Worksheet 10 – Answers to Critical Thinking Questions
The worksheets are available in the tutorials and form an integral part of the learning outcomes and experience
for this unit.
Model 1: Endothermic and Exothermic Processes
1.
(a)
Endothermic
(b)
Endothermic
(c)
Exothermic
Model 2: Enthalpy of Atomization (ΔatomH) and Enthalpy of Atom Combination (ΔacH)
1.
ΔatomH = −ΔacH
2.
ΔH = 0 kJ
Model 3: Enthalpy of Reaction using ΔatomH and ΔacH
1.
It involves bond breaking only, which requires energy.
2.
It involves bond making only, which releases energy.
3.
ΔH = (239 + 435) kJ mol-1 + (2 × -431) kJ mol-1 = -188 kJ mol-1
4.
ΔrxnH = ΔacH (products) – ΔacH (reactants)
5.
ΔrxnH = ΔatomH (reactants) - ΔatomH (products)
6.
ΔrxnH will be positive. If the bonds that need to be broken in the reactants are stronger than those formed
in the products, the reaction is endothermic.
Model 4: Enthalpy of Reaction using ΔfH
1.
ΔfH° (O2(g)) refers to the formation of O2(g) from the element in its standard state (i.e. O2(g)):
O2(g) à O2(g)
ΔfH° (H2(g)) refers to the formation of H2(g) from the element in its standard state (i.e. H2(g)):
H2(g) à H2(g)
Both processes involve no change and so ΔH° = 0.
2.
ΔrxnH° = -892 kJ mol-1
3.
ΔrxnH° = ((-1096) – (-602 + -394)) kJ mol-1 = -100 kJ mol-1
CHEM1101
2003-J-6
June 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.
ΔHr (kJ mol–1)
Reaction
3
1
2B(s) +
2
B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g)
3
H2(g) +
4
H2O(l) → H2O(g)
1
–1273
/2O2(g) → B2O3(s)
–2035
–286
/2O2(g) → H2O(l)
+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:
Δ f Ho (B2O3(s)) = –1273 kJ mol–1 and Δ f Ho (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:
Δ f Ho (H2O (g)) = –286 + 44 = -242 kJ mol–1
Reaction (2) corresponds to the combustion of B2H6(s). The heat of combustion is given by:
Δ rxn Ho = ∑ mΔ f Ho (products) − ∑ nΔ f Ho (reac tan ts)
= [Δ f Ho (B 2 O 3 (s)) + 3Δ f H o (H 2O(g))] − [ Δ f H o (B 2 H 6 (s)) + 3 Δ f H o (O 2 (g))] All of
= −2035kJ mol −1
the Δ f Ho values except that for B2H6(s) are known from above – the Δ f Ho (O2(g)) is zero
because it is an element in its standard state. Substituting these values in gives:
[−1273 + 3 × ( −242)] − [ Δ f H o (B 2 H 6 (s)) + 0] = −2035 kJ mol-1
or
Δ f Ho (B 2 H 6 (s)) = [ −1273 + 3 × ( −242)] + 2035 = +36kJ mol −1
Answer: +36 kJ mol-1
Marks
2
CHEM1101
2007-J-6
Enthalpy Of Reaction (ΔrxnH)
June 2007
• The current “petrochemical economy” is based on the combustion of fossil fuels, of which octane
is a typical example.
2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l)
Calculate the heat of combustion of octane using the supplied heat of formation data.
Data: C8H18(l): –249.9 kJ mol–1; CO2(g): –393.5 kJ mol–1; H2O(l): –285.8 kJ mol–1
Using Δ rxn Ho = ∑ mΔ f Ho (products) − ∑ nΔ f H o (reac tan ts) , the heat of the reaction as
written is:
Δ rxn Ho = [16Δ f Ho (CO 2 (g)) + 18Δ f Ho (H 2O(l))] − [2Δ f Ho (C8 H18 (l))]
= [(16 × −393.5) + (18 × −285.8)] − [(2 × −249.9)]
= −10940kJ mol −1
where ΔfHo(O2(g)) = 0 has been used for the formation of an element in its standard state.
The enthalpy of combustion is defined per mole of fuel. The above reaction is for the
combustion of two moles of C8H18.
Hence, ΔcombHo = ½ × -10940 = -5470 kJ mol-1
Answer: -5470 kJ mol-1
Marks
5