Section A
1.1 Choose the preferred Lewis structure for NOCl.
(1 mark)
a)
c)
Cl
Cl
O
N
N
O
b)
d)
1.2 The following Lewis structure
Cl
N
O
N
Cl
O
C
O
N
represents one of the
resonance forms of the cyanate ion. The formal charges on the N, C and O
atoms, respectively, are:
(1 mark)
a) -2, +4, -1
b) -2, 0, +1
c) 0, 0, -1
d) +2, 0, -1
1.3 Which of the following molecules has the highest boiling point under same
conditions other than temperature?
(1 mark)
a) NH3
b) H2O
c) HF
c) CH4
1.4 Which one of the following properties does not determine the physical state of
a substance?
(1 mark)
a) Pressure
b) Chemical identity
c) Solubility
d) Temperature
1.5 A chemical reaction that absorbs heat from the surroundings is said to be
___________ and has a ___________ ΔH at constant pressure.
(1 mark)
a) endothermic, positive
b) endothermic, negative
c) exothermic, negative
d) exothermic, positive
1.6 The following reaction occurs at room temperature and pressure:
2Cl (g) → Cl2 (g)
ΔH = - 243.4 kJ
Which statement is correct?
(1 mark)
a) 2Cl (g) has the higher enthalpy.
b) Cl2 (g) has the higher enthalpy.
c) 2Cl (g) has the lower enthalpy.
d) the energy of both 2Cl (g) and Cl2 (g) are the same.
1.7 The temperature of a sample of gas is increased from 330 K to 500 K whilst
the pressure is kept constant. Which statement is correct?
(1 mark)
a) the volume of the gas decreases.
b) the volume of the gas increases.
c) the number of moles of the gas decreases.
d) the number of moles of the gas increases.
1.8 Which property listed below is INCORRECT for gases?
(1 mark)
a) pressure
b) volume
c) temperature
d) energy
Section B
2. Draw Lewis structures of the following compound
a) Na2O
(1 mark)
2 Na+
2-
2-
O
Or
2 Na+
+
O
b) H2NOH
(1 mark)
N
O
H
H
H
3. Arrange the following substances in order of increasing magnitude of their
London dispersion forces
(1 mark)
SiCl4, CCl4, GeCl4
soln
CCl4 < SiCl4 < GeCl4
4. a) Which of the following compound would you expect to have the highest
melting point?
KBr, CH3Cl, CH3OH
soln
KBr
(½ mark)
b) Show the trend, from highest to lowest melting point, in question (a) and
explain your choice for the lowest and highest melting point compound
(4 marks)
KBr > CH3OH > CH3Cl
√1 mark
KBr would have the highest melting point because it has the ion-ion
intermolecular force which is much stronger than the dipole-dipole forces and
hydrogen bonding experienced in CH3Cl and CH3OH respectively. (1½ marks)
CH3Cl has the least since the hydrogen bonding in CH3OH is stronger than
dipole-dipole forces experienced in CH3Cl. (1½ marks)
5) Using VSEPR theory, draw the shape and name the molecular geometry of the
following molecules. Include bond angles in your structures.
a) BeH2
(1½ mark)
√ ½ mark
H
Be
H
Linear
√ ½ mark
180° bond angles √ ½ mark
b) PF5
(2 marks)
F
F
P
F
F
F
√ ½ mark
Trigonal bipyramidal √ ½ mark
90°, 120° and 180° bond angles √ 1 mark
6) A sample of a compound (5.04 x 10-3 mole) is dissolved in 41.0 g of water and
then placed into a Styrofoam cup. The temperature of the solution changes
from 23.460 0C to 23.420 0C. What is the heat of solution of the compound
expressed in kJ/mol? Assume the specific heat of the solution is 4.18 J g-1 0C-1.
(1½ marks)
Ans: q = m x s x ΔT
= 41.0 g x 4.18 J g-1 0C-1 x (23.420 – 23.460) 0C √ (1/2 mark)
= -6.86 J
= - 6.86 x 10-3 kJ √ (1/2 mark)
heat per mole = - 6.86 x 10-3 kJ/5.04 x 10-3 mole
= - 1.36 kJ/mol √ (1/2 mark)
7) Use the data given to calculate ΔH for the reaction of ethylene, C2H4, with F2
shown in the following reaction;
C2H4 (g) + 6F2 (g) → 2CF4 (g) 4HF (g)
Data:
H2 (g) + F2 (g) → 2HF (g)
C (s) + 2F2 (g) → CF4 (g)
2C (s) + 2H2 (g) → C2H4 (g)
ΔH = - 537 kJ
ΔH = - 680 kJ
ΔH = + 52.3 kJ
(2 marks)
Soln:
2H2 (g) + 2F2 (g) → 4HF (g)
ΔH = - 537 kJ x 2 = - 1074 kJ √ (1/2
2C (s) + 4F2 (g) → 2CF4 (g)
ΔH = - 680 kJ x 2 = - 1360 kJ √ (1/2
C2H4 (g) → 2C (s) + 2H2 (g)
ΔH = - 52.3 kJ √ (1/2 mark)
mark)
mark)
C2H4 (g) + 6F2 (g) → 4 HF (g) + 2CF4 (g)
mark)
ΔH = - 2.49 x 103 kJ √ (1/2
8) A sample of air occupies 3.82 L at a pressure of 1.25 atm. What volume, in
litres, does it occupy if the pressure is increased to 5016 mmHg?
(1½ marks)
Soln:
1 atm = 760 mmHg
X
= 5016 mmHg
X = (5016 mmHg x 1 atm)/760 mmHg
= 6.60 atm √ (1/2 mark)
PiVi = PfVf
Vf
= PiVi/Pf
= (1.25 atm x 3.82 L)/6.60 atm √ (1/2 mark)
= 0.723 L √ (1/2 mark)
9) Which gas law is used to explain why hot air balloons rise?
Give an
explanation for your answer.
(1 mark)
Soln: Charles’s Law. √ (1/2 mark) Hot air expands as it is heated and causes
the air balloon to rise. √ (1/2 mark)