DUNMAN HIGH SCHOOL
2008 Prelim
H2 CHEMISTRY
Chemical Bonding/Chemical Eqm/Ionic Eqm
2(a) Ammonium cyanate, NH4NCO undergoes thermal decomposition to form a covalent
compound urea, (NH2)2C=O, which is a common fertilizer.
Describe the type of bonding found in ammonium cyanate and hence suggest a
dot–and–cross diagram of cyanate ion, NCO–, stating its shape.
[3]
Ionic bonding between NH4+ and NCO– ions
Covalent bonding between atoms within each ion
Shape:
(b)
linear
On heating urea above its melting point produces an organic cyclic compound N, and
ammonia. N has the composition by mass of C, 28%; N, 33%; H, 2.3%; O, 37% and a
relative molecular mass of 129.
(i)
Determine the molecular formula of N.
Element
C
H
N
O
% mass
28
2.3
33
37
33
= 2.35
14
1
37
= 2.31
16
1
28
= 2.33
12
1
÷ by Ar
Ratio
2.3
1
Empirical formula = CHNO
Let molecular formula be (CHNO)n,
 n=3
n(12+1+14+16) = 129
Molecular formula = C3H3N3O3
(ii)
Suggest a possible structure of N.
O
H
N:
O
C
N
N
C
C
N
H
O
H
(c)
[2]
Cyanic acid (NCOH), with an acid dissociation constant of 3.5 x 10–4 mol dm–3, is the
simplest chemical compound that contains carbon, hydrogen, nitrogen, and oxygen. A buffer
solution can be prepared by mixing 500 cm3 of 0.200 mol dm–3 of ammonium cyanate and
500 cm3 of 0.100 mol dm–3 of cyanic acid solution.
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(i)With the help of an equation, explain how the mixture can act as a buffer on addition of H+.
NCO– + H+  NCOH
H+ is removed as NCOH, by NCO–, hence the pH remains fairly constant.
(ii)Calculate the pH of the prepared buffer solution.
[NCO  ]
pH = pKa + lg
= – lg(3.5  10–4) + lg
[NCOH]
(0.200)
= 3.76
1 (0.100)
2
1
2
(iii)
A 20.0 cm3 sample of 0.100 mol dm–3 of cyanic acid is titrated with
0.100 mol dm–3 aqueous barium hydroxide, Ba(OH)2. Use the data to sketch the pH changes that
occur during the titration when a total of 30 cm3 of Ba(OH)2 is added. Label the following points on
your sketch:
(1)
Initial pH
(2)
End–point
(3)
Point of maximum buffer capacity
pH
>7 –
(2) End–point
(1) Initial pH = 2.22 –
(3) point of maximum
buffer capacity
(need not show value)
0
5
10
Volume of Ba(OH)2 / cm3
(iv)
At certain temperature, cyanic acid is able to form an isomer which is also an acid, via
the migration of a proton. Suggest the structural formula of this isomer.
[8]
Isomer:
(d)
H–N=C=O
Another ammonium salt, ammonium nitrate, sublimes reversibly with heat to produce
ammonia and nitric acid until dynamic equilibrium is reached:
NH4NO3 (s) Ý NH3 (g) + HNO3 (g)
(i)Explain what is meant by the term dynamic equilibrium.
In a reversible reaction, at dynamic equilibrium,
•
Rate of forward reaction = Rate of reverse (backward) reaction
•
The substances are still reacting together although the concentration of the reactants
and products remain constant.
(ii)Write an expression for the equilibrium constant, Kp for the above reaction.
K p  (PHNO3 )(PNH3 )
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(iii)9.00 g of ammonium nitrate is heated in a sealed container and the equilibrium constant is
found to be 15.7 atm2. Calculate the partial pressure of ammonia present at equilibrium.
Since (PNH3 ) = (PHNO3 ) at equilibrium,
(PNH3 )  K p  15.7 = 3.9623 = 3.96 atm
(iv)Hence, determine the percentage of ammonium nitrate which has dissociated, given that one
mole of gas under these conditions exerts a pressure of 50 atm.
No. of mole of NH4NO3 =
No. of mole of NH3 =
% dissociation =
9.00
= 0.1125
80
3.9623
= 0.07924 = No. of mole of NH4NO3 dissociated
50
0.07924
x 100% = 70.4%
0.1125
(v)
At a higher temperature, the equilibrium constant is 16.8 atm 2. Compare this with the
value given in part (iii). Deduce with reasons, whether the sublimation reaction is endothermic or
exothermic.
[7]
When T increases, Kp increases
 (PNH3 ) and (PHNO3 ) increases  equilibrium has shifted right to absorb heat
 (or this) forward endothermic reaction is favoured to absorb heat
 Sublimation reaction is endothermic
[Total: 20]
Chemical Kinetics/Chemical Bonding/Thermochemistry
3(a) The electrolysis of aqueous potassium chloride containing methyl orange indicator is carried
out in the following apparatus using platinum electrodes. The volume of gases liberated at
the electrodes is shown below:
KCl (aq)
platinum
platinum
+
+
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–
+
3
(i)It is found that 12 cm3 of gas is collected at the cathode after passing the current for two minutes
under room conditions. Calculate the size of current used.
Gas collected at cathode is H2:
(n)H2 =
It
2F

12
24000
2H+ + 2e  H2
=
I(2x60)
2(96500)
 I = 0.804 A
(ii)Determine the volume ratio of gases collected at the anode and cathode.
4OH–  O2 + H2O + 2e
At anode:
Overall eqn: H2O  O2 + 2H2
Volume ratio of gases collected:
cathode : anode = 2H2 : O2 = 2:1
(iii)State and explain the colour of the methyl orange indicator around the electrodes.
Around cathode:
colour is yellow
as solution becomes more alkaline when [H+]
decreases as it is being discharged
Around anode:
colour is red
as solution becomes more acidic when [OH–]
decreases as it is being discharged
(iv)The electrolysis is continued over a long time and a greenish–yellow gas is liberated at the
anode. Explain this observation.
Over time, as more OH– gets discharged at the anode, the electrolyte soon becomes
concentrated with Cl–. Since [Cl–] is high or this, greenish–yellow Cl2 gas is evolved,
when Cl– gets preferentially discharged.
[7]
(b)
(i)
When potassium chloride and potassium bromide are separately added to concentrated
sulfuric acid, different observations are noted. Explain the difference in behaviour of
both salts towards concentrated sulfuric acid using relevant data from the Data Booklet.
Write balanced equations for the reactions that occur.
KCl (aq) + H2SO4 (aq)  KHSO4 (aq) + HCl (g)
KBr (aq) + H2SO4 (aq)  KHSO4 (aq) + HBr (g)
2HBr (g) + H2SO4 (aq)  Br2 (g) + 2H2O (l) + SO2 (g)
Cl2 + 2e ⇌ 2Cl–
Br2 + 2e ⇌ 2Br–
E = +1.36 V
E = + 1.07 V
E (Br2 / Br–) is less positive / less than E (Cl2 / Cl–),
Hence, Br– is a stronger reducing agent than Cl–,
Br– can be oxidised to I2 by excess H2SO4 and Cl– cannot be readily oxidised to Cl2.
Or:
Br– can reduce H2SO4 to SO2 but Cl– cannot.
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(ii)
Potassium bromide in concentrated sulfuric acid is used as a reagent in the following
reaction scheme. It is found that organic compound Q produces white fumes when subjected
to PCl5, and has no reaction with alkaline aqueous iodine. Deduce with reasons, the
structures for compounds P – T.
P
H2 / Pt
KBr
Q
c.H2SO4
alcoholic NH3
R
heat
S
C3H7Br
C3H8O
CH3COOH (aq)
T
C5H13O2N
Q produces white fumes when subjected to PCl5 via nucleophilic substitution, and is not
oxidised by alkaline aqueous iodine  Q is an alcohol that does not contain structure
CH3CH(OH)–
P is reduced by H2 to form Q
 P is an alkene, aldehydes or ketone
 Q is an alkane, primary alcohol or secondary alcohol repectively
Q reacts with HBr (from KBr + c.H2SO4) via electrophilic adddion or nucleophilic
substitution to form R  Q is an alkene or alcohol
 R is a halogenoalkane
R undergoes nucleophilic substitution with NH3 to form an amine, S
 R is halogenoalkane
S undergoes neutralisation with CH3COOH to form T
 S is an amine, T is a salt.
CH3CH2CHO
or
CH2=CHCH2OH
CH3CH2CH2OH
CH3CH2CH2Br
CH3CH2CH2NH2
Q
R
S
C3H8O
C3H7Br
CH3COO-NH3+CH2CH2CH3
T
C5H13O2N
P
[10]
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(c)
Give a reagent, other than concentrated sulfuric acid, that can be used to distinguish
between separate solutions of potassium chloride and potassium bromide. State what would
be observed and write an equation for one of the reactions.
Identify another reagent that could be added to the mixtures from the first test to confirm the
identity of the halide ions. State what would be observed in each case.
[3]
Reagent: add dilute HNO3 followed by aqueous AgNO3
For KCl, white ppt will be formed. For KBr, cream ppt will be formed.
Ag+ (aq) + Cl– (aq)  AgCl (s)
To confirm halide ion, add aqueous ammonia to the precipitate.
AgCl ppt dissolves but AgBr will be insoluble / dissolves slightly.
[Total: 20]
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