CHEM1909
2003-N-2
November 2003
 The conversion of hydroquinone (C6H6O2(aq)) to quinone (C6H4O2(aq)) is involved
in many important biochemical reactions. The bombardier beetle, for example, uses
the explosive reaction between hydroquinone and hydrogen peroxide (as described by
the equation below) as a defence mechanism.
C6H6O2(aq) + H2O2(aq)  C6H4O2(aq) + 2H2O(l)
From the following reaction data, calculate Hrxn for the reaction between 1.00 mol of
hydroquinone and 1.00 mol of hydrogen peroxide.
C6H6O2(aq)  C6H4O2(aq) + H2(g)
Hrxn = +177.4 kJ mol–1
O2(g) + 2H2O(l)  2H2O2(aq)
Hrxn = +189.1 kJ mol–1
H2O(l)  H2(g) + ½O2(g)
Hrxn = +285.8 kJ mol–1
O2(aq) and H2O2(aq) and this is related to reaction (1) and the reverse of
reaction (2). As the former generates H2(g) and the latter O2(g) neither of
which are present in the conversion reaction, they are combined with the
reverse of reaction (3).
To ensure a balance equation, the combination is thus (1) – ½ × (2) - (3):
∆Hrxn
(1)
C6H6O2

C6H4O2 + H2
177.4
- ½ × (2)
H2O2

½O2 + H2O
- ½ x 1891.
- (3)
H2 + ½O2

H2 O
- 285.8
C6H6O2 + H2O2

C6H4O2 + 2H2O
-203 kJ mol-1
Hrxn = -203 kJ mol-1
Use the answer you obtained above to calculate the heat liberated (in joules) in the
oxidation of 3.86  10–4 mol of hydroquinone to quinone.
As 203 kJ are liberated by 1 mol, the heat change for 3.86 × 10-4 mol is:
q = (203 kJ mol-1) × (3.86 × 10-4 mol) = 0.0784 kJ = 78.4 J
Answer: 78 J
Calculate the temperature rise of 0.250 g of water for this quantity of heat.
(The heat capacity of water, Cp = 4.184 J K–1 g–1)
Using, q = m × C × ΔT,
ΔT =
q
(78.4 J)

 74.9K
m  C (0.250g)  (4.184 J K 1 g 1 )
Answer: 74.9 K
Marks
8
CHEM1909
2004-N-3
November 2004
Marks
 The final step in the industrial production of urea, CO(NH2)2, is:
CO2(g) + 2NH3(g)  H2O(g) + CO(NH2)2(s)
∆H = -90.1 kJ mol
–1
Using the following data, calculate the standard enthalpy of formation ∆Hf of solid
urea.
4NH3(g) + 3O2(g) → 6H2O(g) + 2N2(g)
∆H = -1267.2 kJ mol–1
C(s) + O2(g) → CO2(g)
∆H = -393.5 kJ mol–1
2H2(g) + O2(g) → 2H2O(g)
∆H = -483.6 kJ mol–1
Using  rxn H o   m f H o (products)   n f H o (reactants) ,  rxn H o for the
industrial production of urea is given by
Δrxn H o =[Δf H o (H2O(l))+Δf H o (CO(NH2 )2 (s)] - [Δf H o (CO2 (g))+2Δf H o (NH3 (g))]
ΔfH(NH3(g)) is not known. NH3(g) is produced in the reverse of reaction (1).
Reaction (1) requires formation of H2O(l) which is given by reaction (3).
Combining these two reactions therefore allows ΔfH(NH3(g)) to be calculated.
-(1)
3×(3)
6H2O + 2N2  4NH3 + 3O2
∆H° = 1267.2 kJ mol–1
6H2 + 3O2  6H2O
ΔH° = 3 x -483.8 kJ mol-1
6H2 + 2N2  4NH3
4 x ΔfH°(NH3)
Therefore, ΔHf°(NH3) = ¼ (1267.2 – 3 × 483.8) kJ mol-1 = -46.05 kJ mol-1.
Hence for the industrial production of urea,
Δrxn H o =([(0.5  -483.6)+Δf H o (CO(NH 2 )2 (s)] - [(-393.5)+(2  -46.05)]) kJ mol-1
= -90.1 kJ mol -1
Hence, Δf H o (CO(NH2 )2 (s) = 333.6 kJ mol-1
∆Hf = -333.6 kJ mol-1
ANSWER CONTIUNES ON THE NEXT PAGE
6
CHEM1909
2004-N-3
November 2004
The formation of urea in this process is only spontaneous above 821 °C. What is the
value of the entropy change ∆S (in J K–1 mol–1) for the reaction?
For a reaction to be spontaneous requires ΔG° = ΔH°  TΔS° < 0.
At 821 °C, ΔH°  TΔS° = 0 so
ΔS° =
H ° (-90.1  103 J mol -1 )

 82.4 J K 1 mol 1
T
(821+273)K
Actually, the question is incorrect. Above this temperature, ΔG° is positive and
the reaction is non-spontaneous. This can be verified by substituting a higher
temperature into ΔG° = ΔH°  TΔS°.
S = -82.4 J K-1 mol-1
Rationalise the sign of ∆S in terms of the physical states of the reactants and
products.
ΔS° is negative as 3 moles of gas react to give 1 mole of gas and 1 mole of solid.
There is a decrease in randomness.
CHEM1909
2005-N-4
November 2005
 The combustion of hydrazine, N2H4, with oxygen is described by the following
equation:
N2H4(l) + O2(g)  N2(g) + 2H2O(l)
Marks
3
∆H = –623 kJ mol–1
Given that ∆Hf of H2O(l) is –286 kJ mol–1, find the standard enthalpy of formation
of N2H4(l).
Using  rxn H o   m f H o (products)   n f H o (reac tan ts) ,
Δrxn H o =[Δf H o (N 2 (g)+2Δf H o (H 2 O(l)] - [Δf H o (N 2 H 4 (l))+Δf H o (O2 (g)]
As  f H o  0 for an element in its standard state, this becomes:
Δrxn H o =[0 + (2  286)] - [Δf H o (N 2 H 4 (l)) + 0] =-623
Hence,
Δf H o (N 2 H 4 (l)) = +51kJ mol -1
∆Hf = +51 kJ mol-1
The combustion of 1.00 mol of N2H4(l) can also be accomplished using N2O4(l) as
the oxidant, whereupon 629 kJ of energy is released at standard temperature and
pressure. What is the standard enthalpy of formation of N2O4(l)?
The chemical equation for the combustion of N2H4(l) with N2O4(l) is:
2N2H4(l) + N2O4(l)  3N2(g) + 4H2O(l)
As written, this reaction corresponds to burning two moles of N2H4(l) and hence
Δrxn H o = 2Δcomb H o  2  ( 629kJ mol 1 )  1258kJ mol 1 . The negative sign
indicates that energy is released during the reaction.
Using  rxn H o   m f H o (products)   n f H o (reac tan ts) ,
Δrxn H o = [3Δf H o (N 2 (g)+4Δf H o (H 2 O(l)]-[2Δf H o (N2 H4 (l))+Δf H o (N2 O4 (l))]
= [(3  0) + (4  -286)] - [(2  +51) + Δf H o (N 2 O4 (l))] = -1258
Hence,
Δf H o (N 2 O4 (l)) = +12kJ mol -1
∆Hf = +12 kJ mol-1
CHEM1909
2006-N-2
November 2006
• High-purity benzoic acid, C6H5COOH, (∆H°comb = –3227 kJ mol–1) is used to
calibrate a bomb calorimeter that has a 1.000 L capacity. A 1.000 g sample of
C6H5COOH is placed in the bomb calorimeter, along with 750 mL of pure H2O(l),
and the remaining 250 mL cavity is filled with pure O2(g) at 10.00 atm. The
C6H5COOH is ignited and completely burned, causing the temperature of the water
and the bomb calorimeter to rise from 27.20 ºC to 33.16 ºC. Write the chemical
equation corresponding to the standard enthalpy of combustion (∆H°comb) of
C6H5COOH.
C6H5COOH(s) +
15
O2(g) → 7CO2(g) + 3H2O(l)
2
Given that H2O(l) has a heat capacity of 4.184 J K–1 g–1 and a density of 0.997 g mL-1,
calculate the heat capacity of the bomb calorimeter itself (in units of J K–1). Ignore
the heat capacity of the gases and of C6H5COOH.
The molar mass of C6H5COOH is (7×12.01 (C))+(6×1.008 (H))+(2×16.00 (O)) =
122.118. Therefore 1.000 g corresponds to:
amount of C6H5COOH =
mass
1.000
= 0.008189 mol
=
molar mass 122.118
As combustion of 1 mol gives out 3227 kJ of heat, this amount gives out:
q = 0.008189 × 3227 = 26.42 kJ = 26.43 × 103 J
This heat leads to a temperature increase of (33.16 – 27.20) = 5.96 °C for the
combination of the calorimeter and the water. As water has a density of 0.997 g
mL-1, 750 mL has a mass of 750 × 0.997 g.
q = (mCwater + Ccalorimeter) × ∆T
= ((750 × 0.997 × 4.184) + (Ccalorimeter)) × 5.96 = 26.42 × 103
Ccalorimeter = 1305 J K-1
Answer: 1305 J K-1
ANSWER CONTINUES ON THE NEXT PAGE
Marks
6
CHEM1909
2006-N-2
November 2006
If 30.0% of the CO2 produced dissolves in the water, calculate the final total pressure
(in atm) inside the 250 mL cavity of the bomb calorimeter. Assume oxygen is
insoluble in water and ignore the vapour pressure of water.
Initially, 10.00 atm of oxygen is present in a volume of 250 mL and a
temperature of 27.20 °C. Using the ideal gas equation, PV = nRT,
PV
(10.00) × (0.250)
initial nO (g) =
= 0.1015 mol
=
2
RT (0.08206) × (273 + 27.20)
15
mol of O2 are required for the combustion of
2
every mole of C6H5COOH. As 0.008189 mol of C6H5COOH is present,
From the chemical equation,
15
= 0.06142 mol
consumed nO (g) = 0.008189 ×
2
2
Therefore,
final nO (g) = 0.1015 – 0.06142 = 0.0400 mol
2
From the chemical equation, 7 moles of CO2 are produced for the combustion
every mole of C6H5COOH. As 0.008189 mol of C6H5COOH is present,
total nCO = 0.008189 × 7 = 0.05732 mol
2
As 30% of this dissolves,
nCO (g) = 0.70 × 0.05732 = 0.04013 mol
2
The total number of moles of O2(g) and CO2(g) is therefore 0.0400 + 0.04013 =
0.0801. This is present in 250 mL at 33.16 °C. Using the ideal gas law,
P=
nRT (0.0801) × (0.08206) × (273 + 33.16)
=
= 8.05 atm
V
0.250
Answer: 8.05 atm
CHEM1909
2006-N-3
November 2006
• The specific heat capacity of water is 4.18 J g–1 K–1 and the specific heat capacity of
copper is 0.39 J g–1 K–1. If the same amount of energy were applied to a 1.0 mol
sample of each substance, both initially at 25 °C, which substance would get hotter?
Show all working.
Using q = C × m × ∆T, the temperature change for a substance of mass m and
specific heat capacity C when an amount of heat equal to q is supplied is given
by:
∆T =
q
C×m
The atomic mass of copper is 63.55. Hence, the temperature change for 1.0 mol
of copper is
∆T (copper) =
q
q
°C
=
(0.39 × 63.55) 24.8
The molar mass of H2O is (2 × 1.008 (H)) + 16.00 (O) = 18.016. Hence, the
temperature change for 1.0 mol of water is
∆T (water) =
q
q
°C
=
(4.18 × 18.016) 75.3
Hence,
∆T (copper) > ∆T (water)
Answer: Copper
Marks
2
CHEM1909
2006-N-5
November 2006
• “Water gas” is a mixture of combustible gases produced from steam and coal
according to the following reaction:
C(s) + H2O(g) → CO(g) + H2(g)
∆H° = 131 kJ mol–1
The equation for the complete combustion of 1 mol of water gas (i.e. 0.5 mol CO(g)
and 0.5 mol H2(g)) can be written as:
½CO(g) + ½H2(g) + ½O2(g) → ½CO2(g) + ½H2O(g)
Calculate the standard enthalpy of combustion of water gas, given the following
thermochemical data.
∆H°vap (H2O) = 44 kJ mol–1
∆H°f (H2O(l)) = –286 kJ mol–1
∆H°f (CO2(g)) = –393 kJ mol–1
Using ∆ rxn H o = ∑ m∆ f H o (products) − ∑ n∆ f Ho (reac tan ts) for the
vaporization of water (H2O(l) H2O(g)) gives
∆ vap H o = [∆ f Ho (H 2 O(g))] − [∆ f H o (H 2 O(l))]
= [∆ f Ho (H 2 O(g))] − ( −286) = +44
Hence ∆ f Ho (H 2 O(g)) = ( +44) + ( −286) = −242 kJ mol −1
Using ∆ rxn H o = ∑ m∆ f H o (products) − ∑ n∆ f Ho (reac tan ts) for the reaction,
C(s) + H2O(g) CO(g) + H2(g) gives
∆ rxn H o = [∆ f Ho (CO(g)] − [∆ f H o (H 2 O(g))]
= [∆ f Ho (CO(g)] − ( −242) = +131
as ∆ f H o (H2(g)) and ∆ f H o (C(s)) are both zero for elements in their standard
states. Hence ∆ f Ho (CO(g) = -111 kJ mol-1
Using ∆ rxn H o = ∑ m∆ f H o (products) − ∑ n∆ f Ho (reac tan ts) for the reaction,
½CO(g) + ½H2(g) + ½O2(g) ½CO2(g) + ½H2O(g) gives
1
1
1
∆ comb H o = [ ∆ f H o (CO 2 (g)) + ∆ f Ho (H 2 O(g))] − [ ∆ f H o (CO(g))]
2
2
2
as the enthalpy of formation of H2(g) and O2(g) are both zero for elements in
their standard states. Hence,
1
1
1
∆ comb H o = [( × −393) + ( × −242)] − [( × −111)] = -262 kJ mol-1
2
2
2
Answer: -262 kJ mol-1
THIS QUESTION CONTINUES ON THE NEXT PAGE.
Marks
3
CHEM1109
2007-N-2
November 2007
• A solution of 2.00 M NaOH (50.0 mL) at 44.9 ºC is added to a constant pressure
(“coffee cup”) calorimeter containing 250.0 mL of 0.70 M HNO3 at 21.5 ºC. The
final temperature of the solution is 29.9 ºC. Calculate the enthalpy of neutralisation
of OH–(aq) and H+(aq) in kJ mol–1. Assume the density of these solutions is
1.000 g mL–1 and the specific heat capacity of the solutions is 4.184 J K–1 g–1.
The final temperature is due to both the mixing of two solutions with different
initial temperatures and the chemical reaction. It is convenient to treat these
two processes separately.
(i) Temperature change due to mixing:
The NaOH(aq) and HNO3(aq) solutions are initially at 44.9 °C and 21.5 °C.
When mixed, heat from the former will warm up the latter to give a solution
with temperature Tm.
50.0 mL of NaOH(aq) corresponds to (50.0 mL × 1.000 g mL-1) = 50.0 g. The
heat lost by this mass is given by:
q1 = mC∆
∆T = 50.0 × 4.184 × (Tm - 44.9)
250.0 mL of HNO3(aq) corresponds to (250.0 mL × 1.000 g mL-1) = 250.0 g. The
heat gained by this mass is given by:
q2 = mC∆
∆T = 250.0 × 4.184 × (Tm - 21.5)
As the heat lost by NaOH(aq) is gained by the HNO3(aq), q1 = -q2 and so:
50.0 × 4.184 × (Tm - 44.9) = -1 × 250.0 × 4.184 × (Tm - 21.5)
so Tm = 25.4 °C
(ii) Temperature change due to reaction:
The final temperature is 29.9 °C so the temperature change due to the reaction
must be (29.9 – Tm) = (29.9 – 25.4) = 4.5 °C. The mixed solution has a total
volume of (50.0 + 250.0) = 300.0 mL.
This volume has a mass of (300.0 mL × 1.000 g mL-1) = 300.0 g. The heat change
corresponding to this mass and temperature increase is therefore:
qr = mC∆
∆T = 300.0 × 4.184 × 4.5 = 5600 J = 5.6 kJ
The reaction is a 1:1 neutralization reaction, OH-(aq) + H+(aq) H2O(aq).
ANSWER CONTINUES ON THE NEXT PAGE
Marks
4
CHEM1109
2007-N-2
November 2007
The number of moles of OH-(aq) present in 50.0 mL of the 2.00 M NaOH
solution is:
number of moles = concentration × volume = 2.00 × 0.0500 = 0.100 mol
The number of moles of H+(aq) present in 250.0 mL of the 0.70 M HNO3
solution is:
number of moles = concentration × volume = 0.70 × 0.2500 = 0.175 mol
The H+ is therefore in excess and the OH- is the limiting reagent.
As 0.100 mol generates a heat change of 5.6 J, the enthalpy of neutralization is:
∆ rH =
−5.6kJ
= −56 kJ mol −1
0.1mol
The reaction increases the temperature and so must be exothermic.
Answer: 56 kJ mol-1
Calculate the pH in the combined solution in the calorimeter at 21.5 ºC.
As 0.100 mol of OH-(aq) reacts with 0.175 mol of H+(aq), the final solution
contains (0.175 – 0.100) = 0.075 mol of unreacted H+(aq).
The final solution has a volume of 300.0 mL so,
[H+(aq)] =
number of moles 0.075mol
= 0.25 M
=
volume
0.3000 L
Hence,
pH = -log10[H+(aq)] = -log10(0.25) = 0.60
Answer: pH = 0.60
CHEM1109
2007-N-5
November 2007
• Ammonium perchlorate mixed with powdered aluminium powers the space shuttle
booster rockets:
2NH4ClO4(s) + 2Al(s) → Al2O3(s) + 2HCl(g) + 2NO(g) + 3H2O(g)
Given the following thermochemical data, how much heat would be released per gram
of Al(s)?
∆H fo (H2O(l)) = –285.1 kJ mol–1
∆H fo (Al2O3(s)) = –1669.8 kJ mol–1
∆H fo (NO(g)) = 90.4 kJ mol–1
∆H fo (NH4ClO4(s)) = –290.6 kJ mol–1
∆H fo (HCl(g)) = –92.3 kJ mol–1
o
∆H vap
(H2O) = 44.1 kJ mol–1
Using ∆ rxn H o = ∑ m∆ f H o (products) − ∑ n∆ f Ho (reac tan ts) :
∆ rxn H o = ∑ ∆ f H°(Al 2 O 3 (s)) + 2∆ f H°(HCl(g)) + 2∆ f H° NO(g)) + 3∆ f H°(H 2 O(g))
− ∑ 2∆ f H°(NH 4 ClO 4 (s)) + 2∆ f H°(Al(s))
= [( −1669.8) + 2( −92.3) + 2(90.4) + 3( −285.1 + 44.1)] − [2( −290.6) + 2(0)]
= −1815.4 kJ mol −1
In this calculation, ∆ f H°(Al(s) ) = 0 for an element in its standard state and
∆ f H°(H 2 O(g) ) = ∆ f H°(H 2 O(l) ) + ∆ vap H°(H 2 O ) have been used.
As written, this enthalpy change is for the reaction of two moles of Al(s).
Therefore, per mole of Al(s), ∆H° = ½ × -1815.4 kJ mol-1 =907.7 kJ mol-1.
As the atomic mass of aluminium is 26.98 g mol-1, the heat released per gram of
Al is:
q=
−907.7 kJ mol −1
26.98 g mol
−1
= 33.64 kJ g −1
Answer: 33.64 kJ g-1
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.
Marks
3
CHEM1109
2008-N-2
November 2008
• Carbon monoxide is commonly used in the reduction of iron ore to iron metal. Iron
ore is mostly haematite, Fe2O3, in which case the complete reduction reaction is:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
∆H° = –25 kJ mol–1
Incomplete reduction, however, results in the formation of magnetite, Fe3O4:
3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g)
∆H° = –47 kJ mol–1
Use these heats of reaction to calculate the enthalpy change when one mole of
magnetite is reduced to iron metal using carbon monoxide.
The required reaction is:
Fe3O4(s) + 4CO(g) 3Fe(s) + 4CO2(g)
The second reaction in the question is reversed so that it leads to loss of Fe3O4(s):
∆H°° = +47 kJ mol–1
2Fe3O4(s) + CO2(g) 3Fe2O3(s) + CO(g)
This reaction is then added to 3 × the first reaction:
3Fe2O3(s) + 9CO(g) → 6Fe(s) + 9CO2(g)
∆H°° = 3 × –25 kJ mol–1
2Fe3O4(s) + CO2(g) 3Fe2O3(s) + CO(g)
∆H°° = +47 kJ mol–1
2Fe3O4(s) + 8CO(g) 6Fe(s) + 8CO2(g)
∆H°° = (-75 + 47) kJ mol–1
The chemical reaction is exactly twice that required, so for one mole of Fe3O4(s),
the ∆H°° = (-75 + 47) / 2 kJ mol–1 = -14 kJ mol-1.
Alternatively, using the data in the next part of the question,
∆rxnH° = Σm∆fH°(products) - Σn∆fH°(reactants),
∆rxnH° = [4∆fH°(CO2(g)] – [∆fH°(Fe3O4(s) + 4∆fH°(CO(g)]
∆fH°(Fe(s)) = 0 as it is an element in its standard state.
Hence using the data in the table below:
∆rxnH° = ([4 × -394] – [-1118 + 4 × -111]) kJ mol-1 = -14 kJ mol-1
Answer: ∆rxnH° = -14 kJ mol-1
ANSWER CONTINUES ON THE NEXT PAGE
Marks
5
CHEM1109
2008-N-2
November 2008
Another iron oxide that can be formed as an intermediate during reduction is FeO.
Use the following table of thermochemical data to show whether the formation of
FeO from Fe3O4 is spontaneous or not at 25 °C.
∆fH° (kJ mol–1)
S° (J K–1 mol–1)
FeO
–272
61
Fe3O4
–1118
146
CO
–111
198
CO2
–394
214
For the reaction,
Fe3O4(s) + CO(g) 3FeO(s) + CO2(g)
∆rxnH° = Σm∆fH°(products) - Σn∆fH°(reactants)
= ([3 × -272 -394] – [-1118 – 111]) kJ mol-1 = +19 kJ mol-1
∆rxnS° = ΣmS°(products) - ΣnS°(reactants)
= ([3 × 61 + 214] – [146 + 146]) J K-1 mol-1 = +53 J K-1 mol-1
Thus,
∆rxnG° = ∆rxnH° -T∆rxnS°
= (+19 × 103 J mol-1) – (298 K)(53 J K-1 mol-1)
= +3200 J mol-1 = +3.2 kJ mol-1
As ∆rxnG° > 0, the reaction is not spontaneous.
CHEM1109
2008-N-3
November 2008
• A 150.0 g block of iron metal is cooled by placing it in an insulated container with
a 50.0 g block of ice at 0.0 °C. The ice melts, and when the system comes to
equilibrium the temperature of the water is 78.0 °C. What was the original
temperature (in °C) of the iron?
Data:
The specific heat capacity of liquid water is 4.184 J K–1 g–1.
The specific heat capacity of solid iron is 0.450 J K–1 g–1.
The molar enthalpy of fusion of ice (water) is 6.007 kJ mol–1.
The heat from the iron is used to melt the ice and to warm the water from 0.0 °C
to 78.0 °C.
The molar mass of H2O is (2 × 1.008 (H) + 16.00 (O)) g mol-1 = 18.02 g mol-1.
Hence 50.0 g of ice corresponds to:
number of moles = mass / molar mass = (50.0 g) / (18.02 g mol-1) = 2.775 mol.
Hence the heat used to melt ice is:
q1 = 6.007 kJ mol-1 × 2.775 mol = 16.67 kJ = 16670 J
The heat used to warm 50.0 g water by 78.0 °C is:
q2 = m × C × ∆T = (50.0 g) × (4.184 J K-1 g-1) × (78.0 K) = 16320 J
Overall, the heat transferred from the iron is:
q = q1 + q2 = 16670 J + 16320 J = 32990 J
This heat is lost from 150.0 g of iron leading to it cooling by ∆T:
q = m × C × ∆T = (150.0 g) × (0.450 J K-1 g-1) × ∆T = 32990 J
∆T = 489 K = 489 °C
As the final temperature of the iron is 78.0 °C, its original temperature was
(78.0 + 489) °C = 567 °C.
Answer: 567 °C
Marks
4
CHEM1109
2009-N-2
November 2009
• The thermite reaction is written below. Show that the heat released in this reaction is
sufficient for the iron to be produced as molten metal.
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l)
Assume that the values in the table are independent of temperature.
Substance
Al
Al2O3
Fe
Fe2O3
Enthalpy of
formation, ΔfHo
kJ mol–1
Molar heat
capacity, Cp
J K–1 mol–1
Melting
point
o
C
Enthalpy
of fusion
kJ mol–1
0
24
660
11
–1676
79
2054
109
0
25
1535
14
–824
104
1565
138
Assume 1 mol of reactants at initial temperature of 25 °C. Need to show that ΔH
for the reaction is greater than the amount of energy required to melt 2 mol of
Fe(s) and heat all the products (2 mol of Fe(s) + 1 mol of Al2O3(s)) to the melting
point of Fe.
ΔH = ∑ΔfH(products) – ∑ΔfH(reactants)
= ΔfH(Al2O3(s)) + 2ΔfH(Fe(s)) – (2ΔfH(Al(s)) + ΔfH(Fe2O3(s)))
= [(–1676 + 2 × 0) – (–824 + 2 × 0)] kJ mol-1
= –852 kJ mol–1
ΔH to heat 2 mol of Fe(s) to its melting point
ΔH = nFe(s) × Cp(Fe(s)) × ΔT
= (2 mol) × (25 J K–1 mol–1) × (1535–25) K = 75.5 kJ
ΔH to heat 1 mol of Al2O3(s) to melting point of Fe(s)
ΔH = ࢔‫ܔۯ‬૛‫۽‬૜ሺ‫ܛ‬ሻ × Cp(Al2O3(s)) × ΔT
= (1 mol) × (79 J K–1 mol–1) × (1535–25) K = 119 kJ
ΔH to melt 2 mol of Fe(s)
ΔH = 2 × ΔfusH° = (2 mol) × (14 kJ mol–1) = 28 kJ
Total energy required to melt the iron = (75.5 + 119 + 28) kJ = +222.5 kJ.
The energy generated by the reaction is more than enough to melt the iron.
Marks
6