Answers to Examination Style Questions

8/16/2019 Answers to Examination Style Questions

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Module 1 (page 74)1 D

2 C

3 A

4 D

5 C

6 D

7 A

8 B

9 A

10 C

11 a An oxidising agent gains electrons and the oxidationnumber of its element decreases. [2]

b i 2MnO4– + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2 [2]

ii Oxidation number of Mn in MnO4– is +7, after it

reacts it decreases to +2. It is reduced (and is therefore the oxidising

agent) [2]

Oxidation number of O in H2O2 is –1, after it reactsit increases to 0. It is oxidised (and is therefore the reducing

agent) [2]

c i I – [1]

ii Br2(aq) + 2KX(aq)→ 2KBr(aq) + X2(aq) [2]

iii Br2 [1] Gains electrons more readily than X [1] Br2 + 2e– → 2Br– [2]

12 a Any THREE of the assumptions below:The molecules are in constant random motion.

The molecules occupy negligible volume whencompared to the volume occupied by the gas itself.The average kinetic energy is proportional to theabsolute temperature.Collisions are perfectly elastic.There are no intermolecular forces between themolecules. [3]

b i Low temperature and high pressure [2] At low temperature, the kinetic energy of the

molecules decreases causing them to slow down,providing the opportunity for intermolecular forcesof attraction to occur. [1]

At high pressure, the molecules are pushed closetogether causing the volume of the molecules tobecome signicant. [1]

ii

P (atm)

P V

CO2

ideal gas

Allow any sort of positive deviation fromhorizontal line [2]

c i Pressure = 3.4 × 1.01 × 105 = 3.434 × 105 Pa [1] Volume = 0.072 × 0.001 = 7.2 × 10–5 m3 [1]

Temperature = 273 + (–48) = 225 K [1] n = PV ___

RT [1]

n = 3.434 ×105 × 7.2 ×10–5 _____________________ 8.31 × 225

= 0.0132 mol [1]

ii Relative molecular mass =mass _____ mole

= 0.94 _______ 0.0132

= 71 [1]

13 a The enthalpy change accompanying a chemical changeis independent of the route by which the chemicalchange occurs. [1]

b i2C

2CO2

�

� 3H2O

2 H f (CO2)Ø 3 H f (H2O)Ø

H f (C2H6)Ø

3H2 C2H6

HC (C2H6)Ø

+1560 kJ mol 1

arrow reversedso sign changes [2]

ii ∆Hf Ø (C2H6) = 2∆Hf

Ø (CO2) + 3∆Hf Ø (H2O) −

∆HcØ (C2H6)

= (2 × –393) + (3 × –286) – (–1560) = –84 kJ mol–1 [2]

c i The polystyrene is a better insulator than theglass. [1]

ii Any ONE assumption:The specic heat capacity of the solution isequal to the specic heat capacity of water.The density of the solution is the same as thedensity of water, 1 g cm–3.Heat is not lost or absorbed from theenvironment. [1]

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iiiH p

Hr

e n e r g y

k J

reaction pathway

NH4NO3(aq)

Hsoln�ve

endothermic

NH4NO3(S)+aq

[4]

iv E = mc∆T = 100 × 4.2 × 12.2 = +5124 J = +5.12 kJ [1] Number of moles of ammonium nitrate = 16 ___

80

= 0.2 mole [1] 0.2 mole produces 5.12 kJ Therefore 1 mole produces 5.12 ____

0.2 = +25.60 kJ mol–1 [2]

14 a To ensure complete combustion of the oxalic acid [1]

b i Empirical formula – the simplest whole numberratio of elements present in a compound [1]

Molecular formula – shows the actual number ofatoms of each type of element present in 1molecule of the compound [1]

ii C H O Mass in 100 g 26.67 2.22 71.11

No. of moles [ 26.67 ______ 12 ] [ 2.22 _____ 1 ] [ 71.11 _____ 16 ] [1] = 2.22 = 2.22 = 4.44 [1]

Mole ratio [ 2.22 _____ 2.22

] [ 2.22 _____ 2.22

] [ 4.44 _____ 2.22

] [1]

1 : 1 : 2 [1] Empirical formula is CHO2 To calculate molecular formula n(CHO2) = 90 n(45) = 90 n = 2 Molecular formula is therefore C2H2O4. [2]

c i Colourless to pink [1]

ii Number of moles of potassium manganate( ):1000 cm3 potassium manganate( )

contains 0.02 mol

Therefore 21.5 cm3 contains 0.02 _____ 1000

× 21.5

= 0.000 43 mole [1] From the mole ratio:

2 moles potassium manganate( )reacts with 5 moles oxalic acid

Therefore 0.00043 moles potassiummanganate( ) reacts with

5 × 0.000 43 ____________ 2

[1]

= 0.0011 mole [1]

iii 25 cm3 contains 0.0011 mole oxalic acid

Therefore 1000 cm3 contains 0.0011 _______ 25

× 1000

= 0.044 mol dm–3 [1]

iv massconcentration = molar

concentration × molarmass

= 0.044 × 90= 3.96 g dm–3 [1]

15 a i The rst ionisation of an element is the energyneeded to convert one mole of an element in itsgaseous atoms into ions with a single positivecharge. [1]

ii Mg(g)→ Mg+(g) + e– [1]

b i The rst ionisation energy increases across theperiod. [1]

This is as a result of the increasing effective nuclearcharge across the period. As a result of this itmakes it more difcult for the outer electrons to beremoved. [2]

ii Between Mg and Al: Electronic conguration of Mg is 1s2 2s2 2p6 3s2. Electronic conguration of Al is 1s2 2s2 2p6 3s2 3p1. [1] The electron to be removed from the Al is in a

higher sub-shell, thus it requires less energy thanthe electron to be removed in Mg. [1]

Between P and S: Electronic conguration of P is 1s2 2s2 2p6 3s2 3p x 1

3p y 1 3p z1

Electronic conguration of S is 1s2

2s2

2p6

3s2

3p x 2

3p y 1 3p z1 [1] In P the three electrons in the 3p orbitals have the

same spin but in S the electron to be removed inthe p orbital is paired. [1]

As a result of this there is electron–electron repulsionand hence it makes it easier to lose this electron thanone of the electrons in the p orbital of P. [1]

c i

i o n i s

a t

i o n

e n e r g y

1

number of electrons

removed

2 3 4 5 6

[1 mark for correctlylabelled axes.

1 mark for correctshape of graph]

ii The sudden jump in values of ionisation energyoccurs when the outer electrons have beenremoved. [1]

The rst and second electrons are easy to removeso must be in group II. [1]

It takes a lot more energy to remove the thirdelectron so this electron must be in a shell closer tothe nucleus. [1]

Answers to exam-style questions



Module 2 (page 132)

1 D

2 A

3 B

4 D

5 D

6 B

7 B

8 C

9 D

10 B

11 a Any strong monoacidic base, e.g. NaOH, KOH [1]

b i Initial pH = 3.0 ∴ at equilibrium [H+] = 1 × 10–3, [acid]

= 0.112 – 1 × 10–3 = 0.111

K a =(1 × 10–3)2 __________

0.111 = 9.0 × 10–6 mol dm–3 [3]

ii pK a = –log 9.0 × 10–6 = 5.0 [1]

c Equivalence volume = 27.5 cm3

∴ molarity of base = 0.102 mol dm–3 [3]

d At the equivalence point, the conjugate base of theweak acid exists. It dissociates in solution to produceOH– ions. [2]

e i pH = 3.5 [1]

ii This is the maximum amount of acid or alkali abuffer can absorb without an appreciable changein pH. [2]

f Yes, it is suitable since the pH range of the indicatorfalls on the pH range of the equivalence point (i.e. thevertical portion of the titration curve). [2]

12 a The electrode potential of a half cell relative to thestandard hydrogen electrode [2]

b iand ii

V

Ag(s)[anode]

Ag� (aq)(1 mol dm 3)

Mn2� (aq)(1 mol dm 3)

Pt(s)[cathode]

H� (aq)MnO4(aq)

(1 mol dm 3) [5]

c EØ

cell = +0.72 V 5Ag(s) + MnO4

–(aq) + 8H+(aq) → 5Ag+(aq) + Mn2+(aq) + 4H2O(l) [2]

d Ag(s) Ag+(aq) [MnO4

–(aq) + 8 H+(aq)],[Mn2+(aq) + 4H2O(l)] Pt(s) [2]

e An increase in the concentration of MnO4–(aq) relative

to Mn2+(aq) and Ag+(aq) would cause the cell e.m.f. toincrease, but a decrease in the concentration of

MnO4–(aq) relative to Mn2+(aq) and Ag+(aq) would

result in a decrease in cell e.m.f. [2]

f Cl– ions from the salt bridge would react with Ag+ ionsin the anode compartment to produce a precipitate ofAgCl, which would affect ionic mobilities and thereforethe cell function. [2]

13 a i For a chemical reaction to occur the particles must

collide. For the collisions to be effective, theparticles must collide with the correct amount ofenergy (energy greater than or equal to theactivation energy) and with the correctorientation. [2]

ii The collision frequency would increase andtherefore the probability of effective collisionswould increase. [2]

b i

time

[ r e a c

t a n

t ]

Draw a tangent to the curve at t = 0, then nd thegradient of the tangent. [2]

ii From comparing Experiments 1 and 2, rst orderwith respect to A. [2]

From comparing Experiments 1 and 3, second orderwith respect to B. [2]

iii Rate = k [A][B]2 [1]

iv Could substitute data from Experiments 1, 2 or 3 in

the rate equation, e.g. k = (1.45 × 10–2 mol dm–3 s–1) ________________________________ (0.125 mol dm–3) (0.125 mol dm–3 )2

= 7.42 dm6 mol–2 s–1 [2]

v Rate = (7.42 dm6 mol–2 s–1)(0.30 mol dm–3)(0.14 mol dm–3)2

= 0.044 mol dm–3 s–1 [2]




Module 3 (page 174)1 D

2 C3 B

4 D

5 A

6 a i A metal which forms at least one ion with apartially lled d sub-shell. [2]

ii Exhibits variable oxidation states. Forms complex ions. Forms coloured compounds. Exhibits catalytic properties. [4] iii Fe3+ [Ar] 3d5; Mn3+ [Ar] 3d4 [2]

b i Fe2+[Ar] 3d6 . The 3d5 conguration of Fe3+ representsa more stable (energetically) arrangement than the3d6 of Fe2+. This is due to the increased repulsion ofelectrons in the sub-shell as well as the stabilitypresent in the half lled sub-shell . Thus the reactionof Fe2+ to Fe3+ will be favoured:

Fe2+ → Fe3+ + e-

3d6 3d5 [2]

ii In this case, Mn2+ with the conguration 3d5 willresist conversion to the less stable conguration3d4 of Mn3+. [2]

c i The hexaaquairon( ) ion of iron( ) chloridebecause of the high charge density of the Fe3+ isable to attract the water molecules stronglythereby weakening the O–H bond. The oxygen ofother water molecules are then able to removethem as protons. The removal of three protonsfrom the hydrated metal ion produces thered-brown precipitate of iron( ) hydroxide. [2]

ii [Fe(H2O)6]3+(aq) + 3H2O(l) → Fe(H2O)3(OH)3(s) + 3H3O+(aq)

Allow: FeCl3(aq) + 3H2O(l) → Fe(OH)3(s) + 3HCl(aq) [1]

7 a i A compound formed when a transition metal atomion is surrounded by species which form coordinatebonds. [2]

ii The species, molecules or ions, which formcoordinate bonds to the metal ion. [2]

b i [Cu(H2O)6]2+(aq) (blue);[CuCl4]2–(aq) (yellow) [2]

ii [Cu(H2O)6]2+(aq) + 4Cl–(aq) → [CuCl4]2–(aq) + 6H2O(l) blue yellow [2]

iii The stability constant indicates the strength of theligand as a Lewis base (lone pair donor).

The Cl– ligand has a larger stability constant thanH2O and hence can displace it from the coppercomplex. [3]

c i Blue solution forms a blue precipitate which

dissolves to form a deep blue solution. [3] ii [Cu(NH3)4(H2O)2]2+(aq) or [Cu(NH3)4]2+(aq) [1]

8 a CCl4 does not react with water while SiCl4 reactsreadily to produce white clouds of hydrogen chloridedue to hydrolysis. The silicon atom of the SiCl4 molecule has available d orbitals of appropriate energywhich enable bonding to occur with the oxygen atomof the water molecules. Carbon does not have similaravailable d orbitals.

SiCl4 + 2H2O → SiO2 + 4HCl Or SiCl4 + 3H2O → H2SiO3 + 4HCl [5]

b Ge4+(aq) + 2e– → Ge2+(aq) ∆E = –1.6 V Sn4+(aq) + 2e– → Sn2+(aq) ∆E = +0.15 V Pb4+(aq) + 2e– → Pb2+(aq) ∆E = +1.69 V From the above electrode reactions, as the group is

descended it becomes easier for M4+ to be converted toM2+ as seen in the increasing positive values for ∆E.Hence the stability of +4 oxidation state decreases infavour of the +2 oxidation state. [3]

c i 3Cl2(g) + 6OH– (aq) → ClO3–(aq) + 5Cl–(aq) + 3H2O(l)

Ox. 0 +5 –1 No. [2]

ii 2OH–(aq) + Cl2(g) → ClO–(aq) + Cl–(aq) + H2O

3ClO–

(aq) → ClO3–

(aq) + 2Cl–

(aq) Disproportionation is a reaction in which a speciesis simultaneously oxidised and reduced. [5]




Answers to exam-style questions9 a i Group II metal sulphates become less soluble as the

group is descended, barium is below magnesium andhence its sulphate is much less soluble. [2]

ii Group II carbonates decompose to the oxide andcarbon dioxide. The smaller oxide ion forms a muchstronger ionic bond with the metal than thecarbonate ion can and therefore is much morestable. As the group is descended the cations getlarger. The larger Sr2+ ion has s smaller chargedensity and a reduced ability to polarise thecarbonate ion thus making strontium carbonatemore resistant to decomposition. [4]

b i Atomic size of Group II elements increase as it isdescended. The presence of additional electronicshells positioned further away from the nuclei withthe resulting reduction in attractive force isresponsible for this trend. [2]

ii In the case of ionic size there is a similar trend, theloss of the valence electrons still leaves eachsuccessive ion with an additional electronic shellwith a noble gas conguration. [1]

c 10.46 g of oxygen atoms combine with 89.54 g of M 10.46/16 mol of oxygen atoms combine with 89.54 g

of M 1 mol of oxygen atoms combines with

89.54 × 16 ______ 10.46

g of M = 136.96 g

Ratio of moles: M : O = 1 : 1, hence rel. atomic mass of Mis 137.

i NO2 [1] ii Atomic mass M = 137 [2]

iii Barium nitrate( ), Ba(NO3)2 [1]

iv Ba(OH)2(aq) + 2HNO3(aq) → Ba(NO3)2(aq) + 2H2O(l) [2]

10 a i state Element/symbol

b.pt. density

[2]

gas F

ClBr

↓ ↓solid IIncreasing

ii Structure: simple molecular to molecular solid Bonding: covalently bonded molecules with

intermolecular van der Waals’ forces. As group is descended the strength of van der

Waals’ forces increase as molecules becomelarger and state changes from simple moleculesto solid. [4]

b i Hydrogen bromide and carbon dioxide [2]

ii Br–, white fuming gas, reddish vapour; and H3O+,white precipitate with calcium hydroxidesolution [2]

iii Br–(aq) + Ag+(aq) → AgBr(s) AgBr(s) + 2NH3(aq) → Ag(NH3)2

+(aq) + Br–(aq) [2]

c Cl2 + 2e– → 2Cl– ∆E = +1.36 V Br2 + 2e– → 2Br– ∆E = +1.07 V I2 + 2e– → 2I– ∆E = +0.54 V From the above electrode potentials, the ease with

which the halogen accepts electrons (oxidising agent)decreases as the group is descended; the +ve characterof ∆E decreases. [3]

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