ANSWERS for ENRICHMENT QUESTIONS Sem 1

CONTENT Page

UJIAN PERTENGAHAN SEMESTER (UPS) 2
UPS 2011/2012 5
UPS 2012/2013 8
UPS 2013/2014 11
UPS 2014/2015 13
UPS 2015/2016
17
PEPERIKSAAN SEMESTER PROGRAM MATRIKULASI (PSPM) 28
PSPM 2012/2013 40
PSPM 2013/2014 52
PSPM 2014/2015 66
PSPM 2015/2016 80
PSPM 2016/2017 95
PSPM 2017/2018
PSPM 2018/2019 106

OBJECTIVE QUESTION
OBJECTIVE (SET 1, SET 2, SET 3)

PREPARED BY,
AZURAH BINTI DAPOR
SHARIFAH NURLIYANA BINTI SYED NASIR

Suggested Answer
Ujian Pertengahan Semester

UPS

1. a) i. SUGGESTED ANSWER UPS 1
ii. SESI 2011/2012

No. of proton = 27
No. of neutron = nucleon number – proton number

= 52 – 27
= 25

∑
∑

u

= 52.11
b)

c) i. (required)
ii Mole O2 available (0.503 mol) > mole O2 required (0.443 mol)
therefore, O2 is excess reactant ad NO is limiting reactant

Mass of NO2 produced = 0.886 x [14+2(16)]
= 40.8 g

2

2. a) i. From n=4 to n=2

ii. -Electron at ground state absorbs energy and excites to higher energy level,
excited state, n=4.

- Electron at excited state is unstable.
- Electrons falls back to lower energy level (n=2) by releasing energy in the form

of light/photon with specific wavelength.

iii.

b) i. 9 orbitals
ii. 1s2 2s2 2p6 3s2 3p1

iii. x

y

x z x
y 3s y

or x z
y 3py
z
or
3pz
z

3px

3. a) Proton number of Q = 11
b) T
c) P<S<R<Q<T

d)  P is smallest since it has only 2 shells.
e)  S, R and Q are in the same shell but S has higher proton number/higher
f)
effective nuclear charge than R and Q.
R3-, S2-, T+
R has half-filled orbital which is more stable than partially-filled orbital in S.
Oxides of P, R and S are acidic.

3

4. a) i.

[Al ]3+ 3[ F ]- Cl
Al Cl

Cl

ii Cl Cl Cl

Al Al

Cl
Cl Cl

b) BeH2 is an incomplete octet molecule due to hydrogen that forms duplet/formal
charge zero on Be.
SF6 is an expanded octet molecule due to the existence of empty d orbitals.

c) O-- 2- 0 2-
S2+ O- -
- O- O

- O S O-

O- - O

0

Structure A Structure B

The most plausible structure is B because it bears lower formal charge on each
atom.

4

SUGGESTED ANSWER UPS 1
SESI 2012/2013

1 a) i. Proton number is the number of protons in the nucleus of an atom.

ii.

b) Assume that,mass of solution = 100 g
mass of solute = 85 g

mass of solvent = 15 g

n of H3PO4 = 85 g
= 98 g mol-1

0.867 mol

Molality = n of solute (mol)

mass of solvent (kg)

= 0.867 mol

0.015 kg

= 57.8 m

c) i. n of C6H12 = 45.0 g
= 84 g mol-1

0.536 mol

From equation,  2 mol H2C6H8O4
2 mol C6H12  0.536 mol H2C6H8O4
0.536 mol C6H12

Mass of H2C6H8O4 = 0.536 mol x 146.0 g mol-1
= 78.3 g

ii. Percentage yield = actual yield x 100
= percentage yield
= 63.5 x 100
78.3
81.1%

5

2 a) i. Infrared region

ii. Line A

() ()

() ()
= 1.097 x 107 (0.0469)
= 1.94 x 10-6 m

b) i. E : 1s2 2s2 2p6 3s2 3p6 4s2 3d1

ii.
z

y or any one of 3d orbitals

x 3dz2
4s

iii. (n, l, m, s) = (4, 0, 0, +½) and (4, 0, 0, -½)

3 a) i. Group 1 Period 3 Block s
ii. P and R
iii. Q
iv. R < S < P < Q

The first ionization energy of P & Q> R&S
Because The number of shell of P & Q< R&S
The shielding effect of P & Q< R&S
The attraction between nucleus and valence electron of P & Q> R&S
The atomic radius of P & Q< R&S

First ionization energy P<Q & First ionization energy R<S
Because effective nuclear charge of P>Q and effective nuclear charge of R>S
The attraction between nucleus and valence electron of P > Q and R>S
The atomic radius of P < Q and R <S

v. Q

6

b) i. valence electron = 4

Drastic increase in ionization energy to remove the fifth electron /
The fifth electron located at inner shell / The ratio IE5 to IE4 is the highest /
The fourth electron is removed from outermost shell

ii. Group 14

iii. 1s2 2s2 2p2

4 a) i. BeCl2 : covalent bond
MgCl2 : ionic bond

ii.

xx xx 2+ xx -

x Cl Be Cl x Mg 2 x Cl x
x xx x x x

xx xx

b) i. Structure I Structure II

xx xx

x N N O x x N N O x
x x x x

xx xx

ii. (-2) (+1) (+1) (0) (+1) (-1)

iii. Structure II.
because the -1 formal charge is on the more electronegative atom, oxygen.

7

1. a) SUGGESTED ANSWER UPS 1
b) SESI 2013/2014

i. Isotope is two or more atoms of the same element that have the same
proton number but different nucleon number.

ii. )
( )(

u

Relative atomic mass  192 u
1
2 x 12 u

 192

()

c) i. 3V2O5(s) + 10Al(s) 6V(s) + 5Al2O3(s)
ii 3V2O5(s) + 10Al(s) 6V(s) + 5Al2O3(s)

8

2. a) i. Paschen series, nf = 3 )
()
ii.
b) i. ( ) ( )(

ii. ∴ The excited state is n = 5
iii. Infrared region.
X: 1s22s22p63s23p64s23d104p1
X3+
1s22s22p63s23p63d10

z

xy z z

4s
z

x yx y xy
4pz
4px 4py

* Choose any one from 4p orbital.

3. a) i. P: Period 3, Group 14
ii. S
iii. R
iv. RS

9

b) i. Y: Group 14, Block P
ii. It is because:
 Drastic increase occur between ionization energy 4 and 5.
 5th electron come from the inner shell.
 Y has 4 valence electron.
Valence electronic configuration: ns2np2

Y: 1s22s22p63s23p2

4. a) i. - - -

O IO O I+ O O IO
O O O

ii -

iii O IO
b) O

This structure has the lowest formal charge on each atom.

IO3- is an expanded octet ion.
Due to existence of empty d orbitals.

ionic bond

+
H

HN H -
Cl

H

dative bond covalent bond

10

SUGGESTED ANSWER UPS 1
SESI 2014/2015

1. a) i. Isotope No. of protons No. of neutrons
107Ag 47 60
109Ag 47 62

ii. Average atomic mass = isotopic mass  abundance
 abundance

= 106.91 51.50  108.90 48.50  107.88 amu
51.50  48.50

b) i. n HCl = 0.08 mol n Mg = 0.06 mol

2 mol HCl ≡ 1 mol Mg
0.08 mol HCl ≡ 0.04 mol Mg

Moles of Mg available (0.06 mol) > moles of Mg needed (0.04 mol)
∴ Mg is excess reactant and HCl is limiting reactant.

ii. 2 mol HCl ≡ 1 mol H2
0.08 mol HCl ≡ 0.04 mol H2

V of H2 at STP = 0.04 mol × 22.4 L mol−1 = 0.896 L

% yield = Actual yield 100% = 672 103 100% = 75%
Theoretical yield 0.896

E   1  1   2.18 1018  1  1  = 2.18 × 10-18 J atom−1
RH    12 2 
2. a) n 2 n 2 
 i f

IE = (6.02 × 1023)( 2.18 × 10-18) = 1312 kJ mol−1

b) i. Heisenberg’s uncertainty principle states that it is impossible to know
simultaneously both the position and momentum of an electron.

ii. Electron moves in circular orbit around the nucleus.

c) i. It does not obey Aufbau Principle;
Electron must be filled into orbital with lower energy (2s) before start filling into
orbital with higher energy (2p).

ii. It does not obey Pauli exclusion principle;
No two electrons can have same set of quantum numbers

11

3. a) i. Group = 16, Period = 3, Block = p

ii. RO3
iii. Acidic
iv. RO3  H2O  H2RO4

b) - Q has higher ionization energy than R.
- because the first electron of Q is removed from more stable half-filled 3p
orbital whereas R is removed from less stable partially filled 3p orbital due to
electrostatic repulsion.

c) - Effective nuclear charge of R is greater than Q.
- Attraction of nucleus towards valence electrons of R is stronger than Q.

4. a) i.

H H+
HNH
H N + +
H
H

H

ii. - - -
b) i. NNN NNN
NNN
AB C

The most stable structure is A.

F F F
FN F FB F F Cl F

ii. BF3 : incomplete octet
ClF3 : expanded octet

12

SUGGESTED ANSWER UPS 1
SESI 2015/2016

1. a) i. Given:

Electrons = 18
Neutrons = 24

Proton number = 18+3 = 21
Nucleon number = 24+21 = 45

ii. Isotope notation,
Y45

21

b) i. Given:

V of Pb(NO3)2 solution needed = 250 mL
[Pb(NO3)2] = 0.25 mol L−1

Mol of Pb(NO3)2 = 0.25 mol L−1×0.25 L
= 0.0625 mol

∴ Mass of Pb(NO3)2 = 0.0625 mol x 331.2 g mol−1
= 20.7 g

ii. Two main steps to prepare solution
 20.7 g of Pb(NO3)2 is dissolved with a small amount of distilled water in a
beaker and transferred into 250 mL volumetric flask.
 Distilled water is added into the flask to the calibrated mark.

c) species that undergo reduction,
MnO4−, MnO2

Explanation: The oxidation number of Mn decrease from +7 to +4.

2. a) i. Given: Paschen series, nf = 3
 = 1094 nm

Region of electromagnetic spectrum: Infrared region

13

ii Energy level of excited state, ni

1  RH  1  1   n2
  n12 n22  , n1
 

1  1 1 
 32 ni2 
   1094109 m  
 1.097 107 m1 

ni = 6

Transition of electron : ni = 6 to nf = 3

b) i. Definition orbital,
Orbital is three-dimensional region around nucleus with high probability of
finding electrons.

ii. Valence electronic configuration for arsenic,
4s24p3

iii. Set of quantum numbers for one valence electron in s orbital
n=4 l=0 m=0 s=+1/2

c) shape of d-orbital that lie on the axes,

z

3. a) i. xy

3dx2  y2

Definition second ionization energy.
The minimum amount of energy required to remove one mole of electrons
from one mole of unipositive ion in the gaseous state.

ii. Electronic configuration of element Z & Group 14
1s22s22p63s23p2

iii. Size of Ca2+ < K+ < Cl- < S2-

 They are isoelectronic species.
 Proton number of Ca2+ > K+ > Cl- > S2-.
 Nuclear charge of Ca2+ > K+ > Cl- > S2-.
 Nucleus attraction towards outer electrons in Ca2+ > K+ > Cl- > S2-

14

4. a) i. Definition of electrovalent bond (ionic bond),
Electrostatic attraction between positively and negatively charge ions.

ii. formation of AlF3, F 3+ -
F Al 3 F
Al +

F

Iii. Type of stability : Noble gas configuration.

b) i. Resonance structure of NCO−

- - -
OCN
OCN OCN

ii. most plausible structure,

-
OCN
-1 0 0

Reason:
Negative formal charge resides on O atom which is more electronegative than
N atom.

15

Suggested Answer

Peperiksaan Semester
Program Matrikulasi

PSPM

SUGGESTED ANSWER PSPM 1
SESI 2012/2013

1. a) TABLE 4 shows the atomic structure of four particles, represented by the

letter A to D. The particles are atoms and ions.

TABLE 4

Particle Electrons Protons Neutrons

A6 66

B 12 12 12

C 10 12 12

D6 68

Use the letter A to D to answer the following questions.
i. Which two particles are atom and ion of the same element? Explain.

B and C
Same number of protons and neutrons but different number of
electrons.

ii. Which two particles are isotopes of the same element? Explain.

A and D
Same number of electrons and protons but different number of neutrons
or nucleon.

b) Compound E is a hydrocarbon, CxHy. When 6.84 g of this compound is burnt

completely in pure oxygen, 21.5 g of CO2 and 8.87 g of H2O are obtained. Determine

the empirical formula of compound E. If the molar mass of the compound is 128.2

gmol-1, what is its molecular formula? [6 m]

Elements C H
Mass(g) 5.86 0.98
Number of mole 0.4887 mol 0.98 mol
Mole ratio
1 2

Empirical formula = CH2 17

[CH2]n = 128
n=9

Molecular formula = C9H18

b) The reaction between the solutions of permanganate ion, MnO4-, and oxalate ion,
C2O42-, in acidic medium is shown below:

MnO4-(aq) + C2O42-(aq) MnO2(s) + CO2(g)

Write:
i. the half-reactions for the reduction and oxidation reactions respectively.

Reduction:

MnO4- + 4H+ + 3e MnO2 + 2H2O

Oxidation:

C2O42- 2CO2 + 2e

ii. the balanced redox equation for the reaction.

2 x [MnO4- + 4H+ + 3e MnO2 + 2H2O]

3 x [C2O42- 2CO2 + 2e]

2MnO4- + 3C2O42- + 8H+ 2MnO2 + 6CO2 + 4H2O

2. a) For each of the molecules given; PCl3, PCl5 and POCl3,
i. draw a Lewis structure and name the shape based on molecular geometry.

Lew is Structure: Lew is Structure: Lew is Structure:
Cl
Cl Cl
Cl P Cl Cl Cl P Cl
Molecular Geometry: O
P Cl
Trigonal pyramidal Cl Cl Molecular Geometry:

Molecular Geometry: tetrahedral

trigonal bipyramidal

ii. deduce the polarity and predict the Cl-P-Cl bond angle of each molecules.
PCl3: less than 109.5o, polar
PCl5: 120o, non-polar
POCl3: 109.5o, polar

18

b) Show the formation of a dative bond in the reaction of;

AlCl3 + Cl- AlCl4-

Cl - Cl -
+Al Cl Cl Al Cl dative bond
Cl
Cl
Cl

3. a) In the offshore base oil drilling operation, methane gas, CH4, was found at the
bottom of the sea at 480oC and a pressure of 12.8 atm.

i. calculate the volume of the gas needed to be transferred to a container in the
refinery mill with a volume of 1.80 x 103 L at a pressure of 1.0 atm and
temperature of 22oC.

( )( )

= 359 L

ii. determine the mass of the gas being transferred.

( )( )( )
( )( )

b) i. define vapour pressure and boiling point.

Vapour pressure:
Pressure exerted by vapour molecules to it liquid at equilibrium in a
sealed container.

Boiling Point:
Temperature at which the vapour pressure of the liquid is equal to the
external atmospheric pressure.

19

ii. based on TABLE 5, arrange the compounds in order of increasing boiling
point. Explain your answer.
TABLE 5
Compound
Propane, CH3CH2CH3
Ethanol, CH3CH2OH
Dimethyl ether, CH3OCH3
Ethanoic acid, CH3COOH

CH3CH2CH3 < CH3OCH3 < CH3CH2OH < CH3COOH
 Strength of intermolecular forces directly proportional to boiling
point.
 Propane is non-polar molecule so it can form weak London
Dispersion forces between its molecules.
 Dimethyl ether is polar molecule, so it can form dipole-dipole
forces between its molecule.
 Dipole-dipole forces stronger than London dispersion forces.
 Both ethanol and ethanoic acid can form hydrogen bond.
 Hydrogen bond stronger than Van der Waals forces.
 Ethanoic acid can form more hydrogen bond compare to ethanol.

4. a) What is meant by end point and equivalence point?
End point:
pH in titration at which the indicator change colour.

Equivalence point:
The pH at which the no. of mole of H+ is stoichiometrically equivalence to the no.
of mole of OH-.

b) A titration of 25.0 mL solution of 0.15 M HNO3 requires 37.5 mL of NaOH to reach the
equivalence point.
i. calculate the pH of the solution before the addition of NaOH.

HNO3(aq) H+(aq) + NO3-(aq)
[ ]i 0.15 --
[ ]f -
0.15 0.15

pH = - log [H+]
= - log (0.15)
= 0.8239

20

iii. calculate the concentration of NaOH solution needed to neutralize the HNO3
solution.

( )( ) ()

c) A solution contains 0.175 M manganese(II) ions and 0.182 M lead(II) ions. The Ksp
for manganese(II) sulphide is 1.4 x 10-15 and for lead(II) sulphide is 8.0 x 10-28.

i. Write the solubility product expression for both sulphides.

[ ][ ] [ ][ ]

ii. Calculate the maximum concentration of sulphide ions present in the solution
without manganese ions being precipitated.

[ ][ ]

PbS(s) Pb2+(aq) + S2-(aq)
ss

[ ][ ]

[ ][ ]
()
M

5. a) Describe the formation of emission spectrum for hydrogen atom. Show and
label the first three series of electron transitions between energy levels.
An electron in a hydrogen atom is transferred from n=5 to n=3. Calculate the energy
of the photon emitted and the wavelength of the spectral line produced.

When energy supplied, the electron at the ground state (n=1) absorbed energy. It
will excite from lower energy level to a higher energy level. Electron at the
excited state is unstable. It will fall back to lower energy level, n=1 and release
specific amount of energy with specific wavelength. Hence the line spectrum
formed.

21

energy

Paschen n=7
n=6
Balmer n=5
n=4
n=3

n=2

Lyman n=1

()
()

Energy emitted is

( )( )
()

b) An element Q has proton number of 8. Draw the orbital diagram of the element.
Explain the two rules applied in arranging the electrons in the orbitals.

8Q: 2p
1s 2s

Aufbau principle:
Electrons are filled into the orbitals according to increase energy level.

Hund’s rule:
The 4 electrons in 2p orbital are spread out with parallel spin before pairing up.

22

d) Given four elements W, X, Y and Z with their proton numbers of 11, 12, 13 and 18
respectively. Identify the group and period of the elements in the Periodic Table.
Arrange the elements according to increasing atomic radii and explain the trend.

Elements Electronic configuration Group Period
W 1s2 2s2 2p6 3s1 1 3
X 1s2 2s2 2p6 3s2 2 3
Y 13 3
Z 1s2 2s2 2p6 3s2 3p1 18 3
1s2 2s2 2p6 3s2 3p6

Increasing order of atomic radii: Z < Y < X < W

W, X, Y, Z are elements located in the same period.
Across period, the proton no. increase from W to Z causes the effective nuclear
charge increases from W to Z, the nucleus attraction towards the valence
electrons increase from W to Z. Therefore, the atomic radii increase from left to
the right across the same period.

6. a) By using Lewis dot symbol, show the formation of the compounds, BCl3 and
LiCl, from their respective atoms. Identify the types of bond in each compound.
Explain the hybridization state of the central atom in the BCl3 molecule. State its
molecular geometry.

B+ Cl Cl
Cl B Cl

Cl

Cl

+Li Cl Li+ -
Cl

Type of bond:
BCl3 : Covalent Bond
LiCl : Ionic bond

BCl3 : There are 3 bonding pairs / 3 σ bonds. Thus, type of hybridization is
sp2 23

B (ground state) : 2p
2s

B (excited state) : 2p
2s

B (hybrid state) :
sp2 2p

Molecular geometry : Trigonal Planar

3p

Cl 

sp2 

sp2 3p
sp2 B Cl



Cl
3p

b) Formic acid is widely used in the preservation of biological specimens. Its chemical
structure is shown below:

O

H C OH

Draw and describe the formation of bonds by involving the overlapping of orbitals.
What is the possible intermolecular forces that exist between these acid molecules?

Type of hybrid: O1 O2: sp3
C: sp2 H COH

2

O1: sp2

C(ground state): 2p O1(ground state): 2p O2(ground state): 2p
2s 2s 2s

C(excited state): 2p O1(excited state): 2p O2(excited state): 2p
2s 2s 2s

C(hybrid state): p O1(hybrid state): O2(hybrid state):
sp2
sp2 p sp3

sp2 

O1 

sp2
sp2

sp2  sp3

sp2 sp3 sp3 sH

C O2 sp3 

sp2



s
H

24

Intermolecular Forces: Hydrogen bond.
7. a) Carbon dioxide, CO2, has a triple triple point at 5.2 atm and -57oC and a critical

point at 72.9 atm and 31oC. Sketch a phase diagram of CO2.

Pressure (atm)

72.9 B C
Solid
liquid
5.2 T

A gas
Temperature (OC)

-57 31

Based on the diagram;
i. explain liquid CO2 can be obtained

Liquid CO2 can be obtained only at pressure higher than 5.2 atm.

ii. the formation of solid CO2 (dried ice) at room condition.
At room condition (1 atm, 28oC), CO2 exist in gas form. Impossible to
form solid at room conditions. Triple point below room temperature and
above 1 atm. To form solid, temperature must be reduced to below 57oC
and pressure must be increased to above 5.2 atm.

b) decomposition of ammonium hydrogen sulphide, NH4HS, on heating in a sealed flask
is an endothermic process.

NH4HS(s) H2S(g) + NH3(g)

If Kp for this reaction is 0.11 at 25oC, determine the partial pressure of NH3
and total pressure in the flask (atm) at equilibrium.
Based on the Kp value, calculate Kc for this reaction at the same temperature.

NH4HS(s) H2S(g) + NH3(g)
n(eq) xx

( )( )

25

Using Le Chatelier’s principle, explain how increasing the temperature would affect
the equilibrium.

When temperature increase, the equilibrium position will shift forward to absorb
more heat in order to overcome the high temperature.

What will happen to the pressure of NH3 if some H2S is removed from the flask?

When some of H2S is removed, equilibrium position will shift forward to produce
more H2S in order to increase the concentration of H2S and pressure of H2S will
increase.

8. a) Define a buffer solution.
Buffer solution is a mixture of salt and weak acid/weak base which can maintain
the pH of the solution when small amount of strong acid or strong base is added
to the solution.

b) Explain how the pH of a buffer solution made from a weak acid and its conjugate
base changes when;
i. the acid dissociation constant, Ka, of the weak acid increases.
When Ka increase, pH will decrease.

ii. the acid concentration is decreased relative to the concentration of its
conjugate base.
The ratio [Conj. Base]/[acid] will increase. The log term increase and pH
should increase.

c) calculate the pH of a buffer solution prepared by adding 5.0 g of sodium acetate,
CH3COONa, to 250 mL of 0.20 M acetic acid, CH3COOH. What is the pH of the
buffer solution after addition of 0.01 M sodium hydroxide, NaOH, solution? [Ka acetic
acid = 1.8 x 10-5]

[ ]

[CH3COOH] = 0.20 M ]
]
[
[

()
()

26

OH- + CH3COOH CH3COO- + H2O
0.2439
[ ]i 0.01 0.20 0.2539

[ ]f 0 0.19

[]
[]

()
()

27

SUGGESTED ANSWER PSPM 1
SESI 2013/2014

1. (a) Vinyl chloride, C2H3Cl, is prepared by the reaction of ethyne gas, C2H2, with
hydrochloric acid, HCl. In preparation of vinyl chloride, 70.0 g of C2H2 is reacted with
102.0 g of HCl.
i. Write a balanced chemical equation for the reaction.

C2H2 (g)  HCl(g)  C2H3Cl(g)
ii. Identify the limiting reagent in the reaction.

n C2H2 = 2.692 mol
n HCl = 2.794 mol

1 mol C2H2 ≡ 1 mol HCl
2.692 mol C2H2 ≡ 2.692 mol HCl

Moles of HCl available (2.794 mol) > moles of HCl needed (2.692 mol)
∴HCl is excess reactant and C2H2 is limiting reactant

iii. Calculate the mass of the product.
1 mol C2H2 ≡ 1 mol C2H3Cl
2.692 mol C2H2 ≡ 2.692 mol C2H3Cl

n C2H3Cl = 2.692 mol
mass C2H3Cl = 2.692 x (12x2+ 3+35.5) = 168.25 g

iv. Calculate the mass of the excess reactant.

n HCl remained = 2.794- 2.692 = 0.102 mol

mass HCl remained =0.102 x 36.5 = 3.72 g

(b) Magnesium is an element with a proton number of 12. In a sample of magnesium, it
was found that magnesium atoms have three different nucleon numbers which are
24, 25 and 26.
i. write the symbol for one of these three atoms, showing its proton number
and nucleon number.

24 Mg
12

28

ii. A sample of magnesium contains 78.6% of atoms with a nucleon number of
24, 10.1% of atoms with a nucleon number of 25 and 11.3% of atoms with a
nucleon number of 26. Calculate the relative atomic mass of magnesium.

Average atomic mass = Isotopic mass  % abundance

% abundance

  24  78.6  2510.1  2611.3 = 24.3u
78.6 10.111.3

relative atomic mass  24.3u
1
12 x 12.00 u

 24.3

iii. Sketch and label the expected mass spectrum of the magnesium.

% abundance

78.6

10.1 11.3

24 25 26 mass

2. (a) Ammonia, NH3, is a colourless gas with a characteristic pungent smell.
i. Using a Lewis dot symbols show the formation of ammonia from the
respective elements.

N + H+ H + H H NH

H

ii. Predict the hybridization of the central atom of ammonia and draw a diagram
of the molecule showing the overlapping of the orbitals.

H

HN

H

 NH3 has 4 electron pair which are 1 lone pair and 3 bonding pair
around central atom.

 Electron pair arrangement : Tetrahedral
 Molecular geometry: Trigonal pyramidal
 Type of hybridisation: sp3

29

Valence orbital diagram: H: sp 3
N(ground state): H 1s
N
2s 2p sp 3

N(excited state):  3 3 
2s s
H sp sp s
H
2p

N(hybrid state):

sp3 s
H

iii. Describe the polarity of the ammonia molecule.

 N is more electronegative than H.

 Bond dipole cannot cancel each other.
 ≠ 0
 Polar molecule

(b) Use the valence shell electron repulsion (VSEPR) theory to explain the difference in

the bond angles between the following pairs of compound/ion.
i. CO2 and CO32-

2-
O

OC O OC

O

CO2 Has 2 electron pair around central atom and all are bonding pair .
 Electron pair arrangement: linear.
 Based on VSEPR theory, the electron pairs are located as far as
 possible in order to minimize the repulsion.
The strength of repulsion between bonding pair-bonding pair
 electrons are equals.

 Molecular geometry: Linear.
 Bond angle: 180o.

CO32-
 Has 3 electron pair around central atom and all are bonding pairs.
 Electron pair arrangement: trigonal planar.
 Based on VSEPR theory, the electron pairs are located as far as
possible in order to minimize the repulsion.

 The strength of repulsion between bonding pair-bonding pair
electrons are equals.

 Molecular geometry: trigonal planar.
 Bond angle: 120o.

30

ii. PCl3 and H2O

Cl H
Cl P OH

Cl

PCl3
 Has 4 electron pair around central atom which are 3 bonding pairs

and 1 lone pair .

 Electron pair arrangement: tetrahedral.

 Based on VSEPR theory, the electron pairs are located as far as

possible in order to minimize the repulsion.

 The strength of repulsion between lone pair-bonding pair >

Bonding pair-bonding pair.

 Molecular geometry: trigonal pyramidal.
 Bond angle: < 109.5o.

H2O Has 4 electron pair around central atom which are 2 bonding pairs
 and 2 lone pairs .
Electron pair arrangement: tetrahedral.
 Based on VSEPR theory, the electron pairs are located as far as
 possible in order to minimize the repulsion.
The strength of repulsion between lone pair-lone pair>lone pair-
 bonding pair > Bonding pair-bonding pair.
Molecular geometry: bent.
 Bond angle: 104.5o.


3. (a) i. Define vapour pressure.

Pressure exerted by vapour molecule in equilibrium with its liquid state.

ii. Explain how temperature affects the vapour pressure of a liquid.

 Temperature increase.
 Average kinetic energy increase.
 More molecule can escape as vapour.
 Vapour pressure increase.

(b) i. State Dalton’s law of partial pressure.

Total pressure of a mixture of non-reacting gases is equal to the sum of
the pressures that each gas would exert if it were present alone.

31

ii. FIGURE 2 shows two connecting vessels containing different gases. When the

valve between the vessels is opened, the gases are allowed to mix. Ignoring

the volume taken by the valve, calculate the partial pressure of each gas and
the final pressure of the mixture at 25 oC.

FIGURE 2

PH2  nRT   0.50  0.08206  298.15  3.1atm
V 4

PCO2  nRT   0.25  0.08206  298.15  1.5 atm
V 4

PT = 3.1 + 1.5 = 4.6 atm

(c) Differentiate between a crystalline solid and an amorphous solid in terms of their
particle arrangement. Draw the structures and state the types of crystal for diamond
and sodium chloride.

Crystalline Solid Amorphous Solid
Have regular 3-D arrangement. Does not have regular 3-D
arrangement.

C

C
C

C
C

Giant Covalent Crystal

Cl Cl- -+ +
Na
Na

Cl Cl- -
+ +
Na Na

Ionic Crystal

32

4. (a) Define buffer solution.

A solution that maintain its pH when a small amount of a strong acid or a strong
base is added to it.

(b) Explain how the pH of a mixture of aqueous ammonium chloride and aqueous
ammonia remains constant when a small amount of strong base is added. Give
suitable equations in your explanation.

The mixture of aqueous ammonium chloride and ammonia form buffer solution.
When small amount of strong base is added, the OH- ion from the strong base is
neutralized by NH4+ of buffer to form NH3.

NH4 (aq)  OH  (aq)  NH3(aq)  H2O(l)

Thus the pH of the mixture is maintained.

(c) A buffer solution of pH 4.84 was prepared by dissolving a certain amount of sodium
ethanoate in 1.0 dm3 of 0.20 M ethanoic acid.

i. Calculate the mass of sodium ethanoate needed to prepare the above buffer

solution.

pH   log Ka  log CH3COO 

CH3COOH 

 4.84  log CH3COO 
1.8 105  log
0.2

[CH3COO-] = 0.252 M

n CH3COO- = n CH3COONa = 0.252 mol

mass CH3COONa = 0.252 x (24+32+23+3) = 20.66 g

ii. Determine the change in pH when 1.0 cm3 of 1.0 M hydrochloric acid is
added to 100 cm3 of the buffer solution.

[Ka for CH3COOH = 1.8 x 10-5M]

CH3COO (aq)  H  (aq)  CH3COOH (aq)

ni 0.252 1 x 10-3 0.2

nc -1 x 10-3 -1 x 10-3 +1 x 10-3

nf 0.251 0 0.201

[ ] 0.251 0.201

 pH  log 0.251
1.8 105  log 0.201 = 4.84

33

5. (a) Bohr used the information from a line spectrum of a hydrogen atom to
explain the electronic structure of a one-electron system. A blue line in the spectrum
of hydrogen atom was observed as a result of a transition of electron from the fourth
to the second shells of an atom.
What is meant by a line spectrum?

Line spectrum is a spectrum that consist of discontinuous and discrete lines with
specific wavelength and frequency.

Calculate the wavelength and energy for this blue line.

ni = 4 nf = 2

1  RH  1  1   n2
  n12 n22  , n1
 

 1.097 107  1  1 
 22 42 

λ = 4.86 x 10-7 m = 486 nm

  E  hc  6.631034 3108 = 4.09 x 10-19 J
  4.86107

Energy emitted = 4.09 x 10-19 J

State two of Bohr’s postulates.

 Electrons in an atom move in certain circular orbits.
 An electron moving in an does not radiate or absorb energy.
 The moving electron has a specific amount of energy and its energy is

quantized.
 Energy is emitted or absorbed only when the electron moves to another

energy level.
 Electron at its excited states is unstable. It will fall back to lower energy

level and emit a specific amount of energy in the form of light.
(any two answer)

34

(b) The first five ionization energy (kJ mol-1) of atoms X and Y in period 3 of the periodic

table are as follows:

TABLE 5

Atom/IE First Second Third Fourth Fifth

X 738 1450 7730 10500 13600

Y 578 1820 2750 11600 14800

Define ionization energy.
Ionization energy is minimum energy required to removed 1 mol electron from 1
mol of gaseous atom to form 1 mol of positively charge gaseous ion.

Explain why the first ionization energy of X is higher than that of Y.
IE1 of X > Y

 No. of proton in X > Y.
 Effective nuclear charge of X > Y.
 Nucleus attraction toward valence electrons in X > Y.
 Size of X < Y.
 Energy needed to remove the valence electrons from X > Y.

Predict the number of electron valence for X and Y and write their respective
electronic configuration. Write the set of four quantum numbers (n,l,m,s) for the
outermost electron(s).

X: Drastic increase from IE1 to IE2.
 The 2nd electron is removed from inner shell.

 There are 1 valence electron.
 1s2 2s2 2p6 3s1

(n: 3, l: 0, m: 0, s:+1/2)

Y: Drastic increase from IE2 to IE3.
 The 3rd electron is removed from inner shell.

 There are 2 valence electron.
 1s2 2s2 2p6 3s2

(n: 3, l: 0, m: 0, s:+1/2) & (n: 3, l: 0, m: 0, s:-1/2)

6. (a) CO2 and BeH2 are triatomic covalent molecules. Describe in detail the
formation of the covalent bonds in these molecules and explain why CO2 obeys the
octet rule while BeH2 does not.

CO2
C has 4 valence electrons, O has 6 valence electrons.
C will shared its valence electrons with O to form covalent bonds.

C +O +O OCO

35

BeH2
Be has 2 valence electrons, H has 1 valence electrons.
Be will shared its valence electrons with H to form covalent bonds.

Be + H + H H Be H

BeH2 has less than 8 electrons around central atom.
CO2 has 8 electrons around central atom.

(b) Formal charge is a useful guide in determining the best or preferred structure.
Explain this statement using [OCN]- ion as example.

00 0 0 -1 +1 0 -2
OCN OCN OCN

AB C

[OCN]- has 3 possible structures.

Among those 3 structures, structure A is the best or preferred structure because

negative charge on structure A lies on the more electronegative atom.

7. (a) State the relationships of the volume of an ideal gas with its pressure,
absolute temperature and the amount of gas respectively.

Based on ideal gas law,

 volume of fixed amount of gas is inversely proportional with its pressure
at constant temperature. (Boyle’s law)

 Volume of fixed amount of gas is directly proportional with its absolute
temperature at constant pressure. (Charles’ law)

 At fixed temperature and pressure, volume is directly proportional with
the amount of gas. (Avogadro’s law)

A student added an amount of hydrochloric acid to a rock sample and observed the
fizzing action indicating a gas being released. At 25 oC, the gas collected in a 0.220 L

gas bulb was 0.300 g and at a pressure of 0.757 atm. Calculate the molar mass and

the density of this gas.

Mr  mRT   0.300  0.08206  298.15  44.1 gmol1
PV  0.757   0.220

  m   0.300  1.36 gL1
V  0.220

36

(b) Sulphur trioxide, SO3, in a 1 L closed container was left to dissociate according to the
equation below:

2SO3(g) 2SO2(g) + O2(g) H = -197.8 kJ

Derive the relationship between Kc and Kp for the above reaction.

SO2 2 O2      Kp PSO22PO2
SO3 2
Kc  2

PSO3

PSO2  SO2  RT PO2  O2  RT PSO3  SO3  RT

Kp  SO2  RT 2 O2  RT   SO2 2 O2    RT 2  RT   Kc  RT 
SO3  RT 2 SO3 2  RT 2

Predict the equilibrium position of the reaction under the following separate
conditions;
temperature is increased,

When T increase, the equilibrium position will move backward in order to lower
temperature. Hence, decrease the value of K.

pressure is decreased,

When pressure is decreased, the equilibrium position will move forward towards
higher number of mole in order to increase the pressure.

and SO2 is removed.

When SO2 is removed, the equilibrium position will move forward in order to
recover the loss of SO2.

8. In a titration experiment at 25 oC, a total of 30 mL of 0.1 M NaOH was added
dropwise into a conical flask containing 25 mL of 0.1 M HCl and a few drops of an indicator.
Show the variation of pH of the solution before the addition of NaOH, at half equivalence
point, at equivalence point and at the final volume of the titration. Sketch a graph of pH
against the volume of NaOH.
[given: Kw at 25 oC = 1 x 10-14]

Before addition of NaOH,
At this point the solution only contain of HCl that dissociate completely in water.

HCl(aq)  H  (aq)  Cl (aq)

pH = - log [H3O+] = - log (0.1 M) = 1

37

At half equivalence point: 12.5 mL

HCl(aq)  NaOH (aq)  NaCl(aq)  H2O(l)

ni 2.5 x 10-3 1.25 x 10-3 - -

nc -1.25 x 10-3 -1.25 x 10-3 +1.25 x 10-3

nf 1.25 x 10-3 0 1.25 x 10-3

[ ] 0.0333

Since NaCl is neutral salt and NaOH is completely neutralized by HCl, the [H3O+] is
mainly come from the dissociation of acid.

pH = - log (0.067) = 1.48

At equivalence point
The solution only contain NaCl salt.
Both Na+ and Cl- ions does not hydrolyzed water.
The salt form is neutral salt.
pH = 7

Final volume
The solution only contain NaOH and NaCl.
Since NaCl is neutral salt. The [OH-] only come from dissociation of NaOH.

pH = 14 – (-log (0.1)) = 13

pH < 13.

pH
13
11

7

3 V NaOH, mL
1

25

38

What is acid-base indicator? By referring to TABLE 6 below, suggest the most suitable

indicator to be used in the titration above and give reasons.

TABLE 6

Indicator pH range

Phenophthalein 8.3-10.0

Cresol red 7.0-8.8

Bromothymol blue 6.0-7.6

Litmus 4.7-8.3

Acid-base indicator is a solution containing weak acid and weak base that can
change colour over range pH value.
Most suitable indicator: Bromothymol blue
 The pH range of indicator cover the pH of end point.
 The pH range start just before pH of end point.

39

SUGGESTED ANSWER PSPM 1
SESI 2014/2015

1 (a) Consider the following reaction.

2NO(g) + 2 CO(g) ⟶ N2(g) +2CO2(g)

If 300.00 g of CO has reacted, calculate the mass percentage, w/w, of CO2 in the
mixture of product.

ANSWER

No. of mol CO  300.00 g
28.0 g mol1

 10.714 mol

From equation, 2 mol CO  1 mol N2 x 1 mol N 2
10.714 mol CO  x mol N 2
x  10.714 mol CO
2 mol CO
 5.3571 mol N 2

From equation, 1 mol N 2  2 mol CO2

5.3571 mol N 2  x mol CO 2

x  5.3571 mol N 2 x 2 mol CO 2
1 mol N 2

 10.714 mol CO 2

Mass of CO2 = 10.714 mol X 44.0 g mol-1
= 471.42 g

Mass of N2 = 5.3571 mol x 28.0 g mol-1
= 150.00 g

%w w  mass of solute x 100
mass of solution

 471.42 g x 100

(471.42 150.00) g

 75.862%

40

(b) A 30.00 mL of 0.025 M sodium dichromate, Na2Cr2O7 solution is titrated with iron(II)

sulphate,Fe2SO4 solution in acidic condition, according to the following reaction:

Cr2O72- + Fe 2+ Cr3+ + Fe 3+

The titration requires 40.00 mL of Fe2SO4 solution to reach the end point.

(i) Balance the redox equation.

ANSWER

6e + 14H+ + Cr2O72- ⟶ 2Cr3+ + 7H2O
6 (Fe2+ ⟶ Fe3+ + e )

Cr2O72- + 6Fe2+ + 14H+ ⟶ 2Cr3+ +6Fe3+ +7H2O

(ii) Calculate the mass of Na2Cr2O7 needed to prepare a 0.025 M solution in a 50 mL
volumetric flask.

ANSWER

Molarity of Na 2Cr2O7  mol of solute

volume of solution L

mol Na 2Cr2O7  (0.025mol L1 )  50  L
 1000 

 1.3 x 103 mol

Mas s Na 2Cr2O7 needed  1.3 x103 mol x 262.0 g mol1
 0.33 g

(iii) Determine the molarity of the Fe2+ solution.

ANSWER

mol of Na 2Cr2O7  0.025 mol L1 x 30.00x103 L
 7.500 x 104 mol

From equation, 1 mol Cr2O7 2  6 mol Fe 2

7.500 x 104 mol Cr2O7 2  x mol

x  7.500 x104 mol Cr2O7 2 x 6mol Fe 2
1 mol Cr2O7 2

 4.500 x 103 mol

41

Molarity of Fe 2 solution  Mole of solute

Volume of solution (L)

 4.500 x 103 mol
40.00x 103 L

 0.1125 mol L1

2 (a) The formula of formic acid is HCO2H.One of the carbon-oxygen bond lengths in this
molecule is 1.36 Å while the other is 1.23 Å. Draw the Lewis structure of this
molecule and labels these bonds.

ANSWER

Formic acid: HCOOH

O

1.23 Å

HCOH

1.36 Å

(b)(i) Xenon can be covalently bonded to fluorine and oxygen to form xenon compounds,
XeF4 and XeO2F2. For both compounds,
(ii) determine the number of bonding electron pair(s) and lone electron pair(s) around
the central atom xenon.
(iii) state the molecular geometry.
(iv) determine the hybridisation of xenon atom.
predict their polarity.

ANSWER

F 3
F Xe F sp

F  P 3 
H sp
3
s sp

3
sp

sH

Hs 

No of bonding pairs XeF4 XeO2F2
electron 4 4
2 1
No.of lone pairs electron
Square planar Distorted tetrahedral @ see
Molecular shape saw

42

XeF4 ↑↓ ↑↓ ↑↓
5p
Xe: ↑↓
5s

Xe excited state: ↑↓ ↑↓ ↑↓ ↑↓ ↑↑
5s 5p 5d

Xe hybrid: ↑↓ ↑↓ ↑ ↑ ↑ ↑
Sp3d2

XeO2F2 ↑↓ ↑ ↑
5p
Xe2+ground state: ↑↓
5s

Xe2+ excited state: ↑↓ ↑ ↑↑ ↑↑
5s 5p 5d

Xe2+ hybrid: ↑↓ ↑ ↑↑ ↑
XeF4 Sp3d

XeO2F2

FF FO

Xe Xe

FF FO

μ = 0 (non polar) μ ≠ 0 (polar)

3 (a) Limestone,CaCO3, decomposed to solid calcium oxide and carbon dioxide gas when
heated at high temperature. At 30 oC, a volume of 107.3 mL of the gas was collected
by displacement of water with a total pressure of 1 atm. Calculate;

the number of moles of carbon dioxide produced.
(i)

ANSWER

CaCO3(s)  CaO(s) + CO2 (g)

43

PH2O  31.8 mmHg
 0.0418 atm

PT  PCO2  PH2O
PCO2  1.0  0.0418

 0.9582 atm

PCO2 V  n CO2 RT

n CO2  (0.9582)(0.1073)
(0.08206) (303.15)

 0.0041 mol

(ii) the mass of limestone decomposed.
[Vapour pressure of water at 30oC is 31.8 mmHg]

ANSWER

1 mol CO2 ≡ 1 mol CaCO3
0.0041 mol CO2 ≡ 0.0041 mol CaCO3

 mass CaCO3 = 0.0041 X 100.1
= 0.4137g

(b) Briefly describe three types of crystalline solids in terms of their interparticle forces.

ANSWER
3 type of crystalline solid

 Metallic crystal/metallic solid
- Electrostatic forces between positively metal ion and the sea of valence
electron.
- Metallic bond @ metal cations with a cloud of delocalized electrons.

 Ionic crystal/ionic solid
- Electrostatic forces between positively ion and negatively ion
- Ionic bond

 Molecular covalent/simple molecular/simple molecular covalent
- Weak van der Waals forces

 Giant covalent
- Covalent bond between atoms.

44

4 (a) HA is a weak acid with an acid dissociation constant,Ka=2.95x10-8.If the
concentration of the acid is 1.12 M, calculate

(i) pH of the solution.
ANSWER

HA(aq) ⇌ H+(aq) + A-(aq)

[ ]o 1.12 0 0
∆[ ]
-x +x +x
[ ]f
1.12-x x x

Ka  [H ][A ]
[HA ]

2.95 x108  (x)(x)
1.12  x

2.95 x108  x 2
1.12

[H ]  x  1.82 x 104 M
pH   log[H ]
  log1.82 x 104
 3.74

(ii) Percent dissociation.

ANSWER

% dissociation  [ ]dissociated x 100%
[ ]initial

 1.82 x 104
x 100%
1.12

 0.0163%

(b) The solubility product constant,Ksp of calcium phosphate,Ca3(PO4)2, in pure water is
1.2 x 10-26 at 25°C.

(i) Write the expression for the solubility product constant.

ANSWER

Ca3(PO 4)2(s) 3Ca2+ + 2PO 3 - (aq)
4

Ksp  [Ca 2 ]3[PO43 ]2

45

(ii) Calculate the molar solubility of calcium phosphate and concentration of each ion.
ANSWER

Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43-(aq)
x 3x 2x

K sp  [Ca2 ]3 [PO 4 3 ]2

1.2 x 1026  (3x)2 2x2

1.2 x 1026  108x5
x5  1.1111x 1028

 2.57 x 106 M
 molar solubility of Ca 3 (PO 4 )2  2.57 x 106 M
[Ca2 ]  7.70 x 106 M
[PO 4 2 ]  5.14 x 106 M

5. The ion of atom X has 8 outermost electrons and 10 inner electrons with a charge of
-1. Discuss all the rule(s) and principle(s) used to fill the electrons in the orbital of
atom X. Explain the change in the radius of atom X as it changes from a neutral
atom to a negatively charged ion.

ANSWER

Aufbau principle:
 electron occupy orbitals in the order of increasing energy level of the
orbitals.
 Orbitals with the lowest energy are always occupied first.

Pauli’s exclusion principle:
 Only two electrons may occupy the same orbital and that these two
electrons must have opposite spins.

Hund’s rule:
 When electrons are placed in a set of orbitals with equal or degenerate
energies, the electrons must occupy them singly with parallel spins before
they occupy the orbitals in pairs.

The extra electron is added to the same shell (n=3). Extra repulsion is produced
by the incoming electron causing the atom to expand. As a result X- ion is
bigger than atom X.

46

Atoms X, Y and Z are in periods n, n+1 and n+2, respectively. These atoms are
also in the same group. Discuss the trend in electronegativity exhibited by these
atoms

Answer

X, Y and Z are located in periods 3, 4 and 5 respectively. As we go down the
group from X to Z, new shell (n) increase and inner electrons are added than
shielding effect increases. Furthermore valence electrons being far away from
the nucleus, so the attractive forces of electrons become smaller. Therefore, the
electronegativities of atom will decrease.

6 Draw and explain the structures of BH3, NH3 and PH3 using valence shell electron
pair repulsion theory. Also, show the overlapping of orbitals in the PH3 molecule.

The boiling points of BH3, NH3 and PH3 are 173 K, 240 K and 185 K, respectively.
Explain why the boiling point of PH3 is greater than BH3 but lower than NH3.

ANSWER

VSEPR theory state that the repulsion of bonding pair-bonding pair < bonding

pair – lone pair < lone pair-lone pair. H

BH3 HB
 Has 3 bonding pairs. H

 Molecular geometry : trigonal planar

 Based on VSEPR theory, the electron pairs are arranged as far as possible

in order to minimize the repulsion. The repulsion of bonding pair-bonding

pair repulsion is equal.

 Bond angle: 120o

NH3 N

 Have 3 bonding pairs and 1 lone pair. HH H
 Molecular geometry: trigonal pyramidal

 Based on VSEPR theory, the electron pairs are arranged as far as possible

in order to minimize the repulsion. The repulsion of lone pair-bonding pair

> bonding pair-bonding pair.

 Bond angle: 107.5o

47

P

PH3 HH H
 Have 3 bonding pairs and 1 lone pair.

 Molecular geometry: trigonal pyramidal

 Based on VSEPR theory, the electron pairs are arranged as far as possible
in order to minimize the repulsion. The repulsion of lone pair-bonding pair

> bonding pair-bonding pair.
 Bond angle: < 109.5o

P(ground state):

3s 3p 3

P(excited state): sp

3s 3p  P sp3 
Hs sH
P(hybrid): sp3
sp3
3
Hs 
sp

 PH3 is a polar molecule, presence of dipole-dipole interaction.
 BH3 is nonpolar molecule, presence of London forces.
 Strength of dipole-dipole interaction > London forces.
 NH3 has hydrogen bonding and has stronger forces than PH3.

7 (a) The melting points of C2H6, CH3OH, NaCl and Si increase in the order as shown
below. Explain the trend.
C2H6 < CH3OH < NaCl < Si

ANSWER

Melting point increases in the order of attraction between molecules or atoms.
 C2H6 is non-polar compound and can form weak Van der Waals forces
between molecules.
 CH3OH has hydroxyl group and can form hydrogen bond between
molecules.
 NaCl is ionic compound and all atoms in the compound are held together
by electrovalent electrostatic forces.
 Si has gigantic covalent network in which Si atom is bonded tetrahedrally
to each other, thus form infinite amount of covalent bond and required
high energy to break all bonds.

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(b) A 10.00-L vessel contains 0.0681 mol phosphorous trichloride,PCl3 and 0.2056 mol
chlorine,Cl2 at 250 °C. If 0.0316 mol phosphorous pentachloride, PCl5, is produced at
equilibrium, calculate Kp for this reaction.

Discuss four factors that can increase the amount of PCl5 produced.

ANSWER

PCl3(g) + Cl2(g) → PCl5(g)

ni 0.0681 0.2056 0

nc -x -x +x

nf 0.0681-x 0.2056-x x = 0.0316

= 0.0365 =0.174

[ ] 0.00365 M 0.0174 M 0.00316 M

Kc = PCl5   =  0.00316 = 49.756
PCl3 Cl2 0.00365  0.0174 

Kp = Kc(RT)∆n = 49.756 (0.08206×523.15)−1 = 1.16

Le Chatelier state that when a system at equilibrium is subjected to change,
the system will adjust to counteract the effect.
 Increase pressure: Reaction will moves forward because the number of

moles of reactants is more than that of the product.
 Increase temperature: Reaction will moves forward in order to counteract

with the added heat by undergo endothermic reaction.
 Add reactants: Reaction will moves forward in order to decrease the

concentration of reactants.
 Remove product: Reaction will moves forward in order to increase the

concentration of product.

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