CHEMI

CALREACTI
ONSI

1.Chemicalequations
Chemicalreacti
onsinvol
vethechemicalpr
opert
iesofasubst
anceand
duri
ngther eact
ion,
thesechemical
propert
ieschange.Chemi
calchangesar
e
di
ffer
entfr
om physicalchanges.

Dur
ingachemi
cal
react
ion,
anewsubst
ancei
sfor
med.

Char
act
eri
sti
csofachemi
calr
eact
ion

i
. Ther
ear
ereact
ant
sandpr
oduct
s.

Hydr
ogen(
gas)+Oxygen(
gas) Wat
er

2H2+ O2 2H2O

 HydrogengasexistsasH2andNOTH.ButyoucanhaveH
exi
sti
ngasHi nothersubstances.
 Oxygengasexist
sasO2
 Thereactant
sarehydrogenandoxygen
 Theproductiswater

i
i. Oneormor
enewchemi
calsubst
ancesar
efor
med.
React
inghydrogenandoxygen(int
heexampl
eabove)givesoutwat
er
asaproduct.Thusi
tisseenthatat
otal
l
ynewsubstancewhichdidnot
exi
sti
nthereactant
sisformed.

i
ii
. Tot
alnumberofat
omsinvolvedasreact
ant
sist
hesameast
het
otal
numberofat
omsinvol
vedintheproduct
s.
 Thereisconservati
onofatomsi nsuchawayt hatthenumberand
typeofatomsinvolvedasreactant
sismaintainedi
nt heproduct
side(t
hisisachievedinbal
ancingtheequat
ionasyouar egoingto
coverinthist
opic)
 Nonewat omsar ecreat
ed(howevernewmol ecul
esar ecr
eated)

i
v. Ener
gyi
stakeni
norgi
venoutdur
ingt
her
eact
ion.
Ener
gycanbet
akeni
norgi
venoutdur
ingchemi
cal
react
ions.

v. Thechangei
susual
lydi
ff
icul
ttor
ever
se
 Tor et
urntotheor
igi
nalreact
antsmayr equi
reanotherchemi
cal
reacti
on(wit
hanexcepti
onofr ever
sibl
eequati
ons).
 Youmaynotnecessar i
l
yr et
urntotheexactquanti
ti
es.

Howt owrit
etheequat
ionforareact
ion
Thefoll
owingar
ethestepsthatar
etobef
oll
owedwhenwr
it
ingachemi
cal
equat
ion
 i
. Writ
et hewor dequati
on.
i
i
. Thenwr it
et heequati
onusingchemi cal symbolsandmakesur eallthe
for
mul aearecor r
ect.
i
ii
. Checkt hattheequati
onisbalanced.i.e.numberofeachat om intur
n
mustbeequal bothinthereact
antsandpr oductsides.(
Not e:
Donot
changeanyf or
mulaotherwiseyoumayendupchangi ngthenameof
thesubstance).
i
v. Addt hestatesymbolsofyourreactant sandpr oducts(i
nwhatst ate
arethey?).Thestatesymbolsusedar e:
 Li
qui
d( l
) Sol i
d( s) Gas(g) Aqueous( aq)

Exampl
es
(
a)Calci
um burnsi
nchlori
netof
orm cal
cium chl
ori
de,
asol
i
d.Wr
it
ean
equati
onforthi
sreact
ion.

1.Wri
tethewordequation.
Cal
cium +Chlor
ine cal
cium chl
ori
de
2.Wri
tetheequati
onusingsymbol
s
Ca+ Cl 2 CaCl
2
3.Bal
ancethenumberofat oms

At
om LHS RHS
Ca 1 1
Cl 2 2

Ca:1at
om ont
heleftandoneat om onther
ight.
:2at
Cl omsontheleftand2at omsontheright
Ther
efor
ethenumberi
sbal ancedhencetheequati
onisbal
anced.

4.Ca(s)+ Cl2(
g) CaCl s)
2(
(
b)I
nindustr
y,hydrogenchl
ori
deisf ormedbybur ni
nghydr
ogeni
n
chlor
ine.Writ
eanequationforthisreact
ion.

1.Hydrogen + chl
ori
ne hydr
ogenchl
ori
de
H2 + Cl2 HCl

2.H:2at omsont helef

tand1at om ontheright.
Cl:2atomsont helef
tandoneat om ontheright.
Equationi sNOTbal anced.Thereisneedforanothermol
ecul
eof
HCl ont herightside
H2 +Cl 2 2HCl
Theequat i
oni snowbalanced.

3.H2(g)+Clg)
2( 2HCl
(
g)

Uni
tsofmeasur
ement
s

Onemol
eofasubst
ancei
sanamountt
hati
sobt
ainedbywei
ghi
ngoutt
he
RAM ( Rel
ati
veAtomi
cMass)orRFM ( Rel
ati
veFormulaMass)ofthe
substanceingr
ams.Onemol eofanysubstancecont
ains6.02x1023
part
icles(
ormolecul
esoratomsorions)ofthesubst
ance.Thenumber6.
02
x1023iscall
edtheAvogadr
osnumberorconst ant
.

TheAvogadr
onumberist
hesamenumberofcarbonat
omscont
ainedi
n
exact
ly12gr
amsoftheCarbon—12i
sot
ope.It
svaluei 02x1023
s6.

Thust
ofindthemassofamol eofasubst
ance;
i
. Wri
tedownt hesymbol
orfor
mulaofthesubst
ance
i
i. Fi
ndoutitsRAM orRFM.
i
ii
. Expr essthatmassingrams.

Forexampl
e:

i
. 1moleofHeli
um (He)
:
RAM ofHe=4
Ther
efor
e1mol eofHeis4grams
i
i
. 1moleofOxygen(O2)
:
RAM ofO=16
Si
ncetherear
e2Oxygenatoms,RFM ofO2=2x16=32
Ther
efor
e1mol eofOxygeni
s32grams

i
i
i. 1moleofEthanol
(C2H5OH)
RAMs: C=12,H=1, O=16.Thereare2C,1Oand6H
RFM ofC2H5OH=(2x12)+(1x16)+(
6x1)=24+16+6=46
Ther
efore1moleofC2H5OHis46grams.

Note:abeam balanceisusedtomeasuremass.
Thenthefol
l
owi ngf or
mulacanbeusedtodeter
minet
henumberof
molesofasubst ancefrom agi
venmass.

Numberofmol
es = massofsubst
ance
massof1mol
eofthesubst
ance

Wecanaswel
luset
hef
oll
owi
ngmol
econver
sioncal
cul
ati
ons:

Numberofmol
esofanat
om (N) =massofsubst
ancei
ngr
ams
Relat
iveAt
omi
cMassingr
ams

Numberofmol
esofmol
ecul
es(
N) =massofsubst
anceingr
ams
Rel
ati
veFor
mul
aMassingrams

Numberofmol
esofmol
ecul
es(N) =massofsubstancei
ngr
ams
Relat
iveMol
ecul
arMassingr
ams

Numberofmol
esofanpar
ti
cles(N) =Numberofpar
ti
cles
Avogadr
osconst
ant
 Example:
1.Howmanymol esofoxygenmol
ecul
esar
ether
ein64gr
amsof
oxygen,
O2.
RFM ofoxygengas=32
So32gr amsofoxygen=1mol
eofoxygen
Numberofmol es= mass
Massof1mol e
= 64g32g
= 2moles
2.Calcul
atet
henumberofmol
esofmagnesi
um oxi
deMgOin
(a)80g(b)10goft
hecompound.
(RAM:Mg=24;O=16)

Sol
uti
on:
(
a)In80g
Numberofmoles = MassofMgO
RFM ofMgO
RFM ofMgO=(
RAM ofMg+RAM ofO)=(
1x24)+(
1x16)=40g

Ther
efor es= 8040 =2mol
eNumberofMol es

(
b)I
n10g
Numberofmoles =MassofMgO
RFM ofMgO
= 1040 =0.
25mol
es

Empi
ri
calFor
mul
aandMol
ecul
arf
ormul
a

Empi
ricalf
ormul
aisthesi
mplestf
orm ofachemicalf
ormul
athatshowst
he
l
owestrati
ooft
heatomsthatmakeupt hecompound.

Molecularformul
aofasubstanceshowstheact
ualnumberofat
omsofeach
el
ementpr esenti
nthechemicalf
ormul
a.Themolecul
arf
ormulamaybeany
multi
pleoftheempiri
calf
ormula.

Exampl
e:

(
a)I
f0.
24gofmagnesi
um bur
nsin0.
16gofoxygent
ofor
m magnesi
um
oxi
de.Wor
kouttheempir
ical
for
mulaofmagnesi
um oxi
de

esofMg=massofMgRAM ofMg
Numberofmol =0.
2424 =0.
01
mol
es

NumberofmolesofO=massofORAM ofO =0.

1616 =0.
01mol
es
Mul
ti
plyby100oneachquant
it
ywi
ll
giveus
- 1mol eofMg
- 1mol eofO
Ther
efor
ethesi
mpl
estf
ormul
aofMagnesi
um oxi
dei
sMgO(
1Mol
eofMgand1
moleofO)

(
b)Letusnowwor kwithanunknowncompound:
Calcul
atetheempir
icalf
ormul
aofanorgani
ccompoundcont
aini
ng92.3%
carbonand7.7%hydrogenbymass.TheRFM oft
hecompoundis78.
Whatisitsmolecul
arfor
mula?(
RAM: C=12,H=1)

Sol
uti
on:
C H
%bymass 92.
3 7.
7
I
n100g 92.
3g 7.
7g
No.ofmol
es
92.
312 7.
71
=7.
7 =7.
7
Rat
ioofmol
es(obtai
nedbydi
vi
ding 1 1
by7.
7oneachside)

Empi
ri
cal
for
mul
a=CH

RFM oft
heempir
ical
for
mul auni
tCH=( 1x12)+(
1x1)=13
Consi
der
ingt
hattheRFM ofthecompoundis78,

NumberofEmpi
ri
calFor
mul a= RFM ofthecompound
RFM oft
heempi r
icalf
ormul
aunit

Therefor
enumberofempi ri
calf
ormulauni
ts= 7813 =6
Andt hemolecul
arfor
mul aoft
heorgani
ccompoundis6xCH=C6H6 and
thi
ssubstanceisbenzene.

Cal
cul
ati
onsoncomposi
ti
on

Thef
ollowingprocedurecanbefoll
owedtocalculat
epercent
ageofan
el
ementinacompound.
 Writedownt hemolecul
arfor
mulaofthecompound
 Usingal i
stofRAMs, workouti
tsfor
mul amass( RFM)
 Writethemassoft heel
ementyouwant, asafract
ionoft
hetot
al(
RFM)
 Multipl
ythefract
ionby100,t
ogiveapercentage.

Int
heexampl
eofbenzeneabove,
cal
cul
atet
heper
cent
agecomposi
ti
onof
car
bon

 t
hemol
ecul
arf
ormul
aofbenzenei
sC6H6.
  RFM ofbenzene=(6x12)+(6x1)=78
 Themassofcar boni
nthef
ormul
a=72
 Thecomposi
ti
onbypercent bon=7278 x100%=92.
ageofcar 3%
Theref
oret
hepercent
agecomposi
ti
onofcar
boninbenzenei
s
92.3%

Concent
rat
ion/Mol
ari
ty

Mol
esandgases
- Gasesar ealwaysli
ghtt
heref
orewhenmeasur i
ng,vol
umeofthegasi
s
consideredrathert
hani
tsmass.
- Thusonemol eofanygasoccupiesavolumeofapproxi
mat
ely24dm3
(24li
tres)atroom t
emperat
ureandpressure
- Ther efore

Numberofmol
esofagas=Vol
umeofthegas(
indm3atr
tp)
3
24dm

- Or

Vol
umeofagas(
indm3atr
t esx24dm3ofgas
p)=numberofmol

Exampl
e:

Cal
culat
ethenumberofmol
esofcarbondioxi
de,i umeof52dm3oft
navol hegas
measuredatr
tp(r
oom t
emperat
ureandpressur
e)

Numberofmol
esofcar
bondi
oxi
de=Vol
umeofcar
bondi
oxi ndm3
dei
24dm3
= 5224 =2. 166mol es
Mol
esandsoluti
ons
 Concent rati
onofasol utionismeasuredi nnumberofmolesperdm3
(moldm-3).
 When1mol eofasubst anceisdissol
vedinwaterandthesolut
ioni
s
3 3
madeupt o1dm ( 1000cm or1l it
re)
,then1molar(1M)soluti
onis
produced.
 Thusconcent r
ation(mol ar
ity)ofasol
utioncanbecalcul
atedas
fol
lows:

Concent
rat
ion= Numberofmol
es
3
Vol
ume(
indm )

Theuni
tsar es/dm3orMol
emol ar(
M)

Example:
Cal
culat
etheconcent
rat
ionofsolutionscont
aini
ng:
(a) 0.2mol
esofNaOHdi ssolvedinwat o100cm3
erandmadeupt
Sol
uti
on:
Concent
rati
on=Numberofmol
es
3
Volume(
indm )
100cm3=1001000 1dm3(
=0. 1000cm3=1dm3)
Concent
rat
ion=0.20.
1 mol dm-3
-3
=2mol dm
=2M
(
b) 9.8gofsulphur
icaci
ddi
ssolvedi
nwat o500cm3
erandmadeupt

Sol
uti
on
 Themolecul
arfor
mulaofsul
phuri
cacidisH2SO4
 (RAM:H=1, S=32,O=16)
 Ther
efor
eRFM ofH2SO4=(2x1)+(1x32)+(
4x16)=98
 NumberofmolesofH2SO4i
n9.8gcanbecalculatedasf
oll
ows

Numberofmoles=Massofsubst
ancei
ngr
ams
RFM i
ngr
ams
=9.
898 =0.
1mol
es

Vol s500cm3 =5001000

umei dm3=0.
5dm3

Ther
efor
econcent
rat
ion=Numberofmol
es
3
Vol
umeindm
=0.10.
5 mol dm-3
=0.
2mol dm-3
=0.
2M

Pr
epar
ati
onofst
andar
dsol
uti
on

ASt andar
dSoluti
onisasolut
ionwhoseconcent
rat
ionisknown.Howeveryou
mayal soberequi
redtopr
oduceasoluti
onofaspecif
icconcent
rati
onfr
om a
givenone.Bothsol
uti
onswill
bestandar
dbecausethei
rconcentr
ati
onsare
known.

Forexample:
1M sol
utonofNaCl
i i
sast
andar
dsol
uti
on.Expl
ainhowyouwoul
dpr
epar
e
1M ofNaClsol
uti
on.

Concentr
ati
on= No.ofmol
es
3
Vol
umeindm

1= No.ofmol
es
 1dm3
1x1=Noofmol es
Noofmol
es=1
1moleofNaClshouldbedissol
vedin1dm3ofwat er
And1mol eofNaClistheRFM ofNaCl measuredi
ngrams
1moleofNaClistheref
ore(16+35.5)g=51.5g
Ther
efor
e51.5gofNaCl shoul
dbedi ssol
vedin1dm3ofwatert
omake1M of
NaCl.

- 4M sol
utionofNaOHi sal
soast
andar
dsol
uti
on.Expl
ainhowyou
wouldprepare1M ofNaOHsol
uti
on.

Concent
rat
ion= No.ofmol
es
3
Volumeindm
4= No.ofmol es
Volumeindm3
4= No.ofmol es
3
1dm

4x1=No.ofmoles
Numberofmol
es=4

Ther
efor
e4mol esofNaOHhastobedi
ssol
vedin1dm3topr
oduce1M of
NaOH.
But1moleofNaOHi s(23+16+1)
g=40g.
And4mol esofNaOHis(4x40)
g=160g
Thus160ghastobedissol n1dm3ofwat
vedi ert
opr
oduce4M ofNaOH.

Di
lut
ingSol
uti
ons

Mostoftheaci
dicoral
kal
iaqueoussol
uti
onsar
eavai
l
ablecommerci
all
yin
concent
rat
edfor
m.Forexamplewemayhave20M HCl ,
15M NaOHetc.

However,i
nourlaborat
ori
esandduri
ngdaytodayuse,t
heconcent
rat
ed
aqueoussolut
ionscanbedil
utedt
odesi
rabl
econcentr
ati
on.

Whendilut
iontakesplace,t
henumberofmol esdoesnotchange,
onlyt
he
concent
rat
iondecreases.Thus,iftheini
ti
alsol
uti
onhadconcentr
ati
onMA,
vol
umeVA andt hedil
utedsolut
ionhaveconcent r
ati
onMBvolumeVB.Remember
for
mulaforconcentr
ation!
Then MAVA =MBVB,( #ofmol es)

Andt
headdedvol
umeVC =VB —VA

Anyuni
tsofvol
umecanbeusedasl
ongasbot
hVA andVBhavet
hesameuni
ts.

Exampl
es:
 1.Ifyoudi
lut
e34mlof1.32M KMnO4(
potassium per
manganat
e)t
o0.28M
KMnO4.Whatist
hefinalvol
umeofthesoluti
on?Whati
stheadded
volume?

Sol
uti
on
MAVA =MBVB
MA =1.32M
VA = 34ml
MB = 0.
28M
Ther
efore
VB=MAVA = 1.
32x34 = 160.
29ml
MB 0.
28

Thef
inalvolumei
s160.
29ml
Theaddedvol
umeVC=VB-VA
160.
29—34=126.
29ml

2. Given5.0mlof14.
8M NH3(
ammonia)
.Whatwil
lthef
inal
vol
umebeaf
tert
his
solut
ionisdi
lut
edwit
hwatert
ogi
ve1.0M NH3?

Sol
uti
on
MAVA =MBVB
VB =MAVA = 14.
8x5 = 74.
0ml
MB 1

3. If100ml i
saddedt
o15mlof0.
5M NaOH,whatwi
l
lbet
hef
inal
concent
rat
ion.
Sol
uti
on:
VB=VC +VA = 100+15=115ml
And MB = MAVA = 0.5x15 =0. 065M
VB 115

Ti
tr
ati
on

Tit
rat
ioni samethodofvolumetr
icanalysi
sinwhi
chavol
umeofoner eagent
(f
orexampl eanacid)i
saddedtoaknownvol umeofanot
herreagent(e.
g.an
al
kali
)slowlyfr
om aburett
e(syri
nge)until
anend-poi
nti
sreached.Ifanacid
andanal kal
iar
eused,thenanindicat
orisusedt
oshowthattheend-pointhas
beenreached.

End-poi
nti
sapoi
ntthati
sreachedwhenenoughaci
dhasbeenaddedt
othe
baset
oneut
ral
i
zei
t,duri
ngti
tr
ati
on.

Inti
tr
ati
onsuchapparatusasbur
ette(
syri
nge),
pipett
e,beaker
sandcanbe
arr
angedasshownbel ow.Thewholeobj
ecti
veoftit
rat
iontofi
ndouthowmuch
vol
umeofonesoluti
onwoul dberequi
redtoneutr
ali
zetheothersol
uti
on.

Thef
oll
owi
ngf
ormul
aisusedi
nti
tr
ati
on:

M1V1 = M2V2
 Macid Malkali
Wher eM1=concent rationoft heaci dinmoles/dm3
M2=concent rati
onoft heal kali(
base)inmol dm3
es/
V1 =vol umeoft heaci di ncm3
V2 =vol umeoft heal kal i incm3
Macid=noofmol esoft heaci dshowni nthebal
ancedchemical
equati
on.
Malkali=noofmol esoft heal kal i(base)showninthebal
ancedchemical
equat i
on
Exampl es:

4.22.4cm3ofasoluti
oncontai
ning0.10mol dm-3ofsulphur
icacidj
ust
3
neutral
i
zes25.0cm ofasodium hydroxi
desol ut
ion.Whatisthe
concentrat
ionoft
hissodi
um hydroxidesolut
ion?

Sol
uti
on:

Usingcal
culat
ionmet
hod
H2SO4 + 2NaOH Na2SO4 + 2H2O

Ther
efor
eMacid =1 andMalkali=2

M1V1 = M2V2
Macid Malkali

Fr
om t
hef
ormul
a,

M2 = M1V1 xMalkali
V2 xMacid
=0.1x22.4x2
25.0x1
M2 =0. 179=0. 18

Ther
efor
etheconcent
rat
ionofNaOH=0.
18M

5.24.2cm3ofasolut
ioncont
aining0.20mol dm-3ofhydrochl
ori
caci
djust
3
neutral
i
zed25.0cm ofapotassium hydroxi
desol ut
ion.Whati
sthe
concentrat
ionoft
hispot
assium hydroxi
desol ut
ion?

Sol
uti
onbycal
cul
ati
on

HCl+ KOH KCl + H2O

HCl KOH
V1 2cm3
24.
V2 0cm3
25.
M1 0. dm-3
20mol
 M2 Tobe
cal
cul
ated
Macid 1
Malkali 1

M1V1 = M2V2
Macid Malkali

M2 =M1V1
V2 si
nceMacid=1andMalkali=1

=24.
2x0.
20
25 194moldm-3
= 0.

194moldm-3
M2 =0.

6.Howmanycm3ofHNO3of1.0M mol
ari
tyi
sneededneut
ral
i
ze
3
i. 1dm of2M KOH
Sol
uti
on:
HNO3 +KOH KNO3 +H2O

Thechemical
equat
ioni
sbal
ancedt
her
efor
ether
eis1M ofHNO3and
1M ofKOH.

AndM1=1. 0M
M2 =2. 0M
V2 =1dm3=1000cm3
Macid =1
Malkali=1

M1V1 = M2V2
Macid Malkali

V1= M2V2 =2x1000 = 2000cm3

M1 1

i
i
. 20cm3of2M Ca(
OH)
2
Sol
uti
on
2HNO3+Ca( OH)2 Ca(
NO3)2 + 2H2O
2mol esofHNO3and1mol eCa(OH)
2
M1V1 = M2V2
Macid Malkali

M1 =1. 0M
M2 =2. 0M
V2 =20cm3
Macid =2
Malkali=1
 V1= M2V2xMacid
M1xMalkali

=2x20x2
1
=80cm3

Usi
ngTitr
ati
onmethod
I
nthispr
ocessanexper
imenti
sconduct
edtodeter
minetheconcent
rat
ionof
onesol
uti
onwhentheconcent
rat
ionoft
heotheri
sknown.

Pr
ocedur
e
 Setupt
heappar
atusasshowni nthediagram below.
 Fi
lli
n0.
5M NaOHinthebeakerwithaknownvol ume(say100cm3)
 Fi
lli
nHCli
ntheburr
ett
euptoaknownval ue( say60cm3)
 Addatl
east2dr
opsofphenolphthal
einindicatori
ntot
hebeaker
(
theNaOHwillt
urntopinkbecausewhenphenol
pht
hal
eini
sadded
t
oabaseitt
urnspink)
 Slowlyr eleasetheacidi ntothebeakerandkeeponshaki ngunt i
l
whent hepi nkcolourjustdisappear s.(
Thi sisyourendpoi nt
),
recordt hevol ume.(t
her eadi ngscanber ecor dedinagiventable)
 Repeatpr ocedure2t o5atl eastthreeti
mest oir
onoutanyer r
ors.
Everyt imeusi ngaf r
eshvol umeofNaOH.
 Getanaver ageoft heaci dused
 Thenusi ngmat hemat i
cal calculat
ion,fi
ndt heconcentr
ationofthe
acid.(youhavet ouset het i
trati
onf or
mul a)
Ti
tr
ati
on

Ini
ti
alVol
ume Fi
nalVolumeof VolumeofHCl
ofHCl(V1) HCl(V2) used(V1—V2)
60 50 10
50 41 9
41 29 11

AveragevolumeofHClused=10+9+11 =30 = 10cm3

3 3
Balancedchemical
equat
ionoft
her
eact
ionofHCl
andNaOHi
s

HCl + NaOH NaCl+H2O

Thenumberofmol
esusedinbot
haci
dandbasei
s1
Usi
ngtheti
tr
ati
onf
ormula

M1V1 =M2V2 (
sinceMacid =Malkali=1)

AndM1 =M2V2 = 0.
5x100 =5M
V1 10
 Not
ethat
:
 i
tisadvisabletotakeyourlastr
eadingfr
om theburet
teasini
ti
alr
eadi
ngfor
thesubsequentobservati
on.
 makesur ethatthebeakerisri
nsedbeforet
henextsetup.
 Readyourbur ett
efrom t
hebot t
om ofthemeniscus
 Theshakingoft hebeakerasyouaddt heaci
distoavoidgett
ingbi
ased
resul
ts.

AMolarsol
uti
onsisasoluti
onthatcont
ains1moleper1dm3.I
tisact
ual
l
ya1M
sol
uti
on.I
tishoweveraStandar
dsoluti
onbecausei
tsconcent
rat
ioni
sknown.

Met
hodsofder
ivi
ngconcent
rat
ionofasol
uti
on.

(
a)Massperuni
tvol
ume

Concent
rat
ion= massofsol ut
e
Vol
umeofsol
uti
on ( uni
ts:
g/l
, cm3,
g/ g/dm3)

(
b)Massperuni
tmass

Concent
rat
ion= massofsolut
e
Massofsol
uti
on(
orsol
ventexpr
essedasper
cent
age)

(
c)Mol
espervol
ume
Concent
rat
ion= Numberofmol esofsolut
e
3
Vol
umeofsol
uti
onindm
3 -
3
(
unit
s:molperdm ormoldm )
3
Recal
lthat1dm =1l
i
tret
her
efor
econcent
rat
ioncanal
sobeexpr
essedi
n
moles/li
tre

Youmayal soberequir
edtoproducesolut
ionsofknownconcentrat
ionfr
om
standardsol
uti
ons.Normall
yacidscomeinconcentrat
edform andhaveto
bedi l
utedt
orequi
redconcentr
ationsandt
histopi
cenablesyoutomanage
this.

Pr
act
icepr
obl
em:

Youar
egivena0.5M NaOHsolut
ionwhosevol s250cm3.Expl
umei ainhow
3
youwoul
dprepar
efrom i
t500cm of0.
1M NaOH.

Mol
arvol
ume

a.Gases
Molarvol
umeforagasisthevol
umeoccupiedby1mol eofagas.I
tis
24dmᶟatroom t
emper
atur
eandpr essur
e(RTP)or22.
4dmᶟatst
andard
 t
emper
atur
eandpr
essur
e(STP)

b.Sol
uti
ons
Forsol
uti
ons,mol
arvol
umei
sthevol
umeoccupi
edby1mol
eofa
sol
uti
on.I
tis1dmᶟ

Heat
sofr
eact
ion

Mostchemical
react
ionsi
nvolvesomeheatener
gy.Whenheati
sgivenout
,it
wil
lcauset
hecontai
nertogethotandwhenheatist
akeni
nthecontai
nerwil
l
getcol
d.

Exother
mi creact
ions
Arechemicalreact
ionst
hatgi
veoutheatt
othesur
roundi
ngs(
maki
ngt
he
cont
ainerheatup)

Examplesofexot her micreact ionsare:

 sulphuricaci dreact ingwi t
hwat er
 Sodium hydr oxi
der eacti
ngwi thhydr
ochl
ori
cacid.
Endothermicreact ions
Arereact
ionst hattakei nheatf rom thesurr
oundi
ngs.Thust
hecont
ainer
becomescoldt hanbef ore.
Examplesofendot hermicr eactionsare:
 Dissolvingammoni um ni tr
ateinwater
 Evapor ationofwat ert owatervapour.

Bondbr eakingandbondf orming

 i
nt hereactantssideoft hechemi cal
equat i
on, bondsar ebr okenandenergy
i
sr equir
edt obreakt hebond( thusener gyist akenin)
 i
nt heproductssideoft hechemi calequation, newbondsar eformedand
energyisgivenout .
 Ift
heener gyoutisgr eatert
hant heener gyin, t
henther eactionisexot
hermi
c
 Ift
heener gyoutislesst hantheener gyin,thent hereactionisendothermi
c.
Theener gyrequi redtobreakonemol eofbondsi st hesameast he
energyrequiredt omaket hesamemol e.Thisenergyi scaledbond
l
energy.

Thef
oll
owi
ngar
esomeval
uesofbondener
giesi
nkJ/
Mol
e

Typeofbond Bondenergy
(kJ/
Mole)
C-C 346
C=C 612
C-O 358
O=O 498
O-H 464
H-H 436
N≡ N 946
 N-H 391

Example
I
fst
eam i
sblownt hr
oughwhi te-hotcoke,thecarbongetsoxidi
zed.
a.Calcul
atehowmuchener gyisrequiredt
obr eakthebonds
b.Theamountofener gyr equir
edt oformthebonds.
c.Calcul
atetheoverall
ener gyinthechemi calreact
ion.
d.I
st hereact
ionexothermicorendot hermic
e.Stateareasonforyouransweri n(d)above.

Sol
uti
on:
Usi
ngthebondener
gyt
abl
eabove

Carbon+ Watervapour Carbonmonoxi

de+ Hydr
ogen
C(s)+H2O(g) CO(
g) + H2(
g)

a./Energyneededtobr eakt
hebonds
1mol eofcarbon 717kJ
1moleofwat er 928kJ
Totalenergyin(needed) 1645kJ

b.Energyoutfrom f
ormi
ngbonds
1mol eofcarbonmonoxi
de 1077kJ
1mol eofhydrogen 436kJ
Totalener
gyout 1513kJ
c.Overall
Energy=Energyi
n—Energyout=1645—1513=132kJ

d.Ther
eact
ioni
sendot
her
mic

e.Ther
eact
ionshowst
hati
ttakesi
nmor
eener
gyt
hani
tgi
vesout
.

Ener
gyDi
agr
ams

Anener gydiagr
am i sbasicall
yanener gyprofi
leofr eacti
ons.Inachemi cal
reacti
on, bondsarebr okeni nthereactant
sandbondsar eformedint he
product s.Whenpr esentedasavect or
,energyinisdr awnasavect orpointing
upwar ds( posi
ti
vevect or)andener gyoutispresentedasavect orpoi nti
ng
downwar ds(negativevect or
).Thebiggerthevectoris,thebiggerthe
magni tude.Thust hevect orsoftheenergydiagram shouldbepr opor t
ional to
magni tudesofthevect ors.Intheexampl eofcarbonr eacti
ngwi t
hst eam, the
energydi agram canbedr awnasf oll
ows: Bonds
broken
 Thepr oductsareatahigherenergylevel(fr
om theref
erencepoi
nt)t
hanthe
react
ant s.Therefor
ethereact
ionisendot hermi
c.Ther eact
ionhastakeni
n
heatener gy.
Letusconsi derthechemicalr
eacti
ont hattakeplacewhenhydrogenand
chlor
ineexpl odi
nginthesun.

Hydrogen + Chl
ori
ne HydrogenChl
ori
de
g) + Cl
H2( 2(
g) 2HCl
(g)

 Itt
akes436kJofener gytobreakthebondsinonemol eofhydr
ogen
molecul
es.
 And242kJofener gytobreakthebondsi nonemoleofchlor
inemolecul
es.
 431kJofenergyisrequir
edtof or
m onemol eofhydr
ogenchlori
de
Ener
gyintobreakbonds
Foronemol eofhydrogenmol ecul
e 436kJ
Foronemol eofchlor
inemol ecul
e 242kJ
TotalEner
gyIN 678kJ

Energyoutf
rom f
ormingbonds
for2molesofhydrogenchl
ori
demol
ecul
es 862kJ

Ener
gyout—Ener
gyi
n = 862—678 =184kJ

Ener
gydi
agr
am f
ort
her
eact
ioni
s
 Thepr
oduct
sareatal
owerener
gyl
evel
thant
her
eact
ant
s,t
hust
her
eact
ion
i
sexot
hermi
c.

Acti
vati
onener gyist hemi nimum ener gyneededt obr eakenoughbondst o
gett
her eacti
onst arted.Activati
onener gycanbesour cedfrom sunlight,ali
t
match,spir
itburner ,
etc.
I
fweanal yzetheener gydiagr am usingt hevectorprincipl
e,avectorpoi nti
ng
upispositi
veandt heonepoi nti
ngdowni snegati
vet hen
 i
ftheresultantisnegat ive,
theover al
lener gyi
sener gyOUT.Ther eactionis
EXOTHERMI C
 i
ftheresultantisposi t
ive,t
heover all
ener gyisenergyI N.Thereactioni s
ENDOTHERMI C.