Assessment Schedule 2013 for AS 91390 (Chemistry 3.4) Demonstrate understanding of thermochemical principles and the properties of particles and
substances.
Question
number
1 (a)
1 (b) (i)
Answer
Ti+2 1s2 2s2 2p6 3s2 3p6 3d2
Ga 1s2 2s2 2p6 3s2 3p6 3d104s24p1
O-2 1s2 2s2 2p6
Br has a larger first ionization energy than arsenic but a smaller first
ionization energy than chlorine.
IE represents the amount of energy to remove the outermost electron. Br is
in the same period as As and has more protons than As. Because the outer
electrons of both are similarly shielded by the inner electrons it is the large
positive nucleus that is able to attract Br’s electrons with more force which
makes it harder to remove the outermost electron.
Bromine has a smaller first ionization energy than chlorine because although
it has more protons, the outermost electrons are in the next energy level and
are shielded from the nuclear charge by 1 extra energy level of electrons. Due
to the increased shielding the attraction to the nucleus is less and the I.E> is
less.
1 (b) (ii) Sulfur ion is larger than sulfur atom because it gains electrons to become an
ion. When sulfur gains two electrons there are still 16 protons while there
are now 18 electrons still in 3 energy levels. The extra 2 electrons increases
electron-electron repulsion, causing increase in size, when sulfur is an ion the
outer shell will be full which will result in the maximum electron-electron
repulsion.
Magnesium ions are smaller than magnesium atoms because the magnesium
ions have lost two electrons while keeping the same number of protons.
These are the only two electrons in the outer energy level so it has indeed lost
an entire energy level making it significantly smaller.
Achievement
Merit
Excellence
2 correct configurations
In (b)
A correct statement
comparing the IE of Br to
As
OR
A correct statement
comparing the IE of Br to
Cl
OR
A correct statement
regarding the size of S ion
vs. S atom
OR
A correct statement
regarding the size of Mg
ion vs. Mg atom.
In (b) correct explanation
of comparison of IE of Br
to As
OR
correct explanation of
comparison of IE of Br to
Cl
AND
A correct explanation of the
size of S ion vs. S atom
OR
A correct explanation of the
size of Mg ion vs. Mg atom
All correct
1 (c) (i)
1 (c)
(ii)




N0
N1
N2
A3
A4
M5
M6
E7
E8
Bent
Square-planar
SF4 has five regions of electron density and four atoms connected so it is a
see-saw shape.
SF6 has six regions of electron density around the central atom so its
shape is an octahedral shape
SF4 is polar. S and F have different electronegativities so the S–F bond is
polar covalent. The shape of the molecule is non-symmetrical about the
central S atom therefore the bond dipoles do not cancel, resulting in a
net/overall molecular dipole.
SF6 is non-polar. S and F have different electronegativities so the S–F
bond is polar covalent. The shape of the molecule is symmetrical about
the central S atom, so bond dipoles cancel / add to zero to give no net
dipole/ molecular dipole
No relevant response
One correct response at the A level
Two correct responses at the A level
Three correct responses at the A level
Four correct responses at the A level
Two correct responses at the M level
Three correct responses at the M level
Minor errors/omissions in responses at the E level
Two correct responses at the E level
In (c) (i) Both Lewis
diagrams correct
OR
Both shape diagrams
correct
OR
Both names correct
Both columns correct
In (c) (ii) Correct polarities
stated for both.
OR
Correct polarity for one
with some correct
explanation.
Correct polarities for both
stated, with an explanation
regarding shape or polarity
for both.
Comprehensive
discussion of both
molecules in terms
of shape and
polarity
Question
number
Answer
Achievement
2 (a) (i)
6C(s) + 3O2(g) + 6H2(g) → C6H12O6(s)
In (a)
Correct equation for ∆fH
2 (a) (ii)
Endothermic reactions are reactions where energy is absorbed by the
reactants to form the products. An endothermic reaction can be
spontaneous when, at a given temperature, the increase in entropy of the
reaction is greater than the decrease in enthalpy. (A reaction is spontaneous
when ∆G is negative and ∆G = ∆H - T∆S).
The enthalpy of fusion is the energy needed to melt 1 mole of sugar (at
1atm/100 kPa at its melting point
C6H12O6 (s) → C6H12O6 (l) ΔfusH°= 31.4 kJ mol-1
ΔvapH is energy needed to vapourise 1 mole of a substance. Vaporisation
means breaking all og the forces between particles, whereas melting only
means overcoming enough of the forces to form a liquid. Therefore ΔvapH
will always be higher than ΔfusH.
3 x C(s) + O2(g) → CO2(g)
(3x -394)
4x H2(g) + ½ O2(g) → H2O(g)
(4x -286)
-1 x C3H8(g) + 5 O2(g) → 4 H2O(l) + 3 CO2(g)
(-1x -2220)
Adding and cancelling gives ∆rHo = -106 kJ mol-1
Correct statement about
endothermic reactions
2 (b)
(i)
(ii)
2 (c)
N0
No relevant responses
N1
One correct response at the A level
N2
Two correct responses at the A level
A3
Three correct responses at the A level
A4
Four correct responses at the A level
M5
Two correct responses at the M level
M6
Three correct responses at the M level
E7
Minor errors/omissions (e.g. units) in both responses at the E level
E8
All correct responses at the E level
Definition OR equation
Merit
Both parts of (a) correct
(∆G statement and equation
are not required)
Definition of ΔfusH
including conditions
AND equation
Discusses the
difference between
ΔfusH and ΔvapH
in terms of forces
between particles
Minor error in calculation
or method
Correct calculation
and units
AND
ΔvapH described
Correct methodology
Excellence
Question
number
Answer
3 (a) (i)
Entropy will increase because H2O gas is less ordered than H2O liquid
3 (a) (ii)
Entropy will decrease because there are three moles of reactant gas and
only two moles of product gas. Less moles of gas would generally be more
ordered than a larger number of moles of gas so the entropy would increase.
Achievement
Merit
Two entropy changes
correct
Two correct changes with
correct explanations
One correct explanation
Two correct explanations
One correct comparison
Two correct comparisons
Excellence
3 (a) (iii) Entropy will decrease because there are three moles of reactant gas and
3 (b)
3 (c)
N0
N1
N2
A3
A4
M5
M6
only one mole of product gas along with two moles of liquid. The overall
reaction went from highly disordered (three moles of gas) to more ordered
(one mole of gas and two moles of liquid). Since it is more ordered the
entropy decreased.
Entropy will generally increase as temperature increases because the
particles are moving faster. When particles increase their movement, they
will become less ordered and entropy will increase.
Entropy will generally decrease as pressure increases because the particles
will be pushed closer together. As the particles are moved closer together,
they will become more ordered so the entropy will decrease.
Generally as the molar mass increases there are more electrons and more
instantaneous dipole-dipole interactions so all things being equal, the
boiling points would increase. Ammonia has the lowest molar mass and the
highest boiling point of the three. This is because ammonia has hydrogen
bonds. Hydrogen bonds are very strong intermolecular forces. The
difference in boiling point between nitrogen and hydrogen sulphide is very
large even though the molar mass difference is very small. This is due to
the fact that hydrogen sulfide is polar and will have permanent dipoledipole interactions. These will increase the boiling point more than just the
increase in instantaneous dipole-dipole interactions with strong attractive
forces causing an increase in the boiling point. The stronger the
intermolecular forces the greater the energy needed to overcome them,
therefore the higher the b.p.
No relevant responses
One correct enthalpy response in question 3a
One correct response at the A level
Two correct responses at the A level
Three correct responses at the A level
One correct responses at the M level
Two correct responses at the M level
Complete
discussion of all
three substances
with comparisons
made
E7
E8
Minor errors/omissions in responses at the E level (e.g) link not made between energy and bp)
All correct responses at the E level
Judgement statement:
NA: 0 – 7
A:
8 – 13
M:
14 – 19
E:
20 – 24