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Transcript of New Way Chemistry for Hong Kong A-Level Book 41 1 Nitrogen and its Compound 43.1Introduction...
1New Way Chemistry for Hong Kong A-Level Book 4
1
Nitrogen and its Compound43.143.1 IntroductionIntroduction
43.243.2 Unreactive Nature of NitrogenUnreactive Nature of Nitrogen
43.343.3 Direct Combination of Nitrogen and Oxygen Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxidesleading to Formation of Nitrogen Oxides
43.443.4 AmmoniaAmmonia
43.543.5 Nitric(V) AcidNitric(V) Acid
43.643.6 Nitrates(V)Nitrates(V)
Chapter 43Chapter 43
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43.1 Introduction (SB p.110)
Nitrogen (first member of Group VA):
• Electronic configuration: 1s22s2
2p3
• Complete octet by forming diatomic molecules N N
• Non-metal, colourless and odourless gas
• Very low melting and boiling points
• Slightly soluble in water and does not support combustion
Covalent radius (nm) 0.074
Melting point (°C) –210
Boiling point (°C) –196
Bond enthalpy (kJ mol–
1)+944
First ionization enthalpy (kJ mol–1)
+1 400
Electron affinity (kJ mol–1)
+3
Electronegativity 3.0
Some information about nitrogen
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43.1 Introduction (SB p.110)
Nitrogen
• Mainly as free N2 molecules in the atmosphere (78%
by volume)
• Combine with other elements in the form of proteins in
all living things
• Liquid N2 is used as coolant
• Raw material for Haber process
(manufacture of ammonia)
• Ammonia is the major component
of nitrogenous fertilizers
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Nitrogen in gaseous state
• As diatomic molecules (N2) which are held by weak van der Waals’ forces
• 2 atoms are joined by extremely strong triple covalent bonds
• Bond enthalpy of the triple bond = +944 kJ mol–1
• Due to extremely strong covalent bonds and absence of bond polarity
Nitrogen molecule is very unreactive
43.2 Unreactive Nature of Nitrogen (SB p.111)
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43.2 Unreactive Nature of Nitrogen (SB p.111)
Bond Bond
enthalpy (kJ mol–1)
Bond Bond
enthalpy (kJ mol–1)
N N
O = O
H – H
C – C
+944
+496
+436
+348
S – S
Cl – Cl
P – P
F – F
+264
+242
+172
+158
Bond enthalpies of some common covalent bonds
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• Reactions involving nitrogen usually have high activation en
ergies and unfavourable equilibrium constants
e.g. At 25°C
N2(g) + O2(g) 2NO(g) Kc = 4.5 10–31
• The presence of catalyst and high temperature and
pressure may be required for nitrogen to react
N2(g) + 3H2(g) 2NH3(g)400 – 500°C, 300 – 1000 atm
Fe as catalyst
43.2 Unreactive Nature of Nitrogen (SB p.111)
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43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.112)
• N2 will not react at room temperature due to high bond
enthalpy
• At high temperature, N2 shows some reactions with
other elements
∵ sufficient energy to break N N triple bond
• At high temperature,
N2(g) + O2(g) 2NO(g)
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43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.112)
• The electric discharge in lightning provides sufficient e
nergy to break the N N triple bond and then react
with O2
N2(g) + O2(g) 2NO(g)
2NO(g) + O2(g) 2NO2(g)
lightning
colourless Reddish brown(poisonous)
colourless
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43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.112)
• The above reactions are very important in nature
• The NO2 formed dissolves in rainwater to produce nit
ric(V) acid and nitric(III) acid
2NO2(g) + H2O(l) HNO3(aq) + HNO2(aq)
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43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.112)
• At high temperatures in car
engines, N2 & O2 react to form NO(g)
which emitted into air with
exhausted gas
• The NO formed will be oxidized to
NO2
• The NO2 absorbs sunlight and breaks down
into NO and O atom
NO2(g) NO(g) + O(g)
• These leads to formation of photochemical smog
sunlight
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43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.113)
• In laboratory, we use
the apparatus shown
on the right to convert
N2 into NO2
• When current is switched on, electric discharges occur in
the gap between the electrodes
• NO is formed and followed by NO2
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43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.113)
• Other than NO and NO2, N2 can form other oxides
e.g. 2 NO2 molecules (brown) can combine to form a
N2O4 molecule (yellow)
• NO2 & N2O4 exist in equilibrium in gas phase
2NO2(g) N2O4(g) H = –58 kJ mol–1brown yellow
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• The formation of N2O4 is exothermic
N2O4 predominantes at low
temperatures
NO2 predominantes at high
temperatures
the colour of mixture fades on
cooling, darkens on warming
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.113)
NO2 (left) and N2O4 (right) predominate in hot water and ice water respectively
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Check Point 43-1 Check Point 43-1
(a) Draw the structures of the following compounds.
(i) Dinitrogen monoxide
(ii) Nitrogen monoxide
(iii) Dinitrogen trioxide
(iv) Nitrogen dioxide
(v) Dinitrogen tetraoxide
(vi) Dinitrogen pentaoxide Answer
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.114)
(a) (i) Dinitrogen monoxide
(ii) Nitrogen monoxide
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43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.114)
(iii) Dinitrogen trioxide
(iv) Nitrogen dioxide
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43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.114)
(v) Dinitrogen tetraoxide
(vi) Dinitrogen pentaoxide
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Check Point 43-1 Check Point 43-1
(b) Arrange N2, O2 and F2 in an ascending order of reactivity. Explain the order briefly.
Answer(b) The ascending order of reactivity is: N2 < O2 < F2.
The reactivity of diatomic molecules depends on the bond enthalpy of covalent bonds. The bond enthalpy of N N is greater than that of O = O, which in turn is greater than that of F – F. Therefore, the breakage of N N bond requires the greatest amount of energy, whereas the breakage of F – F bond requires the least amount of energy.
43.3 Direct Combination of Nitrogen and Oxygen leading to Formation of Nitrogen Oxides (SB p.114)
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43.4 Ammonia (SB p.114)
Ammonia
• colourless, pungent gas
• polar molecules
• trigonal pyramidal shape with a lone pair of electrons on nitrogen
• extremely soluble in water and easy to condense to liquid due to hydrogen bonds
• good solvent for ionic compounds
• weakly alkaline
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
Kb = 1.8 10–5 mol dm–3
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43.4 Ammonia (SB p.114)
Ammonia
• one of the most fundamental raw materials for modern i
ndustries
• important source of fertilizers and 85% of ammonia i
s used to make nitrogenous fertilizers (e.g. (NH4)2SO4,
NH4NO3)
• making fibres and plastics (rayon, nylon)
• making nitric(V) acid (used to make fertilizers, dyes)
• making household cleaners
• making detergents
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43.4 Ammonia (SB p.115)
Percentages of ammonia used in different industries
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Manufacture of Ammonia by the Haber ProcessManufacture of Ammonia by the Haber Process
43.4 Ammonia (SB p.115)
• NH3 is manufactured industrially by t
he Haber Process, named after the G
erman chemist Fritz Haber
• The process involves direct combina
tion of N2 and H2 under special condi
tions
N2(g) + H2(g) 2NH3(g)
H = –92 kJ mol–1
Fritz Haber (1868 – 1934)
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43.4 Ammonia (SB p.116)
Flow diagram for the Haber process
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Process of Haber Process:
• N2 is obtained from fractional distillation of liquid air
• H2 is obtained from methane, naphtha or mixture by ste
am reforming
CH4(g) + H2O(g) CO(g) + 3H2(g)
CH4(g) + air CO(g) + 2H2(g) + N2(g)
Ni900°C
Ni900°C
43.4 Ammonia (SB p.116)
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• Mixture of CO & H2O is mixed with steam and passed over a heated catalyst
CO(g) + H2O(l) CO2(g) + H2(g)
The CO2 formed is dissolved in water under pressure
• The gases (N2 & H2) are purified before proceeding to the next stage
∵ Compounds of oxygen and sulphur will poison the catalyst
43.4 Ammonia (SB p.116)
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• Purified N2 & H2 are mixed in ratio of 3 : 1 by volume
Compressed to 200 – 1000 atm and heated in the heat exchanger
Hot gaseous mixture is passed over iron in the catalytic chamber
Gases contain 10 – 15% of NH3 and unreacted N2
and H2 when leaving the chamber
The gases are cooled after passing through the heat exchanger
NH3 is liquefied under pressure and unreacted gas
es are recycled
43.4 Ammonia (SB p.116)
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43.4 Ammonia (SB p.117)
Physico-chemical principles:
• Synthesis of ammonia is an exothermic and reversible
reaction
N2(g) + H2(g) 2NH3(g) H = –92 kJ mol–1
• According to Le Chatelier’s principle,
(1) high pressure will increase the yield
(2) low temperature will increase the yield
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43.4 Ammonia (SB p.117)
• Apart from increasing yield, the reaction rate should be
fast
Low temperatures would lower the rate of reaction
∴ optimum temperature is around 500°C which is high
enough for reaction to proceed quickly but low enough
to give satisfactory yield
• Catalyst is used to increase the reaction rate
poisoned by CO, CO2, H2S
Gases entering the catalytic chamber should have
high purity!!
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Chemical Properties of AmmoniaChemical Properties of Ammonia
43.4 Ammonia (SB p.118)
As a base
• NH3 partly ionizes in water to give NH4+ and OH– ion
s
∴ NH3(aq) is alkaline
NH3(g) + H2O(l) NH4+(aq) + OH–(aq)
Kb = 1.8 10–5 mol
dm–3
∴ NH3(aq) is a weak base
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43.4 Ammonia (SB p.118)
Reaction with Acids
• NH3 neutralizes acids to give ammonium salts
e.g. 2NH3(aq) + H2SO4(aq) (NH4)2SO4(aq)
NH3(aq) + HNO3(aq) NH4NO3(aq)
ammonium sulphate(VI)
ammonium nitrate(V)
Filter paper soaked with
NH3
Filter paper soaked with
HCl
• Formation of NH4Cl by reacting NH3 with HCl
NH3(aq) + HCl(aq) NH4Cl(s)
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43.4 Ammonia (SB p.118)
Reaction with Metal Salts
• NH3 precipitates the hydroxide of many metals from solutio
ns of their salts
CaSO4(aq) + 2NH3(aq) +2H2O(l) Ca(OH)2(s) + (NH4)2SO4(aq)
ZnSO4(aq) + 2NH3(aq) +2H2O(l) Zn(OH)2(s) + (NH4)2SO4(aq)
Pb(NO3)2(aq) + 2NH3(aq) +2H2O(l) Pb(OH)2(s) + 2NH4NO3(aq)
CuSO4(aq) + 2NH3(aq) +2H2O(l) Cu(OH)2(s) + (NH4)2SO4(aq)
FeSO4(aq) + 2NH3(aq) +2H2O(l) Fe(OH)2(s) + (NH4)2SO4(aq)
Fe2(SO4)3(aq) + 6NH3(aq) +6H2O(l) 2Fe(OH)3(s) + 3(NH4)2SO4(aq)
white
white
white
blue
dirty green
reddish brown
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43.4 Ammonia (SB p.119)
Pb(OH)2(s) Cu(OH)2(s) Fe(OH)2(s) Fe(OH)3(s)
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43.4 Ammonia (SB p.119)
• Some metal hydroxides (e.g. Zn(OH)2 & Cu(OH)2) redis
solve in excess NH3 solution and form complex compound
s
Zn(OH)2(s) + 4NH3(aq) [Zn(NH3)4]2+(aq) + 2OH–(aq)
Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 2OH–(aq)
colourless
deep blue
A solution containing Cu2+(aq)
Cu(OH)2(s)
[Cu(NH3)4]2+(aq)
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43.4 Ammonia (SB p.120)
AgCl(s) dissolves
Addition of excess NH3(aq)
water
AgCl(s)
• Silver(I) ions also form a complex with ammonia
• AgCl is insoluble in water and acids, but dissolves in excess
NH3 forming soluble complex ion [Ag(NH3)2]+(aq)
AgCl(s) Ag+(aq) + Cl–(aq)
Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq)
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• NH3 does not burn in air or support combustion
• It burns in O2 with a yellow flame, forming N2 and water vapour
43.4 Ammonia (SB p.120)
As a Reducing Agent
Laboratory set-up for oxidation of ammonia
Reaction with Oxygen
4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)
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• In the presence of catalyst (red hot spiral coil of platinum
at 800 – 900°C), NH3 is oxidized to NO by O2
43.4 Ammonia (SB p.120)
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
• This is called catalytic oxid
ation of ammonia
• Key reaction in the preparat
ion of HNO3
Pt
Laboratory set-up for catalytic oxidation of ammonia
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43.4 Ammonia (SB p.121)
Reaction with Copper(II) Oxide
• When dry NH3 is passed over heated black CuO, NH3 i
s oxidized to N2 and H2O
• The CuO turns from black to reddish brown as it is r
educed to Cu
2NH3(g) + 3CuO(s) 3Cu(s) + N2(g) + 3H2O(g)
Laboratory set-up for oxidation of ammonia by copper(II) oxide
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Check Point 43-2 Check Point 43-2
(a) Write chemical equations to show how hydrogen is produced from
(i) the reaction of natural gas (mainly methane) with water;
(ii) the reaction of coal (mainly carbon) with water.
Answer
43.4 Ammonia (SB p.121)
(a) (i) CH4(g) + H2O(g) CO(g) + 3H2(g)
(ii) C(s) + H2O(l) CO(g) + H2(g)
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Check Point 43-2 (cont’d) Check Point 43-2 (cont’d)
(b) Consider the following reversible reaction:
N2(g) + 3H2(g) 2NH3(g) H = –92 kJ mol–1
Discuss how each of the following factors affects the above equilibrium:
(i) increase in temperature
(ii) decrease in pressure
(iii) addition of a suitable catalyst Answer
43.4 Ammonia (SB p.121)
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(b) (i) The forward reaction is exothermic. According to Le Chatelier’s principle, exothermic reactions are favoured at low temperatures. Therefore, an increase in temperature will favour the backward reaction, and thus decrease the yield of ammonia.
(ii) According to Le Chatelier’s principle, a high pressure will increase the yield of ammonia as the forward reaction is accompanied by a decrease of volume from four to two volumes of the gas. Therefore, a decrease in pressure will decrease the yield of ammonia.
(iii) Addition of a suitable catalyst will increase the rate of both forward and backward reactions to the same extent. As it does not change the position of the equilibrium, the yield of ammonia remains constant.
43.4 Ammonia (SB p.121)
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Check Point 43-2 (cont’d) Check Point 43-2 (cont’d)
(c) Ammonia reacts with oxygen in two different ways. Give equations for both of these reactions and explain how one of them is used industrially to produce nitric(V) acid.
Answer
43.4 Ammonia (SB p.121)
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(c) In the absence of catalyst, ammonia burns to give molecular nitrogen and water vapour.
4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)
Industrially, in the presence of red hot platinum-rhodium at about 850°C, ammonia is catalytically oxidized to nitrogen monoxide.
4NH3(g) + 5O2(s) 4NO(g) + 6H2O(g)
The nitrogen monoxide formed then reacts with oxygen from the air to give nitrogen dioxide.
2NO(g) + O2(g) 2NO2(g)
The nitrogen dioxide reacts with excess air and water to produce aqueous nitric(V) acid.
4NO2(g) + O2(g) + 2H2O(l) 4HNO3(aq)
H2O(l) + 3NO2(g) 2HNO3(aq) + NO(g)
The NO(g) is recycled and subsequently combines with more oxygen and water to give more nitric(V) acid. Finally, the product is distilled to give concentrated nitric(V) acid (containing 68% HNO3).
43.4 Ammonia (SB p.121)
Pt – Rh
850°C
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Nitric(V) acid
• a very strong acid
• turns yellow on storage as the formation
of dissolved NO2 from decomposition of s
ome acid
4HNO3(l) 4NO2(aq) + 2H2O(l) +O2(g)
• keep in brown bottles as light will speed
up decomposition
• used to make explosives, nylon, fertilizers
and dyestuff synthesis
43.5 Nitric(V) Acid (SB p.121)
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Manufacture of Nitric(V) Acid from the Catalytic Oxidation of Ammonia
Manufacture of Nitric(V) Acid from the Catalytic Oxidation of Ammonia
43.5 Nitric(V) Acid (SB p.122)
• Most of the ammonia formed is converted to nitric(V) ac
id by Ostward process
• Ostward process is divided into 3 stages:
1. Mixture of ammonia and excess air is passed over
Pt-Rh catalyst at around 700-800°C under low pres
sure
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)Pt-Rh
850°C
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2. The NO formed then reacts with O2 to form NO2
2NO(g) + O2(g) NO2(g)
3. The NO2 reacts with excess air and water to give aqueo
us HNO3
4NO2(g) + O2(g) + 2H2O(l) 4HNO3(aq)
43.5 Nitric(V) Acid (SB p.122)
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Nitric(V) Acid as an Oxidizing AgentNitric(V) Acid as an Oxidizing Agent
• HNO3 is a strong oxidizing agent, especially when
concentrated
• NO3– acts as an electron acceptor when H+ ions are present
• HNO3 can be reduced to different nitrogen compounds
with different oxidation states, depending on
1. the conc. of HNO3
2. nature of substance being oxidized
43.5 Nitric(V) Acid (SB p.122)
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• If dilute or moderately concentrated HNO3 is reduce
d, NO will be formed
4HNO3(aq) + 3e– 3NO3 –(aq) + 2H2O(l) + NO(g)
or NO3–(aq) + 4H+(aq) + 3e– NO(g) + 2H2O(l)
• If concentrated HNO3 is reduced, NO2 will be formed
2HNO3(aq) + e– NO3 –(aq) + NO2(g) + H2O(l)
or NO3–(aq) + 2H+(aq) + e– NO2(g) + H2O(l)
• The electrons are supplied by the reducing agent in t
he reaction
43.5 Nitric(V) Acid (SB p.122)
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• Cu reacts with warm dilute HNO3 to give NO
3Cu(s) + 8HNO3(aq)
3Cu(NO3)2(aq) + 4H2O(l) + 2N
O(g)
• The NO formed reacts with atmospheric O2 to give
NO2
2NO(g) + O2(g) 2NO2(g)
43.5 Nitric(V) Acid (SB p.123)
Reaction with Copper
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• Conc. HNO3 (~14 M) reacts with Cu to give NO2 and a blue s
olution of Cu(NO3)2
Cu(s) + 4HNO3(aq) Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)
43.5 Nitric(V) Acid (SB p.123)
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• Conc. HNO3 oxdizes green Fe2+ ions to brown Fe3+ io
ns while itself reduced to NO
3Fe2+(aq) + NO3–(aq) + 4H+(aq)
3Fe3+(aq) + NO(g) + 2H2
O(l)
• The NO formed reacts with atmospheric O2 to form N
O2
2NO(g) + O2(g) NO2(g)
43.5 Nitric(V) Acid (SB p.123)
Reaction with Iron(II) Ion
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• Hot concentrated HNO3 oxidizes sulphur to give sulph
uric(VI) acid and brown fumes of NO2
S(s) + 6HNO3(aq)
H2SO4(aq) + 6NO2(g) + 2H2
O(l)
43.5 Nitric(V) Acid (SB p.123)
Reaction with Sulphur
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Check Point 43-3 Check Point 43-3
Account for the following observation by giving a balanced equation.
(a) Nitrogen monoxide turns brown when exposed to air.
Answer
43.5 Nitric(V) Acid (SB p.123)
(a) Nitrogen monoxide reacts with atmospheric oxygen to give brown nitrogen dioxide gas.
2NO(g) + O2(g) 2NO2(g)
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Check Point 43-3 (cont’d) Check Point 43-3 (cont’d)
Account for the following observation by giving a balanced equation.
(b) Nitric(V) acid turns yellowish brown on standing.Answer
43.5 Nitric(V) Acid (SB p.123)
(b) Nitric(V) acid turns yellowish brown on standing because of the dissolved nitrogen dioxide formed from the decomposition of some of the acid.
4HNO3(aq) 4NO2(aq) + 2H2O(l) + O2(g)
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Check Point 43-3 (cont’d) Check Point 43-3 (cont’d)
Account for the following observation by giving a balanced equation.
(c) Silver dissolves in dilute nitric(V) acid, yielding a colourless gas.
Answer
43.5 Nitric(V) Acid (SB p.123)
(c) Dilute nitric(V) acid is reduced by silver to form colourless nitrogen monoxide gas.
3Ag(s) + 4HNO3(aq)
3Ag+(aq) + 3NO3–(aq) + 2H2O(l) + NO(g)
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Check Point 43-3 (cont’d) Check Point 43-3 (cont’d)
Account for the following observation by giving a balanced equation.
(d) The nitrate of a metal ion decomposed on heat to give the metal.
Answer
43.5 Nitric(V) Acid (SB p.123)
(d) Both mercury nitrate(V) and silver nitrate(V) decompose on heating to give the corresponding metal.
Hg(NO3)2(s) Hg(s) + 2NO2(g) + O2(g)
2Ag(NO3)2(s) 2Ag(s) + 2NO2(g) + O2(g)
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43.6 Nitrates(V) (SB p.124)
• Metal nitrates(V) can be prepared by reacting very dilute nit
ric(V) acid with metals, metal oxides, hydroxides or carbon
ates
e.g. Mg(s) + 2HNO3(aq) Mg(NO3)2(aq) + H2(g)
CuO(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l)
NaOH(aq) + HNO3(aq) NaNO3(aq) + H2O(l)
Na2CO3(aq) + 2HNO3(aq)
2NaNO3(aq) + H2O(l) + CO2(g)
very dilute
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43.6 Nitrates(V) (SB p.124)
• Metal nitrates(V) can be prepared by reacting metals wit
h concentrated nitric(V) acid
e.g. Mg(s) + 4HNO3(aq)
Mg(NO3)2(aq) + 2H2O(l) + 2N
O2(g)
concentrated
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Action of Heat on Nitrates(V)Action of Heat on Nitrates(V)
When metal nitrates(V) in solid form are strongly heated, they decompose differently according to their thermal stability
43.6 Nitrates(V) (SB p.124)
Metal oxide, nitrogen dioxide and oxygen
2Ca(NO3)2(s) 2CaO(s) + 4NO2(g) + O2(g)
2Mg(NO3)2(s) 2MgO(s) + 4NO2(g) + O2(g)
4Al(NO3)3(s) 2Al2O3(s) + 12NO2(g) + 3O2(g)
2Zn(NO3)2(s) 2ZnO(s) + 4NO2(g) + O2(g)
2Fe(NO3)2(s) 2FeO(s) + 4NO2(g) + O2(g)
2Pb(NO3)2(s) 2PbO(s) + 4NO2(g) + O2(g)
2Cu(NO3)2(s) 2CuO(s) + 4NO2(g) + O2(g)
Calcium
Magnesium
Aluminium
Zinc
Iron
Lead
Copper
Metal nitrate(III), oxygen
2KNO3(s) 2KNO2(s) + O2(g)
2NaNO3(s) 2NaNO2(s) + O2(g)
PotassiumSodium
ProductReaction Nitrate of
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43.6 Nitrates(V) (SB p.124)
Dinitrogen oxide and water
NH4NO3(s) N2O(g) + 2H2O(l)Ammonium ion
Metal, nitrogen dioxide and oxygen
Hg(NO3)2(s) Hg(s) + 2NO2(g) + O2(g)
2AgNO3(s) 2Ag(s) + 2NO2(g) + O2(g)
Mercury(II)Silver
ProductReaction Nitrate of
Cont’d
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Brown Ring Test for Nitrate(V) IonsBrown Ring Test for Nitrate(V) Ions
• The brown ring test is used to detect nitrate(V) ions in
aqueous solutions
43.6 Nitrates(V) (SB p.125)
Procedure:
1. Mix a freshly prepared FeSO4 solution with a solution suspected of containing nitrate(V) ions in a test tube
2. Conc. H2SO4 is added carefully along the side tothe bottom of the test tube with the test tube tilted
Laboratory set-up for brown ring test
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• Formation of a brown ring confirms the presence of
nitrate(V) ions in the solution
43.6 Nitrates(V) (SB p.125)
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43.6 Nitrates(V) (SB p.125)
Reactions involved in the brown ring test:
• Nitrate(V) ions react with conc. H2SO4 to give HNO3
NO3–(aq) + H2SO4(l) HNO3(aq) + HSO4
–(aq)
• The nitric(V) acid oxidizes some FeSO4 to Fe2(SO4)3 and is itself reduced to NO
HNO3(aq) + 3Fe2+(aq) + 3H+(aq)
NO(g) + 3Fe3+(aq) + 2H2O(l)
• Finally, NO reacts with unreacted FeSO4 to form a brown complex
FeSO4(aq) + NO(g) FeSO4 • NO(aq)Brown complex
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Check Point 43-4 Check Point 43-4
Give the name of the ion responsible for the following observation.
(a) An ion produces a blue precipitate with ammonia solution. The blue precipitate redissolves in excess ammonia solution to give a clear deep blue solution.
Answer
(a) Copper(II) ion
43.6 Nitrates(V) (SB p.125)
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Check Point 43-4 (cont’d) Check Point 43-4 (cont’d)
Give the name of the ion responsible for the following observation.
(b) An ion produces a dirty green precipitate with ammonia solution.
Answer
(b) Iron(II) ion
43.6 Nitrates(V) (SB p.125)
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Check Point 43-4 (cont’d) Check Point 43-4 (cont’d)
Give the name of the ion responsible for the following observation.
(c) An ion gives a positive result in the brown ring test.
Answer
(c) Nitrate(V) ion
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The END