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CHEM1901/3
2013-J-8
June 2013
• The atmosphere of Venus contains 96.5 % CO2 at 95 atm of pressure, leading to an
average global surface temperature of 462 °C. The energy density of solar radiation
striking Venus is 2625 J m–2 s–1. The radius of Venus is 6052 km, and the average
albedo (the fraction of solar radiation reflected back into space) of its surface is 0.90.
Calculate the magnitude of the greenhouse effect on Venus.
An albedo of 0.90 means that 90% of the solar energy is reflected and only 10%
is available to heat the planet. If Venus has a radius R, the energy input from the
Sun is:
Ein = (2625 × πR2 × 0.10) J
The energy output from the surface area 4πR2 is:
Eout = (4πR2 × 5.67 × 10-8 × T4) J
At equilibrium, the energy input and output are equal and so the temperature
can be calculated:
2625 × πR2 × 0.10 = 4πR2 × 5.67 × 10-8 × T4
T4 = 0.10 × 2625 / (4 × 5.67 × 10-8) K
so T = 184 K
As the actual temperature on Venus is 462 °C = 735 K, the magnitude of the
greenhouse effect is:
greenhouse effect = (735 – 185) K = 551 K
Answer: 551 K
The main absorption bands of CO2 lie in the energy range 600 – 750 cm–1. What
range of wavelengths (in nm) corresponds to this energy range?
600 cm-1 corresponds to a wavelength of (1/600) cm = 0.00167 cm. As 1 cm = 0.01
m and 1 nm = 10-9 m, this corresponds to 16700 nm.
750 cm-1 corresponds to a wavelength of (1/750) cm = 0.00133 cm. As 1 cm = 0.01
m and 1 nm = 10-9 m, this corresponds to 13300 nm.
ANSWER CONTINUES ON THE NEXT PAGE
Marks
5
CHEM1901/3
2013-J-8
June 2013
Sketch the emission spectrum of Venus on the axes below. Note the wavelength of
maximum intensity, and point out any other important features.
The wavelength corresponding to the intensity maximum is given by Wien’s
displacement law:
T λ max ≈ 2.898 × 106 K nm
Using T = 735 K, λ max ≈ (2.898 × 106 K nm) / (735 K) = 3940 nm
Intensity
y
λmax = 3940 nm
CO2 absorption
bands
4000
(nm)
13,000 17,000
wavelength
CHEM1901/3
2013-J-12
June 2013
• Differential scanning calorimetry (DSC) is an experimental technique that measures
the temperature of a sample as a function of the heat supplied to it. Negative or
positive peaks on a DSC curve therefore indicate endothermic or exothermic
processes respectively. The figure below shows a series of DSC curves collected for
methane at different pressures. The scales of all the heat flow curves are the same,
but they have been offset from zero for clarity. Clearly identify the type of phase
change associated with every peak in the DSC curve.
s→l
s→l
s→l
l→g
l→g
s→g
On the 0.1, 1 and 10 bar curves, there are two phase transitions observed so
melting and boiling occurs. From the size of the peak is consistent with the single
peak in the 100 bar being melting. The size of the peak in the 0.01 bar curve is
not consistent with melting or boiling so it must correspond to sublimation.
ANSWER CONTINUES ON THE NEXT PAGE
Marks
8
CHEM1901/3
2013-J-12
June 2013
Use the DSC data shown to sketch a pressure-temperature phase diagram on the graph
below (note that pressure is on a log scale). Label all the important regions of the
phase diagram.
supercritical
fluid
solid
critical point
liquid
gas
triple point
CHEM1901
2012-J-8
June 2012
• Explain the following phenomena.
(a) When O2 dissolves in water at room temperature, the total entropy of the system
decreases.
The decrease in entropy of O2 going from (g) → (aq) outweighs the increase in
entropy of pure water going to water with O2 dissolved in it.
(b) When MgSO4 is dissolved in water, there is a very small but measurable decrease
in volume.
The hydration spheres around the dissolved ions (especially the small Mg2+ ion)
create a denser structure than that of random H2O molecules in pure water.
(c) A crystal (e.g. of NaCl) in its lowest energy configuration (thermodynamic ground
state) will always contain defects at finite temperatures.
Defects increase enthalpy, but they also increase entropy. So a small number of
defects will lower the total free energy (except at T = 0 K).
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY
Marks
6
CHEM1901/3
2010-J-10
June 2010
• Iron oxide that has been exposed to the atmosphere for any length of time will
generally contain a mixture of magnetite, Fe2O3, and haematite, Fe3O4. This mixture
can be converted to pure Fe3O4 by heating it under an excess of flowing hydrogen gas:
3Fe2O3(s) + H2(g) → 2Fe3O4(s) + H2O(g)
∆H° = –11.0 kJ mol–1
If 0.364 kJ of heat is liberated during the conversion of a 7.18 g sample of iron oxide
(consisting only of magnetite and haematite) to pure Fe3O4, what was the initial mass
percentage of haematite in the sample?
From the chemical equation, 1 mol of H2 liberates 11.0 kJ of heat. Hence, the
number of moles of H2 that must be reacting to produce 0.0364 kJ of heat is:
number of moles of H2 = (0.0364 / 11.0) mol = 0.00331 mol
From the chemical equation, 1 mol of H2 reacts with 3 mol of Fe2O3 and so:
number of moles of Fe2O3 = (3 × 0.00331) mol = 0.00993 mol
The molar mass of Fe2O3 is (2 × 55.85 + 3 × 16.00) g mol-1 = 159.7 g mol-1. Hence,
this amount of Fe2O3 corresponds to:
mass of Fe2O3 = number of moles × molar mass
= (0.00993 mol) × (159.7 g mol-1) = 1.59 g
The mass percentage of Fe2O3 in the initial 7.18 g sample is therefore:
mass percentage of Fe2O3 = (1.59 / 7.18) × 100% = 22.1 %
The mass percentage of haematite is therefore (100.0 – 22.1)% = 77.9%.
Answer: 77.9%
Fe3O4 can be further reduced to FeO under flowing hydrogen.
Fe3O4(s) + H2(g) → 3FeO(s) + H2O(g)
Based on the following table of thermodynamic data, what is the minimum
temperature at which this would be a spontaneous reaction?
Fe
∆fH° (kJ mol–1)
S° (J K–1 mol–1)
27
FeO(s)
Fe3O4(s)
–272
–1118
61
146
H2(g)
H2O(g)
–242
131
Using ∆rxnH° = ∆fH°(products) - ∆fH°(reactants):
∆rxnH° = [3∆fH°(FeO) + ∆fH°(H2O] – [∆fH°(Fe3O4)]
= ([(3 × -272) + (-242)] – [-1118]) kJ mol-1 = 60. kJ mol-1
ANSWER CONTINUES ON THE NEXT PAGE
189
Marks
4
CHEM1901/3
2010-J-10
June 2010
Using ∆rxnS° = S°(products) - S°(reactants):
∆rxnS° = [3S°(FeO) + S°(H2O] – [S°(Fe3O4) + S°(H2)]
= ([(3 × 61) + (189)] – [(146 + 131)]) J K-1 mol-1 = 95 J K-1 mol-1
The reaction is spontaneous if ∆G° < 0. As ∆G° = ∆H° - T∆S°, this condition
becomes:
∆H° - T∆S° < 0
Hence:
(60. × 103 J mol-1) – T × (95 J K-1 mol-1) < 0
T > (60. × 103 J mol-1) / (95 J K-1 mol-1)
T > 630 K
Answer: T > 630 K
CHEM1901/3
2010-J-11
June 2010
Would the resulting FeO be stable under those conditions, or would it be reduced
further to Fe metal by the following reaction?
FeO(s) + H2(g) → Fe(s) + H2O(g)
Explain, with calculations as appropriate.
For this reaction, using ∆rxnH° = ∆fH°(products) - ∆fH°(reactants):
∆rxnH° = [∆fH°(H2O] – [∆fH°(FeO)]
= ([-242] – [-272]) kJ mol-1 = 30. kJ mol-1
Using ∆rxnS° = S°(products) - S°(reactants):
∆rxnS° = [S°(Fe) + S°(H2O] – [S°(FeO) + S°(H2)]
= ([27 + 189] – [61 + 131]) J K-1 mol-1 = 24 J K-1 mol-1
At 630 K,
∆G° = ∆H° - T∆S° = (30. × 103) – (630 K) × (24 J K-1 mol-1) = 15000 J mol-1
As ∆G° > 0, the reaction is not spontaneous and FeO is stable.
Marks
3
CHEM1901/3
2009-J-8
June 2009
• A new process has been developed for converting cellulose from corn waste into the
biofuel butanol, C4H9OH. A bomb calorimeter with a heat capacity of 3250 J K–1 was
used to determine the calorific value by burning 5.0 g of butanol in excess oxygen.
Write a balanced reaction for the combustion of butanol in oxygen.
C4H9OH(l) + 6O2(g) → 4CO2(g) + 5H2O(g)
Calculate the heat released from this combustion if the temperature of the calorimeter
increased from 23.0 to 78.6 °C during the test.
The heat, q, required to change the temperature of the bomb calorimeter by ∆T
is given by:
q = C ∆T
The temperature increases from 23.0 to 78.6 °C corresponding to ∆T = 55.6 °C =
= 55.6 K. Hence:
q = (3250 J K-1)(55.6 K) = 181000 J = 181 kJ
Answer: 181 kJ
Use this value to determine the calorific value and molar enthalpy of combustion of
butanol.
The calorific value is the energy content per gram. As 5.0 g of butanol releases
181 kJ, the calorific value is:
calorific value =
. = 36 kJ g-1
The molar enthalpy of combustion is the energy released by combusting one
mole. The molar mass of C4H9OH is:
molar mass = (4×12.01 (C) + 10×1.008 (H) + 16.00 (O)) g mol-1 = 74.12 g mol-1
5.0 g of butanol thus corresponds to:
number of moles =
. . = 0.067 mol
As 0.067 mol generates 181 kJ when combusted, the molar enthalpy of
combustion is negative and is given by:
∆H =
. = -2.7 × 103 kJ mol-1
Marks
5
CHEM1901/3
2008-J-8
June 2008
• Write a balanced equation for (i) the explosive decomposition, and (ii) the combustion
in air, of TNT, C7H5N3O6(s).
(i) 2C7H5N3O6(s) → 12CO(g) + 3N2(g) + 5H2(g) + 2C(s)
(ii) 2C7H5N3O6(s) +
21
/2O2(g) → 14CO2 (g) + 5H2O(g) + 3N2(g)
What is the essential difference between these two processes?
In the combustion reaction, the oxidant comes from an external source (air). In
the explosion, the oxidant is contained in the explosive material.
What is the increase in the number of moles of gas (per mole of TNT consumed) for
each of these two processes?
For process (i), the change in the number of moles of gas, ∆n, is:
∆n = n(moles of gaseous products) – n(moles of gaseous reactants)
= ((12 + 3 + 5) – (0)) mol = 20 mol
As (i) involves 2 mol of TNT, ∆n = 10 mol per mole of TNT
Answer: (i) 10 mol
For process (ii), the change in the number of moles of gas, ∆n, is:
∆n = n(moles of gaseous products) – n(moles of gaseous reactants)
= ((14 + 5 + 3) – (21/2)) mol = 23/2 mol
As (i) involves 2 mol of TNT, ∆n = ½ × 23/2 mol per mole of TNT
Answer: (ii) 23/4 mol
Which of these two processes releases more energy into the surroundings?
Data:
∆fH°(TNT) = 6.9 kJ mol–1
∆fH°(CO2(g)) = –393 kJ mol–1
∆fH°(CO(g)) = –111 kJ mol–1
∆fH°(H2O(g)) = –242 kJ mol–1
Using ∆rH° = Σ m∆fH°(products) - Σ m∆fH°(reactants), the enthalpy change for
reaction (i) can be written as
∆rH° = [12∆fH°(CO(g))] – [2∆fH°(C7H5N3O6(s) )]
= ([12 × -111] – [2 × 6.9]) kJ mol-1 = --1346 kJ mol-1
ANSWER CONTINUES ON THE NEXT PAGE
CHEM1901/3
2008-J-8
June 2008
Similarly. the enthalpy change for reaction (ii) can be written as
∆rH° = [14∆fH°(CO2(g)) + 5∆fH°(H2O(g))] – [2∆fH°(C7H5N3O6(s) )]
= ([14 × -393 + 5 × -242] – [2 × 6.9]) kJ mol-1 = -6726 kJ mol-1
∆fH°(N2(g)), ∆fH°(H2(g)) and ∆fH°(C(s) all represent formation of the elements
from their standard states and are all zero.
The combustion reaction releases more energy.
Answer: combustion reaction
CHEM1901/3
2008-J-10
June 2008
• A calorimeter containing 300.0 mL of water at 25 °C was calibrated as follows.
A 1000.0 W heating coil was run for 10.0 s, after which time the temperature had
increased by 7.5 °C. Calculate the heat capacity of the empty calorimeter. The
specific heat of water is 4.184 J K–1 g–1.
With a power of 1000.0 W = 1000.0 J s-1, the amount of heat generated by the coil
is:
q = (1000.0 J s-1) × (10.0 s) = 1.00 × 103 J
As the density of water at 25 °C is 0.997 g mL-1, 300.0 mL of water corresponds
to:
mass of water = m = density × volume = (0.997 g mL-1) × (300.0 mL) = 299 g
The heat required to heat this quantity of water by 7.5 °C is:
q = × × ∆T = (4.184 J K-1 g-1) × (299 g) × (7.5 K) = 9380 J
The remaining heat, (1.00 × 103 – 9380) J = 620 J, is used to heat the calorimeter.
This also increases in temperature by 7.5 °C so the heat capacity of the
calorimeter is:
ccalorimeter =
=
.
= 82 J K-1
Answer: 82 J K-1
A solution containing 0.040 mol Ag+(aq) was mixed with a second solution containing
0.050 mole Br–(aq) in this calorimeter, causing AgBr(s) to precipitate. The
temperature increased by 2.4 °C. Given the solubility product constant is
Ksp(AgBr) = 5 × 10–13 M2 , calculate the equilibrium concentrations of Ag+(aq) and
Br–(aq) present in the final solution of volume 320 mL.
To calculate the equilibrium constants of Ag+(aq) and Br-(aq), a reaction table
can be used, with x representing the number of moles which do not precipitate:
initial (mol)
final (mol)
Ag+(aq)
Br-(aq)
AgBr(s)
0.040
x
0.050
0.010 + x
0
0.040 - x
The concentrations in the final solution of volume 320 mL are therefore:
[Ag+(aq)] =
.
and
[Br-(aq)] =
.
.
Hence, the solubility product is:
ANSWER CONTINUES ON THE NEXT PAGE
Marks
6
CHEM1901/3
2008-J-10
Ksp = [Ag+(aq)][Br-(aq)] = .
June 2008
.
.
= 5 × 10-13
Ksp is very small, x is tiny and so 0.010 – x ~ 0.010. This approximation gives:
.
.
.
= 5 × 10-13
or
= (5 × 10-13) .
M = 1.6 × 10-11 M
[Ag+(aq)] = .
.
[Br-] =
.
.
M = 0.031 M
[Ag+(aq)]: 1.6 × 10-11 M
[Br–(aq)]: 0.031 M
Calculate the enthalpy of solution of AgBr(s).
As the final solution has a volume of 320 mL, its mass is:
mass = density × volume = (0.997 g mL-1) × (320 mL) = 319 g
The heat produced causes the temperature of this mass of solution and the
calorimeter to rise by 2.4 °C:
q = ( × × ∆T) + ( !"! × ∆T)
= ((4.184 J K-1 mol-1) × (319 g) × (2.4 K)) + ((82 J K-1) × (2.4 K))
= 3400 J
This corresponds to the heat given out in the precipitation of 0.040 mol of
AgBr(s). As the precipitation is exothermic, the heat of solution is endothermic:
#
heat of solution = +
= +85 kJ mol-1
.
#
Answer: +85 kJ mol-1
CHEM1901/3
2008-J-11
June 2008
• Use the figure below to help answer the following.
following
Write a balanced equation for the smelting of one of these metal oxides with coke in
which a major product is CO2. Give the approximate temperature range over which
this reaction is spontaneous and state what happens outside this temperature range.
Between 400 – 950 K, the Ni – NiO lines is below the C – CO2 line and hence the
oxide will be reduced by coke to produce CO2:
2NiO(s) + C(s) 2NI(l) + CO2(g)
Below 400 K, it appears that the Ni – NiO will be above the C – CO2 line so this
reduction will not occur.
Above 950 K, the C – CO line is higher than the C – CO2 line so the reduction
produces CO2. (Any CO2 produced would be reduced by C to produce CO.)
Over what temperature range can ZnO be reduced by Fe? What other metal could be
used instead to increase the temperature range in which metallic Zn was produced?
The Zn – ZnO line falls below the Fe – FeO line at approximately 1450 K and
reduction of ZnO by Fe will thus occur above this temperature.
The Zn – ZnO line is below the Ca – CaO, Al – Al2O3 and Mn – MnO lines over
the entire temperature range of the diagram and so Ca, Al or Mn could be used
too reduce ZnO at these temperatures.
THIS QUESTION CONTINUES ON THE NEXT PAGE.
Marks
4
CHEM1901/3
2008-J-12
June 2008
Estimate the partial pressure of CO that would be expected at equilibrium in the
smelting of ZnO by coke at 1500 K.
At 1500 K,
C + ½O2 CO
Zn + ½O2 ZnO
lnKp(1) ~ 21
ln
ln p(2) ~ 14
lnK
Hence, for:
ZnO + C Zn + CO
Kp = p(CO) = Kp(1) / Kp(2)
or
lnKp = ln(Kp(1)/Kp(2)) = lnK
ln p(1) – lnKp(2) = 21 – 14 = 7
Kp ~ 103 and hence p(CO)
(CO) ~ 1000 atm
Metallic copper is produced by smelting chalcopyrite, CuFeS2(s), directly in oxygen
to produce iron oxides and SO2. Write a balanced equation for this reaction, and
sketch the lnKp versus temperature curve for Cu-CuO
Cu CuO on the diagram on page 24.
Clearly label the curve you have drawn.
CuFeS2(s) + 5/2O2(g) → Cu(s) + FeO(s) + 2SO2(g)
The lnKp curve for Cu – CuO lies below Fe – FeO at all temperatures. A sketch
(in bold / red) is shown below.
Marks
5
CHEM1901/3
2008-J-14
June 2008
• State the Second Law of Thermodynamics, and explain how an exothermic process in
a closed system changes the entropy of the surroundings.
In any spontaneous process, the entropy of the universe always increases.
A closed system can exchange heat or work with its surroundings. An exothermic
reaction releases heat, q, into the surroundings: ∆surrS = q / T > 0
The enthalpy of reaction defines the heat released into or absorbed from the
surroundings at constant temperature and pressure.
The standard enthalpy and entropy of solution of poly(oxyethylene) in water are
∆H° = –7.8 kJ mol–1 and ∆S° = –31 J K–1 mol–1. Use these data to predict whether the
solubility of poly(oxyethylene) in water increases or decreases when the solution is
warmed.
The surroundings receive 7.8 kJ mol-1 from this exothermic reaction. Hence:
∆surrS° = (7.8 × 103 kJ mol-1) / T
As ∆sysS° = –31 J K–1 mol–1, the total change in entropy is:
∆univS°° = (–31 + 7800/T) J K–1 mol–1
Spontaneous process occur when ∆univS°° is positive:
•
At low T, the second term is large and dominates the first term. This leads
to ∆univS°° > 0 and dissolution being favoured.
•
At high T, the second term is small and the first term dominates. This leas
to ∆univS°° < 0 and dissolution being unfavoured.
Hence, the solubility decreases with temperature.
4
CHEM1901/3
2007-J-7
June 2007
• Write a balanced equation for the combustion of methylhydrazine, CH3NHNH2(g).
CH3NHNH2(g) + 5/2O2(g) CO2(g) + N2(g) + 3H2O(g)
Using bond enthalpies, estimate the enthalpy of combustion of methylhydrazine.
Bond enthalpies:
H
Bond
∆H / kJ mol–1
Bond
∆H / kJ mol–1
C–N
285
O–H
464
N–N
159
O=O
498
C–H
416
C=O
806
N–H
391
N≡N
945
H
O
O
H
C
N
N
+
5
N
/2
C
+
O
H
H
H
+3
N
O
H
H
O
∆atomH(CH3NHNH2) = 3∆H(C-H) + ∆H(C-N) + 3∆H(N-H) + ∆H(N-N)
= (3 × 416) + (285) + (3 × 391) + (159) = 2865 kJ mol-1
∆atomH(O2) = ∆H(O=O) = 498 kJ mol-1
∆atomH(CO2) = 2 × ∆H(C=O) = (2 × 806) = 1612 kJ mol-1
∆atomH(N2) = ∆H(N≡
≡N) = 945 kJ mol-1
∆atomH(H2O) = 2∆H(O-H) = (2 × 464) = 928 kJ mol-1
∆combH = ∆atomH(reactants) – ∆atomH(products) =
= [∆atomH(CH3NHNH2) + 5/2∆atomH(O2)]
– [∆atomH(CO2) + ∆atomH(N2) + 3∆atomH(H2O)]
= [(2865) + (5/2 × 498)] – [(1612) + (945) + (3 × 928)] = −1231 kJ mol-1
Answer: −1231 kJ mol-1
ANSWERS CONTINUES ON THE NEXT PAGE
Marks
6
CHEM1901/3
2007-J-7
June 2007
Liquid methylhydrazine and liquid oxygen can be used as a rocket fuel. Calculate the
calorific value (in kJ g–1) of this fuel.
The molar mass of O2 is (2 × 16.00) = 32.00. The molar mass of CH3NHNH2 is:
(12.01 (C)) + (6 × 1.008 (H)) + (2 × 14.01 (N)) = 46.078
To burn one mole of methylhydrazine requires 5/2 moles of liquid O2 to be
carried so the combined mass is (46.078) + (5/2 × 32.00) = 126.078.
The calorific value is therefore:
calorific value =
∆ comb H
1231
=
= 9.8kJ g −1
molar mass 126.078
Answer: 9.8 kJ g-1
When methylhydrazine is used as a rocket fuel, the usual oxidant is dinitrogen
tetroxide rather than liquid oxygen. Why?
N2O4 has a boiling point of 21 °C, so there are no problems associated with
handling cryogenic liquids. N2O4 and methylhydrazine are hypergolic - the
reaction occurs spontaneously on contact.
CHEM1901/3
2007-J-8
June 2007
• Estimate the average temperature of Mercury given the solar power density at its
surface of 9150 J m–2 s–1, and assuming an average albedo of 6% and zero Greenhouse
effect.
With an albedo of 6%, 94% is not reflected. Hence,
Eout = 4πr2 × 5.67 × 10-8 × T4 J s-1
Ein = πr2 × 9150 × 0.94 J s-1
These balance at a temperature:
T=
4
( πr 2 ) × (9150) × (0.94)
= 441 K
(4πr 2 ) × (5.67 × 10 −8 )
Answer: 441 K
Marks
2
CHEM1901/3
•
2007-J-10
June 2007
Use the figure below to help answer the following.
Write a balanced equation for the smelting of NiO by coke. In what temperature
range will this process be spontaneous?
NiO(s) + C(s)
Ni(l) + CO(g)
The reaction is spontaneous at temperatures where the C-CO line is above the
Ni-NiO line.
As marked on the diagram, this occurs at temperatures above ~620 K.
Why are (a) aluminium and (b) tungsten not recovered from their oxides by smelting
with coke? What alternative processes are used and why?
Tungsten forms tungsten carbide if smelted with coke. The oxide is usually
reduced using another reductant such as H2.
Aluminium is usually recovered electrolytically as smelting only occurs at
temperatures well above 1800 K, which are not readily achieved.
Marks
4
CHEM1901/3
2007-J-13
June 2007
• State the Second Law of Thermodynamics, and explain how the enthalpy of reaction
is related to the entropy change of the surroundings.
In any spontaneous reaction, the entropy of the universe always increases.
The enthalpy of reaction defines the heat released into or absorbed from the
surroundings at constant temperature and pressure. The entropy change of the
surroundings arises from the heat released.
By definition, ∆surrS =
q surr
T
As the heat from the reaction is lost to the surroundings, qsurr = –∆rxnH at
constant P.
−∆ rxn H
Hence, ∆surrS =
T
Formic acid HCOOH, can dimerise in the gas phase according to the reaction
2HCOOH
(HCOOH)2
with a standard enthalpy and entropy of dimerisation of ∆H° = –62 kJ mol–1 and
∆S° = –150 J K–1 mol–1 respectively. Predict the temperature-dependence of the
dimerisation reaction.
∆universeS = ∆sysS + ∆surrS = ∆sysS −
∆ rxn H
T
Assuming that ∆S° and ∆H° do not change greatly with temperature, this
becomes:
∆universeS = (-150) −
( −62 × 103 )
(62 × 103 )
= -150 +
J K-1 mol-1
T
T
(62 × 103 )
Dimerization is spontanteous if ∆universeS > 0. This occurs when T <
.
(150)
Hence, dimerization is spontaneous at T < 413 K and the reverse process is
spontaneous at T > 413 K. More dimer is formed at low temperature.
Draw a structure that shows the bonding in the dimer.
H-bond
O
H O
H C
C H
O H
O
Marks
5
CHEM1901/3
2006-J-7
June 2006
• Write the equation whose enthalpy change represents the standard enthalpy of
formation of hydrazine, N2H4(g).
N2(g) + 2H2(g) N2H4(g)
Write the equation whose enthalpy change represents the enthalpy of combustion of
hydrazine, N2H4(g) to produce water vapour.
N2H4(g) + O2(g) N2(g) + 2H2O(g)
Given the following data, calculate the standard enthalpy of formation of N2H4(g).
∆H°f (H2O(g)) = –242 kJ mol–1
∆H°comb (N2H4(g)) = –580 kJ mol–1
Using ∆ rxn H o = ∑ m∆ f Ho (products) − ∑ n∆ f Ho (reac tan ts) , the enthalpy
change for the combustion reaction is:
∆ comb H o = [2∆ f H o (H 2 O(g))] − [∆ f Ho (N 2 H 4 (l))] as the remaining
compounds are elements in their standard states.
Hence:
[2 × −242] − [∆ f H o (N 2 H 4 (l))] = −580kJ mol −1
∆ f Ho (N 2 H 4 (l))] = +96kJ mol −1
Answer: +96 kJ mol-1
Marks
4
CHEM1901/3
2006-J-12
June 2006
• State the Second Law of Thermodynamics, and explain how the enthalpy of reaction
is related to the entropy change of the surroundings.
The entropy of the universe always increases: ∆univS = ∆sysS + ∆surrS > 0
The surroundings act as a reservoir of heat for the system: heat gained or lost in
a reaction is taken or given to the surroundings, respectively.
As ∆sysH = -qsurr at constant pressure:
∆surrS =
qsurr -∆sys H
=
T
T
Give an example of a chemical reaction or a chemical process that corresponds to each
of the following.
∆S > 0, ∆H > 0
-∆sys H
< 0, such processes will only occur if the temperature is
T
sufficiently high that ∆surrS is smaller in magnitude than ∆sysS.
As ∆surrS =
Examples include:
•
Melting of ice at 300 K: H2O(s) H2O(l)
•
Boiling of water at 400 K: H2O(l) H2O(g)
•
NH4NO3(s) NH4+(aq) + NO3−(aq)
∆S > 0, ∆H < 0
-∆sys H
> 0 and ∆sysS > 0, such processes will occur at all
T
temperatures. Examples include:
•
Burning of ethane: 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)
•
CaCl2(s) Ca2+(aq) + 2Cl−(aq)
As ∆surrS =
∆S < 0, ∆H < 0
-∆sys H
> 0, such processes will only occur if the temperature is
T
sufficiently low that ∆surrS is larger in magnitude than ∆sysS.
Examples include:
•
Freezing of ice at 263 K: H2O(l) H2O(s)
Condensing of steam at 300 K: H2O(g) H2O(l)
•
As ∆surrS =
Marks
5
CHEM1901/3
2005-J-9
June 2005
 In an experiment, 50.0 mL of 1.00 M HNO3 was combined with 50.0 mL of 0.540 M
NaOH in a calorimeter. Give an equation for the reaction that took place.
H3O+(aq) + OH(aq)  2H2O(l)
The temperature of the solution was found to increase by 2.98 C. If the heat capacity
of the calorimeter was 80.0 J K–1 and the heat capacity of the final solution was
426 J K–1, determine the molar heat of reaction.
As q = C  ΔT for both the calorimeter and the solution,
q = ((80.0  2.98) J) + ((426  2.98) J) = 1510 J
As the reaction is a 1:1 reaction and fewer moles of OH are present, it is rate
limiting. The number of moles of OH is:
number of moles of OH = concentration × volume
= 0.0500 L × 0.540 mol L-1= 0.027 mol
This number of moles generates 1510 J of heat. The reaction is exothermic as the
temperature increases. The enthalpy change for one mole is therefore:
ΔrH =
−𝟏𝟓𝟏𝟎 𝐉
𝟎.𝟎𝟐𝟕 𝐦𝐨𝐥
= 55800 J mol-1 = 55.8 kJ mol-1
Answer: 55.8 kJ mol-1
The average bond enthalpy of the O-H bond is 463 kJ mol–1. Explain briefly why the
heat of neutralisation calculated in the first part of this question differs significantly
from this value.
The bond enthalpy refers to half the enthalpy change for the reaction H2O(g) 
O(g) + 2H(g) rather than the transfer of a proton from H3O+ to OH in aqueous
solution.
Marks
4
CHEM1901/3
2004-J-9
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 SiO2(s) to silicon by Al(s).
3SiO2(g) + 4Al(s)  2Al2O3(s) + 3Si(s)
Given the following thermochemical data, evaluate the enthalpy change per gram of
reactants for the SiO2 and Fe2O3 reactions above.
Compound
Hf (kJ mol–1)
Fe2O3
-821
Al2O3
-1668
SiO2
-877
Using ΔrxnH° = Σ mΔfH°(products) - Σ mΔfH°(reactants), for the SiO2 reaction
with Al:
ΔrxnH° = [2ΔfH°(Al2O3)] – [3ΔfH°(SiO2)]
= {[2 × -1668] – [3 × -877]} kJ mol1 = -705 kJ mol-1
This is the enthalpy change for 3 mol of SiO2 and 4 mol of Al. The combined
mass of the reactants in these amounts is:
mass of reactants = (3 × (28.09 (Si) + 2 × 16.00 (O)) + 4 × 26.98 (Al)) g mol-1
= 288.16 g mol-1
The enthalpy change per gram of reactants is therefore:
enthalpy change =
−𝟕𝟎𝟓 𝐤𝐉 𝐦𝐨𝐥−𝟏
𝟐𝟖𝟖.𝟏𝟔 𝐠 𝐦𝐨𝐥−𝟏
= -2.45 kJ g-1
For the Fe2O3 reaction with Al:
ΔrxnH° = [ΔfH°(Al2O3)] – [3ΔfH°(Fe2O3)]
= {[-1668] – [-821]} kJ mol-1 = -847 kJ mol-1
This is the enthalpy change for 1 mol of Fe2O3 and 2 mol of Al. The combined
mass of the reactants in these amounts is:
mass of reactants = ((2 × 55.85 (Fe) + 3 × 16.00 (O)) + 2 × 26.98 (Al)) g mol-1
= 213.66 g mol-1
ANSWER CONTINUES ON THE NEXT PAGE
Marks
4
CHEM1901/3
2004-J-9
June 2004
The enthalpy change per gram of reactants is therefore:
enthalpy change =
−𝟖𝟒𝟕 𝐤𝐉 𝐦𝐨𝐥−𝟏
𝟐𝟏𝟑.𝟔𝟔 𝐠 𝐦𝐨𝐥−𝟏
Answer (SiO2): -2.45 kJ g-1
= -3.96 kJ g-1
Answer (Fe2O3): -3.96 kJ g-1
Which set of reactants would make the better rocket fuel on the basis of most energy
provided per mass of fuel (i.e. biggest “bounce per ounce”)?
Fe3O3/Al is better fuel than SiO2/Al
CHEM1901/3
2003-J-2
June 2003
 Use the thermochemical data provided to calculate the heat of reaction of the
following reaction:
Marks
2
PCl3(l) + Cl2(g)  PCl5(s)
Data:
P4(s) + 6Cl2(g)  4PCl3(l)
H = –1280 kJ mol–1
P4(s) + 10Cl2(g)  4PCl5(s)
H = –1774 kJ mol–1
The reactions given in the data correspond to the enthalpy of formation of PCl 3(l)
and PCl5(s). The reactions as written correspond to the formation of 4 moles of
PCl3(l) and of 4 moles of PCl5(s).
The fHo values for these two substances are therefore:
fHo (PCl3(l)) = (-1280 × ¼) kJ mol-1
fHo (PCl5(l)) = (-1774 × ¼) kJ mol-1
As ΔrH° = Σ mΔfH°(products) - Σ mΔfH°(reactants) and fHo (Cl2(g)) = 0,
ΔrH° = ((-1774 × ¼) – ((-1280 × ¼)) kJ mol-1 = -124 kJ mol-1
ANSWER: 124 kJ mol1
 Identify one property of a molecule necessary for it to be considered a “greenhouse
gas”.
To be a greenhouse gas, a molecule must be able to absorb infrared light: it
must undergo a vibration which causes a change in the dipole moment.
1
CHEM1901/3
2003-J-6
June 2003
 The normal boiling point of chloroform is 61.7 C and its enthalpy of vaporisation
is 31.4 kJ mol–1. Calculate the entropy of vaporisation for chloroform at 1 atm.
The entropy change for vaporisation is:
ΔvapS =
∆𝐯𝐚𝐩 𝑯
𝑻𝐯𝐚𝐩
=
𝟑𝟏.𝟒 ×𝟏𝟎𝟑 𝐉 𝐦𝐨𝐥−𝟏
𝟔𝟏.𝟕+𝟐𝟕𝟑.𝟎 𝐊
= 93.8 J K-1 mol-1
Answer: 93.8 J K-1 mol-1
Marks
2
CHEM1901/3
2003-J-10
June 2003
 If wet silver carbonate is dried in a stream of hot air, the air must have a certain
concentration level of carbon dioxide to prevent decomposition by the reaction
Ag2CO3(s)
Ag2O(s) + CO2(g)
The enthalpy change, H, for this reaction is 79.14 kJ mol–1 in the temperature
range of 25 to 125 C. Given that the partial pressure of CO2 in equilibrium with
solid Ag2CO3 is 8.20  10–6 atm at 25 C, calculate the partial pressure of CO2
necessary to prevent decomposition of Ag2CO3 at 110 C. Assume that S does
not change over this temperature range.
As Ag2CO3 and Ag2O are present as solids, they are not involved in the
equilibrium expression which is simply, Kp = pCO = 8.20 × 10-6 atm = (8.20 ×
2
10-6 × 1.01325 × 105) Pa = 0.831 Pa at 298 K.
This is related to universeS by:
universeS = RlnKp = (8.314 J K-1 mol-1) × ln(0.831) = -1.54 J K-1 mol-1
 rxn H o
= rxnS T
79.14  103
= {rxnS } J K-1 mol-1
298
Hence, rxnS = 264 J K-1 mol-1.
If rxnH and rxnS are the same at 110 C (378 K), universeS becomes:
 rxn H o
T
= 57.4 J K-1 mol-1
universeS = rxnS -
= {264 -
79.14  103
} J K-1 mol-1
383
The equilibrium constant at 110 C (383 K) is then:

S
Kp = exp( universe )
R
𝟓𝟕.𝟒 𝐉 𝐊 −𝟏 𝐦𝐨𝐥−𝟏
= exp(
) = 992 Pa = 0.00979 atm = 𝑷𝐂𝐎𝟐
𝟖.𝟑𝟏𝟒 𝐉 𝐊 −𝟏 𝐦𝐨𝐥−𝟏
(Kp increases with temperature for an endothermic reaction.)
Answer: 0.00979 atm
Marks
6