CHEM1001/CHEM1101 combined
2003-N-12
November 2003
Celcius)
(deg. (C)
pointPoint
Boiling
Boiling
 The figure below shows the boiling points of Group 14 and 17 hydrides as a function
of the period (row) of the periodic table.
0
HF
-50
HI
-100
-150
HBr
HCl
CH4
SiH4
2
3
GeH4
Period
Period
4
SnH4
5
A number of trends are apparent from this figure, including:
- the tetrahydrides have lower boiling points than the monohydrides,
- the boiling point increases with period, with the exception of HF.
Explain these two trends, and the reason that HF is exceptional.
The tetrahydrides are non-polar, so intermolecular attraction is due to
dispersion forces only. As the period increases, the central atom gets bigger
(more electrons) and its polarisability increases - hence the dispersion forces and
boiling points increase.
The monohydrides are polar, so they have dipole - dipole attractions as well as
dispersion forces. Hence they have higher boiling points than corresponding
tetrahydride from same period.
HF is anomalous as the F atom is very small and very electronegative. HF is
therefore able to form strong H-bonds, which are relatively strong
intermolecular attractions. This results in an exceptionally high boiling point for
HF.
Marks
3
CHEM1001/CHEM1101 combined
2003-N-9
November 2003
 Diborane (B2H6) is a highly reactive compound, which was once considered as a
possible rocket fuel for the US space program. Calculate the heat of formation of
diborane at 298 K from the following reactions.
Hr (kJ mol–1)
Reaction
3
/2O2(g)  B2O3(s)
–1273
1
2B(s) +
2
B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g)
3
H2(g) +
4
H2O(l)  H2O(g)
1
/2O2(g)  H2O(l)
–2035
–286
+44
The heat of formation of a compound refers to its formation from the elements in
their standard states.
In the reactions above, (1) and (3) correspond to the formation of B2O3(s) and
H2O(l) from their elements so:
ΔfH°(B2O3(s)) =–1273 kJ mol–1 and ΔfH°(H2O(l)) = –286 kJ mol–1
Formation of H2O(g) from its elements corresponds to formation of H2O(l) [-286
kJ mol-1] followed by vaporization of H2O(l) [+44 kJ mol-1 (reaction (4))]. The heat
of formation of H2O(g) is therefore:
ΔfH°(H2O (g)) = (–286 + 44) kJ mol-1 = -242 kJ mol–1
Reaction (2) corresponds to the combustion of B2H6(g). The heat of combustion is
given by:
ΔrxnH° = ΣmΔfH°(products) - ΣnΔfH°(reactants)
= [ΔfH°(B2O3(s) + 3ΔfH°(H2O(g))] – [ΔfH°(B2H6(g) + 3ΔfH°(O2(g))]
= -2035 kJ mol-1
All of the ΔfH° values except that for B2H6(g) are known from above – the
ΔfH°(O2(g)) is zero because it is an element in its standard state. Substituting these
values in gives:
([-1273 + 3 × -242] – [ΔfH°(B2H6(g)) + 0] kJ mol-1 = -2035 kJ mol-1
or
ΔfH°(B2H6(g)) = [-1273 + 3 × -242 + 2035] kJ mol-1 = +36 kJ mol-1
Answer: +36 kJ mol-1
Marks
2
CHEM1001
2004-J-7
June 2004
 Aluminium metal is a very effective agent for reducing oxides to their elements. For
example, it is used as a component of the solid fuel in the space shuttle, and in the
thermite reaction shown in lectures:
Fe2O3(s) + 2Al(s)  Al2O3(s) + 2Fe(s)
Write a balanced equation for the reduction of CuO(s) to the base metal by Al(s).
3CuO(s) + 2Al(s)  Al2O3(s) + 3Cu(s)
Given the following thermochemical data, evaluate the enthalpy change per gram of
reactant for the CuO and Fe2O3 reactions above.
Compound
Hf (kJ mol–1)
Fe2O3
-821
Al2O3
-1668
CuO
-157
Using ΔrxnH° = ΣmΔfH°(products) - ΣnΔfH°(reactants), for the reaction of CuO
and Al, ΔrxnH ° = ([-1668] – [3 × -157]) kJ mol-1 = -1197 kJ mol-1.
The formula mass of CuO is (63.55 (Cu) + 16.00 (O)) g mol-1 = 79.55 g mol-1. The
atomic mass of Al is 26.98 g mol-1. The enthalpy change is for the reaction of 3
moles of CuO and 2 moles of Al, corresponding to a total mass of (3 × 79.55 + 2 ×
26.98) g = 292.61 g.
The change per gram of reactants is therefore ΔH° =
−𝟏𝟏𝟗𝟕 𝐤𝐉
𝟐𝟗𝟐.𝟔𝟏 𝐠
= -4.09 kJ g-1
For the reaction of Fe2O3 and Al, ΔrxnH ° = ([-1668] – [-821]) = -847 kJ mol-1.
The formula mass of Fe2O3 is ((2 × 55.85) + (3 × 16.00)) g mol-1 = 159.7 g mol-1.
The enthalpy change is for the reaction of 1 mole of Fe2O3 and 2 moles of Al,
corresponding to a total mass of (159.7 + 2 × 26.98) g = 213.66 g.
The change per gram of reactants is therefore =
−𝟖𝟒𝟕 𝐤𝐉
𝟐𝟏𝟑.𝟔𝟔 𝐠
= -3.96 kJ g-1
Answer: CuO/Al: 4.09 kJ g-1, Fe2O3/Al: 3.96 kJ g-1
ANSWER CONTINUES ON THE NEXT PAGE
Marks
5
Which would make the best rocket fuel on the basis of most energy provided per mass
of fuel (i.e. biggest “bounce per ounce”)?
The CuO / Al system is a better fuel on the basis of its “bounce per once”.
CHEM1001
2005-J-9
June 2005
 Consider the following reaction.
2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(l)
E = –10909 kJ mol–1
A mixture of C8H18 (10.00 g) and O2 (30.00 g) is allowed to react. Assuming that the
reaction goes to completion, how much energy will be produced?
The molar mass of C8H18 is (8 × 12.01 (C) + 18×1.008 (H)) g mol-1 = 114.224 g
𝟏𝟎.𝟎𝟎 𝐠
mol-1. The number of moles of C8H18 =
= 0.08755 mol
𝟏𝟏𝟒.𝟐𝟐𝟒 𝐠 𝐦𝐨𝐥−𝟏
The molar mass of O2 is (2 × 16.00) g mol-1 = 32.00 g mol-1. The number of moles
𝟑𝟎.𝟎𝟎 𝐠
of O2 is therefore
= 0.9375 mol.
𝟑𝟐.𝟎𝟎 𝐠 𝐦𝐨𝐥−𝟏
25 moles of O2 is required for every 2 moles of C8H18 so 12.5 mol of O2 is
required for every 1 mol of C8H18.
The ratio of O2 : C8H18 is actually
0.9375
 10.71 so O2 is the limiting reagent.
0.08755
10909 kJ is produced for every 25 mol of O2 that reacts. As 0.9375 mol of O2 are
present, the energy produced is:
ΔE =
𝟎.𝟗𝟑𝟕𝟓
𝟐𝟓
× 10909 kJ = 409.1 kJ
Answer: 409.1 kJ
Marks
4