7-2.indd 202 7/12/2007 8:25:2
-
8/2/2019 Chapter 07 Equilibrium
23/26
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!)
070803 Chem Chap 7-2.indd 203 7/12/2007 8:25:2
-
8/2/2019 Chapter 07 Equilibrium
24/26
CHAPTER 7
204
EXTEns
iOn
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)
he equbrum consan s gven by
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?
070803 Chem Chap 7-2.indd 204 7/12/2007 8:25:2
-
8/2/2019 Chapter 07 Equilibrium
25/26
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
070803 Chem Chap 7-2.indd 205 7/12/2007 8:25:2
-
8/2/2019 Chapter 07 Equilibrium
26/26
CHAPTER 7
EXTEns
iOn
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 )