1) Atoms, Molecules and Stoichiometry

Q. 150cm3 of oxygen were added to 20cm3 of a gaseous hydrocarbon D. After explosion and cooling, the gaseous mixture occupied 130cm3 and, after absorption by potassium hydroxide, 90cm3 of oxygen remained. (All volumes being measured at r.t.p.). Find D.

C x H y+(x+ y4 )O2 → xC O2+

y2

H 2O

20cm3 60cm3 40cm3

1 mol 2 mol 2 mol

D is C2H4

Q. A gardener fumigates his greenhouse (volume = 24m3) by burning a sulfur candle. A gaseous concentration of sulfur dioxide of 50ppm by volume is found to be effective in getting rid of pests and moulds. Determine the mass of sulfur the gardener must burn to obtain a concentration of 50ppm. (All volumes being measured at r.t.p.).

V S O2=50 ppm×V greenhouse=50 ×1 0−6× 24=1.2×10−3m3

V m=Vn

n= VV m

=1.2 ×1 0−3

24 ×1 0−3 =0.050 mol

S+O2 → S O2

Mass of S=n× A r=0.050 ×32.1=1.61 g

Q. 23.00cm3 of H2X was diluted to 250cm3 with distilled water. 25.0cm3 of this diluted solution was found to require 22.40cm3 of 0.100 M NaOH for complete neutralisation. Calculate the concentration (in M) of H2X.

nNaOH=22.40 ×1 0−3× 0.100=2.24 ×10−3 mol

H 2 X+2 NaOH → N a2 X+2 H 2O

∴nH 2 X=12

nNaOH =1.12× 10−3 mol(¿25.0 c m3)

¿250 c m3 , nH 2 X=0.0112 mol

c H2 X=nV

= 0.0112

23 × 10−3=0.487 M

Q. Baking powder causes cake to rise as a result of the reaction between cream of tartar (potassium hydrogen tartrate, KHC4H4O6) and baking soda (NaHCO3).

KH C4 H 4 O6+NaHC O3 → KNaC4 H 4O6+C O2+H 2O

(a) What mass of cream of tartar is needed for the complete reaction of 1.00g of baking soda?

(b) A sample of cheap self-rising flour contains 10.0g of cream of tartar and 5.0g of baking soda. What volume of CO2 (at r.t.p.) would be released when the reaction between the two components proceeds to completion?

(a) nNaHC O3=1.00

84.0=0.0119mol

¿nKH C4 H 4 O6

mass ( KH C4 H 4O6 )=0.0119× 188.1=2.24 g

(b) nNaHC O3= 5.0

84.0=0.0595 mol

nKH C4 H 4 O6= 10.0

188.1=0.0532 mol (Limiting reagent)

∴nC O2=0.0532 mol

V C O2=0.0532× 24.0=1.28d m3

Q. 10.0cm3 of a sample of the spirit was diluted to 250cm3. To a 25.0cm3 portion of the diluted solution, 25.0cm3 of 0.156M K2Cr2O7 (aq) and excess dilute H2SO4 were added. Ethanol reacts with acidified Cr2O7

2- as follows:

3 C H 3 C H 2OH +2C r2 O72−¿+16 H+¿→ 3C H 3C O2 H+4C r

3+ ¿+11 H2

O ¿¿ ¿

After allowing the mixture to stand for about an hour at room temperature, the excess K2Cr2O7 in the mixture was then titrated with 0.118M (NH4)2Fe(SO4)2 (aq) using an appropriate indicator. In the titration, 12.25cm3 of the Fe2+ (aq) solution was used. What is the concentration of ethanol, in M, in this brand of spirit?

C r 2O72−¿+6 F e2+¿+14H +¿→ 2C r

3+¿+6 Fe3+¿+7H

2O ¿

¿ ¿¿ ¿

nFe2+¿=12.25 ×1 0−3× 0.118=1.446 ×10−3 mol ¿

∴nC r2 O7

2−¿=16

nF e2+¿=2.409 ×1 0−4 mol¿

¿

Amount of total dichromate added=25.0 ×1 0−3× 0.156=3.90 ×1 0−3mol

Amount of dichromatereacted with EtOH=3.90 ×1 0−3−2.409× 10−4=3.66 ×10−3 mol

nEtOH=32

nC r2 O7

2+¿=5.489× 10−3 mol(¿25.0 c m3)¿

nEtOH=0.0549 mol (¿250.0 c m3 )

c EtOH=nV

= 0.0549

10.0 × 10−3=5.49M

Q. A sample of household bleach contains sodium chlorate (I), NaClO, as the active ingredient. The following experiment was carried out to determine the concentration of NaClO in a brand of household bleach.

A sample of the bleach was diluted 50 times using deionised water. To 25.0cm3 of the diluted bleach, excess KI (aq) and H2SO4 (aq) were added. The mixture was then titrated against 0.0552M Na2S2O3 (aq). 16.00cm3 of the Na2S2O3 (aq) was required to reach the end-point.

(a) Describe how the sample of bleach can be diluted using apparatus available in a school laboratory.

(b) Suggest how the end-point of the titration can be detected.(c) Under acidic conditions, ClO- (aq) reacts with I- (aq) as follows.

ClO−¿ (aq )+2 I−¿ (aq )+2H +¿ ( aq ) →C l−¿ ( aq) +I

2( aq )+ H

2O ( l)¿

¿¿ ¿What is the concentration of ClO- (aq), in M, in this brand

of bleach?

(a) 5.0cm3 of the sample can be pipetted into a 250cm3 volumetric flask and diluted with deionised water.

(b) When the solution has a pale yellow colour, starch is added to give a deep blue colour. Titration can then be continued and the end-point is reached when the solution turns colourless.

(c) 2S2O32−¿+I 2→ S 4 O6

2−¿+2 I−¿¿¿¿

n I2=1

2×16.00 × 10−3× 0.0552=4.416 ×1 0−4 mol=nClO−¿¿

¿250 c m3 , nClO−¿=4.416 ×1 0−3 mol ¿

c

ClO−¿=nV

=4.416 ×1 0−3

5×10−3 =0.883 M ¿

2) Atomic Structure

Q. Write the electronic configuration and hence deduce the number of unpaired electrons for Fe3+ (g) at its ground state.

Electronic configuration of Fe (Z=26): 1s2 2s2 2p6 3s2 3p6 3d6 4s2

Electronic configuration of Fe3+ : 1s2 2s2 2p6 3s2 3p6 3d5

Thus there are 5 unpaired electrons in the 3d orbitals.

Q. State and explain which noble gas (He, Ne, Ar, Kr, Xe) has the smallest first I.E.

Xe has the smallest first I.E. because it has the largest atomic size and so, the outermost electron is furthest away from the nucleus and better shielded by the inner shells of electrons. Hence the electrostatic attraction experienced by its outermost electron is the smallest.

Q. The table below gives the first ionisation energies (I.E.) of seven consecutive elements (T, U, V, W, X, Y, Z) in the 2nd and 3rd periods of the Periodic Table.

Element T U V W X Y Z1st I.E. /kJmol-1 1090 1400 1310 1680 2080 500 740

(a) Deduce from the above information what these elements may be.(b) Explain why the first ionisation energy of V is smaller than that of U.

(a) The very large drop in I.E. value from X to Y suggests a change in quantum shell. Among the 7 elements, X has the highest first I.E., suggesting that it might be a noble gas, Ne. Hence, T, U, V, W, X, Y, Z, are C, N, O, F, Ne, Na and Mg respectively.

(b) U is nitrogen and V is oxygen. Electronic configuration of N: 1s2 2s2 2p3, electronic configuration of O: 1s2 2s2 2p4

In oxygen, the two electrons occupying the same orbital give rise to inter-electron repulsion. Thus less energy is required to remove an electron from the paired 2p electrons in O.

Q. The diagram shows the first 8 ionisation energies (I.E.) of an element R.

1 2 3 4 5 6 7 80.91.11.31.51.71.92.12.32.5

lg I.E. against number of e- removed

(a) To which group in the Periodic Table does R belong?(b) Write down the electron arrangement in the outer shell of R.(c) Can R be nitrogen? Explain.(d) Explain why the ionisation energy increases as electrons are removed.

(a) R belongs to Group 15 in the Periodic Table.

(b) The electron arrangement in the outer shell is ns2 np3

(c) R cannot be nitrogen because nitrogen only has 7 electrons, but the diagram shows that R has at least 8 I.E.

(d) Ionisation energy increases as electrons are removed because the remaining electrons are attracted more strongly by the constant positive charge on the nucleus.

3) Chemical Bonding

Q. Although Kr and Rb have nearly the same relative atomic mass, the boiling point of Kr is -152°C whereas that of Rb is 686°C. Explain this observation, as far as you can, in terms of structure and bonding.

Kr has a simple atomic structure. It consists of a lattice of atoms held together by weak van der Waals forces, which arises due to induced dipole dipole attraction.

Rb has a metallic structure and consists of a lattice of positive ions with the outer electrons forming a sea of delocalised electrons. Strong electrostatic attractions (metallic bonds) exist between the delocalised electron cloud and the metal ions.

Much more energy is required to separate the Rb atoms due to a larger force of attraction. Hence, Rb has a much higher boiling point than Kr.

Q. Selenium dioxide, SeO2, is a solid that melts at 315°C and does not conduct electricity when molten. Explain this observation in terms of structure and bonding.

SeO2 has a giant molecular structure (like diamond) with strong covalent bonds between Se and O atoms. Hence, a large amount of energy is required to break the strong covalent bonds and this accounts for SeO2 being a solid with high melting point. SeO2 does not conduct electricity when molten because there are no free electrons/ions to carry the current.

Q. (a) Boron trifluoride, BF3, and aluminium fluoride, AlF3, have very different physical properties, e.g. melting point of BF3 is -144°C, while that of AlF3 is 1291°C. What type of bonding is present in each of these compounds? Draw ‘dot-and-cross’ diagrams to illustrate this bonding. (b) Draw a diagram to show the shape of, and bonding in, the product formed when boron trifluoride reacts with ammonia. Describe the type of bond formed during this reaction.

(a) BF3 is simple molecular. It has strong covalent bonds within the molecule (between B and F), and weak van der Waals forces between BF3 molecules. AlF3 is ionic. The oppositely charged Al3+ and F- ions are held together by strong electrostatic forces of attraction (ionic bonds)

(b) A dative bond is formed. The lone pair of electrons on the N atom is donated into the vacant orbital in B.

Q. Describe the type of bonding present in each of the given compounds, relating the bonding to boiling point.

Compound Formula Mr Boiling point /°CW CH3(CH2)3CH3 72 36X (CH3)4C 72 10Y CH2ClO(CH2)2CH3 108.5 54Z CH3CH2CO2H 74 141

W and X are isomers. They consist of discrete covalent molecules held together by weak van der Waals forces. They have very low boiling points, as little energy is required to break the weak van der Waals forces.

X has a lower boiling point than W because its molecule is highly branched, and so, has smaller van der Waals forces.

In Y, the discrete covalent molecules are held together by dipole-dipole forces which are stronger than van der Waals forces in W and X. Thus, more energy is required to overcome the stronger attraction and this accounts for its higher boiling point.

In Z, the discrete covalent molecules are held together by hydrogen bonds, which are stronger than both dipole-dipole forces in Y and van der Waals forces in W and X. Hence, Z has the highest boiling point as a large amount of energy is required to overcome the stronger intermolecular hydrogen bonds.

Q. Electric cables used in fire alarm systems has copper wire surrounded by MgO (which acts as an insulator) and the whole cable I encased in thin copper tubing.(a) Describe the bonding in MgO and explain why it acts as an insulator.(b) Explain why MgO is preferred to pvc (polyvinyl chloride) as an insulator.(c) Suggest a reason why copper is suitable for encasing the magnesium oxide.

(a) MgO has ionic bonding, which are strong electrostatic forces of attraction between Mg2+

and O2- ions in the lattice. In the solid state, the ions are fixed and so cannot carry current. Thus MgO acts as an insulator.

(b) MgO is preferred over pvc because MgO has a higher melting point, thus can withstand high temperature whereas pvc will not only melt but give off toxic fumes of HCl.

(c) Copper is malleable, ductile, resistant to corrosion. It also has high tensile strength and so, acts as a support for the powdery MgO.

4) The Gaseous State

Q. At 433K and 80.0kPa, 0.569g of aluminium chloride produces a vapour which occupies a volume of 96.0cm3. [1kPa = 1 x 103Nm-2](a) Assuming that the aluminium chloride vapour behaves as an ideal gas, what is its molar mass?(b) Hence deduce the molecular formula of aluminium chloride and suggests a structure consistent with this formula.

(a) pV=massM r

RT

(80.0 × 103) ( 96.0× 1 0−6 )=0.569M r

×8.31 × 433

M r=266.6 gmo l−1

(b) Let the molecular formula be AlnCl3n

27 n+ (3n × 35.5 )=267

n=2, Al2Cl6

Q. In an experiment to determine the SO2 level in air, 200dm3 of an air sample, measured at 298K and 101kPa was bubbled into excess H2O2 (aq) to convert the SO2 to SO4

2- (aq). The resulting solution was then treated with excess BaCl2 (aq) to give 0.0583g of BaSO4 (s). Assuming that the air sample behaves ideally, what is the concentration of SO2 in the sample? Express your answer in μgm-3.

S O2+ H 2O2 → S O42−¿+2 H +¿¿ ¿

S O42−¿+BaC l2 → BaSO 4+2 C l−¿¿ ¿

nBaSO 4=0.0583

233.1=2.501 ×1 0−4 mol=nSO 2

∈200 d m3

cSO 2=2.501× 10−4

200=1.251 ×10−6 mold m−3

¿1.251 ×1 0−6 ×64.1=8.02× 10−5 gd m−3

¿ 8.02× 10−5 gd m3 ×

106 μg1 g

×1 d m3

10−3 m3 =8.02 ×1 04 μg m−3

V=nRTp

=2.501 ×1 0−4 ×8.31×298101 ×1 03 =6.132 ×10−6m3

Q. A food chemist used cylinders of pressurised CO2 to make carbonated drinks. One such cylinder (with internal volume of 2.5dm3) contains 2.3kg of CO2.(a) Assuming the gas behaves ideally, determine the pressure (in Pa) exerted by CO2 inside the cylinder at 298K. (b) Under these conditions, the actual pressure inside the cylinder is 2.2 x 107Pa. Suggest why this value differs from the pressure calculated in (a).

(a) p=nRTV

=52.273 × 8.31× 298

2.5 × 10−3=51.78 MPa

(b) The actual pressure inside the cylinder is lower than that calculated in (a) because CO2 is not an ideal gas. CO2 has significant size (Mr = 44) and significant intermolecular van der Waals forces of attraction holding the molecules together. Thereby reducing the number of collisions per unit area exerted on the walls of the cylinder, i.e. pressure decreases.

Q. Calculate the total pressure in a 2dm3 vessel when 4dm3 of O2 at a pressure of 400kPa and 1dm3 of N2 at a pressure of 200kPa are introduced.

When dioxygen contracts from 4dm3 to 2dm3, the pressure increases.

p1V 1=p2V 2

p2=p1V 1

V 2

=400 ×1 03 × 42

=800 kPa

When dinitrogen expands from 1dm3 to 2dm3, the pressure decreases.

p2=p1V 1

V 2

=200 ×1 03 ×12

=100 kPa

Total pressure = 900kPa

Q. Argon is an inert gas which exists in the atmosphere as single atoms. (a) Give two assumptions of the kinetic theory as applied to inert gas.(b) How many atoms of argon are there in one mole of the gas?(c) To calculate the percentage of the volume occupied by the atoms themselves in one mole of argon at room temperature and pressure: (i) Calculate the volume of one atom of argon. [Given radius of Ar = 1.92Å] (ii) Hence calculate the volume of one mole of argon atoms. (iii) Assuming that argon behaves as an ideal gas, what is the volume occupied by one mole of argon at room temperature and pressure? (iv) Determine the percentage of this volume occupied by the atoms themselves.(d) How does your answer to (c)(iv) justify one of your assumptions in (a)?(e) Electric light bulbs are filled with argon, and have a fine metal filament which glows white hot from its electrical resistance to the current. Suggest why argon, rather than air, is used to fill the light bulbs.

(a) Two assumptions are gas particles have negligible size/volume, and gas particles have negligible intermolecular forces of attraction.

(b) Number of Ar atoms in 1 mol = 6.02 x 1023 atoms

(c i) V= 43

π r3=43

π (1.92× 10−10 )3=2.96 × 10−29m3

(c ii) V=2.96× 1 0−29× 6.02× 1023=1.782× 10−5 m3=1.782× 1 0−2 dm3

(c iii) At r.t.p. V=24.0 dm3

(c iv) % occupied by Ar=1.782 ×1 0−2

24.0× 100 %=0.0742 %

(d)The very small percentage (<1%) of space occupied by Ar atoms themselves shows that Ar atoms have negligible size/volume.

(e) Argon is used because Ar is inert and will not react. If air is used, the hot metal will react with dioxygen in air to form oxides. In addition, the high temperatures of electric bulbs, dioxygen and dinitrogen will react to give NOx.

5) Chemical Energetics

Q. (a) When 0.10g of magnesium was added to an excess of dilute hydrochloric acid (in a polystyrene cup of negligible heat capacity), a maximum temperature rise of 4.3°C was recorded. What is the enthalpy change for the reaction? [Heat capacity of acid = 494 JK-1]

(b) In a similar experiment, the enthalpy change for the reaction of MgCO3 (s) with hydrochloric acid was found to be -90kJmol-1. What is the enthalpy change of formation of MgCO3 (s) under the conditions of the experiment? [Given ΔHf

°/kJmol-1: H2O (l) = -285, CO2 (g) = -393, HCl (aq) = -167]

(a) Heat evolved, Q=C ×∆ T=494 × 4.3=2.124 kJ

nMg=0.1024.3

=4.12 mmol

∆ H r=Qn

=−2.124 ×1 03

4.12 ×1 0−3 =−516 kJmo l−1

(b) MgC O3 (s )+2 HCl (aq ) → MgC l2 ( aq )+H 2O ( l )+C O2 (g ) , ∆ H r=−90 kJmol−1

By Hess’s law, ∆ H f +2 (−167 )+(−90 )=(−516 )+ (−285 )+(−393 )

∆ H f=−770 kJmol−1

Q. In an experiment, the maximum change in temperature was measured when 0.025mol of CuSO4 (s) and 0.025mol of CuSO4.5H2O (s) were dissolved separately in 50.0cm3 of distilled water in a polystyrene cup. The maximum change in temperature recorded was +7.7°C for CuSO4 (s), and -1.7°C for CuSO4.5H2O (s).

(a) Calculate the enthalpy change of solution of CuSO4 (s) and of CuSO4.5H2O (s). [Assume that CuSO4 (aq) formed has a specific heat capacity of 4.2Jg-1K-1 and a density of 1.0gcm-3; and that heat capacity of polystyrene cup is negligible.]

(b) Hence calculate the enthalpy change for the reaction:

CuS O4 (s )+5 H 2O (l )→ CuS O4 .5 H 2O(s)

And suggest why this value cannot be determined directly.

(a) QCuSO 4=mc ∆ T=50.0× 4.2 ×−7.7=−1617 J

∆ HCuS O4=−1617

0.025=−64.7kJmo l−1

QCuSO 4 .5H 2 O=mc ∆ T=50.0× 4.2× 1.7=+357J

∆ HCuS O4 .5 H 2O=+357

0.025=+14.3 kLmo l−1

(b) By Hess’s law, ∆ H r +14.3=−64.7

∆ H r=−79.0kJmo l−1

This value cannot be determined directly because it is difficult to measure the temperature of a solid.

Q. (a) Calculate the standard enthalpy change of formation of cyclopropane, C3H6 (g), using the following standard enthalpy changes of combustion.

ΔHc° of C (graphite) = -394kJmol-1

ΔHc° of H2 (g) = -286kJmol-1

ΔHc° of C3H6 (g) = -2090kJmol-1

(b) Estimate the C-C bond enthalpy in cyclopropane given that the enthalpy change of atomisation of graphite is +715kJmol-1, and the average C-H and H-H bond enthalpies are +413kJmol-1 and +436kJmol-1 respectively.

(a) 3 C ( s)+3 H2(g)→ C3 H 6(g)

By Hess’ law, 3 (−394 )+3 (−286 )=∆ H r+(−2090 ) , ∆ H r=+50 kJmo l−1

(b) Formation of C3H6 involves the atomisation of 3 C atoms, breaking 3 H-H bonds and formation of 3 C-C bonds along with 6 C-H bonds.

Let the C-C bond enthalpy be x kJmol-1, thus

∆ H f=3 (+715 )+3 (+436 )+3 (−x )+6 (−413 ) , x=+308 kJmol−1

Q. Rocket propellants are made up of two classes of substances – fuels and oxidisers. When hydrazine is used as a fuel, fluorine or oxygen can be used as the oxidiser. The reactions involved are as follows:

( I ) H 2 NN H 2 ( g )+2 F2 (g )→ 4 HF (g )+N 2(g)( II ) H 2 NN H 2 ( g )+O2 (g )→ 2 H 2O (g )+N2(g)

(a) Calculate the enthalpy changes, in kJmol-1, for reactions (I) and (II). [Bond energies / kJmol-1: N-N, 163; N≡N, 945; N-H, 390; F-F, 158; H-F, 565; O=O, 498; O-H, 464]

(b) Suggest, with reasons, which of the two compounds (fluorine or oxygen) is a better choice as an oxidiser in powering rockets.

(a I)

∆ H=∑ (bonds formation )+∑ (bondsbroken)=(−945 )+4 (−565 )+4 (+390 )+(+163 )+2 (+158 )=−1166 kJmo l−1

(a II) ∆ H=4 (−464 )+ (−945 )+4 (+390 )+(+163 )+( 498 )=−580 kJmol−1

(b) While reaction (I) is more exothermic than (II), it produces HF which is toxic and highly corrosive. Reaction (II), however, gives harmless products. Besides, it is difficult to store F2 (g) as it is highly reactive. Hence, O2 (g) is a better choice as oxidiser in powering rockets.

Q. (a) Calculate the lattice energy of CaO given that the energy required to form one mole of O2- ion from gaseous oxygen atom is 704kJ, and the following data:

ΔH /kJmol-1

Enthalpy change of atomisation of calcium +177Enthalpy change of atomisation of oxygen +250First ionisation energy of calcium +587Second ionisation energy of calcium +1156Enthalpy change of formation of calcium oxide -637

(b) How would you expect the numerical magnitude of the lattice energy of MgO to compare with that of CaO?

(a) Ca (s )+ 12

O 2(g)−637→

CaO (s )

By Hess’ law, ∆ H r +177+250+587+704+1156=−637 , ∆ H r=−3511kJmo l−1

(b) Lattice energy of MgO is expected to be larger (more exothermic) than that of CaO due to the smaller ionic radius of Mg2+ which makes a stronger lattice with O2- ions. Hence, more energy is released when 1 mol of Mg2+O2- (s) is formed from its gaseous ions.

ΔHr

+704

+1156

+587

+250+177

O2−¿(g)¿C a2+¿(g )¿

C a+¿(g)¿

O(g)Ca(g)

Q. The dimerization of NO2 (g) is found to take place as follows:

2 N O2 ( g ) → N2 O4 ( g ) , ∆ H θ=−57.2 kJmo l−1; ∆ Sθ=−176 J K−1 mol−1

(a) State how the values of ΔHθ and ΔSθ relate to the changes that occur at the molecular level.

(b) Estimate the temperature at which the reaction is at equilibrium.

(a) ΔHθ is exothermic because a new bond (N-N) is being formed, but none other bonds were broken. It is only slightly exothermic as the new N-N bond formed is weak. ΔSθ is negative because in the reaction only 1 gas molecule is formed from 2 gas molecules; i.e. a decrease in the number of particles. This results in a loss of translational degrees of freedom (the system becomes more orderly) and hence, entropy decreases.

(b) ∆ G=∆ H−T ∆ S, at equilibrium, ΔG = 0, thus ∆ H=T ∆ S

T=∆ H∆ S

=−57.2× 103

−176=325 K

Q. Use the standard enthalpy change of formation, ΔHfθ, and the standard free energy of

formation, ΔGfθ, given in the table to calculate the standard entropy change of formation,

ΔSfθ, of CO (g) and of CO2 (g).

CO (g) CO2 (g)ΔHf

θ /kJmol-1 -110.5 -393.5ΔGf

θ /kJmol-1 -137.2 -394.4

(i) Why are the two values so different?

(ii) Deduce the feasibility of the reaction: C ( s )+C O2(g)→ 2CO (g) at 298K and at 1000K.

∆ G=∆ H−T ∆ S⇒∆ S=∆ H−∆ GT

∆ S fθ (CO )=(−110.5+137.2)×1 03

298=+89.6 J K−1mo l−1

∆ S fθ (C O2 )=

(−393.5+394.4 )×1 03

298=+3.02 J K−1 mol−1

(i) The formation of CO (g), C ( s )+ 12

O2(g)→CO (g), involves an increase in the amount of

gaseous particles, thus change in entropy is large and positive. On the other hand, the

formation of CO2 (g), C ( s )+O2(g)→C O 2(g), there is no change in the amount of gaseous

particles, thus change in entropy is close to zero.

(ii) C ( s )+ 12

O2 (g )→ CO ( g ) , ∆ G fθ=−137.2 kJmol−1

-------(1)

C ( s )+O2 ( g ) →C O2 ( g ) , ∆ Gfθ=−394.4 kJmol−1-------(2)

2x(1) – (2): C ( s )+C O2(g)→ 2CO (g)

∆ Grθ=2 (−137.2 )−(−394.4 )=+120 kJmo l−1

As the Gibbs free energy is positive at 298K, the reaction is not spontaneous at this temperature.

At 1000K,

∆ H r=2 ∆ H f (CO )−∆ H f (C O2 )=2 (−110.5)− (−393.5 )=+172.5 kJmo l−1

∆ Sr=2∆ S f (CO )−∆ S f (C O2 )=2 (+89.6 )−(+3.02 )=+176.2 Jmo l−1

∆ Gr=∆ H r−T ∆ Sr=172.5 × 103−(1000 ) (+176.2 )=−3.70 kJmo l−1

The reaction is spontaneous at 1000K.

Q. Suggest why ammonium nitrate is soluble in water at room temperature despite the fact that the enthalpy change of solution of ammonium nitrate is endothermic. Suggest also why it becomes very soluble at 75°C.

∆ H sol=∆ H hyd−∆ H latt

ΔHsol (NH4NO3) is endothermic indicates that the lattice energy of NH4NO3 is much more exothermic than ΔHhyd (NH4NO3).

When NH4NO3 (s) is added to water, the disorder achieved in breaking up the lattice is greater than the ordering of water molecules around the ions. ΔS is therefore positive as the system becomes less orderly, thus ΔG is negative; i.e. the reaction is feasible and proceeds spontaneously. Hence NH4NO3 dissolves in water at room temperature.

As temperature T increases, -TΔS becomes more negative. Hence the Gibbs free energy becomes more negative, and the reaction is more feasible and spontaneous.

Q. Comment on the following in terms of free energy and/or entropy changes.(a) At 273K, ice and water are in equilibrium but at 298K, ice spontaneously changes into water. [Given: enthalpy change of fusion of ice = +6.0kJmol-1](b) The polymerisation of ethane is a spontaneous process.

(a) At 273K, ice and water are in equilibrium because a phase change occurs at that temperature and the process of ice melting to form water is reversible.

At equilibrium, ∆ G=0⇒∆ S=∆ HT

=6.0 × 103

273=21.98 Jmol−1

At 298K, ∆ Gr=∆ H r−T ∆ Sr=6.0× 103−(298 ) (21.98 )=−0.550 kJmo l−1

As the Gibbs free energy is negative, ice spontaneously changes into water at 298K.

(b) When ethane polymerises, the number of molecules is reduced and ΔS is negative. However, in the polymerisation process, a stronger C-C bond is formed for every weak C-C π bond is broken. Hence, the enthalpy change of polymerisation is very exothermic, making ΔG to be negative.

6) Electrochemistry

Q. Given the following:

Y 3+¿+ e−¿→Y 2+¿ ,Eθ=−3.1V ¿ ¿¿

Y 2+¿+2 e−¿→ Y , Eθ=−2.0 V ¿ ¿

Predict, based on Eθ values given, whether an aqueous solution of Y2+ would exist under room conditions in the absence of oxygen.

Y 2+¿→Y 3+¿+e−¿ ,Eθ=+3.1 V ¿¿¿

Y 2+¿+2 e−¿→ Y , Eθ=−2.0 V ¿ ¿

3 Y 2+¿ →2 Y 3+¿+Y ,E cellθ =+1.1 V ¿¿

Since Eθcell is positive, the reaction is feasible. Hence, Y2+ (aq) would not exist under room

conditions since it is unstable and readily undergoes disproportionation to give Y3+ and Y.

Q. (a) Calculate the e.m.f. of the cell: Co (s) | Co2+ (aq) // Ag+ (aq) | Ag (s).(b) Write a balanced equation for the cell reaction.(c) How would you expect the e.m.f. of this cell to change, if at all, when the concentration of Ag+ ion is doubled?

(a) From Data booklet, C o2+¿+2 e−¿⇌ Co, Eθ=−0.28 V ¿¿

A g+¿+e−¿⇌ Ag, Eθ=+0.80V ¿ ¿

Ecellθ =ERed

θ −EOxθ =(+0.80 )−(−0.28 )=+1.08 V

(b) The cell equation is Co+2 A g+¿→ 2 Ag+C o2+¿¿¿

(c) When the concentration of Ag+ is doubled, the equilibrium shifts to the right and more electrons are removed from the Ag electrode. Hence Eθ of Ag+/Ag half-cell becomes more positive, thus the e.m.f. of the cell to be greater than +1.08V.

Q. Chrome yellow (PbCrO4), a yellow pigment used in oil paintings, is found to change colour when the painting is exposed to an atmosphere containing sulfur dioxide.

(a) Using the data below, suggest a reason for this colour change.

S O42−¿+4H +¿+2 e

−¿⇌ SO2+2 H

2O , E

θ=+0.17V ¿¿ ¿

Cr O 42−¿+8 H+¿+3e−¿⇌ C r

3+¿+4 H2

O , Eθ=+1.33V ¿

¿¿ ¿

(b) Why does this colour change take a long time?

(c) State what colour change takes place.

(a) 2 Cr O42−¿+4 H +¿+3S O

2→2C r

3+¿+ 2H2

O+ 3SO42−¿¿¿

¿ ¿

Ecellθ =ERed

θ −EOxθ =(+1.33 )−(+0.17 )=+1.16 V

Since Eθcell > 0, the reaction is feasible. Hence SO2 would reduce the yellow CrO4

2- to green Cr3+, which explains why the yellow pigment in oil paintings changes colour when exposed to an atmosphere containing SO2.

(b) This colour change takes a long time due to the low concentration of SO2 in the atmosphere and hence, slow rate of reaction.

(c) The yellow pigment turns green.

Q. An electrochemical cell containing an oxygen cathode and a hydrogen anode is shown below. The pistons above the gas chambers are frictionless.

(a) Write balanced equations for the half reactions and for the overall reaction in the cell.

(b) What is the e.m.f. of the cell?

(c) How does the concentration of H2SO4 affect the equilibria of the half reactions?

(d) If weights are added to the pistons of both chambers, how would the reading of the voltmeter change? Explain your answer.

(a) At the anode: 2 H 2 ( g )⇌ 4 H+¿ (aq )+4 e−¿ ,Eθ=0.00 V ¿¿

At the cathode: O2 (g )+4 H+¿ (aq )+4e−¿ ⇌2 H 2O (l ) ,Eθ=+1.23 V ¿¿

Overall: O2 (g )+2H 2(g)→2 H 2O( l)

(b) Ecellθ =(+1.23 )−0.00=+1.23 V

(c) At the anode, an increase in [H2SO4] shifts the equilibrium to the left, thus less H2 will be oxidised. At the cathode, an increase in [H2SO4] shifts the equilibrium to the right, allowing more O2 to be reduced.

(d) If the weights are added to the pistons of both chambers, the pressure of the system is increased. At the anode, an increase in the pressure of H2 (g) shifts the equilibrium position to the right, and Eθ becomes more negative as more H2 (g) are oxidised. At the cathode, an increase in the pressure of O2 (g), shifts the equilibrium position to the right, and Eθ becomes more positive as more O2 (g) are being reduced.

Hence the e.m.f. of the cell will be greater than +1.23V.

Q. In many of the early attempts to produce fluorine by the electrolysis of aqueous solutions of metallic fluorides, oxygen was produced at the anode instead.

(a) In the electrolysis of NaF (aq), hydrogen is produced at the cathode and oxygen at the anode. Write the half equation for each electrode reaction.

(b) Use relevant Eθ values from the Data Booklet to explain why fluorine is not produced in the electrolysis of NaF (aq).

(a) At cathode: 4 H+¿+4 e−¿→ 2H 2¿ ¿

At anode: 4 O H−¿→O 2+2 H 2 O+4 e−¿¿ ¿

(b) F2+2 e−¿⇌2F−¿ ,Eθ=+2.87 V ¿ ¿

O2+2H 2O+4 e−¿⇌ 4 OH−¿ ,Eθ=+0.40 V ¿¿

Oxidation occurs at the anode. Since F2/2F- has a very high positive reduction potential, F2 has a high tendency to remain in the reduced state, fluoride ions.

Q. The given diagram shows the apparatus used to obtain strontium metal by the electrolysis of molten strontium bromide, SrBr2.

(a) Write equations, with state symbols, for the two electrode reactions.

(b) Why is an atmosphere of argon used around the cathode?

(c) Suggest why it is not possible to extract strontium by electrolysis of aqueous strontium bromide.

(a) Reduction at cathode: S r2+¿(aq)+2e−¿→Sr (s )¿ ¿

Oxidation at anode: 2 B r−¿ ( aq) → Br2 ( g)+2 e−¿ ¿¿

(b) An atmosphere of argon is used around the cathode to prevent the oxidation of molten strontium to strontium oxide.

(c) If aqueous strontium bromide is used, H+ (aq) ion would be discharged at the cathode instead of Sr2+, then H2 (g) would be obtained instead of Sr metal.

Q. Contacts of electronic circuit boards are often coated with gold. An electrolytic bath for gold-plating is found to contain the following substances: K[Au(CN)2], KCN, K2CO3, K2HPO4, KOH.

(a) Why are the contacts of electronic circuit boards coated with gold?

(b) Give two reasons for using the complex K[Au(CN)2] as a source of gold (I) ions in the electroplating process.

(c) Suggest why the electroplating process should not be carried out under acidic conditions.

(d) The total surface area of the contacts of an electronic circuit board is 1.0cm2. It is gold plated using a current of 0.50A. Assuming 100% current efficiency, calculate the time taken to deposit a layer of gold of 5 x 10-4cm thick on the contacts of the circuit board. [Density of gold = 19.3gcm-3]

(a) This is done to prevent oxidation or corrosion of the contacts.

(b) ¿

The CN- complexes with Au+ and hence prevents the precipitation of Au (s) by OH- (aq). It also helps to provide a steady supply of Au+ (aq) ions.

(c) Acid is not used as it will form the highly toxic HCN (g).

(d) V Au=1.0 ×5× 10−4=5.0× 10−4 c m3

Mas s Au=V Au× ρ=5.0 ×1 0−4 ×19.3=9.65 ×1 0−3 g⇒ nAu=4.90× 10−5 mol

To deposit 1 mol of Au (s), 1 mol of electrons are needed;

F=N A e=96500Cmol−1

Q=4.90 ×1 0−5 ×96500=4.73C

Q=¿⇒ t=QI=4.73

0.50=9.46 s

Q. In the electrolysis of aqueous sulphuric acid using graphite anode, the gas liberated is a mixture of O2, CO and CO2. In an experiment, 35cm3 of gas formed above the cathode and 20cm3 of gas above the anode.

(a) Give the ion-electron equation for the reaction at the cathode.

(b) Explain why oxides of carbon are produced at the anode.

(c) Explain, with the aid of an equation, why the volume of gas collected at the anode is larger than expected.

(d) Calculate the volume of oxygen produced at the anode if an inert electrode is used instead.

(a) Reduction occurs at the cathode: 4 H+¿ ( a q)+4 e−¿→2 H2(g )¿ ¿

(b) Oxidation occurs at the anode: 4 O H−¿→O 2+2 H 2 O+4 e−¿¿ ¿

The dioxygen produced at the anode reacts with the graphite electrode to produce CO and CO2.

(c) 2 C+O2 →2 CO

This is because each volume of O2 gives two volumes of CO gas.

(d) From the half equations, every mol of dioxygen will produce 2 mol of dihydrogen. Thus the volume of dioxygen formed will be 17.5cm3.

7) Equilibria

Q. A 2:1 mixture of SO2 and O2 was allowed to reach equilibrium at 500°C and a total pressure of 5atm. The partial pressure of SO3 at equilibrium was found to be 4.7atm. Write an expression for Kp, and calculate its value.

2 SO 2 ( g )+O2(g)⇌2 S O3(g)

At equilibrium, pS O2+ pO2

=5.0−4.7=0.3atm

Since the mole ratio of SO2:O2 is 2:1, then pS O2=0.2atm , pO2

=0.1atm

K p=( pS O3 )

2

( pS O2)2 ( pO2

)at m−1

K p=( 4.7 )2

¿¿

Q. (a) Identify the two acids and the two bases present in each of the reactions.

I N H 3+H 2O⇌ N H 4+¿+O H−¿ ,Kc=1.8 ×10−5mol d m−3 ¿¿

II C6 H 5O−¿+C H 3 C O2 H ⇌C6 H 5OH +C H 3 C O2−¿ ,Kc=1.3 ×1 06 mold m−3¿ ¿

(b) For each of the above reactions, suggest which ion (or molecule) is the stronger acid and which, the stronger base. Give your reasoning.

(a) I: the acids are H2O and NH4+; bases are NH3 and OH-.

II: the acids are CH3CO2H and C6H5OH; bases are C6H5O- and CH3CO2-.

(b) I: NH4+ is the stronger acid (or OH- is the stronger base) because Kc is small (<1)

suggesting that the position of equilibrium is more to the left. II: CH3CO2H is the stronger acid (or C6H5O-) is the stronger base) because Kc is large (>1) suggesting that the position of equilibrium is more to the right.

Q. Benzoic acid, C6H5CO2H, is used as a preservative in fruit juices as it inhibits the growth of micro-organisms. It is found to be more effective if the pH of the fruit juice is below 5. Benzoic acid reacts with water as follows:

C6 H 5 C O2 H+ H 2O⇌C6 H 5C O2−¿+H 3 O+¿ ¿¿

(a) Write an expression for the acidity constant, Ka, for benzoic acid.

(b) Calculate the pH of a solution of benzoic acid of concentration 0.010moldm-3. [Ka for benzoic acid is 1.0 x 10-4 moldm-3]

(c) Using the expression in (a), calculate the ratio [C6H5CO2-]:[C6H5CO2H], in

(i) a fruit juice of pH 6; (ii) a fruit juice of pH 4. Hence, comment on which is the better preservative, benzoic acid or the benzoate ion.

(d) At which of the two pH values would the better ‘buffering’ occurs? Give a reason for your answer.

(a) Ka=¿¿

(b) Since ¿

pH=−lg ¿¿

(c i) pH=6⇒ ¿

ratio¿

(c ii) pH=4⇒¿

ratio¿

Benzoic acid is the better preservative, since there is very little of this present at pH6

(where preservation is poor).

(d) Better buffering occur at pH4, where the ratio of [C6H5CO2-]:[C6H5CO2H] is near to 1.

Q. The solubility of lead sulfate at 25°C is 1.26 x 10-4 moldm-3. Calculate its Ksp value at 25°C.

PbS O4 (s )⇌P b2+¿ ( aq)+ S O42−¿(aq)¿ ¿

In a saturated solution, ¿

K sp=¿

Q. Silver carbonate has a solubility product of 8.0 x 10-4 moldm-3. What is the solubility in gdm-3? [Mr of Ag2CO3 = 276]

Let the solubility of Ag2CO3 be x.

A g2C O3⇌ 2 A g+¿+C O 32−¿¿ ¿

K sp=¿

x=3√ 8.0× 1 0−4

4=1.26 × 10−4 M

Hence solubility of Ag2CO3 is 1.26 ×1 0−4 ×276=0.0348 gd m−3

x 2x x

Q. (a) Will a precipitate form if equal volumes of 3 x 10-5 moldm-3 nickel (II) sulfate and

4 x 10-5 moldm-3 sodium hydroxide are mixed?

[Ksp of nickel (II) hydroxide = 6.0 x 10-18 mol3dm-9]

(b) Calculate the volume of water required to dissolve 1 mol of nickel (II) hydroxide if pH5.5 is maintained.

(a) Since the volume of resulting mixture is doubled after mixing, the concentration of each solution is halved.

¿¿

ionic product ,¿

Since the ionic product > Ksp, Ni(OH)2 will precipitate.

(b) pOH=14−pH =14−5.5=8.5

pOH=−lg¿¿

To dissolve Ni(OH)2, ionic product must not be larger than Ksp.

¿

¿

To dissolve 1 mol of Ni(OH)2, V= 10.60

=1.67 d m3

Q. A photographic company found that the wash water had a dissolved silver ion concentration of 2.5 x 10-3moldm-3.

(a) What is the minimum [Cl-] needed to recover silver chloride as a precipitate?

(b) What mass of AgCl will be precipitated from 1dm3 of wash water if [Cl-] is 0.10moldm-3? [Ksp of AgCl = 2.0 x 10-10mol2dm-6]

(a) For precipitation to occur, ionic product > Ksp

¿

(2.5 ×1 0−3 )¿

¿

(b) Since [Cl-] = 0.10M > 8.0 x 10-8M, practically all of Ag+ is precipitated as AgCl.

mass of AgCl=n M r=2.5 ×1 0−3×143.5=0.359 g

Q. To calculate the pH of buffer solution as follows: 50cm3 of 0.1moldm-3 CH3CO2H mixed with 150cm3 of 0.3moldm-3 CH3CO2Na. [Ka of CH3CO2H = 1.7 x 10-5moldm-3]

Total volume after mixture = 200cm3

[ acid ]=50 × 0.1200

=0.025 M

[ salt ]=150 × 0.3200

=0.225 M

pH=p Ka+lg[ salt ][ acid ]

=−lg (1.7×10−5 )+ lg0.2250.025

=5.72

Q. To calculate the composition of the following buffer solution: An alkaline buffer of pH9.75 made by mixing aqueous NH3 and NH4Cl, both of the same concentration. [pKb (NH3) = 4.74]

pOH=14−pH =14−9.75=4.25

pOH=p K b+ lg[salt ][base ]

⇒ 4.25=4.74+lg[ salt ][base ]

⇒ [salt ][base ]

=0.324

Hence, the buffer is made by mixing aqueous NH3 and NH4Cl in the ratio 1:0.324, e.g. adding 100cm3 of aqueous NH3 to 32.40cm3 of NH4Cl of the same concentration.

Q. Barium salts are poisonous. If a solution containing barium ions with concentration about 10-3moldm-3 is drunk, it will cause stomach upsets. Yet, patients are given a ‘barium meal’ of barium sulfate, BaSO4, before doctors take stomach X-rays.

(a) (i) Calculate the concentration of barium ions in a saturated solution of BaSO4. [Ksp of BaSO4 = 1.0 x 10-10mol2dm-6]

(ii) Would this concentration of barium ions pose a threat to patients taking a barium meal? Explain.

(b) A patient has taken a barium meal shortly after taking some ‘Epsom salts’, MgSO4, and is worried about the effects of mixing the two. A doctor estimates that the sulfate ion concentration in the stomach is about 10-2moldm-3.

(i) Calculate the concentration of barium ions in the patient’s stomach.

(ii) Is the patient’s fear justified?

(c) Suggest why BaSO4 is used for stomach X-rays rather than a compound of a less potentially toxic cation.

(d) Explain why barium carbonate (another insoluble barium salt) could not be used as an alternative barium meal.

(a i) K sp=¿

(a ii) Since [Ba2+] is less than 1.0 x 10-3M, it will not pose a threat.

(b i) ¿

(b ii) Since [Ba2+] is still less than 1.0 x 10-3M, it will not pose a threat.

(c) BaSO4 has suitable heavy nuclei, which is needed to absorb the X-rays and allow the shape of the stomach to be seen.

(d) BaCO3 reacts with HCl in the stomach to form BaCl2, which is insoluble.

Q. To calculate the solubility of silver chloride in the following solutions: [Solubility product of silver chloride = 2.0 x 10-10mol2dm-6] (i) water (ii) 0.15moldm-3 sodium chloride

(i) Let the solubility of AgCl to be x. then K sp=¿

K sp=x2⇒ x=√ K sp=1.41 ×1 0−5 M

(ii) Let the solubility of AgCl to be y.

¿

K sp=¿

8) Chemical Kinetics

Q. Given the variation of initial reaction rate with [NO2] for the reaction:

N O2 ( g )+CO (g )→ N O ( g )+C O2(g)

Experiment Rate /moldm-3s-1 [NO2] /moldm-3

1 0.010 0.152 0.040 0.303 0.160 0.604 0.360 0.90

Let rate equation be: rate=k ¿

From experiment 1, 0.010=k ¿From experiment 2, 0.040=k ¿

Then 0.0400.010

=( 0.300.15 )

n

⇒ n=2

The reaction is therefore second order with respect to NO2.

k= rate¿¿

Q. For the reaction I−¿+C H 3 Cl→ C H 3 I +C l−¿ ¿¿, the rate equation is found to be

rate=k ¿. What information can be deduced about the mechanism of this reaction?

The rate equation shows that the reaction is 1st order with respect to I-, and 1st order with respect to CH3Cl. Hence, the slow step in the mechanism involves only one molecule of CH3Cl and one I- ion.

Q. Consider the reactions:

I C3 H 7 Cl ( aq )+NaOH (aq )→ C3 H 7 OH (aq )+NaCl (aq )

II HCl ( aq )+NaOH (aq )→ H 2O (l )+NaCl(aq )

(a) Reaction I must be heated for some time before any reaction occurs, but reaction II takes place almost instantaneously at room temperature. Explain this observation as far as you can.

(b) Reaction I is a second order reaction. Suggest how the rate of this reaction would be affected when the solution is diluted with an equal volume of the solvent.

(c) State and explain how the rate of reaction between C3H7Br (aq) and NaOH (aq) would compare to that of reaction I.

(a) Reaction I has a higher activation energy than reaction II.

In I, C3H7Cl is an organic compound and energy is required to break the C-Cl bond before a reaction can occur.

In II, the reaction does not involve any breaking of covalent bonds since both NaOH and HCl

are completely ionised in aqueous solution. The reaction is basically H+¿+O H−¿→ H2 O ¿¿. The

oppositely charge ions combine readily (low activation energy) and hence, instantaneous at room temperature.

(b) Rate equation, rate=¿

When the solution is diluted with an equal volume of solvent, [reactant] is halved. Hence the rate of reaction would be decreased to a quarter of the original rate.

(c) The reaction would be faster because the C-Br bond is weaker than C-Cl bond.

Q. Nitrosyl fluoride, ONF, can be produced from nitrogen monoxide and fluorine.

2 NO ( g )+F2(g)→2 ONF (g)

The rate equation for this reaction is rate=k [ NO ][ F2]. The mechanism has two steps, one of

which produces ONG and the free radical F· in equimolar amounts.

(a) Suggest equations for the two steps of the mechanism, indicating clearly the slow step.

(b) By considering the relative reactivities of the species involved, justify your answer to (a).

(a) The rate equation shows that one molecule of NO and one molecule of F2 are involved in the slow step. Hence a possible mechanism is:

NO+F2 →ONF+F ∙(slow )

NO+F ∙→ ONF

(b) The slow step involves breaking the F2 molecule into highly reactive fluorine free radicals. The free radicals will react quickly because they have high energies. Besides, NO has an odd number of electrons and can react with the fluorine radical. Hence, the second step is a fast step.

Q. Dinitrogen pentoxide (in CCl4) decomposes as follows:

2 N2O5 (¿CC l4 ) → 4 N O2 (¿CC l 4 )+O2(g)

(a) Suggest an experimental method that can be used to follow the progress of the decomposition, and state the underlying principle of the method.

(b) An experiment was carried out to study the kinetics of the decomposition of dinitrogen pentoxide at 318K. The results obtained are given in the table.

Time /mins 0 20 40 60 80[N2O5] /10-2 moldm-3 10.00 5.49 3.02 1.65 0.91

Plot a suitable graph to determine the rate equation for the decomposition, and hence the rate constant at 318K.

(c) If the rate constant for the decomposition at 332K was found to be 5 times that at 318K, what is the activation energy for the decomposition?

(a) The progress of the decomposition can be followed by monitoring the pressure of the system since both N2O5 and NO2 are soluble in CCl4 while O2 is insoluble.

(b) Plot graph of [N2O5] against time.

Since t½ is constant ( = 23mins), the reaction is first order with respect to N2O5.

rate=k [ N2 O5]

t 1/2=ln 2k

= ln223

=0.0301 mi n−1

(c) Using Arrhenius equation, k=A e−E a/RT

k2

k1

=5=eEa ( 1

318 R− 1

322 R )⟹ Ea=100.9kJmo l−1

9) Inorganic Chemistry

9.1) The Periodic Table: Chemical Periodicity

9.2) Group II

Q. Beryllium oxide is found to react with aqueous sodium hydroxide as follows:

BeO+2NaOH → N a2 Be O2+H 2O

Discuss whether this behaviour of beryllium oxide is what you would expect from the position of the element in the Periodic Table.

This behaviour of beryllium oxide is not expected.

Be (Group 2) is metallic in character and is expected to form basic oxide. However, due to the high charge density of Be2+ ion, BeO shows a high degree of covalency and is amphoteric.

Q. On strong heating, CaSO4 decomposes into CaO and SO3. The compound CaCO3 decomposes at a lower temperature than CaSO4.

(a) Explain why this is so.

(b) Write a balanced equation and hence, the Kp expression, for the decomposition of calcium carbonate.

(c) State and explain how you would expect the value of Kp to change if BaCO3 is heated instead of CaCO3 (at the same temperature).

(a) CaCO3 decomposes at a lower temperature than CaSO4 because the CO32- ions are more

easily polarised than SO42- ions, hence a greater covalent character in CaCO3.

(b) CaC O3 (s )⇌CaO ( s)+C O2(g), K p=pCO 2

(c) The value of Kp would decrease because BaCO3 is more stable to thermal decomposition than CaCO3, hence less CO2 would be produced.

Q. Group II peroxides, on heating, decompose to the oxide and oxygen as follows:

BaO2 ∆→

BaO+ 12

O2

Predict how the temperature of decomposition of magnesium peroxide would differ from that of barium peroxide. Give your reasoning.

MgO2 would decompose at a lower temperature than BaO2 due to the smaller cation size, and hence greater polarising power of Mg2+ ion, resulting in greater extent of covalency; lower temperature.

Q. Explain why magnesium hydroxide is less soluble in water than barium hydroxide.

Given that ∆ H sol=∆ H hyd−∆ H lattice and ∆ H lattice∝q+¿

q−¿

r+¿r−¿¿¿¿¿

Mg2+ is smaller than Ba2+ ion. Hence, ΔHhyd and ΔHlatt are greater in Mg(OH)2 than in Ba(OH)2.

However, owing to the small size of OH-, the difference in ΔHlatt between Mg(OH)2 and Ba(OH)2 is much greater than the difference in ΔHhyd between Mg2+ and Ba2+ ions. This results in ΔHsol being less exothermic in Mg(OH)2, which accounts for its lower solubility in water.

9.3) Group VII

Q. A weedkiller can be prepared by heating sodium chlorate (I), a household bleach.

3 NaClO ∆→

2 NaCl+NaCl O3

(a) What are the oxidation states of chlorine in these three compounds? Hence, identify the type of reaction occurring.

(b) Describe the reaction you could use in the laboratory to make a solution that contains sodium chlorate (I). Write an equation for the reaction involved.

(a) The oxidation state of chlorine is +1 in NaClO, -1 in NaCl and +5 in NaClO3. This is a disproportionation reaction. NaClO is simultaneously reduced to NaCl and oxidised to NaClO3.

(b) A solution of sodium chlorate (I) could be prepared by passing Cl2 (g) into cold, dilute aqueous NaOH.

C l2+2 NaOH → NaCl+NaClO+H 2O

Q. Potassium chlorate (V), KClO3, is widely used in fireworks and match heads. It decomposes when heated with a catalyst to form potassium chloride and oxygen.

(a) Write a balanced equation for this decomposition chlorate (V), and suggest why it is used in fireworks and match heads.

(b) Suggest how potassium chlorate (V) could be obtained from chlorine.

(a) 2 KCl O3∆→

2 KCl+3O2

When heated, KClO3 produces oxygen which supports combustion.

(b) KClO3 could be obtained by bubbling Cl2 (g) into hot, aqueous KOH.

6 KOH+3 C l2 →5 KCl+KCl O3+3 H 2O

Q. Astatine is a radioactive element which is below iodine in Group VII of the Periodic Table. Ignoring the radioactive nature of astatine, predict what happens when:

(a) aqueous silver nitrate is added to aqueous sodium astatide, followed by aq. NH3;

(b) astatine is mixed with aqueous potassium chloride;

(c) concentrated sulfuric acid is added to solid potassium astatide.

Write equations for any reactions which take place.

(a) When aqueous silver nitrate is added to aqueous sodium astatide, silver astatide is formed, which is insoluble in aq. NH3.

A g+¿ ( aq) +At−¿(aq)→ AgAt( s) ¿¿

(b) When astatine is mixed with aqueous potassium chloride, there is no reaction. Only a higher dihalogen can displace a halide lower in the Periodic Table.

(c) When conc. H2SO4 is added to KAt, HAt is produced which is readily oxidised by conc. H2SO4 to give At2.

KAt+H 2 S O 4→ KHSO 4+¿̂

8 ^+H 2 S O 4→ 4 A t2+H 2 S+4 H 2 O

9.4) An Introduction to the Chemistry of Transition Elements

Q. Explain the following colour changes as fully as you can.

(a) When thiocyanate ions are added to aqueous (III) ions, a deep blood-red colour is obtained, which disappears on adding aqueous sodium fluoride.

(b) The yellow solution obtained when concentrated hydrochloric acid is added to blue aqueous copper (II) sulfate reverts to a light blue colour when an excess of water is added.

(a) The SCN- ligand displaces water ligands to produce the deep blood-red colour due to the formation of [Fe(SCN)(H2O)5]2+ ions.

When NaF is added, the colour disappears due to the formation of a stable complex, [FeF6]3-; the strong F- ligand displaces all ligands.

(b) When concentrated HCl is added to aqueous CuSO4, a yellow solution is obtained due to the formation of [CuCl4]2- complex.

The yellow solution reverts to a light blue colour when an excess of water is added due to the gradual replacement of Cl- ligands by aqua ligands to form the blue [Cu(H2O)6]2+ ion.

Q. By quoting relevant Eθ values from the Data Booklet, explain the following observations:

(a) The green precipitate obtained when aqueous sodium hydroxide is added to aqueous iron (II) sulfate rapidly turns brown on exposure to the air.

(b) The blue solution, containing Cr2+ (aq), contained when chromium metal is dissolved in dilute sulfuric acid slowly turns green in the absence of air.

(a) Fe ¿

O2+2H 2O+4 e−¿→ 4 O H−¿Eθ=+0.40 V ¿¿

4 Fe ¿

Ecellθ =+0.40+0.56=+0.96 V

Since Eθ > 0, the reaction is feasible.

(b) C r3+¿+e−¿→C r 2+¿Eθ=−0.41 V ¿¿¿

2 H+¿+2 e−¿→H 2 Eθ=0.00 V ¿¿

2 C r2+¿+2 H +¿→ 2C r3+ ¿+ H

2¿¿¿

Ecellθ =0.00−(−0.41 )=+0.41V

Reaction is feasible.

Q. By quoting relevant Eθ values from the Data Booklet, predict any chemical changes involving chromium that would occur when

(i) excess chromium granules are added to deoxygenated dilute hydrochloric acid at room temperature;

(ii) the solution obtained in (i) is acidified and is then allowed to stand in air;

(iii) excess zinc granules are added to the solution obtained in (ii).

(i) C r 2+¿+2e−¿→Cr Eθ=−0.91 V ¿¿

2 H+¿+2 e−¿→H 2 Eθ=0.00 V ¿¿

Cr+2 H+¿→C r 2+¿+H 2 Ecellθ =+0.91V ¿ ¿

(ii) C r3+¿+e−¿→C r 2+¿Eθ=−0.40 V ¿¿¿

O2+4 H+¿+4 e−¿→2 H 2O Eθ=+1.23 V ¿¿

4 C r2+¿+O 2+4 H +¿ →4 Cr3+¿+2H

2O E

cellθ =+1.63 V ¿

¿¿

(iii) C r3+¿+e−¿→C r 2+¿Eθ=−0.40 V ¿¿¿

Z n2+¿+2e−¿→Zn Eθ=−0.76 V ¿¿

2C r3+¿+Zn→ 2C r 2+¿+Z n2+¿E

cellθ =+ 0.36V ¿

¿ ¿

Q. Explain the following observations.

(a) A green complex cation Y with formula [Co(NH2CH2CH2NH2)2Cl2]+ is produced by passing air through an aqueous solution containing CoCl2, H2NCH2CH2NH2 and HCl.

(b) A red complex cation Z (same formula as Y) is produced by evaporating an aqueous solution of Y at 90°C. Y has no dipole moment, whereas Z does.

(a) When air is passed through, Co2+ is oxidised to Co3+.

(b) Y and Z are cis-trans isomers. Y is the trans isomer, and Z is the cis isomer.

Q. In the haemoglobin molecule, the iron atom is surrounded by six ligand atoms – five are nitrogen atoms from the globin protein, and one is an oxygen atom from a water molecule. This molecule is replaced by an oxygen molecule in oxyhaemoglobin as shown by the following equilibrium:

(a) Describe the bonding between the oxygen and iron atoms in haemoglobin and in oxyhaemoglobin.

(b) The iron atom in haemoglobin contains six 3d electrons.

(i) What oxidation state of iron does this corresponds to?

(ii) When haemoglobin is converted to oxyhaemoglobin, what change of oxidation number, if any, occurs? Explain your answer.

(iii) Suggest, with explanation, the likely magnetic property of iron in haemoglobin.

(a) Dative covalent / coordination bond is formed between O and the Fe atoms in haemoglobin and oxyhaemoglobin. The lone pair of electrons on oxygen atom is donated into the vacant orbital in iron atom to form the dative bond.

(b i) Oxidation state of iron = +2

(b ii) There is no change in oxidation number.

(b iii) Iron in haemoglobin is likely to be paramagnetic since it contains unpaired electrons. In the presence of a magnetic field, these odd electrons aligned themselves in such a way that they are attracted to a magnetic field.

Q. Explain briefly how EDTA could be used to treat hard water.

Hard water contains Ca, Mg ions which can complex with EDTA, and renders them ineffective as hard water agents.

10) Organic Chemistry

10.1) Introductory Topics

Q. State the total number of stereoisomers shown by compound X, which is used as an optical brightener in detergents. Draw the structures of these isomers.

There are 4 isomers, zero chiral centres. Products are cis-cis, trans-trans, cis-trans and trans-cis.

Q. Give the structural formulae of all possible isomers shown by a compound with molecular formula C4H8.

Q. State the type of isomerism shown by CH3CH=CHCHClCH3 and draw diagrams to represent the isomers.

10.2) Hydrocarbons

Q. A hydrocarbon X (molecular formula C8H8) burns with a smoky flame, and 1 mol of X is found to react with 1 mol of bromine molecules.

(a) Draw the displayed formula of (i) X, and (ii) the product formed between X and bromine.

(b) What organic product would you expect if X is heated under reflux with concentrated alkaline potassium manganate (VII)? Suggest the colour change observed.

(a)

(b) The organic product expected is benzoic acid, which would react with the base (in alkaline KMnO4) to give benzoate salt. The purple KMnO4 solution is decolourised and a brown precipitate of MnO2 is formed.

Q. The given diagram shows the apparatus used in the laboratory preparation of propene by the dehydration of propan-1-ol: C H 3 C H 2 C H 2OH →C H 3 CH=C H 2+H 2 O

(a) Pumice (or a ceramic such as broken crockery) may be used as material Y. Give the chemical name of a substance that might be present in one of these materials.

(b) Why must material Y be heated strongly?

(c) Material Y becomes black in colour. Name the substance responsible for this black colour, and suggest how it arises in the reaction.

(d) At the end of the reaction, when no more propene comes over, the apparatus is left to cool with the delivery tube out of the water. Why is this done?

(a) Alumina, Al2O3, SiO2 or Fe2O3 may be used.

(b) Y needs to be heated strongly to break the bonds.

(c) The black solid is carbon, which is always formed when organic molecules are subjected to strong heat (thermal decomposition).

(d) This is done so that cold water would not be sucked back into the hot test tube, thereby causing it to crack.

Q. Hydrocarbon A reacts with HBr to give B as the major product.

(a) Give the structure of B and its systematic name.

(b) Outline the mechanism of the reaction and explain why B is the major product.

(c) A student commented that B obtained from the above reaction is optically active. State and explain whether or not you agree with the student.

(a) B is 3-bromo-3-methylhexane

(b) Mechanism: electrophilic addition.

B is the major product because a tertiary carbocation is more stable.

(c) No. The carbocation intermediate has a planar structure and so, bromide can attack from either side of the plane. Hence a racemic mixture is formed. B is therefore not optically active as the rotating power of one enantiomer exactly cancels that of the other.

10.3) Halogen Derivatives

Q. Chlorine atoms are formed in the stratosphere by the homolytic fission of CFC molecules such as CCl2F2 (a volatile material used as a refrigerant).

(a) Describe how the formation of chlorine atoms in the stratosphere affects the ozone layer.

(b) The compound CF3CH2F is proposed as a replacement for CCl2F2 as a refrigerant. Suggest what physical property of CH3CH2F makes it an appropriate choice as a refrigerant. Give your reasoning.

(a) Chlorine atoms are generated in a chain reaction and each Cl atom can destroy many ozone molecules. This eventually creates a hole in the ozone layer.

(b) The low boiling point of CH3CH2F makes it an appropriate choice because it is necessary for a refrigerant to vaporise rapidly at a low temperature.

Q. Fluothane, CF3CHBrCl, is a volatile liquid commonly used as an anaesthetic in hospitals. To confirm that Fluothane does not undergo hydrolysis, a mixture of Fluothane and sodalime (which contains NaOH) was stirred for an hour. It was then acidified and tested with (i) aqueous silver nitrate, and (ii) aqueous chlorine.

(a) Describe what would be observed if bromide ions had been produced, and write an equation in each case.

(b) The attempted hydrolysis did not break the C-Br bond in Fluothane. Hence it may be assumed that the other two types of C-Hal bond in Fluothane also remain intact after hydrolysis. Give a reasoning for this assumption.

(a i) Cream ppt of AgBr

(a ii) orange/brown Br2 solution. Cl2 + 2Br- -> Br2 + 2Cl-

(b) The other two types of C-Hal bond in Fluothane are C-Cl and C-F bonds, both which are stronger than C-Br bond.

Q. When hydrogen chloride is eliminated from 2-chlorobutane, three isomeric alkenes with the formula C4H8 are produced. Draw the structures of the three butenes produced.

Q. Outline a synthetic route, in not more than five steps, for each of the following conversions. For each step, give the reagent(s), conditions and the structure of the organic product.

(a) C H 3 C H 2 C H 2Cl →C H 3 C H 2 C H 2C O2 H

(b) C H 3 C H 2 C H 2Cl →¿

(c) C H 3 CH=CH →¿

10.4) Hydroxy Compounds

Q. Arrange compounds K, L and M in order of increasing pKa values with reasons.

The smaller the pKa value, the stronger is the acid.

The order is: M < K < L

L is the weakest acid as intramolecular hydrogen bond is formed due to the close proximity of the –OH group and –NO2 group. Hence proton is less readily lost.

K is a stronger acid than L due to the presence of an electron withdrawing NO2 group, which attracts electrons away from the oxygen atom and so stabilises the phenoxide ion, reducing its tendency to attract protons.

M is a strong acid as it has 2 electron withdrawing groups.

Q. Triphenylmethanol is converted to triphenylmethoxymethane by the following synthetic route:

(a) Give the structural formula of compound W, and outline a mechanism for the reaction of W with CH3I. Hence explain why this reaction is extremely slow.

(b) Triphenylmethoxymethane can be synthesized by heating triphenylmethanol in methanol with a catalytic amount of HCl (aq). Given that the reaction mechanism involves the protonation of the OH group of triphenylmethanol followed by an SN1 pathway, outline a mechanism for the conversion of triphenylmethanol to triphenylmethoxymethane.

(a) W is Ph3CO-Na+. Reaction is slow due to the bulky nucleophile.

10.5) Carbonyl Compounds

Q. Menthol and methone are the major constituents of peppermint oil. Describe one simple test-tube reaction to distinguish menthol from methone.

Warm each compound separately with 2,4-dinitrophenylhydrazine. Methone gives an orange precipitate but menthol does not.

Other tests: Menthol gives off HCl fumes with reaction with PCl5, SOCl2, or effervescence of H2 with Na metal.

Q. Describe one simple chemical test to distinguish propanone, CH3COCH3, from

(i) methyl ethanoate, CH3CO2CH3, and

(ii) propanal, CH3CH2CHO

(i) Add 2,4-DNPH to each and warm. Ketone will give orange precipitate.

(ii) Add Tollen or Fehling’s solution, the aldehyde will react only.

Q. Devise a 3-stage synthesis of compound X starting from a suitable carbonyl compound, and using a cyanohydrin as an intermediate.

Q. Give the structure of the organic product B in the following reaction, and label all chiral centres with asterisks. Will the product obtained rotate the plane of polarisation of a beam of plane-polarised light? Explain.

The product obtained will not rotate the polarised light, because the carbonyl group is planar, hence the cyanide can attack on either sides, giving a racemic mixture.

10.6) Carboxylic Acids& Derivatives

Q. Suggest structures for compounds J, K, L and M.

Under acidic conditions, K reacts with butan-1-ol to give M (C16H22O4).

Q. Compound X (Mr 58) is neutral and water soluble. X does not react with sodium metal nor with Fehling’s solution but it does react with alkaline aqueous iodine. Suggest a structural formula for X. Explain.

X is neutral, so it is not a carboxylic acid.X is water soluble showing it is polar.X does not reat with Na, it cannot be alcohol.X is not aldehyde as it does not react with Fehling’s solution.X reacts with alkaline aqueous iodine giving a yellow precipitate of MeI showing that it contains the Ac group. Hence X is acetone.

Q. Outline a synthetic route, in not more than five steps, to accomplish the following conversion. For each step, give the reagents, conditions and structure of the organic product.

Q. Three chlorine-containing compounds are prepared starting from ethanoic acid (A):

(a) What are the reagents and conditions for reactions I and II?

(b) Compound D reacts with ethanol to give steamy fumes and a fruity-smelling liquid. Write a balanced equation for this reaction, giving the structural formula of the organic product.

(c) Compounds A, B and D (not necessarily in that order) are added to separate portions of water to give solutions with pH values of 0.5, 2.5 and 3.0. On adding aqueous silver nitrate to these solutions, only one of the solution gives a thick white precipitate.

(i) Suggest which pH value is associated with ach of A, B and D. Explain.

(ii) Account for the formation of the white precipitate.

(d) What is the likely pH value of an aqueous solution of compound C? Explain.

(a) I: Cl2, hv II: PCl5

(b) MeCOCl+EtOH → MeC O2 Et+ HCl

(c i) A has pH 3.0, B has pH 2.5 and D has pH 0.5.

Ethanoic acid (A) is a weak acid. B is more acidic than A due to electron withdrawing Cl. D is the most acidic because it undergoes hydrolysis to give HCl, a strong acid.

(c ii) The free Cl- released upon hydrolysis can react with silver nitrate to form silver chloride.

(d) An aqueous solution of C is likely to have a pH value of 1.5, due to the presence of 2 electron withdrawing groups.

10.7) Nitrogen Compounds

Q. Paracetamol, a common analgesic drug, can be synthesized by reacting 4-aminophenol with compound X.

(a) Identify compound X.

(b) Suggest a reagent that could be used to convert ethanoic acid, CH3CO2H into X.

(a) MeCOCl

(b) SOCl2 or PCl3/PCl5

Q. Aspartame, a synthetic sweetening agent, has the structure shown:

(a) Which two functional groups in aspartame contribute to its solubility in water? Explain.

(b) Give the conditions and reagents that could be used to hydrolyse aspartame into its constituent parts, and draw the structural formulae of all the organic molecules obtained.

(a) Amino and the carboxylic acid group, by forming hydrogen bonds with water.

(b) Reagent: dilute acid for hydrolysis, reflux.

Q. When heated, benzoic acid reacts with butan-1-amine to give compound Y.

(a) Write the systematic name of Y, and state the type of reaction involved.

(b) At room temperature, benzoic acid reacts with butan-1-amine to give another product Z. Give the structure of Z.

(c) Using benzoic acid as the starting material, devise another route to synthesize Y without involving reactions that take place at temperatures much higher than room temperature.

(a) Name is N-butylbenzamide, condensation reaction.

(b) Z is an ionic salt.

(c) Change to acid chloride, then add in the amine.