St. Andrew's Junior CollegeH2 Chemistry 2014/2015

SOLUTIONS to Revision Package for Common Test

Section Topic

I Atoms, Moles & Stoichiometry and Redox

II Atomic Structure

III Chemical Bonding with summary notes

IV The Gaseous State

V Chemical Energetics

1

I: Atoms, Moles & Stoichiometry and Redox

1 (i) 3As2O3 + 4NO3- + 4H+ + 7H2O 6H3AsO4 + 4NO

(ii) 2MnO4- + 3NO2

- + H2O 2MnO2 + 3NO3- + 2OH-

(iii) 5C2O42- + 2MnO4

- +16H+ 10CO2 + 2Mn2+ + 8H2O

(iv

)

5H2XeO4 + 6Mn2+ + 18OH- 6MnO4- + 5Xe + 14H2O

(v) 5SO2 + 2IO3- + 4H2O 5SO4

2- + I2 + 8H+

2 2S2O32- + I2 S4O6

2- + 2I-

No. of moles of S2O32- = 16.80/1000 x 0.1 = 1.680 x 10-3 mol

No. of moles of I2 in 25 cm3 = 1.680 x 10-3 / 2 = 8.400 x 10-4 mol

No. of moles of I2 in 250 cm3 = 8.400 x 10-4 x 250/25 = 8.400 x 10-3 mol

No. of moles of Cu2+ in 250 cm3 = 8.400 x 10-3 x 2 = 0.01680 mol

Mass of Cu2+ in sample = 0.01680 x 63.5 = 1.0668 g

% by mass of Cu = 1.0668/9.40 x 100 % = 11.3 %

3 M2O3 + 6 HCl 2 MCl3 + 3 H2O

Total no. of moles of HCl = 80/1000 x 1.5 = 0.1200 mol

2 HCl + Na2CO3 2 NaCl + CO2 + H2O

No. of moles of Na2CO3= 22.50 / 1000 x 0.125 = 2.8125 x 10-3 mol

No. of moles of HCl in 25 cm3 = 2.8125 x 10-3 x 2 = 5.625 x 10-3 mol

No. of moles of excess HCl in 400 cm3 = 5.625 x 10-3 x 400 / 25 = 0.09000 mol

No of moles of HCl reacted in 400 cm3 = 0.1200 – 0.09000 = 0.03000 mol

No. of moles of M2O3 = 0.03000 / 6 = 5.000 x 10-3 mol

Mr of M2O3 = 0.829 / 5 x 10-3 = 165.8

Ar of M = [165.8 – (3 x 16)]/2 = 58.9

Metal is Cobalt.

4 (a) H2O2 + 2H+ +2e 2H2O

Fe2+ Fe3+ + e

Overall: H2O2 + 2H+ + 2Fe2+ 2H2O + 2Fe3+

(b) MnO4- + 8H+ + 5e Mn2+ + 4H2O

Fe2+ Fe3+ + e

Overall: MnO4-+ 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+

(c) No. of moles of MnO4- = 13.5/1000 x 0.02 = 2.700 x 10-4 mol

No. of moles of Fe2+ in 10 cm3 = 2.700 x 10-4 x 5 = 1.350 x 10-3 mol

No. of moles of excess Fe2+ in 65 cm3 = 1.350 x 10-3 x 65/10

1

= 8.775 x 10-3 mol

Total no. of moles of Fe2+ = 40/1000 x 0.25 = 0.01000 mol

No. of moles of Fe2+ reacted = 0.01000 – 8.775 x 10-3 = 1.225 x 10-3 mol

No. of moles of H2O2 = 1.225 x 10-3 /2 = 6.125 x 10-4 mol

[H2O2] = 6.125 x 10-4 / (25/1000) = 0.0245 mol dm-3

5 (a) Cr2O72- + 14H+ + 6e 2Cr3+ + 7H2O

MnO4- + 8H+ + 5e Mn2+ + 4H2O

(b) (i) No of moles of KMnO4 = 18/1000 x 0.2 = 0.00360 mol

(ii) No of moles of electrons gain by 0.00360 moles of KMnO4

= 0.0360 x 5 = 0.0180 mol

(c) No of moles of electrons gain by K2Cr2O7 = 0.0180 mol

[Note: No. of moles of e- gained by Cr2O72- is the same as MnO4

- for same

vol of FA 1 used]

No of moles of K2Cr2O7 = 0.0180/6

= 0.003000 mol

Volume of K2Cr2O7 = 0.003000 /0.15

= 20.0 cm3

6 (a) Cr2O72- + 14H+ + 6e 2Cr3+ + 7H2O

(b) No of moles of Cr2O72- = 22.55/1000 x 0.0325 = 0.0007329 mol

No of moles of Xn+ = 25/1000 x 0.044 = 0.001100 mol

Mole ratio of Cr2O72- : Xn+ = 2 : 3

(c) 2 moles of Cr2O72- gain 12 moles of electrons

3 moles of Xn+ loss 12 moles of electrons

1 mole of Xn+ loss 4 moles of electrons

(d) Final oxidation state of X in XO3- = + 5

Hence, initial oxidation state of X in Xn+ = + 5 - 4

n = +1

7 (a) No of moles of Cu+= 25/1000 x 0.2 = 0.005 mol

No of moles of K2Cr2O7 = 12.5/1000 x 0.1 = 0. 00125 mol

No. of moles of Cu+ that reacts with 1 mole of K2Cr2O7 = 0.005/0.00125

= 4

(b) Cu+ Cu2+ + e

4 moles of Cu+ loses 4 moles of electrons

4 moles of electrons are gained by 1 mole of K2Cr2O7

1 mole of K2Cr2O7 contains 2 moles of Cr

2

1 mole of Cr gains 4/2 = 2 moles of electrons

Initial O.S of Cr is +6

New O.S of Cr = +6 -2

= +4

8 (a) 2ClOn- + 4nH+ + (4n-2) e- Cl2 + 2nH2O

(b) No. of moles of I2 = 10/2 = 5 mol

2I- I2 + 2e-

Mole ratio of ClOn- : I2 = 2 : 5

5 moles of I2 gain 10 moles of electrons

2 moles of ClOn- gain 10 moles of electrons

4n – 2 = 10

n = 3

9 Fe2+ + C2O42- Fe3+ + 2 CO2 + 3 e-

No. of moles of FeC2O4 = 20/1000 x 0.02 = 4.000 x 10-4 mol

No. of moles of MnO4- = 15/1000 x 0.02 = 3.000 x 10-4 mol

Mole ratio of FeC2O4 : MnO4- = 4 : 3

4 moles of FeC2O4 lose 12 moles of e-

12 moles of e- is gained by 3 moles of MnO4-

4 moles of e- is gained by 1 mol of MnO4-

O.N of Mn in MnO4- = +7

New O.N of Mn = +7 - 4 = +3

10 No of moles of Fe2+ = 15/1000 x 0.1 = 0.001500 mol

Since initial oxidation state of X in XO42- = + 6

Final oxidation state of X is = + 3

Change in oxidation state = 6 - 3 = 3

Hence 1 mole of XO42- gain 3 moles of electrons from Fe2+

Fe2+ Fe3+ + e

Mole ratio of XO42- : Fe2+ = 1: 3

No of moles of XO42- in 10 cm3 = 0.001500/3 = 0.0005000 mol

No of moles of XO42- in 1 dm3 = 0.0005000 x 1000/10 = 0.05000 mol

Molecular mass of XO42- = 6.80 / 0.05000

= 136

Hence, relative atomic mass of X = 136 -4(16) = 72.0

3

II: Atomic Structure

1. (a) (i) Number of protons : ……31………

Number of neutrons: ……37……..(ii) – – – – – – – –

+ + + + + + + +

Sourceα

Neutrons

The deflection remains the same at 3o as the alpha particle and deuterium isotope

have the same charge/mass ratio.

2. There is a big increase from 2 nd ionisation energy to 3 rd ionisation energy . The

3rd electron to be removed is in the inner shell and Z has 2 valence electrons. Hence

it is in Group II.

3. (a) Group 0

(b) G2O3, expected nature: ionic

(c) Covalent

(d) Down the group, I has more protons , hence nuclear charge increases.

I has more inner shell electrons than A, screening effect increases significantly.

The increase in screening effect outweighs the increase in nuclear charge, the

effective nuclear charge decreases. Valence electrons are less strongly attracted to the nucleus. Hence, less energy is required to remove the valence

electrons and the first ionisation energy of I is lower than that of A.

4. (i) M (g) M+ (g) + e-

(ii) Iodine has a higher nuclear charge as it has more protons but it also experiences

a greater screening effect from the increased number of inner shell electrons.

Hence, the valence electrons in iodine experience similar effective nuclear charge

as those in phosphorus.

5. (i) 1s22s22p63s23p63d10

(ii) Ga3+ has a larger ionic radius than Ge4+. They are isoelectronic hence same screening effect. Ge4+ has more protons so higher nuclear charge. Thus, Ge4+ has

4

higher effective nuclear charge and the attraction between the nucleus and

outermost electrons are stronger.

6. (a) S+ (g) S2+ (g) + e-

(b) (i) Element D. The sharp drop from G to H indicates that the second electron in G

is removed from the inner shell and it only has one valence electron hence G

belongs to Group I. Hence, since they are consecutive elements, D is in group VI.

(ii) The second ionisation energy generally increases from A to G because there is an

increase in nuclear charge while shielding effect is almost constant due to the

electrons being added to the same outer shell. Hence, effective nuclear charge

increases and more energy is required to remove the 2nd outermost electron from A

to G.

Second ionisation energy of B is lower than A because the electron is removed from a

3p-subshell which is further away from the nucleus compared to the 3s-subshell in A and there is also additional screening effect from the two 3s electrons. This effect

outweighs the increase in nuclear charge resulting in a lower effective nuclear charge and less energy required to remove an electron from the 3p than the 3s

orbital.

Second ionisation energy of E does not increase as much because the electron is

removed from a doubly-filled 3p orbital which experiences inter-electronic repulsion. This effect outweighs the increase in nuclear charge, resulting in a lower effective nuclear charge and less energy is required to remove the paired 3p

electrons.

5

III: Chemical Bonding

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10

D C D D B C C C D B

Q11 Q12 Q13

D C B

Essay Questions1 (a) -bond is a covalent bond formed when atomic orbitals overlap in a head-on

manner.

-bond

-bond is a covalent bond formed when two orbitals overlap in a side-on manner.

-bond

(b) CO2 has a simple molecular structure consisting of non-polar molecules

held together by induced dipole-induced dipole interactions.

SO2 has a simple molecular structure consisting of polar molecules held

together by permanent dipole-permanent dipole interactions.

Since induced dipole-induced dipole interactions are weaker than the

permanent dipole-permanent dipole interactions, less energy is required to

separate the CO2 molecules. Hence CO2 has a lower boiling point compared to SO2.

(c) Both catechol and hydroquinone are polar simple covalent molecules

capable of forming intermolecular hydrogen bonds. However, catechol is

also able to form intramolecular hydrogen bonds due to the proximity of the 2 –OH groups. (can use diagram below to explain)

Formation of intramolecular hydrogen bonds in catechol reduces the

6

extent of intermolecular hydrogen bonding. This leads to weaker intermolecular hydrogen bonding in catechol and therefore less energy is

required to overcome these forces. Hence catechol has lower melting point.

(d)

In liquid state, molecules are close together, able to dimerise.

Chlorine from neighbouring molecule donates its lone pair of electrons to

form dative bond with the electron deficient boron.

2 (a) (i) 2AlP + 3H2SO4 2PH3 + Al2(SO4)3

(ii)

AlP does not conduct electricity due to absence of delocalised electrons or free mobile ions.

(iii) The energy released when permanent dipole-permanent dipole interactions are formed between PH3 and H2O is not enough to overcome the hydrogen bonds between H2O molecules.

The energy released when hydrogen bonds are formed between NH3

and H2O is enough to overcome the hydrogen bonds between NH3

7

molecules and the hydrogen bonds between H2O molecules. Hence

PH3 is less soluble in water and NH3 is more soluble in water.

(b) MgCl2 has a giant ionic lattice structure with strong electrostatic forces of attraction between the oppositely charged ions. A lot of energy is

required to overcome the strong electrostatic forces.

AlCl3 and PCl5 are both simple covalent non-polar molecules with weak induced dipole-induced dipole interactions between the molecules.

AlCl3 exists as dimer (Al2Cl6) and has stronger induced dipole-induced

dipole interactions due to a greater number of electrons. Hence more

energy is required to overcome the induced dipole-induced dipole interaction

in AlCl3 than PCl5.

3 (a)

Lone pair-lone pair repulsion > Lone pair-bond pair repulsion > Bond pair-bond pair repulsionThe lone pair of electrons exerts the greatest extent of repulsion, the bond angle decreases with the increase in number of lone pairs. Hence the

order of increasing H-N-H bond angle is NH2- < NH3 < NH4

+.

(b) Ethylene glycol and acetone are both polar simple covalent molecules.

Ethylene glycol has intermolecular hydrogen bonds while acetone has

permanent dipole-permanent dipole interactions between its molecules.

As hydrogen bonds are stronger than permanent dipole-permanent dipole

interactions, more energy is required to overcome the intermolecular forces

of attractions between ethylene glycol molecules than acetone molecules.

Hence ethylene glycol has a higher boliing point and is less volatile.

When the intermolecular forces of attractions are strong, the individual

molecules lose their mobility, i.e., ability to move freely. Since ethylene

glycol possesses stronger intermolecular forces of attraction than

acetone, it has a higher viscosity.

8

4 (a) (i) 2NH4ClO4 N2O + Cl2 + 4H2O + 3/2O2

(ii)

(iii) NO2 has an unpaired electron and readily reacts with another NO2 to

form the more stable dimer N2O4 in which all the electrons are paired.

(iv) Both N atoms acquire a partial positive charge when attached to the

electronegative oxygen atoms. The slightly positively charged N atoms

are in close proximity and when heated, the N-N bond cleaves easily

and the expected products are NO and NO2.

(b) (i) Sn4+ has a high charge density. It has a high polarising power and

polarises the large electron cloud of C l - . Hence SnCl4 is covalent.

(ii) Element Sn C H

% by mass 50.64 40.90 8.46

No. of moles 0.42555 3.4083 8.46

Simplest ratio 1 8 20

Empirical formula: SnC8H20

Molecular formula: SnC8H20

Structural formula: Sn(CH2CH3)4

5 (a) (i)

(ii) 1. 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s1

2. Caesium has a giant metallic structure. It can enhance the electrical

9

conductivity of boron nitride with its sea of delocalised electrons.

(iii) It can be used as a cutting tool.

(b) Both the cis and trans isomers are simple molecular compounds with

intermolecular hydrogen bonds. However, the cis isomer is also able to

form intramolecular hydrogen bonds due to the proximity of the –COOH and –NO2 groups. Formation of intramolecular hydrogen bonds in the cis

isomer reduces the extent of intermolecular hydrogen bonding. This

leads to weaker intermolecular hydrogen bonding in the cis isomer and

therefore less energy is required to overcome these forces. However in the

trans isomer, it has more extensive intermolecular hydrogen bonding.

Hence more energy is required to overcome them and thus has a higher melting point

6 (a)

(b) (i) 12

(ii) n = 1

Oxidation number = +4

10

IV : The Gaseous State

1.(i)

(ii)

(iii)

11

-273 = 0 K

T/ºC

V

V

p

T/K

p

(iv)

(v)

2(a)(i) At relatively high P, the HCl molecules are closer together and this causes the

attractive forces between them to become more significant. Hence, there is

greater deviation from ideality which results in the PV values to decrease with

increase in P.

(Note: For an ideal gas, PV = constant applies.)

aii) The conditions needed for HCl to approach ideal behavior are low pressure

and high temperature.

At low pressure, the gas particles are far apart such that that the attractive

forces between them are negligible. In addition, the volume of each particle is

negligible as compared to the total volume occupied by the gas. Hence, at

lower pressure, the real gas displays more ideal behavior.

At high temperature, the particles possess high kinetic energy and move about

quickly. The particles have sufficient energy to overcome the forces of

attraction between them. This causes the attractive forces to become less

significant and hence, there is lesser deviation from ideality at high

temperature.

12

1V

p

1/V

p

p

pV

4. (i) V1/T1 = V2/T2 => V2 = 115/298 x 986 = 381 cm3

(ii) pV=nRT, p = 12.5 atm = 1266.25 x 103 Pa, V = 380.5 x 10-6 m3

n = pV/RT = [(1266.25)(380.5) x 10-3]/[8.314(986)]= 0.0588 mol

(iii) Final volume is 400 cm3,

Final n = pV/RT = [(1266.25)(400) x 10-3]/[8.314(986)] = 0.06179 mol

Increase in amount of gas, n = 0.06179 – 0.05877 = 3.02 x 10-3 mol

5(a) pV = nRT

pV= m M

RT

ρ=mV=pM

RT

=(1 . 01×105)(64 . 1)

(8 .31 )(298 )

= 2614 g m-3

= 2.61x10-3 g cm-3 (3 s.f.)13

(b) At a higher temperature (500C), the NH3 molecules have greater kinetic energy

and they move faster. The molecules have sufficient energy to overcome the

forces of attraction between each other and this causes the attractive forces

between them to become less significant. Hence, the behavior of the NH3

molecules deviates less from that of an ideal gas at higher temperature.

14

V: Chemical Energetics

15

3(b) The temperature rise for neutralisation involving CH3COOH will be lower as

CH3COOH is a weak acid which only partially dissociates. Part of the heat

energy evolved from the neutralisation is expended to further dissociate

CH3COOH into CH3COO- and H+, resulting in lesser energy evolved during

neutralisation to form H2O.

16

By Hess’ Law,

2 x Electron affinity (EA) =

– (+ 738) – (+ 1451) – (+ 244) – (+ 148) + (– 641) – (– 2526 )

= – 696 kJ mol-1

Electron affinity = – 696/2 = - 348 kJ mol-1

(b) Enthalpy change of hydration of CaCl2 is less exothermic than that of MgCl2.

Although Ca2+ has the same charge as Mg2+, it is larger and hence has a lower

charge density than the latter. Hence, It attracts H2O molecules less strongly

(or forms weaker ion-dipole interactions with H2O molecules) and so, gives out

less heat during hydration.

17

8a)(i) Standard enthalpy change of combustion (Hc) of hydrazine is the enthalpy

change when one mole of the hydrazine is completely burnt in excess oxygen at 298K and 1 atm.

(ii) Amount of heat absorbed by water, q = 200 x 4 x 4.18 = 3344 J

Amount of heat released by reaction, q’ = q / 0.8 = 4180 J

No of moles of N2H4 = 0.210 / (28.0 + 4.0) = 6.563 x 10-3 mol

Standard enthalpy change of combustion of hydrazine

= - q/n = - 4180 / (6.563 x 10-3) = - 637 kJ mol -1

18

(iii)

O2 (g) + N2 (g) + 2H2 (g) N2H4 (l) + O2 (g)

N2(g) + 2H2O (g) N2(g) + 2H2O (l)

By Hess’ law,

∆Hf (N2H4) = 2(-242) – (-637) – 2(+44) = + 65.0 kJmol -1

(b)(i)

By Hess Law,

160 + 4 x B.E(N-H) + 235 = 994 + 2(436)

B.E (N-H) = + 368 kJ mol -1

(ii) The bond energy values obtained from the Data Booklet are average values

and would differ from the experimental values.

9 (i) Ag(s) + 1

2 N2(g) + 3

2 O2(g) AgNO3(aq)

19

∆Hc(N2H4)2 x ∆Hf(H2O(g))

2 x ∆Hv(H2O)

N2H4 (g)

N2 (g) + 2H2 (g)

∆Hf (N2H4) = + 235 kJmol-1

B.E(N-N) + 4 x B.E(N-H) = + 65 kJmol-1 B.E(N N) +

2x B.E(H-H) = + 65 kJmol-1

2N (g) + 4H (g)

(ii) Hr = ∑nHfo(products) - ∑nHf

o(reactants)

= [4(+106) + 4(207) + 90 + 33 + 3(286)] 6(207)

= +103 kJ mol 1

10

11. (i) At 298K, iodine is a solid. Hence, there is an increase in the number of gaseous

particles during the reaction and the entropy change is more positive.

At 500K, iodine is a gas. Hence, there is very little entropy change as the

number of gaseous reactants and products are the same.

(ii) At 298K, G = 53.0 – (298)(165 x 10-3) = +3.83 kJ mol-1

At 500K, G = -10.4 – (500)(22 x 10-3) = -21.4 kJ mol-1

The reaction is more feasible at 500K because G is negative at 500K whereas

G is positive at 298K.

12. (i) ∆Hr = - 2(90.3) + 2(33.2) = - 114 kJ mol-1

20

(ii)

(iii) There is a decrease in disorder in the system as the number of

particles decrease as the reaction proceeds. The forward reaction

produces 2 moles of gaseous products from 3 moles of gaseous

reactants.

(iv) As temperature increases, ∆G values becomes more positive. As

reaction is spontaneous when ∆G < 0, the reaction is spontaneous at

low temperature.

13.

14.

(i) CH4 (g) + H2O (g) CO (g) + 3H2 (g)21

S is positive because there is an increase in number of gaseous molecules produced.

Since S is positive, -T S is negative . At high temperatures, the -T S

term dominates over the H term, resulting in a overall negative value of G since G = H - TS. Hence reaction becomes spontaneous.

(ii) Since H is negative and S is positive, G is always negative at all temperatures. Hence reaction is feasible at all temperatures.

However, the activation energy of the reaction is very large hence the

reaction does not occur at room temperature.

(iii) CaO(s) + CO2(g) CaCO3(s)

At Ts, G = 0 H - TsS = 0

-178 x 103 - Ts (-164) = 0

Ts = 1085 K

22