BENGKELBENGKEL

KECEMERLANGAN 2015 KECEMERLANGAN 2015

STPM CHEMISTRY STPM CHEMISTRY

STPM 2013

Semester 1962 / 1

Chemistry Paper 1

1 Tetraethyllead(IV), (CH3CH2)4Pb, is added to petroleum as an anti-

knock agent. (CH3CH2)4Pb is prepared from ethane, CH3CH3,

according to the following reactions:

CH3CH3 + C12 � CH3CH2Cl + HCl

4 CH3CH2C1 + 4NaPb � (CH3CH2)4Pb + 3 Pb + 4 NaCl

Which statement about the reactions is not true?

A One mole of Cl2 produces one mole of CH3CH2Cl.

B Four moles of Cl2 produce one mole of (CH3CH2)4Pb.

C One mole of HCl is formed for each mole of (CH3CH2)4Pb C One mole of HCl is formed for each mole of (CH3CH2)4Pb

produced.

D Four moles of CH3CH3 is required for each mole of (CH3CH2)4Pb

produced.

Since 1 mol of HCl = 1 mol of CH3CH2Cl ; and 4 mol CH3CH2Cl @

4 mol of HCl 1 mol of Pb(CH3CH2)4.

2 Gas X is collected in a 250 cm3 volumetric flask at 25°C until the

pressure of the gas is 73.5 kPa. The mass of X in the flask is 0.118

g at 250C. What is the mass of 1 mol of X?

[The gas constant, R, is 8.31 J K-1 mol-1.]

A 7.42 g B 11.3 g C 14.6 g D 15.9 g

3 An atom M has seven valence electrons and forms a stable M2+ ion

PV = (m/Mr) RT ;

Mr = (0.118)(8.31)(25+273) / (73.5 x 103)(250 x 10-6)

Mr = 15.9

3 An atom M has seven valence electrons and forms a stable M2+ ion

in an aqueous solution. What is electronic configuration of atom M?

A 1s22s22p63s23p5 B 1s22s22p63s23p63d6

C 1s22s22p63s23p63d54s2 D 1s22s22p63s23p63d14s2

Since M has 7 valence electrons and form +2 oxidation

state, M is probably a d-block element with valence

electrons of 3d54s2

4 Which compound has an ionic bond?

A HCl B BeCl2 C AlBr3 D C6H5NH3Br

5 The Lewis structures of molecules HF, H2O and NH3 are shown

respectively below. Which statement about the molecules is true?

A HF and H2O molecules are planar.2

B H2O molecule is a stronger Lewis base than NH3 molecule.

C The boiling points of the molecules increase in the order of H2O,

HF, NH3.

D The hydrogen bonds formed by the molecules decrease in the

order of HF, H2O, NH3.

The strength of hydrogen bond increased from N-H < O-H < F-H,

since the electronegativity increased from N, O, F. Therefore,

hydrogen bond increased accordingly

6 The atomic number of several elements represented by letters which are

not the usual symbols for the element concerned in the Periodic Table is

shown in the table below.

Element W X Y Z

Atomic number 3 11 13 26

Which element has the strongest metallic bond?

A W B X C Y D Z

Since Z is a transition metal, so metallic bond is the strongest

( as they can delocalised more electrons)

7. The graph of pV / RT against p for one mole of gases L and M at 298 K is

shown below.

Which statement is true of gases L and M?

A The compressibility of gas M is more than that of gas L at pressures lower

than x atm.than x atm.

B The repulsive forces between L molecules exist at pressures greater than x

atm.

C The collisions between L molecules are elastic at pressures lower than 1.0

atm.

D The attractive forces between M molecules are significant

Since gas M is lighter, therefore harder to compress (A wrong).

Furthermore repulsive forces between L exist at pressure greater than x

atm, as it give a positive deviation (B right). Elastic collision will only

occur for ideal gas (C wrong). M only have repulsion forces (D wrong)

8 Fullerene, C60, is one of carbon allotropes found in soot. Which statement

about fullerene is true?

A Its carbon atom has sp3 hybrid orbitals.

B Its molecule has delocalised electrons.

C Its bond angle is smaller than that of graphite.

D Its density is higher than that of diamond.

Each carbon in fullerene is sp2 hybridised (A wrong), which is the same with

graphite (hence angle is the same [C wrong]), hence can delocalised electrons

and conduct electricity (B right). Diamond, due to its sp3 hybridised, has very

high density (D wrong)

9 The reaction of acidified aqueous potassium iodide with aqueous

hydrogen peroxide is as follows:

2 I– (aq) + H2O2 (aq) + 2 H+ (aq) → I2 (aq) + 2 H2O (1)

The following steps may be involved in the mechanism of the reaction.

H2O2 + I– H2O + OI– OI– + H+ HOI

HOI + H+ + I– I2 + H2O

What could be the unit of the rate constant, k?

A s–1 B dm3 mol–1 s–1 C dm6 mol–2 s–1 D mol dm-3 s–1

Since slow step is rate determining step, therefore overall order is 2nd

order of reaction. Therefore, unit of rate constant is dm3 mol-1 s-1.

10 Which statement about the activation energy of an endothermic

reaction is true?

A The activation energy of the reaction varies with temperature.

B The rate of the reaction is higher if the activation energy of the

reaction is larger.

C The presence of a catalyst has no effect on the activation energy of

the reaction.

D The activation energy for the forward reaction is larger than that of

the backward reaction.the backward reaction.

For endothermic reaction, activation energy of forward reaction is

higher than that of activation energy of backward reaction.

11 The pH of blood of a healthy person is in the range of 7.35 to

7.45. Which statement explains how the human blood system

maintains its pH within that range if there is an excess of H+ ions?

A They are eliminated by sweating.

B They are neutralised by H2CO3 in the blood.

C They are neutralised by OH– ions in the blood.

D They are neutralised by HCO3– ions in blood.

Blood can act as buffer according to equation

H2CO3 + H2O ↔ HCO3– + H3O

+

Therefore, if H+ is in excess, it will shift equilibrium to left, hence

neutralise HCO3– in the blood.

12 The reforming of methane is the principal commercial source of

hydrogen gas. The equation for the endothermic reaction

involved is as follows:

CH4 (g) + H2O (g) ↔ CO (g) + 3 H2 (g)

At equilibrium, the quantity of hydrogen gas formed can be

increased by

A adding a suitable catalyst

B increasing the temperature

C adding a dehydrating agent

D reducing the volume of the vessel

To caused equilibrium shift to right, it can be done by increasing

temperature, as forward reaction is endothermic, therefore

equilibrium shift to right.

13 The liquid-vapour phase diagram of a mixture of liquids X and 7 is

shown below. What could be deduced from the phase diagram?

A Liquid X could be ethanol and liquid Y could be methanol.

B The attraction between X — Y is stronger than X—X and Y — YB The attraction between X — Y is stronger than X—X and Y — Y

C The liquid mixture of composition Q can be separated into liquids X

and Y by fractional distillation.

D When the liquid mixture of composition P is fractionally distilled, the

distillate is the liquid mixture of composition Q and the residue is

liquid XSince this deviation is a positive deviation, therefore X-Y is weaker

than X-X and Y-Y, and X and Y cannot be separated using fractional

distillation, since the first distillate will be azeotropic mixture, leaving X

as residue

14The buffer capacity of a solution is a measure of the amount of acid

or base that the solution can absorb without a significant change in

pH. What are the concentrations of NH3 (aq) and NH4+ (aq) that will

produce the maximum buffer capacity?

[Base dissociation constant, Kb, of NH3 is 1.8 x 10-5 mol dm-3.]

A B C D

[NH3] / mol dm-3 0.1 0.5 1.0 2.0

[NH4+] / mol dm-3 0.2 0.5 2.0 1.0

To form a basic buffer solution with maximum buffer capacity,

concentration of base and and its salt must have equal

concentration, as

where pOH = pKb

)salt(]acidconjugate[

]baseweak[lgpKpOH b −=

15 The solubility product, Ksp, for silver carbonate is 6.30 x 10-12

mol3 dm-9 at 298 K. What is the concentration of silver ions, in

mol dm-3, for a saturated solution of silver carbonate at that

temperature?

A 1.77 x 10-6 B 2.51 x 10-6

C 1.16 x 10-4 D 2.32 x 10-4

When Ag2CO3 � 2 Ag+ + CO32–When Ag2CO3 � 2 Ag+ + CO32–

x � 2x x

Ksp = [Ag+]2 [CO32–] 6.30 x 10–12 = (2x)2 (x)

x = 1.16 x 10–4 mol dm-3

Since [Ag+] = 2 x @ 2 x 1.16 x 10–4 mol dm-3

= 2.32 x 10–4 mol dm-3