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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
TOPIC (1A): INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS1
1
Define
(a) functional group.
The atom or group of atoms that gives the organic compound
its characteristic properties.
(b) molecular formula.
Formula that shows the number of atoms present in one
molecule of compound
(c) structural formula.
Formula that shows the arrangement of atoms in a molecule
2
Complete the table below.
(a)
No. Group
Functional Molecular
Structural Formula
Group
Formula
IUPAC NAME
A
Alkenes
C=C
C4H8
CH3CHCHCH3
2-Butene
B
Carbonyl
–C=O
C4H8O
CH3COC2H5
Butanone
C
Alcohols
–OH
C3H8O
CH3CH(OH)CH3
2-Propanol
D
Carboxylic acid
–COOH
C4H8O2
CH3CH2CH2CO2H
Butanoic Acid
E
Halogenoalkane –X, –Cl
C5H11Cl
CH3CH2CHClCH2CH3
3Chloropentane
(b)
No.
General Formula
A
Homologous
series
Alcohol
CnH2n+2O
Molecular
Formula
C4H10O
B
Cycloalkane
CnH2n
C5H10
C
Alkene
CnH2n
C5H10
1
Structural Formula
CH3CH2CH2CH2OH
Or
CH3CH2CH2(OH)CH3
CH2=CHCH2CH2CH3
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Write the expanded structural formula or full structural formula for
each of the following condensed structural formula.
(a)
2-pentene
(b)
H
H
H
Br Br H
C
C
C
C
C
H
H
H
H
H
H
H
H C C C H
H H H
(c) 2,2-dichloro-5-methylhexane
H
H
Cl H
H
H
H
C
C
C
C
C
C
H
Cl H
H
H
C
H
H
1,2-dibromopropane
(d)
3-bromo-2-pentene
H
H
H
H
Br H
H
C
C
C
C
C
H
H
H
H
H
(d)
3-ethyl-2-methyl-1-hexene
(e)
3-chloro-1-butene
H
H
H
H
H
H
C
H
H
H
C
C
C
C
C
C
H
H
H
H
H
C
H
H
H
H
2,2,4-trimethylheptane
H
H
H
H
H
C
H
H
H
H
H
C
C
C
C
C
C
C
H
C
H
C
H
H
H
H
H
H H
H
C
C
C
H
Cl
H
H
C
(f)
H
H
H
2
H
H
C
H
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Write the IUPAC name for each of the following compounds.
(a)
CH3CH(CH3)CH2CH3 :
2-methylbutane
(b)
CH3C(CH3)ClCH2C(CH3)2CH3:
(c)
CH3(CH2)2CH(C2H5)NH2 :
4-chloro-2,2,4trimethylpentane
(d)
CH3CH2CH(C2H5COOH)CH3 :
3-hexylamine
4-methylhexanoic acid
Draw five possible structural isomers of C5H10 which contain a double
bond. Give the IUPAC names.
H
C
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
C
C
C
C
H
H
H
C
H
H
H
H
H
A: 2-pentene
H
H
H
H
C
C
C
B: 1-pentene
H
H
H
C
C
H
H
C
C
C
H
C
H
H
H
C
H
H
H
H
H
H
H
C: 3-methyl-1-butene
H
H
C
C
C
C
H
H
H
D: 2-methyl-2-butene
H
C
H
H
H
H
E: 2-methyl-1-butene
(a)
Identify two molecules which are positional isomers.
A and B or C and D or C and E or D and E
(b)
Identify two molecules which are chain isomers.
A and D or B and C or A and C or A and E or B and E
(c)
Identify one molecule which shows stereoisomerism and state
the type of stereoisomerism shown. Draw and give the IUPAC
name of the two stereoisomers.
3
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
2-pentene shows geometrical isomerism
CH3
CH2CH3
CH3
CH2CH3
Cis-2-pentene
6
trans-2-pentene
Identify which of the following molecules display geometrical
isomerism and, if they do, draw and name the two isomers:
(a)
1-pentene
No geometrical isomerism
(b)
2-pentene
H
H
C
C
CH2
CH3
CH3
2-methyl-1-butene
No geometrical isomerism
(d)
3-hexene
C
C
CH2
H
CH3
CH3
trans-3-hexene
7
CH3
H
H
H
C
CH2
trans-2-pentene
(c)
CH2
C
H
cis-2-pentene
CH3
H
CH3
C
C
CH2
CH2
CH3
cis-3-hexene
Name the type of reaction or product in the following chemical
reactions.
(a)
C2H4
(b)
CH3CH2Cl 
(c)
Propanal
(d)
2-bromobutane 

C2H6

ADDITION
CH3CH2OH
Propanoic Acid
2–butene
4
SUBSTITUTION
OXIDATION
Elimination
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Name all the functional groups, organic compound groups and the
possible types of reaction in the compound below. Mark any chiral
carbon with an (*)
H
H
H
C
O
C
Cl
H
H
No.
Functional Group
Type of reaction
1
C=O, carbonyl group
Oxidation,
addition
2
C–Cl, carbon-halogen
Substitution
3
9
C
C*
reduction
C
=
C,carbon-carbon Addition, oxidation
double bond
Complete the table below.
Fission
Reaction
Chemical Equation (1 each)
Homolytic
fission in Cl2
Heterolytic
fission
in
CH3Cl
Examples (2 each)
Electrophiles
H+, NO2+, CH3+
Nucleophiles
OH–, Cl–, H2O, NH3
Free Radicals

H , CH3 , Cl , CH2Cl
5
or
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Methane CH4 reacts with Cl2 in the presence of UV/sunlight. The
reaction occurs through 3 stages.
(a)
Write the name of each stages and steps in each stage of the
reaction.
Step I : Initiation
Cl–Cl   Cl
+

Cl
Step II : Propagation
CH4

CH3
+
+
Cl
Cl2 

CH3 + HCl
CH3Cl + Cl

Step III : Termination
(Show only 1 step) Or any other step between 2 radicals.

CH3 +  Cl

CH3Cl


CH3 +
CH3  C2H6
(b)
What is the name of the above mechanism?
(Free)-radical substitution
(c)
Why is UV/sunlight important in the reaction?
To break the Cl–Cl covalent bond.
11
State the type of reaction for halogenation of ethene with bromine in
inert solvent. Show the mechanism of the reaction.
Electrophilic addition/addition
6
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13
14
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Write the reactions of complete combustion for the following alkanes
and alkenes:
(a)
ethane
C2H6 + 7/2O2  2CO2 + 3H2O
(b)
propane
C3H8 + 5O2  3CO2 + 4H2O
(c)
ethene
C2H4 + 3O2  2CO2 + 2H2O
Cracking is used to break large hydrocarbon molecules into smaller
molecules.
(a)
State the conditions for cracking reaction
High temperature and pressure if without catalyst. Low T
& P if with solid catalyst (aluminium oxide or silicon(IV)
oxide).
(b)
A hydrocarbon, C15H32 is cracked to form 3 products. Ethene and
propene are two of the products. What is the molecular formula
for the other product?
C10H22
(a)
State Markovnikov’s Rule
According to Markovnikov’s rule, the more positive
reagent is added to the carbon of the double bond with
higher number of hydrogen atoms.
(b)
The reaction between 1-butene and HBr give two products in
unequal amounts. Identify the two products, state which is the
major product. Explain in term of Markovnikov’s Rule why it is
the major product.
2-bromobutane and 1-bromobutane.
2-bromobutane is the major product. According to
Markovnikov’s rule, the more positive reagent (in this
reaction is H+) is added to the carbon of the double bond
with higher number of hydrogen atoms. Hence Br- is
added to the carbon of the double bond with lower
number of hydrogen atoms.
7
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Complete the following reaction by providing the major product or
reagents needed.
(a)
H2
Pt
Ans:
H
H
(b)
Br
Br
Br2
Ans:
CH2Cl2
(c)
OH
H+
Ans:
H2O
8
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
16
Complete the chart below stating the reagents (R), conditions (C) and
structural formulae where appropriate.
17
Reactions of halogenoalkanes (R–X) with CN– to produce nitrile (R–CN)
compounds are important in organic chemistry. State two
importance of this reaction.
Increases the number of carbon atom in the compound.
Nitriles are useful compounds that can be converted into
amines (by reduction) and carboxylic acids (by hydrolysis).
18
The structures of two iodo-compounds, A and B are shown below.
(a)
Complete the table.
Class of R–X
(1o, 2o or 3o)
Compound
Systematic Name
A
CH3CI(CH3)2
2–iodo–2–methyl propane
3o
1–iodobutane
1o
B
CH3(CH2)3I
(b)
Both A and B undergoes hydrolysis reactions via nucleophilic
substitution mechanism. What is the nucleophile of the reaction?
OH- or H2O
9
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Show how the following conversions can be achieved.
R: dilute
acid
(a)
CH3(CH2)2Cl
(b)
Cl
CH3–(CH2)2–CN
CH3(CH2)2CO2H
Br
R:
R:
C:
Br
R: NaOH
(ethanol)
C; heat
20
R: Br2 in
CCl4
Predict the products , reagents and conditions wherever appropriate
for the following reactions :
(a)
(b)
1-Chloropropane
NaOH in alcohol , boil
CH3CH=CH2
I
Bromocyclopentane
I : NaOH in ethanol/ heat @ KOH in ethanol
(c)
Chloroethane
NH3 , alcohol
CH3CH2NH2
heat
21
Rank
CH3CH2CH2CH2I,
CH3CH2CH2CH2Br,
CH3CH2CH2CH2F
and
CH3CH2CH2CH2Cl in ascending order of reactivity of haloalkane towards
nucleophilic substitution. Explain your answer.
CH3CH2CH2CH2I>
CH3CH2CH2CH2Br>
CH3CH2CH2CH2F
C-X bond becomes stronger from I to F.
10
CH3CH2CH2CH2Cl
>
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Describe the
iodomethane.
SN2
reaction
mechanism
in
the
hydrolysis
+ INu-
Electrophile
transition state
11
product
leaving group
of
SHE1325
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
TOPIC 1 (B) : HYDROXY, CARBONYL, CARBOXYL AND
AROMATIC
COMPOUNDS
HYDROXY COMPOUNDS
1
Name the following compounds according to IUPAC rules :
2
Write reaction equations that can produce ethanol from alkene,
haloalkanes and aldehyde. State reagents and conditions in each
reaction.
(a)
(b)
(c)
CH2=CH2 + H2O (g)  CH3CH2OH
Condition : H2SO4
CH3CH2Cl + NaOH in aqueous  CH3CH2OH
Condition : Heat
CH3CHO + LiAlH4 in ether  CH3CH2OH
Condition : Room temperature.
12
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Below are isomers that have the molecular formula C4H10O
(i)
(ii)
(iii)
(iv)
(a)
(i)
(ii)
(iii)
(iv)
C3H7-CH2OH
(CH3)2CH-CH2OH
C2H5-CH(OH)CH3
(CH3)3COH
State the class of each alcohol above.
Primary
Primary
Secondary
Tertiary
(b)
Name the reagent that can be used to identify the classes of the
alcohol above. State the observation of the reaction for each
class of alcohols.
Lucas reagent : ZnCl2 + HCl (aq)
1° ROH : no ppt formed
2° ROH : white ppt formed within 5 minutes
3° ROH : white ppt formed immediately.
(c)
State the product of reaction between all of the isomers above
with K2Cr2O7.
(i)
C3H7-COOH
(ii)
(CH3)2CH-COOH
(iii)
C2H5-CO-CH3
(iv)
No reaction
(d)
One of the isomers gives a positive observation in iodoform test.
Identify the isomer, write the equation and state the
observation.
C2H5-CH(OH)CH3 + I2 + NaOH  CHI3 + C2H5-COONa + NaI +
H2O
Observation : yellow ppt formed with antiseptic smell.
4
Complete the following equation.
(a)
rt
CH3CH2CH2OH + Na (s)
13
CH3CH2CH2ONa
+ H2 (g)
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(b)
O
CH 3 CH 2 CH 2 CH 2 OH + CH 3 CH 2 CH COOH
CH 3 CH 2 CH 2 CH 2 O C CH CH 2 CH 3
CH 3
CH 3
(c)
OH
CH3CH2CH2CCH 3
170°C
+ H2SO4
CH2
CH3CH2CH2C
CH3
CH3
and
C H 3C H 2C H
C
CH3
CH3
CARBONYL COMPOUNDS
1
Draw full structural formula of following compounds
(a)
3-methyl-2-pentanone
O
CH3
H3C
CH3
(b)
Phenylethanal and phenylethanone
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(c)
Phenylethanone
(d)
2,3-dimethylpentanal
CH 3
CH 3 CH 2 CHCHCH
CH 3
(e)
O
4-methyl-3-hexanone
CH3CH2CH(CH3)COCH2CH3
2
Fill in the empty boxes with products of reaction and the observations.
Reagent
A
B
CH3COCH3
CH3CH2CHO
C
CH3
C
O
2,4-DNP
CH3
CH3 C
N
CH2CH3
NO2
NH
H
NO2
NO2
C
N
NH
NO2
CH3 C
N
NH
Orange ppt
Orange ppt
No silver mirror
CH3CH2COOH
Silver mirror
No silver mirror
No brick red ppt
CH3CH2COOH
Brick red ppt
No brick red ppt
Tollen’s
Fehlings
NO
2
15
Orange ppt
NO2
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
I2 +
NaOH
CH3COONa
Yellow ppt &
antiseptic smell
ONa
No Yellow ppt &
antiseptic smell
C
O
Yellow ppt &
antiseptic smell
3
Show the mechanism involved in the reaction between propanone and
HCN.
O
CH3
C CH3 +
-
CN
cold NaOH
-
O
CH3
+
H
C CH3 +
OH
cold NaOH
CH3 C CH3
CN
CN
4
Show how the following conversions can be carried out:
(a)
butanone to butane-1,2-diol
O
OH
conc H2SO4
KMnO4/ OHL
i
A
l
H
CH2 CHCH2CH3
CH3CCH2CH3
4 CH3CHCH2CH3
CH2( OH) CH( OH) CH2CH3
dry ether
c
ol
d
0
160-180 C
(b)
3-pentanone to 3-chloropentane
O
C H 3C H 2C C H 2C H 3
(c)
dry ether
K 2 C r2 O 7 /H
OH
LiAlH 4
C H 3C H 2C H C H 2C H 3
w arm
+
Cl
C H 3C H 2C H C H 2C H 3
2-bromopropane to propanone
Br
NaOH(aq)
OH
CH3CHCH 3
heat
CH3CHCH 3
16
K2Cr2O7 / H+
warm
O
CH3C CH3
SHE1325
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Compounds W and X have the molecular formula of C8H8O. W reacts
with hot Fehling’s solution producing brick red precipitate. X gives
negative result towards Fehling’s test but it reacts with LiAlH4 (dry
ether) to form alcohol Y. Both X and Y react with iodine in NaOH
solution to give a yellow precipitate with antiseptic smell.
(a)
6
Determine the structural formula of W, X and Y.
C H2 C HO
O
C C H3
OH
C HC H3
W
X
Y
Given the following compounds :
G
CH3COCH2CH2CH3
Q
C6H5CH2COCH3
J
C6H5CHO
L
C6H5CH2CH2CHO
(a)
Which of these compounds are aldehydes and which are
ketones?
Aldehydes – J and L
(b)
Ketones – G and Q
Which of the above compounds can be transformed to the
following compounds? Write the complete equation involved.
(i)
(ii)
C6H5CH2CH2CH2Cl
C6H5CH2C(OH)(CH3)COOH
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(i)
L
LiAlH 4 / ether
C6H5CH2CH2CHO
PCl 5
C6H5CH2CH2CH2OH
rt
C6H5CH2CH2CH2Cl
rt
+ POCl3 + HCl
(ii)
O
Q
CN
HCN in alkaline
C6H5CH2C(OH)CH
C6H5CH2CCH3
H2SO4 (dilute)
3
cold
boil
COOH
C6H5CH2C(OH)CH
CARBOXYLIC ACID
1
Compound Y is given below :
CH3CH(OH)CH2COOH
(a)
Write the IUPAC name for compound Y.
3-hydroxybutanoic acid
(b)
Complete the table below :
Reagent
Structural formula of organic
Observation
product
(i) PCl5 (s)
Cl
White fume
CH3CHCH 2COCl
(ii) Na(s)
+ HCl(g)
ONa
CH3CHCH 2COONa + H2(g)
18
Gas bubbles of
H2
3
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(iii) LiAlH4 (dry)
OH
-
CH3CH CH2CH2OH
(iv) Na2CO3(aq)
Gas bubbles of
OH
CO2 @ lime
CH3CHCH 2COONa + CO2
solution turns
cloudy
(v)
OH
O
CH3CH2CH2CH2NH2
-
CH3CCH2CNHCH 2CH2CH2CH3
(vi) CH3CH2OH
OH
Sweet smell of
O
ester
CH3CHCH 2COCH 2CH3 + H2O
AROMATIC COMPOUNDS
1
Draw the structural formulae and give the IUPAC names for the
isomers of C8H10 containing a benzene ring.
CH3
CH3
CH3
CH3
1,2-dimethylbenzene
1,3-dimethylbenzene
CH3
CH3
CH2CH3
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
1,4-dimethylbenzene
CH2CH3
ethylbenzene
2
Three reactions of methylbenzene can be summarized on the flow
sheet as shown below;
(a)
State the reagents and conditions for step I and II
I
Reagents : Concentrated HNO3
Condition : Conc. H2SO4, T=50-60°C
II
Reagent : Br2
Condition : FeBr3, rt
(b)
Draw the structural formula of compound S
CH 2 Br
(c)
Write the IUPAC nomenclature for compound Q
1, 4-nitromethylbenzene
20
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Show the conversions for the following compound;
(state the reagent, conditions and intermediate compound involved)
(a)
COOH
NO2
(b)
Br
NO2
NO2
Br
Br2, FeBr3
HNO3, H2SO4
50 - 60°C
warm
21
NO2
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Draw the mechanism of nitration of benzene.
Step 1 :
HO-SO3-H+ HO-NO2  H2O + NO2+ + HSO4Step 2 :
Step 3 :
5
State the relative reactivity of benzene and methylbenzene in the
nitration reaction. Explain.
Relative reactivity : benzene < methylbenzene
Methyl is activating groups which activate the benzene ring (by
making it more partially negative) / methyl donating electron
to the benzene ring, makes it more nucleophilic.
6
Benzene, methylbenzene and phenol form different products in the
nitration reaction. Explain why this occurs.
Benzene has no substituent, only one product formed.
Methylbenzene has CH3 group which is ortho-para director.
Phenol has OH group which is ortho-para director. Both will
produce 2 products.
22
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7
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Complete the reaction below :
CH3
Cl
CH3
AlCl 3/FeCl3
+
+ Cl2
Cl
uv/sunlight
CH2Cl
8
CH3
Complete the reaction below :
23
SHE1325
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
TOPIC 2 : CHEMICAL THERMODYNAMICS
1
Define the following terms:
(a) Energy: the ability to do work
(b) Heat, q: Energy transfer/exchange due to a difference in
temperature between system and surroundings. (Heat
flows from region of high temperature to lower
temperature and does not belong to the system.)
(c) Heats of rxn, qrxn: quantity of heat exchanged between a
reaction (system) and its surroundings when the reaction
occurs at a fixed temperature.
(d) State function: A property of a system that is determined
only by the present state/condition of the system and not
by how the state is reached. e.g. P, T, V, E, H, S and G
(e) Hess’s Law: The enthalpy change for an overall process is
the sum of the enthalpy changes of the individual steps of
the process.
(f) Heat capacity: the measurable physical quantity of heat
energy required to change the temperature of an object or
body by a given amount
(g) Specific heat capacity: the heat required to raise unit mass of
a substance by unit temperature interval under specified
conditions, such as constant pressure
2
Describe the transfer of heat and matter in an open, closed and isolated
system.
Heat and matter are exchanged in an open system. Only heat is
exchanged in a closed system. No heat or matter is exchanged
in an isolated system.
3
Consider following phenomenon.
“ Calcium hydroxide solid was dissolved with hydrochloric acid in
250 ml beaker and stirred using glass rod.”
Identify which is universe, surrounding and system.
System: Neutralization reaction between calcium hydroxide
and hydrochloric acid
Surrounding: Beaker, glass rod, table, water
24
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Universe: Other than system and surrounding
4
Describe exothermic and endothermic reactions.
An exothermic reaction releases heat to the surroundings such
that the energy level of the reactant is higher than its product.
A negative sign is used to indicate the drop in energy level. An
endothermic reaction absorbs heat from the surroundings such
that the energy level of the product is higher than its reactant.
A positive sign is used to indicate the rise in energy level.
5
How much heat is released when 7.1 g of an alloy is cooled from
100°C to room temperature? Specific heat capacity of that alloy is
0.460 J/g.°C.
ΔT = 25°C -100°C = -75°C
q=mcΔT
q= 7.1 g (0.460 J/ g.°C)( -75°C)
= -244.95 J
Hence, heat released is 244.95 J
6
25 mL of 1 M NaOH is placed into a calorimeter at 25.0 oC. 25 mL of
1.5 M HCl is added into the calorimeter at the same temperature. The
temperature raised by 3.0 oC. Calculate the heat of reaction, qrxn and
enthalpy of reaction, ΔHrxn. (Assume negligible heat loss. Density of
water = 1.00 g/mL and Specific heat of water, c = 4.18 Jg-1C-1)
qrxn = - qsoln = - (50 mL)(1 g/mL)( 4.18 Jg-1C-1)(3.0oC) = - 627 J
0.025L x 1 M = 0.025 mol NaOH reacts with 0.025 L x 1.5 M =
0.0375mol HCl. Thus, NaOH is the limiting reactant while HCl is
in excess.
∆Hrxn = qrxn /mol water produced = - 627 J/0.025 mol
= - 25 kJ/mol
Check: Temperature rise indicates heat released by the system,
thus exothermic reaction.
7
5 grams of potassium chloride is dissolved in 50 mL of water at 28.0
o
C such that it reaches a final temperature of 32.0 oC. Calculate the
heat of reaction, qrxn and enthalpy of reaction, ΔHrxn. (Assume
negligible heat loss. Density of water = 1.00 g/mL and Specific heat of
water, c = 4.18 Jg-1C-1)
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
qsoln = mcΔT = (50 mL)(1.00 g/mL)( 4.18 Jg-1C-1)(32.0 28.0)C
= 836 J
qrxn = - qsoln = - 836 J
ΔHrxn = qrxn/mol of solute = -836 J/(5g/74.55 g/mol) = -12.5
kJ/mol
8
When a 1.23 g sample of propene, C2H4, is burned in a bomb
calorimeter, the temperature of the surrounding bath rises by
18.22°C. Calculate the standard enthalpy of combustion of propene.
(Heat capacity of bomb calorimeter = 2.12 kJ/°C)
qrxn
= -qcal
= - (heat capacity x ΔT)
= -(2.12 kJ/°C x 18.22°C) = -38.6264 kJ
Moles of C2H4
= mass/molar mass
= 1.23g / 28.06 g/mol
= 0.0438 mol
ΔH°c = -38.6264 kJ/0.0438 mol
= -881.88 kJ/mol
9
10
Identify the type of standard enthalpies for the following equations:
(a)
Li(s) → Li(g)
Standard enthalpy of atomization
(b)
NaCl(s) → Na+(g) + Cl-(g)
Lattice energy
(c)
Cs(g) → Cs+(g) + e
Ionization energy
Calculate the enthalpy of formation of butane (C4H10) from the
following data:
Enthalpy of Combustion of graphite = -393.6kJmol-1
Enthalpy of Combustion of hydrogen = -285.9kJmol-1
Enthalpy of Combustion of butane = -2877.1kJmol-1
4C(s) + 5H2(g) → C4H10(g)
ΔH°rxn
= Σm ΔH°c (reactant) - Σn ΔH°c (product)
= (4ΔH°c C + 5 ΔH°c H2) - ΔH°c C4H10
=-126.8 kJmol-1
26
SHE1325
12
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Given the following data, find the heat of formation for benzene using
the
∆Hoc (C6H6(l)) = - 3267 kJ/mol
∆Hoc (C(s))
= - 394 kJ/mol
∆Hoc
= - 286 kJ/mol
(H2(g))
(a)
algebraic method
(b)
energy cycle method
(a)
algebraic method
6 CO2(g) + 3 H2O(l) ⟶ C6H6(l) + 𝟏𝟓⁄𝟐O2 (g)∆Hoc=+3267 kJ/mol
6 C(s) + 6 O2(g)⟶ 6 CO2(g)
∆Hoc= 6(-394) kJ/mol
3 H2(g) + 𝟑⁄𝟐O2(g) ⟶3 H2O(l)
∆Hoc= 3(-286) kJ/mol
6 C(s) + 3 H2(g) ⟶C6H6(l)
(b)
∆Horxn = +45 kJ/mol
energy cycle method
∆Hof (C6H6 (l))= 6[∆Hoc (C(s))]+ 3[∆Hoc (H2(g))]- ∆Hoc (C6H6 (l))
= 6(-394) + 3(-286) + 3267
= + 45 kJ/mol
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SHE1325
13
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(a)
Write a balanced equation for complete combustion of
methanol, CH3OH.
CH3OH (l) + 3/2 O2 (g) → CO2 (g) + 2 H2O (g)
(b)
Calculate the ΔH°f of methanol, using following data:
ΔH°c of CH3OH
ΔH°f of CO2
ΔH°f of H2O
= -638.5 kJ
= -393.5 kJ
= -241.8 kJ
ΔH°c or ΔH°rxn
= Σm ΔH°f (product) – Σn ΔH°f (reactant)
-638.5 kJ = (ΔH°f CO2 + 2 ΔH°f H2O) - ΔH°f CH3OH
ΔH°f CH3OH = -238.6 kJ/mol
(c)
Calculate the ΔH°f of methanol, using energy cycle, based on
the data given at 7(b)
ΔH°f
C(s) + 2 H2(g) + ½ O2(g)
+ O2
(-393.5)
+ O2
2(-241.8)
CH3OH(l)
+ 3/2 O2
(-638.5)
CO2(g) + 2 H2O(l)
ΔH°f = (-393.5) + 2(-241.8) + (638.5)
= -238.6 kJ/mol
28
SHE1325
14
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
The complete combustion of propane is represented by the following
reaction,
CH3CH2CH3 (g) + 5 O2 (g) → 3 CO2 (g) + 4H2O (l)
(a)
Calculate the ΔH°rxn using enthalpy of formation.
Enthalpy of formation of propane= −104.7 kJ/mol
Enthalpy of formation of carbon dioxide= -393.5 kJ/mol
Enthalpy of formation of water= -285.8 kJ/mol
ΔH°rxn = Σm ΔH°f (product) - Σn ΔH°f (reactant)
= (3ΔH°f CO2 + 4 ΔH°f H2O) - (ΔH°f CH3OH + 0)
= -2219 kJ/mol
(b)
Calculate the ΔH°rxn using average bond enthalpy data:
Bond
C-H
C-C
O=O
C=O
H-O
(c)
ΔH°BE (kJ/mol)
+413
+347
+498
+805
+464
Bonds broken
: 8(C-H) = 8(+413) = +3304
: 2(C-C) = 2(+347) = +694
: 5(O=O) = 5(+498) = +2490
Bonds formed
: 3(2)(C=O) = 6(+805) = +4830
: 4(2)(O-H) = 8(+464) = +3712
ΔH°rxn
ΔH°BE(bond broken) - ΔH°BE(bonds formed)
(3304+694+2490) – (4830+3712)
-2054 kJ/mol
=
=
=
Explain why the enthalpy combustion of propane calculated from
ΔH°f is different than using ΔH°BE.
Enthalpy combustion of propane calculated from ΔH°BE is
less accurate because using average data.
29
SHE1325
15
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Construct Born-Haber cycle for reaction of formation of RbCl(s) by
using the data given:
Equation
RbCl(s) →Rb+(g) + Cl-(g)
ΔH° (kJ/mol)
696
+403
Rb(g) → Rb+(g) + e-
+86
Rb(s) → Rb(g)
Cl2(g) → 2 Cl(g)
+242
Cl(g) + e- → Cl-(g)
-349
=-696
∆Hof=
(d)
Calculate the lattice enthalpy
ΔH°rxn= ΔH°atm Rb + 1/2ΔH°atm Cl + ΔH°IE Rb + ΔH°EA Cl +
(-ΔH°LE)
= -435 kJ/mol
30
SHE1325
16
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Using the data given below, construct a Born-Haber cycle (enthalpy
diagram) for the formation of CsF and calculate the lattice enthalpy of
CsF. Label all heats of reaction involved and its values.
Enthalpy of atomization of Cs (s)
First ionization energy of Cs (g)
Enthalpy of atomization of F2 (g)
Electron affinity of F (g)
Enthalpy of formation of CsF (s)
∆Hof CsF (s)
=
=
=
=
=
+ 76.5 kJ/mol
+ 375.7 kJ/mol
+ 79.4 kJ/mol
- 328.2 kJ/mol
- 553.5 kJ/mol
= ∆Hoatom Cs(s) + ∆HoIE1 Cs(g) + ∆Hoatom F2 +
∆HoEA F(g)+ (- ∆Holattice CsF (s))
- 553.5 kJ/mol = (76.5+375.7+79.4+ -328.2 – x ) kJ/mol
x = +756.9 kJ/mol
31
SHE1325
17
(a)
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Define standard lattice enthalpy
The enthalpy change when 1 mole of solid ionic
compound breaks into its gaseous ions under standard
states.
(b)
Explain how the following factors affect lattice energy.
(i)
ionic radius
In a lattice, each ion is surrounded by a number of
ions of opposite charge, resulting in strong forces
of attraction. the smaller the radius, means the
greater attraction, therefore the greater the lattice
energy.
(ii)
ionic charge
The higher the charge on an ion the stronger the
attractive force that will result in an ionic lattice.
the greater the lattice energy.
18
Write an equation to show the relationship between lattice enthalpy,
enthalpy change of hydration and enthalpy change of solution.
ΔH°solution= ΔH°lattice + ΔH° hydration
19
Spontaneity of a reaction is driven by two factors. State the factors.
∆H and ∆S
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INTRODUCTION TO ORGANIC CHEMISTRY,
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TOPIC 3 : CHEMICAL EQUILIBRIUM
1
2
Describe the following terms.
(a)
Reversible reaction
(b)
Dynamic equilibrium
(a)
A reversible reaction is a reaction that is able to proceed
in the forward and reverse directions
(b)
A state of balance in which the rate of forward reaction is
equal to the rate of reverse reaction and the
concentrations of reactants and products remain constant
with time
Write the Kc and Kp expression for the following reactions.
(a)
CO(g) + Cl2(g) ⇌ COCl2(g)
(b)
SnO2(s) + 2 H2(g) ⇌ Sn(s) + 2 H2O(g)
(c)
NaHCO3 (s) ⇌ NaOH (s) + CO2 (g)
(d)
CH3COOH(l) + CH3CH2OH(l) ⇌ CH3COOCH2CH3(l) + H2O(l)
a
Kc
Kc = [COCl2] / [CO][Cl2]
Kp
Kp = PCOCl2 / (PCO)(PCl2)
b
Kc = [H2O]2 / [H2]2
Kp = (PH2O)2 / (PH2)2
c
Kc = [ CO2]
Kp =
d
Kc= [CH3COOCH2CH3][ H2O]/
[CH3COOH][ CH3CH2OH]
33
(PCO2)
No Kp
SHE1325
3
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
The following reactions yield Nitrogen monoxide gas. Based on the Kc
values provided, explain which reaction would yield the greater
amount of Nitrogen monoxide.
Reaction I
Reaction II
N2(g) + O2(g) ⇌ 2 NO(g)
2 NOBr(g) ⇌ 2 NO(g) + Br2(g)
Kc = 4.8 x 10-31
Kc = 0.50
Reaction II because it has a large Kc which indicates a high
yield of the products NO(g) and Br2(g)
4
Given the following information, calculate the Kc for the desired
reactions.
(a)
Br2(g) + Cl2(g) ⇌ 2 BrCl(g)
Kc1=58.0
BrCl(g) ⇌ ½ Br2(g) + ½ Cl2(g)
Kc2 = ?
Kc2= [Br2]1/2 [Cl2]1/2/BrCl
= (1/58)1/2
= 0.13
5
Given the following two equilibrium:
Equilibrium
PbI2(s) ⇌ Pb2+(aq) + 2 I-(aq)
PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq)
(a)
Write the equilibrium constant expression for each reaction.
Kc1
Kc2
(b)
Kc
8.7 × 10-9
1.8 × 10-8
=
=
[Pb2+][I-]2
[Pb2+][SO42-]
Calculate the equilibrium constant for the following reaction.
PbI2(s) + SO42-(aq) ⇌ PbSO4(s) + 2 I-(aq)
Kc3
=
=
=
=
[I-]2 /[SO42-]
K1/K2
8.7 × 10-9/1.8 × 10-8
0.483
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SHE1325
6
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(a)
Calculate the value of Kp for the following equilibrium.
2 SO3(g) ⇌ 2 SO2(g) + O2(g)
Kc = 0.02 at 1000K
Kp = Kc(RT)n; n = 3-2 = 1
= 0.02 [(0.0821 Latm/molK) x 1000K]
= 1.642
(b)
Calculate the Kc for the following reaction.
2 NO(g) + Cl2(g) ⇌ 2 NOCl(g)
Kp = 6.5  10-4 at 308 K
Kc = Kp /(RT)n gas ; ngas = 2-3 = -1
= 6.5  10-4 / [(0.0821 L atm/mol K) x 308K]-1
= 0.0164
7
Hydrogen and Iodine gas react together to form hydrogen iodide gas.
The equation for this reaction is
H2(g) + I2(g) ⇌ 2 HI(g)
The equilibrium constant for this reaction is 7.1 x 102 at 25 °C. The
initial concentration of gases are
[H2]0 = 0.81 M
[I2]0 = 0.44 M
[HI]0 = 0.58 M
In whicht direction will the reaction proceed to reach equilibrium?
Calculate Q
Q
Q
Q
Q
=
=
=
=
[HI]02/[H2]0·[I2]0
(0.58 M)2/(0.81 M)(0.44 M)
0.34/.35
0.94
Compare Q to K
K = 7.1 x 102
Q = 0.94
Q < K, therefore the reaction will shift to the right.
The reaction will shift to the right to produce more hydrogen
iodide gas to reach equilibrium.
35
SHE1325
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
0.50 moles of N2 gas is mixed with 0.86 moles of O2 gas in a 2.00 L
tank at 2000 K. The two gasses react to form nitric oxide gas by the
reaction
N2(g) + O2(g) ⇌ 2 NO(g).
Calculate the equilibrium concentrations of each gas.
(Kc for the reaction is 4.1 x 10-4 at 2000 K)
Conc
(M)
I
N2(g)
0.50 mol/2.00L
=0.25 M
-x
0.25 M - x
C
E
O2(g)
2 NO(g)
0.86 mol/2.00 L
= 0.43 M
-x
0.43 M - x
0
+2x
2x
K = [NO]2/[N2][O2]
[2x]2/(0.25-x)(0.43-x)= 4.1 x 10-4
assume X is negligible,
0.25-x≈ 0.25 and 0.43-x≈ 0.43
[2x]2/(0.25)(0.43)= 4.1 x 10-4
x = 3.32 x 10-3
Therefore [N2]= 0.25 M, [O2]= 0.43 M, [NO]=2x = 6.64 x 10-3 M
9
Nitrogen dioxide decomposes according to the following reaction.
2 NO2(g) ⇌ 2 NO(g) + O2(g)
Kp = 4.48 x 10-13
0.55 atm of NO2 is introduced in a container and allowed to reach
equilibrium. Calculate the equilibrium partial pressures of NO, O2 and
NO2 gases.
Atm
I
C
E
2 NO2(g)
0.55
- 2x
0.55 – 2x
2 NO(g)
0
+2x
2x
O2(g)
0
+x
x
Kp = 4.48 x 10-13 = (P NO)2(P O2) / (PNO2)2
= (2x)2 (x) / (0.55 – 2x)2
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Assume x is small since 0.55/4.48 x 10-13 >> 400,
then (0.55 – 2x)2 ≈ (0.55)2
4x3 / (0.55)2 = 4.48 x 10-13
;
x = 3.24 x 10-5
P NO = 2x = 2(3.24 x 10-5) = 6.48 x 10-5 atm
P O2 = x = 3.24 x 10-5atm
P NO2 = 0.55 – 2x = 0.55 atm
10
Consider a reaction at equilibrium. Initially, the concentrations are
4.00 M CH4, 8.00 M H2S, 4.00 M CS2 and 8.00 M H2. At equilibrium,
[CS2] = 2.44 M. Calculate the equilibrium concentration of H2.
CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g)
M
I
C
E
CH4(g)
4.00
+x
2 H2S(g)
8.00
+2x
CS2(g)
4.00
-x
4.00-x
= 2.44
4 H2(g)
8.00
-4x
8.00 – 4x
x can only have a positive value. Thus, CS2 is a reactant.
4.00 - x = 2.44 ; x = 1.56 M
[H2] = 8.00 – 4x = 1.76 M
11
The enthalpy change for the following reaction at standard conditions
is +92.2 kJ.
2 NH3(g) ⇌ N2(g) + 3H2(g)
Kp = 6.25 at 250C
State the effect (direction) of the equilibrium and the amount of
nitrogen (increase, decrease or unchange).
(a)
Adding more NH3(g)
Shift to the right , increase N2
(b)
Removing some H2(g)
Shift to the right, increase N2
(c)
Decreasing the temperature of system
Shift to the right, decrease N2
(d)
Decreasing the volume of the container
Pressure increase, therefore Shift to the left, decrease N2
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(e)
Adding Krypton gas
No effect, N2 remain unchanged
(f)
Adding catalyst
No effect because catalyst will only increase the rate of
reaction
(g)
What is the effect of temperature increase towards the value of
Kp?
Magnitude Kp increases.
TOPIC 4 : ACID-BASE EQUILIBRIA
1
2
(a)
Define acid and base, based on Bronsted-Lowry theory.
Acid-a substance that can donate a proton to another
substance
Base-a substance that can accept a proton to another
substance
(b)
Determine whether these species are acids and/ or bases
according to Bronsted- Lowry theory:
(a)
CO32(b)
HBr
(c)
H 2O
BL base
BL acid
BL acid and base
HClO, CH3NH2 and H2SO3 are some weak acids and base. For each of
them
(a)
classify as a Bronsted-Lowry acid or base.
(b)
write the equation for dissociation in water.
(c)
identify all conjugate acid-base pairs.
(d)
write the Ka or Kb expression.
(a)
HClO is a BL acid
HClO (aq) + H2O (l) ⇌ ClO- (aq) + H3O+ (aq)
Conjugate acid-base pairs: HClO/ClO- and H3O+/H2O
Ka = [ClO-][ H3O+]/[ HClO]
(b)
CH3NH2 is a BL base
CH3NH2 (aq) + H2O (l) ⇌ CH3NH3+ (aq) + OH- (aq)
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Conjugate acid-base pairs: CH3NH3+/CH3NH2 and H2O/OHKb = [CH3NH3+][OH-]/[CH3NH2]
(c)
H2SO3 is a BL acid (diprotic)
H2SO3 (aq) + H2O (l) ⇌ HSO3- (aq) + H3O+ (aq)
1st dissociation
Conjugate acid-base pairs: H2SO3/HSO3- and H3O+/H2O
Ka = [HSO3-][H3O+]/[H2SO3]
HSO3- (aq) + H2O (l) ⇌ SO32- (aq) + H3O+ (aq)
2nd dissociation
Conjugate acid-base pairs: HSO3-/SO32- and H3O+/H2O
Ka = [SO32-][H3O+]/[ HSO3-]
3
Explain why hydrochloric acid is a strong acid while HF is a weak acid.
HCl
is
a
strong
acid
because
of
its
complete
dissociation/ionization in water, while HF is a weak acid
because of its partial dissociation/ionization in water. This is
supported by the large Ka value for HCl and small value (6.3 x
10-4)
for
HF.
Ka
value
shows
the
extent
of
acid
dissociation whereby the larger the Ka value, the greater the
dissociation and the stronger the acid.
4
Complete the following equations and list down all conjugate acid-base
pairs that are present :
(a)
NH3 (aq)+ CH3COOH (aq) ⇌ NH4+ (aq) + CH3COO- (aq)
NH4+/NH3 and CH3COOH/CH3COO-
(b)
H2PO4- (aq) + CO32- (aq) ⇌ HPO42- (aq) + HCO3- (aq)
H2PO4-/HPO42- and HCO3-/CO32-
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INTRODUCTION TO ORGANIC CHEMISTRY,
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(c)
H2PO4- (aq) + HF (aq) ⇌ H3PO4 (aq) + F- (aq)
H3PO4/H2PO4- and HF/F-
5
Determine if the following solutions are acidic or basic. Explain.
(a)
A solution with [OH-] = 6.3 x 10-12
A solution is basic when [OH-] > 1 x 10-7. Thus, this
solution is acidic since [OH-] < 1 x 10-7.
(b)
A solution with [H3O+] = 2.4 x 10-4
A solution is acidic when [H3O+] > 1 x 10-7. Thus, this
solution is acidic since [H3O+] > 1 x 10-7.
(c)
A solution with pH of 7.1
A solution is basic when pH > 7.0. Thus, this solution is
basic.
(d)
A solution with pOH of 13.5
A solution with pOH = 13.5 has pH = 0.5. This solution is
acidic since pH < 7.0
6
Which of the following is more acidic? Explain.
(a)
A 0.1M acid solution with Ka = 5.5 x 10-9 or 0.1M acid solution
with Ka = 3.4 x 10-6
(b)
0.1M acid solution with pKa=7.38 or 0.1M acid solution with
pKa=4.56
(c)
An acid solution with concentration of 0.2M or 0.4M
(d)
0.1M of a weak acid or 0.1M of a strong acid
(e)
0.1M of a weak acid or 0.1M of weak base
(a)
The solution with Ka = 3.4 x 10-6 because it has greater
Ka
(b)
The solution with pKa=4.56 because it has lower pKa
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(greater Ka)
(c)
The solution with concentration of 0.4M because it is
more concentrated (more H3O+ is present)
(d)
The strong acid because it dissociates completely in
water (more H3O+ is present)
(e)
The weak acid because a solution of weak acid contain
more hydronium ion than a solution of weak base
7
Calculate the Ka for the following solutions:
(a)
0.05M HA with [A-] = 8.2 x 10-9 M
(M)
HA
H2O
A-
H3O+
I
0.05
-
0
0
C
-x
-
+x
+x
E
0.05-x
-
x
x
From the table, x = [A-] = [H3O+] = 8.2 x 10-9
Ka = [A-][ H3O+]/[HA] = x2/(0.05-x) = 1.34 x 10-15
(b)
1.6M HF, given that pH of the solution is 1.48.
(M)
HF
H2O
F-
H3O+
I
1.6
-
0
0
C
-x
-
+x
+x
E
1.6-x
-
x
x
Because pH = 1.48, [H3O+] = x = 10-pH = 3.29 x 10-2
Ka = [F-][ H3O+]/[HF] = x2/(1.6-x) = 6.9 x 10-4
[Theoretical Ka = 6.8 x 10-4]
41
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8
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Calculate the [H3O+], [OH-], pH and pOH for the following solutions.
For (c) and (d), calculate the pKa and pKb as well.
(a) 1M HNO3
(b) 1M LiOH
(c) 2.5 M CH3COOH [Ka = 1.8 x 10-5]
(d) 5M NH3 [Kb = 1.8 x 10-5]
(a)
1M HNO3 makes [H3O+] = 1M ; pH = - log [H3O+] = 0
[OH-] = (1 x 10-14)/1 = 1 x 10-14
;
pOH= - log [OH-] = 14
OR
pOH = 14 – pH = 14
;
[OH-] = 10-pOH = 1 x 10-14
(b)
1M LiOH makes [OH-] = 1M
; pOH = - log [OH-] = 0
[H3O+] = 1 x 10-14/1 = 1 x 10-14
;
pH = - log [H3O ] = 14
+
OR
pH = 14 – pOH = 14
;
[H3O+] = 10-pH = 1 x 10-14
(c)
2.5 M CH3COOH
(M)
CH3COOH
H2O
CH3COO-
H3O+
I
2.5
-
0
0
C
-x
-
+x
+x
E
2.5-x
-
x
x
Ka = 1.8 x 10-5 = x2/(2.5-x) ≈ x2/2.5 ;
can be simplified since 2.5/Ka >>400
Thus, x = 6.71 x 10-3 = [H3O+]
;
pH = - log [H3O ] = 2.17
+
[OH-] = (1 x 10-14)/6.71 x 10-3 = 1.49 x 10-12
pOH= - log [OH-] = 11.827
42
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
OR
pOH = 14 – pH = 11.827
;
[OH-] = 10-pOH = 1.49 x 10-12
pKa = - log (1.8 x 10-5) = 4.74
; pKb = 14 - pKa = 9.26
(d)
5M NH3
(M)
NH3
H2O
NH4+
OH-
I
5
-
0
0
C
-x
-
+x
+x
E
5-x
-
x
x
Kb = 1.8 x 10-5 = x2/(5-x) ≈ x2/5 ;
can be simplified since 5/Ka >>400
Thus, x = 9.49 x 10-3 = [OH-]
;
pOH = - log [OH-] = 2.02
[H3O+] = (1 x 10-14)/ 9.49 x 10-3 = 1.05 x 10-12 ;
pH= - log [H3O+] = 11.98
OR
pH = 14 – pOH = 11.98
[H3O ] = 10
+
-pH
= 1.05 x 10
pKb = - log (1.8 x 10-5) = 4.74
9
;
-12
; pKa = 14 - pKa = 9.26
Determine and explain if the following salt is acidic, basic or neutral.
Arrange these salts in the order of increasing pH.
(a)
NH4Br
(b)
CH3COOK
(c)
Zn(NO2)2 [Ka (Zn2+) = 1.0 x 10-9 ; Ka (HNO2) = 7.1 x 10-4]
(a)
[Kb (NH3) = 1.8 x 10-5]
Acidic salt. NH4+ is acidic because it is derived from a
weak base while Br- is neutral because it is derived
from a strong acid. NH4+ will hydrolyze to form H3O+
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(b)
Basic salt. CH3COO- is basic because it is derived from a
weak acid while K+ is neutral because it is derived from a
strong base. CH3COO- will hydrolyze to form OH-
(c)
Acidic salt. Ka (Zn2+) > Kb (NO2-)
Order of increasing pH: Zn(NO2)2 < NH4Br < CH3COOK
10
For each of the following, write the relevant acid-base equilibrium
equation and predict if pH increases or decreases with each change.
Explain your prediction.
(a)
KCN is added into HCN
(b)
NH4Cl is added into NH3
(c)
NaOH is added into LiOH
(d)
NaCl is added into HCl
(a)
HCN (aq) + H2O (l) ⇌ CN- (aq) + H3O+ (aq)
Addition of CN- will shift equilibrium to the left (reactant
side) and
produce more HCN which leads to lesser
H3O+. Thus, pH will increase as acidity is reduced.
(b)
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Addition of NH4+ will shift equilibrium to the left (reactant
side) and produce more NH3 which leads to lesser OH-.
Thus, pH will decrease as basicity is reduced.
(c)
Both NaOH and LiOH will dissociate completely in
water. Thus, addition of NaOH only adds more OH-,
increasing the pH of solution.
(d)
11
NaCl is a neutral salt. Thus, the addition of NaCl has no
effect whatsoever on the pH of solution.
Define buffer.
A solution of weak acid and its conjugate base or weak base
and its conjugate acid that can resist pH changes when a small
amount of strong acid or base is added.
44
SHE1325
12
13
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Predict the pH at equivalence point when the following titration is
carried out:
(e)
CH3COOH with LiOH
Basic
(f)
Ca(OH)2 with HClO3
Neutral
(g)
Sr(OH)2 with CH3COOH
Basic
(h)
HClO4 with NH3
Acidic
A 1.0 L buffer solution is made up of 0.5M HCOOH and 0.4M HCOONa.
[Ka = 1.8 x 10-4]
(a)
Calculate the initial pH
(b)
Calculate the pH if 0.01 mol of KOH is added into the buffer
solution
(c)
Calculate the pH if 0.01 mol of HCl is added into the buffer
solution
(a)
pH
= pKa + log [A-]/[HA]
= - log (1.8 x 10-4) + log (0.4)/(0.5)
= 3.648
(b)
0.01 mol of OH- will react with 0.01 mol of HCOOH to
form 0.01mol of HCOO-.
pH
= pKa + log [A-]/[HA]
= - log (1.8 x 10-4) + log (0.4+0.01 )/(0.5-0.01)
= 3.667
(c)
0.01 mol of H+ will react with 0.01 mol HCOO- to form
0.01 mol HCOOH
pH
= pKa + log [A-]/[HA]
= - log (1.8 x 10-4) + log (0.4-0.01)/(0.5+0.01)
= 3.628
45
SHE1325
14
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Listed below is some acid, base and salt.
Acids (Ka)
Bases (Kb)
Salts
HCOOH (1.8 x 10-4)
CH3NH2 (4.4 x 10-4)
NaF
HF (6.8 x 10-4)
C6H5NH2 (4.0 x 10-10)
CH3NH3Br
(a)
How many pairs of buffer can you make from this list?
(b)
Suggest the pH in which each buffer works best.
(a)
Two. HF and NaF. CH3NH2 and CH3NH3Br
(b)
Best at pH = pKa.
Thus, HF is best at pH = - log Ka = 3.17
Thus, CH3NH2 is best at pH = - log Kw/Kb = 10.64
15
20 mL of 0.30M HF is titrated with 0.10 M KOH. [Ka (HF) = 6.8 x 10-4]
(a)
Predict whether pH at the equivalence point will be acidic, basic
or neutral.
(b)
Determine a suitable indicator for this reaction.
(c)
Calculate the initial pH.
(d)
Calculate the pH after 10 mL of KOH is added
(e)
Calculate the pH after 60 mL of KOH is added
(f)
Calculate the pH after 70 mL of KOH is added
(g)
Sketch the titration pH curve based on your calculations in c, d,
e and f.
(a)
Basic pH due to the presence of F- ion which will
hydrolyze to form
OH-
(b)
Phenolphthalein
46
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(c)
Initial pH
(M)
HF
H2O
F-
H3O+
I
0.30
-
0
0
C
-x
-
+x
+x
F
0.30 - x
-
x
x
0.30M/Ka >> 400, so the Ka expression can be simplified
into Ka = x2/0.30 ; x = [H3O+] = 1.43 x 10-2 ;
pH = - log 1.43 x 10-2 = 1.85
(d)
10 mL KOH
(mol)
HF
KOH
KF
H2O
I
0.020Lx0.30M
0.010Lx0.10M
0
-
= 6 x 10-3
= 1 x 10-3
C
- 1 x 10-3
- 1 x 10-3
+ 1 x 10-
-
3
F
5 x 10-3
0
1 x 10-3
-
From the table, you can see that this is prior to equivalence
point. Also, acid and its conjugate base are present.
pH
= pKa + log [A-]/[HA]
= - log (6.8 x 10-4) + log (1 x 10-3)/(5 x 10-3)
= 2.47
47
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(e)
60 mL KOH
This is at the equivalence point whereby 6 x 10-3 mol of
KOH reacts with 6 x 10-3 mol of HF to form 6 x 10-3 mol of
KF. KF will hydrolyze in water as such:
(mol)
F-
H2O
HF
OH-
I
6 x 10-3
-
0
0
C
-x
-
+x
+x
F
6 x 10-3 - x
-
x
x
Kb = Kw/Ka = 1.47 x 10-11 = x2/(6 x 10-3-x) ≈ x2/(6 x 10-3) ; can
be simplified since 0.30/Ka >>400
x = [OH-] = 2.97 x 10-7
;
pH = 14 – pOH = 14 – (- log 2.97 x 10-7) = 8.12
(f)
70 mL KOH
This is beyond the equivalence point whereby an excess
of 0.010 L x 0.10M = 1 x 10-3 mol KOH is present.
[OH-] = (1 x 10-3 mol)/(0.020+0.070)L = 0.0111 MH = 14
– pOH = 14 – (- log 0.0111) = 12.05
(g)
Sketch of pH-volume curve
Titration of HF with KOH
14
12
10
8
pH
6
4
2
0
0
20
40
48
60
Volume of KOH (mL)
80
SHE1325
16
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
30 mL of 0.15M NH3 is titrated with 0.10 M HBr.
[Kb (NH3) = 1.8 x 10-5]
(a)
Predict whether pH at the equivalence point will be acidic, basic
or neutral.
(b)
Determine a suitable indicator for this reaction.
(c)
Calculate the initial pH.
(d)
Calculate the pH after 10 mL of HBr is added
(e)
Calculate the pH after 45 mL of HBr is added
(f)
Calculate the pH after 60 mL of HBr is added
(g)
Sketch the titration pH curve based on your calculations in c, d,
e and f.
(a)
Acidic pH due to the presence of NH4+ ion which will
hydrolyze to form H3O+
(b)
Methyl red
(c)
Initial pH
(M)
NH3
H2O
NH4+
OH-
I
0.15
-
0
0
C
-x
-
+x
+x
F
0.15 - x
-
x
x
0.15M/Kb >> 400, so the Kb expression can be simplified
into
Kb = x2/0.15
;
x = [OH-] = 1.64 x 10-3 ;
pH = 14 - pOH = 14 – (-log 1.64 x 10-3) = 11.21
49
SHE1325
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(d)
10 mL HBr
(mol)
NH3
HBr
NH4+
Br-
I
0.030Lx0.15M
0.010Lx0.10M
0
-
= 4.5 x 10-3
= 1 x 10-3
C
- 1 x 10-3
- 1 x 10-3
+ 1 x 10-
-
3
F
3.5 x 10-3
0
1 x 10-3
-
From the table, you can see that this is prior to
equivalence point. Also, base and its conjugate acid are
present.
pH = pKa + log [A-]/[HA]
= - log [(1 x 10-14)/(1.8 x 10-5)] + log [(3.5 x 10-3)/(1 x 10-3)]
= 9.80
(e)
45 mL HBr
This is at the equivalence point whereby 4.5 x 10-3 mol of
HBr reacts with 4.5 x 10-3 mol of NH3 to form 4.5 x 10-3
mol of NH4Br. NH4Br will hydrolyze in water as such:
(mol)
NH4+
I
4.5 x 10-
H2O
NH3
H3O+
-
0
0
3
C
-x
-
+x
+x
F
4.5 x 10-3 - x
-
x
x
Ka
= Kw/Kb = 5.56 x 10-10
= x2/(4.5 x 10-3-x) ≈ x2/(4.5 x 10-3) ;
can be simplified since 0.15/Ka >>400
x = [H3O+] = 1.58 x 10-6
pH = - log 1.58 x 10-6 = 5.80
50
;
SHE1325
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(f)
60 mL HBr
This is beyond the equivalence point whereby an excess
of
0.015 L x 0.10M = 1.5 x 10-3 mol HBr is present.
[H3O+] = (1.5 x 10-3 mol)/(0.045+0.060)L = 0.0143 M
pH = - log 0.0143 = 1.84
(g)
Sketch of pH-volume curve
12
Titration of NH3 with HBr
10
8
pH
6
4
2
0
0
10
20
30
40
50
Volume of HBr (mL)
51
60
70
SHE1325
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
TOPIC 5 : ELECTROCHEMISTRY
1
Balance the following redox equations by using the half-reaction
method.
(a)
Cr2O72- + C2O42-  Cr3+ + CO2
[acidic]
Cr2O72- + 14 H+ + 3C2O42-  2Cr3+ + 6CO2 +7H2O
(b)
Mn2+ + H2O2  MnO2 + H2O
[basic]
Mn2+ + H2O2 +2OH-  MnO2 + 2H2O
(c)
Bi(OH)3 + SnO22-  SnO32- + Bi
[basic]
2Bi(OH)3 + 3SnO22-  3SnO32- + 2Bi + 3H2O
2
State two differences between voltaic cell and electrolytic cell.
Voltaic cell
1. Chemical energy to
electrical energy
2. Spontaneous chemical
reaction occur
3. Anode – negative
terminal
Cathode – positive
terminal
3
Electrolytic cell
1. Electrical energy to
chemical energy
2. Non-spontaneous
reaction is forced to occur
3. Anode – positive
terminal
Cathode – negative
terminal
Consider the cell notation below
ZnZn2+(aq, 1M) Fe3+(aq, 1M), Fe2+(aq, 1M)Pt
(a)
Write the overall cell reaction.
Anode : Zn (s)  Zn2+ (aq) + 2ē
Cathode : 2Fe3+
E0red = -0.763
(aq) + 2ē  2Fe2+ (aq)
E0red = +0.771
Overall: Zn (s) + 2Fe3+ (aq)  Zn2+ (aq) + 2Fe2+ (aq)
(b)
Sketch a cell. Label the anode and cathode, salt bridge and
direction of electron flow.
52
SHE1325
(b)
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Calculate the E0cell.
E0cell = E0red (cathode) - E0red (anode)
= 0.771 – (-0.763)
= 1.534 V
(c)
Calculate the Ecell if [Zn2+]=0.01 M, [Fe3+]=0.01 M and
[Fe2+]=1.5 M.
Ecell = E0cell– 0.0592 log Q
n
= 1.534 – 0.0592 log [Zn2+] [Fe2+]2
2
[Fe3+]2
= 1.534 - 0.0592 log (0.01) (1.5)2
2
(0.01)2
= 1.464 V
(d)
State two purpose of salt bridge. What might it contains?
-
(i) It prevents the electrolytes in two half-cells from
mixing
(ii) It maintains ionic balance in the cell
Inert electrolyte eg. KNO3,KCl
53
SHE1325
4
(a)
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Define the term Standard Hydrogen Electrode (SHE).
Standard electrode potential of an element is defined
as the potential difference between SHE half-cell and
a half-cell of the element in a solution of its ions at
1M, 298 K and 1 atm.
(b)
5
Draw labeled diagrams that could be used to measure the
standard electrode potential of Cu.
Given a selected standard reduction potentials (298 K)
V2+ (aq) + 2 e → V (s)
Pb2+ (aq) + 2 e → Pb (s)
l2 (s) + 2 e → 2 l- (aq)
Cl2 (g) + 2 e → 2 Cl- (aq)
(a)
E°
E°
E°
E°
=
=
=
=
-1.18V
-0.13V
+0.54V
+1.36V
Determine:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
the strongest reducing agent
V (s)
the ion/s or element/s could be reduced by Pb
Cl2 (g), I2 (g)
the ion/s or element/s could be oxidized by Pb2+
V (s)
the ion/s or element/s could be reduced by V
Cl2 (g), I2 (g), Pb2+ (aq)
the ion/s or element/s could be oxidized by V2+
none
Arrange the appropriate species in the order of increasing
oxidizing strength
V2+< Pb2+< I2 < Cl2
54
SHE1325
6
7
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Calculate whether the following reactions would occur spontaneously in
aqueous solution at 25°C. Assume that the initial concentrations of
dissolved species are 1.0 M.
(a)
Ca(s) + Cd2+ (aq)  Ca2+(aq) + Cd(s)
E° = E°red (cathode) - E°red (anode)
= -0.40-(-2.76)
= 2.36 V (spontaneous)
(b)
2Br- (aq) + Sn2+(aq)  Br2(l) + Sn(s)
E° = E°red (cathode) - E°red (anode)
= -0.14-1.07
= -1.21 V (nonspontaneous)
(a)
Calculate the standard potential of a cell consisting of Zn/Zn2+
half cell and the Standard Hydrogen Electrode.
Zn (s) + 2H+  Zn2++ H2 (g)
E° = E°red (cathode) – E°red (anode)
=0.00 V – (-0.76 V)
= 0.76V
(b)
What is the cell potential (Ecell) if the concentration of Zn2+ is
0.45 M, pressure of H2 is 2.0 atm and the concentration of H+ is
1.8 M.
E = E° - (0.0592 / n) log Q
= 0.76 V – (0.0592/2) log [(0.45 x 2.0)/(1.82)
= 0.78 V
8
Consider the following galvanic cell set up at 298 K:
Al(s) l Al3+(aq) ll Ni2+(aq)l Ni(s)
(a)
Calculate Eocell.
Eocell= -0.25V – (-1.66V)
= +1.41V
55
SHE1325
(b)
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Calculate Ecell when water is added into Ni2+(aq)l Ni(s) half cell
such as Ni2+(aq) decreasing ten times. Explain how does cell
potential change.
𝟎.𝟎𝟓𝟗𝟐
𝟏
) 𝐥𝐨𝐠 𝟏
𝟔
( )𝟑
Ecell=1.41-(
=+1.38V. Cell potential decreases.
𝟏𝟎
9
(a)
Sketch a labeled diagram of electrolysis of diluted NaCl solution
using inert electrode.
(b)
Write the reactions at anode and cathode and the observations.
Anode
Reaction
Observation
Cathode
Reaction
Observation
:
:
2H2O (l)  O2 (g) + 4H+ + 4eBubbles of gas evolved.
:
:
2H2O (l) + 2e-  H2 (g) + 2OHBubbles of gas evolved
56
SHE1325
(c)
10
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Write the reactions at anode and cathode when concentrated
NaCl is used as electrolyte. Explain.
Anode
Reaction
Observation
:
:
2Cl-(aq) Cl2(g) + 2eBubbles of yellow-green gas
evolved.
Cathode
Reaction
Observation
:
:
2H2O (l) + 2e-  H2 (g) + 2OHBubbles of gas evolved
Plastic object can be chromium-plated by coating them with a thin
layer of graphite paste first and then placing them in a bath of
aqueous chromium (III) sulphate and electroplating.
(a)
Draw a fully labeled diagram of the electrical circuit for the
electroplating.
Plastic object
coated with
graphite
(b)
(a)
Cr2(SO4)3(aq)
Write the half reactions at both the electrodes.
Anode
Cathode
11
chromium or
graphite
: Cr(s)  Cr3+(aq) + 3e: Cr3+(aq) + 3e- Cr(s)
Write all the possible half reactions occur at the anode and
cathode when the following substances are electrolysed using
inert electrodes:(i)
AgNO3(aq)
Anode: NO3- cannot be oxidized.
: 2H2O(l)  O2(g) + 4H+(aq) + 4eCathode: Ag+(aq) + e-  Ag(l)
: 2H2O + 2e-  H2(g) + 2OH-(aq)
57
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INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(ii)
Molten NaBr
Anode: 2Br-(l)  Br2(g) + 2eCathode: Na+(l) + e-  Na(l)
(iii)
K2SO4(aq)
Anode: SO42- cannot be oxidized.
: 2H2O(l)  O2(g) + 4H+(aq) + 4eCathode: K+(aq) + e-  K(l)
: 2H2O + 2e-  H2(g) + 2OH-(aq)
(iv)
Concentrated MgCl2(aq)
Anode: Cl-(aq)  Cl2(g) + 2e: 2H2O(l)  O2(g) + 4H+(aq) + 4eCathode: Mg2+(aq) + 2e-  Mg(s)
: 2H2O + 2e-  H2(g) + 2OH-(aq).
(b)
What are the products formed at the anode and cathode
in(a)(i)-(iv)? Give your reasoning.
(i)
Anode: oxygen gas because only water can undergo
oxidation. No competition
Cathode: silver molten because of its more positive
reduction potential
(ii)
Anode: bromine gas
Cathode: sodium molten
Only sodium ion and bromide ion are present.
(iii)
Anode: oxygen gas because only water can undergo
oxidation. No competition.
Cathode: hydrogen gas because water has a more
positive reduction potential.
(iv)
Anode: chlorine gas because of its high
concentration. The formation of oxygen needs an
‘overvoltage’.
Cathode: hydrogen gas because Mg has less
positive reduction potential.
58
SHE1325
12
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
(a)
Sketch a labeled diagram of electrolysis of purification of
copper.
(b)
Describe the industrial processes of purification of copper.
The impure copper from a smelter is cast into a block to
form the positive anode. The cathode is made of
previously purified copper. These are dipped into an
electrolyte of copper(II) sulphate solution.
When the d.c electrical current is passed through the
solution electrolysis takes place. The copper anode
dissolves forming blue copper(II) ions Cu2+.
These positive ions are attracted to the negative cathode
and become copper atoms. The mass of copper dissolving
at the anode exactly equals the mass of copper deposited
on the cathode. The concentration of the copper(II)
sulphate remains constant.
Any impurities present in the impure copper anode fall to
the bottom of the electrolysis cell tank. This 'anode
sludge' is not completely mineral waste, it can contain
valuable metals such as silver.
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SHE1325
13
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
Consider the electrolysis of molten barium chloride.
(a)
Write the half- reactions.
Anode : 2Cl-(aq) Cl2(g) + 2eCathode : Ba2+ (aq) +2e-  Ba (s)
(b)
14
How many grams of barium metal can be produced by supplying
0.50 A for 30 min?
Q = It
= 0.50 x 30 x 60
= 900 C
Mass of I2
= 900 C x 1 mol e/96500 C x 1 mol Ba/2 mol e x 137.3
g/1 mol Ba
= 0.62 g
When an aqueous solution of KI is electrolyzed, the following halfreaction occurs.
Anode : 2I-(aq)  I2(aq) + 2e
Cathode : 2H2O (l) + 2e  H2 (g) + 2OHIn the electrolysis, a current of 8.52 x 10-3 A flows through the cell for
2 hours. Calculate
(a)
The amount of charge required to produce 1 mole of I2.
Q = It
= 8.52 x 10-3 x 2 x 60 x60
= 61.344 C
(b)
Calculate the mass of iodine and hydrogen produced.
Mass of I2
= 61.344 C x 1 mol e/96500 C x 1 mol I2/2 mol e
x253.8g/1 mol I2
= 0.079 g
Mass of H2
61.344 C x 1 mol e/96500 C x 1 mol I2/2 mol e x 2.016g/
1 mol H2
= 6.408 x 10-4 g
60
SHE1325
15
INTRODUCTION TO ORGANIC CHEMISTRY,
HYDROCARBON AND HALOGEN COMPOUNDS
A constant current deposits 365 mg of Ag in 216 min from aqueous
silver nitrate solution. Calculate the current used.
Q =365 mg Ag x 1 g Ag/1000 mg Ag x 107.9 g Ag/ 1 g Ag
x 1 mol e/1 mol Ag x 96500 C/1mol e
= 326.4 C
A = Q/t
= 326.4/12960
= 0.0252 A.
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