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Topic 15 Energetics
15.1 Exercises
1. What is meant by the terms standard conditions and standard state. Are they the same?
Use examples in your answer.
The standard conditions correspond to the conditions used in determining thermochemical data,
i.e. a temperature of 298 K and a pressure of one atmosphere. These conditions are designated
by °, e.g. ΔH°.
Standard state is the stable state of an element or compound, under standard conditions, e.g. the
standard state of CO2 is a gas, while water a liquid.
No they are not the same. Conditions describe circumstances, e.g. the temperature, pressure and
concentration. State, in chemistry, refers to physical state, i.e. (s), (l), (g).
2. Define, with an example, standard enthalpy change of formation.
Is the difference (change) in enthalpy between 1 mol of a compound in its standard state when it is
formed from its constituent elements in their standard state; shown as ΔHf°, e.g.
C(s) + O2(g) → CO2(g)
ΔH°formation = +393.5 kJ
3. Define, with an example, standard enthalpy change of combustion.
The enthalpy change when 1 mol of a substance is completely oxidised under standard conditions,
e.g. CH4(g) + 2O2(g) → 2H2O(g) +
ΔH°combustion = –889 kJ
CO2(g)
4. State Hess’s law. Why is this law so useful?
The enthalpy change for a reaction is independent of the reaction pathway. It depends only upon
the difference between the enthalpy of the products and the enthalpy of the reactants. It is used to
determine enthalpy changes for reactions that cannot be measured directly.
5. Define, with an example, standard enthalpy change of reaction.
Consider the reaction:
2H2(g) + O2(g) → 2H2O(g)
The enthalpy change for this reaction is when the molar amounts shown in the equation react, e.g.
2 mol H2(g) react with 1 mol O2(g) under standard conditions.
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Note that the standard enthalpy of reaction applies to the reaction given. The standard enthalpy of
formation for water, by definition, uses this equation:
H2(g) + ½ O2(g) → H2O(g)
6. The standard enthalpies of formation of SO2 and SO3 are - 297 kJ mol-1 and - 395 kJ mol-1
respectively.
(a) Write the thermochemical equations for these reactions.
S(s) + O2(g) → SO2(g)
S(s) +
/2O2(g) → SO3(g)
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ΔHf° = –297 kJ mol‾1
ΔHf° = –395 kJ mol‾1
(b) Calculate the enthalpy of reaction for SO2(g) + ½O2(g)  SO3(g)
Rearrange given enthalpy equations to fit reactants / products. When reversing a reaction, change
the ΔH° sign.
SO2(g) → S(s) + O2(g)
S(s) +
/2O2(g) → SO3(g)
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ΔH° = +297 kJ mol‾1
ΔH° = –395 kJ mol‾1
Add equations, cancel duplications and add enthalpies:
SO2(g) + ½O2(g) → SO3(g)
ΔH° = – 98 kJ mol‾1
7. (a) Calculate the enthalpy of formation of propane, C3H8(g), given the enthalpy of
combustion of C(s) is –393 kJ mol-1 , of H2(g) is –285 kJ mol -1and of propane is –2217 kJ
mol-1.
We are given:
C(s) + O2(g) → CO2(g)
ΔH° = –393 kJ mol‾1
H2(g) + ½O2(g) → H2O(g)
ΔH° = –285 kJ mol‾1
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
ΔH° = –2217 kJ mol‾1
And we require:
3C(s) + 4H2(g) → C3H8(g)
ΔH° = ?
3C(s) + 3O2(g) → 3CO2(g)
ΔH° = 3x –393 kJ mol‾1
4H2(g) + 2O2(g) → 4H2O(g)
ΔH° = 4x –285 kJ mol‾1
3CO2(g) + 4H2O(g) → C3H8(g) + 5O2(g)
ΔH° = +2217 kJ mol‾1
Therefore,
3C(s) + 4H2(g) → C3H8(g)
ΔH° = +102 kJ mol-1
(b) Why is it necessary to calculate the enthalpy of formation of propane in this way?
Propane, C3H8, cannot be made from the direct combination of C(s) and H2(g).
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8. The standard enthalpy of combustion of ethanol, C2H5OH, is –1364 kJ mol-1, and for
ethanal, CH3CHO, –1191 kJ mol-1. Calculate standard enthalpy of reaction for the oxidation
of ethanol to ethanal.
We are given:
C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)
ΔH° = –1364 kJ mol‾1
CH3CHO(g) + 2O2(g) → 2CO2(g) + 2H2O(g)
ΔH° = –1191 kJ mol‾
We require:
C2H5OH(g) + ½O2(g) → CH3CHO(g) + H2O(g)
ΔH° = ?
Rearrange:
C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)
2H2O(g) + 2CO2(g)
ΔH° = –1364 kJ mol‾1
→ CH3CHO(g) + H2O(g)
ΔH° = +1191 kJ mol‾1
Therefore:
C2H5OH(g) + ½O2(g) → CH3CHO(g) + H2O(g)
ΔH° = –173 kJ mol‾1
9. (a) Using a thermochemical cycle, calculate the enthalpy of formation of methane given
the following ∆H combustion data:
C(s) = –393 kJ; H2(g) = –285 kJ; CH4(g) = –892 kJ
C(s)
+
+O2
2H2(g)
→
+O2
ΔH2
ΔH3
CH4(g) ΔH1 = ?
+2O2
ΔH4
CO2(g) + 2H2O
Therefore, according to Hess’s law, ΔH2 + ΔH3 = ΔH1 + ΔH4
Therefore, –393 + 2 (–285) = ΔH1 + (–892)
Therefore, ΔH1 = –71 kJ mol-1
(b) In (a) the value for carbon is that of the allotrope, graphite. Would the ∆H value be
the same if diamond were used? Give your reasons for your choice.
Graphite and diamond are allotropes. These different physical forms have different energies
so the values for ΔH°combustion would be different.
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10. Compound A has a ∆H formation of –1786 kJ mol-1, and compound B a ∆H formation
of –178 kJ mol-1. Explain why A is considered to be the more stable compound.
The more heat released in forming the compound A (ΔH negative and large) the less
energetic that compound is, because more energy is required to decompose it back to its
constituent elements. Therefore, as more heat is required to decompose Compound A, it is
more stable than Compound B.
Wind power is one of the oldest sources of energy. Windmills were used in the then Persia in
the 7th century. At the end of the 19th century there were about 100 000 windmills on the North
Sea coast between Denmark and Holland. These pumped water, drove sawmills, paper mills,
olive oil presses and ground corn. The 21st century is seeing resurgence in the use of wind
power for generating electricity.
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15.2 Exercises
1. Define the term lattice enthalpy and explain how the sign of the lattice enthalpy can tell if
the lattice is being formed, or being broken.
The standard enthalpy change when 1 mole of an ionic compound is formed from its free gaseous
ions. The negative (–) shows that energy is released as charges come together (i.e. to form a
lattice). A positive sign (+) shows the energy to be added in order to break the lattice. An example:
Na+(g) + Cl–(g)  Na+ Cl–(s) ΔH °lattice = –787 kJ mol-1
2. The following enthalpies are included in a Born-Haber cycle for an ionic lattice. Define,
and give an example of each.
(a) The standard enthalpy change of formation.
Is the enthalpy change when 1 mol of a compound is formed from its constituent elements under
standard conditions, e.g.
Na(s) + ½Cl2(g) → NaCl(s)
ΔH°f = –411 kJ mol‾1
(b) The standard enthalpy of atomisation.
Is the enthalpy change required to produce one mole of gaseous atoms from the element in its
standard state, e.g.
Na(s) → Na(g)
ΔH°atom = –107 kJ mol‾1
(c) The first ionisation energy.
Is the enthalpy change required to remove 1 mole of electrons from 1 mole of a gaseous atom, e.g.
Na(g) → Na+(g) + e‾
ΔH°ion = –496 kJ mol‾1
(d) The first electron affinity.
Is the enthalpy change when 1 mole of gaseous atoms gains 1 mole of electrons, e.g.
O(g) + e‾ → O‾(g)
ΔH°ea = –141 kJ mol‾1
3. Describe and explain the factors that determine lattice energy.
Lattice energy gives a measure of the strength of the ionic bond present in a lattice. This is
determined by the charge on the two ions because when the ion charge increases, there is an
increase the bond strength (force of attraction is directly proportional to charges). Also the size of
the ions is important. An increase the size of the ion decreases the bond strength (attraction is
inversely proportional to the distance squared). Ion size also changes how the ions pack together
in the lattice.
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Ionic bonding results from the electrostatic attraction of opposite charges, i.e. the ions, have for
each other. Therefore, lattice energy increases when the charges on the ions present increases,
and when the ions get smaller and, as a result, pack closer together.
4. Given the following thermochemical data:
A. Enthalpy change of formation of sodium bromide = –361 kJ mol-1
B. First electron affinity for bromine = –325 kJ mol-1
C. Enthalpy change of atomisation of sodium = +107 kJ mol-1
D. First ionisation energy of sodium = + 496 kJ mol-1
E. Enthalpy change of atomisation of bromine = +112 kJ mol-1
(a) Write thermochemical equations for A to E.
A Na(s) + ½Br2(g) → NaBr(s)
ΔHf°
B Br(g) + e → Br‾(g)
ΔH°ea = – 325 kJ mol‾1
C Na(s) → Na(g)
ΔHa°
D Na(g) → Na+(g) + e
ΔH°ion = + 496 kJ mol‾1
E ½Br2(g) → Br(g)
ΔHa°
= – 361 kJ mol‾1
= +107 kJ mol‾1
= +112 kJ mol‾1
(b) Construct a Born-Haber cycle for NaBr(s).
(c) Calculate the lattice energy for NaBr(s)
Using Hess’s law:
A + C + E + D + B + F = 0 (You start with NaBr and finish with NaBr,
therefore no overall enthalpy change)
F = – A – C – E – D + B = –361 –107 –112 –496 + 325 = –751 kJ mol-1
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(d) There is a difference between the value for the lattice energy of NaBr using
thermochemical data and the value using theoretical data. Explain why.
Experimental data (i.e. that used in the Born Haber cycle) assumes 100% ionic character,
Theoretical values are calculated from the charge on the ions, the ion sizes and how the ions are
packed together gives the actual value. This difference gives a measure of the ionic (or covalent)
character present in the lattice.
Note that the higher the difference between the experimental values for a lattice energy (i.e. the
Born-Haber value) and the theoretical values (the real value) the higher the percentage of covalent
character present. Silver fluoride, AgF, is an example. The experimental value is –960 kJ mol-1.
The theoretical value is –870 kJ mol-1. The difference is due some covalent character being
present in AgF. The values for sodium chloride, NaCl, show only a small difference; experimental
–770 kJ mol-1, theoretical –766 kJ mol-1, therefore mainly ionic, very little covalent character.
Differences in the electronegativities of the atoms combing also indicate degree of ionic character
present (or covalent). The difference in values for sodium chloride, NaCl, Na = 0.9 and Cl = 3.0
gives a different of 2.1 indicating high ionic character. In H2 there is no difference, each H = 2.1,
therefore 100% covalent character; the bonding electrons are shared equally between each H.
Use this aid for solving Born-Haber lattice energy problems.
Intermediate
Adding
Energy
(+)
List all
positive
enthalpy
changes
List all
negative
enthalpy
changes
Removing
Energy
(-)
Product e.g. NaCl
Total Energy Increase (+) + Total Energy Decrease (-) = 0
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5. (a) Here are some enthalpy values which can be used to determine the lattice energy for
sodium chloride. Next to each write what the equation is representing:
Na(s) + ½Cl2(g)
 NaCl(s) ∆HO = - 410 kJ mol-1;
Standard enthalpy of formation for NaCl(s)
Na(s) 
Na(g) ∆H = +107 kJ mol-1;
Standard enthalpy change of atomisation of Na(s)
½Cl2(g)  Cl(g) ∆H = +122 kJ mol-1;
Standard enthalpy of atomisation of ½Cl2(g)
Na(g)  Na+(g) + e– ∆H = + 495 kJ mol-1;
First ionisation energy of Na(g), i.e the energy to remove 1 mole of electrons from 1 mole of
gaseous Na atoms.
Cl(g) + e–  Cl– (g) ∆H = - 107 kJ mol-1;
First electron affinity for Cl(g), i.e. energy released when 1 mole electrons added to 1 mole gaseous
Cl atoms.
(b) Now use this information to construct a Born-Haber cycle for sodium chloride, Na+Cl-(g).
(Work this out. DO NOT COPY.)
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(c) The lattice energy for Na+Cl- is –1027 kJ mol-1
A + B + C + D + E + F = 0, therefore +410 +107 +122 +495 –107 –F = 0,
therefore F = –1027 kJ mol-1
6. What is the explanation for the following trends in lattice energies?
(a) NaF: –1088 kJ mol-1; NaCl: –787 kJ mol-1; NaBr: –741 kJ mol-1
The charges are the same for Na+ and X‾, but the halide ion increases in size (F < Cl < Br < I),
therefore the lattice energy decreases.
(b) NaCl: –787 kJ mol-1; KCl: –707 kJ mol-1; CsCl: –623 kJ mol-1
The Cl‾ is the common ion, but the cation (M+) increases in size in going down a group; therefore
the lattice energy decreases. Na < K < Cs, therefore Na+ < K+ < Cs+.
7. Discuss how the difference between the theoretical and experimental lattice enthalpy
values for ionic character can be used to determine the degree of covalent character
present.
The degree of covalent character is usually expressed as the difference between experimental and
theoretical lattice enthalpy values divided the experimental value. For example the experimental
value for AgCl(s) is –905 kJ mol‾1. The theoretical value is –770 kJ mol‾1. The difference is
therefore 135 kJ mol‾1. Therefore the covalent character is 135/905 = 14.9%.
8. Supply the missing words:
Lattice enthalpy becomes more negative i.e. more energy is given out when charge of ions
increases and ion size decreases.
9. In silver iodide, Agl, the experimental lattice energy calculated from a Born-Haber cycle is
- 876 kJ mol-1 (using the assumption that the only bonding is ionic). The theoretical lattice
energy is calculated to be - 736 kJ mol-1. Explain why there is a difference.
The difference in values is due to the covalent character present. There is a smaller
electronegativity difference between Ag and I compared to that between Na and Cl. Therefore AgI
is more covalent in nature than NaCl, hence the difference.
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10. In going across period 3 in the periodic table a change in both ionic size and ionic
charge occurs.
(a) Consider the cations Na+, Mg2+ and Al3+. Discuss how these changes affect the
magnitude of their lattice enthalpies.
In going from left to right, there is an increase in charge, and provided the anion remains constant,
an increase in lattice energies results. A decrease in ionic radii (there is an increase in nuclear
charge, which pulls in the outer electrons). This also increases the lattice energy.
(b) Consider the anions P3–, S2– and Cl–. Discuss how these changes affect the magnitude of
their lattice enthalpies.
In going from left to right the ionic radii decrease due increasing nuclear charge; therefore the
lattice energies increase. This, however, is countered, by a decrease in ion charge which reduces
lattice energy.
11. Here is a not to scale way of showing the stages in the Born- Haber cycle for the
formation of sodium chloride from its constituent elements:
(a) Now complete this table using the letters A to F above next to the appropriate definition
below.
ΔH
Definition
Letter
1st ionisation energy of sodium
B
+ 494
1st electron affinity of chlorine
F
- 364
The enthalpy of atomisation of sodium
C
+109
The enthalpy of atomisation of chlorine
A
+121
The lattice enthalpy of sodium chloride
E
- 770
10
kJ mol-1
(b) Calculate the enthalpy of formation of sodium chloride from the data given.
We need to determine D which is the enthalpy of formation for Na(s) + ½Cl2(g) → NaCl(s)
From the Born-Haber cycle:
D=C+B+A+F+E
= 109+ 494+121–363–770
= +724–1134
= –407 kJ mol‾1.
(c) The lattice enthalpies calculated theoretically sometimes differ from those determined
by experiment. Here are two examples:
Experimental ΔH
Theoretical ΔH
Sodium chloride
- 770 kJ mol-1
- 766 kJ mol-1
Silver chloride
- 889 kJ mol-1
- 778 kJ mol-1
Why is there such a big difference in the values for silver chloride in comparison to those
for sodium chloride?
Sodium has a much lower electronegativity value than silver; hence the electronegativity difference
is greater for NaCl than for AgCl (2.1 and 1.1 respectively). Hence, there is more covalent
character in AgCl than in NaCl.
12. (a) Using the following enthalpy changes construct a Born-Haber cycle and calculate the
lattice energy for magnesium oxide.
Reaction
∆H kJ mol-
∆H formation MgO(s)
- 606
∆H atomisation Mg(s)
+150
∆H sum of 1st and 2nd ionisation energies for Mg(g)
+ 2240
∆H atomisation O2(g)
+ 250
∆H sum first two electron affinities for O2(g)
+ 650
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(b) In the Born-Haber cycle for barium oxide, BaO, what values would be different to those
given for MgO in the above table? Explain why they differ and how they differ (i.e. larger or
smaller).
Firstly the ionisation energy for barium would be much less than that for magnesium since barium
is in period 6 while magnesium is in period 3 (i.e. ionisation energy reduces down a group)
Secondly the enthalpy of atomisation is greater for Ba than for Mg since barium has a higher
boiling point than does magnesium.
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15.3 Exercises
1. If a simple answer to the question, ‘What is enthalpy?’ is, ‘the heat content of a substance
at a given temperature and pressure, symbol H’, what is the simple answer to, ‘What is
entropy?’
The measure of the disorder of a system at a given temperature and pressure, symbol S.
2. State and explain the factors which increase the entropy of a system.
•
Changing the state, e.g. (s) → (g), increases movement of particles, hence disorder.
•
Mixing different types of particles, e.g. water and salt, more spread;
•
Increasing the T, particles move more quickly;
•
Increasing the number of particles, e.g. A2 → 2A results in more disorder
3. Which of the following systems has the higher entropy? Include a reason for your choice.
(a) A teaspoon of sugar in the teaspoon or when dissolved?
When dissolved, the molecules are spread out through the solution, hence more disordered.
(b) An unopened pack of playing cards, or cards in use?
Cards in use now spread over the table, therefore greater disorder, therefore greater entropy.
(c) An electric jug of water before or after it is switched on?
After – increasing the temperature increases the kinetic energy hence the molecules move faster,
therefore more collisions, therefore move around in a more disordered fashion.
4. State and explain whether the entropy change S is positive or negative for each of the
following processes:
(a) Freezing of water
(–) Less disorder, less entropy, so a loss, therefore the sign is negative.
(b) Evaporating water
(+) More disorder, more entropy, therefore the sign is positive.
(c) When sugar dissolves in water
(+) The order in the salt crystal is broken down in the solution, more disorder, therefore the sign is
positive.
(d) When dry ice sublimes.
(+) CO2(s) → CO2(g)
solid to gas, less condensed, more entropy, therefore the sign is positive.
(e) When the following reaction occurs: 3H2(g) +
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N2(g)  2NH3(g)
(–) More order (4 mol → 2 mol) therefore, less entropy, therefore the sign is negative.
The more disordered a system becomes the more positive the change in entropy (ΔS)
becomes.
5. What characteristics of an explosion would suggest that an increase in entropy has
occurred?
Explosions create large volumes of gases and very high temperatures. Both increase disorder and
therefore entropy.
6. Ammonium nitrate can decompose explosively. The equation for this reaction is:
2NH4NO3(s)  2N2(g) + 4H2O(g) + O2(g)
Predict the entropy sign and give reasons for your choice.
ΔS is large and (+) as 2 mol → 7 mol and, also, there is a change of state, (s) → (g).
7. Define standard entropy change.
The change in entropy (ΔS) of a system at standard temperature (293 K) and pressure (1
atmosphere)
8. State and explain whether the value for ∆S is positive or negative for the following
reactions:
(a) Ag+(aq) + Cl-(aq)  AgCl(s)
(–) ΔS; less disorder as 2 ions become one species
(b) NH4Cl(s)  NH3(g) + HCl(g)
(+) ΔS; the solid becomes a gas and 1 mol becomes 2 moles, therefore an increase in entropy.
(c) H2(g) + Cl2(g)  2HCl(g)
(–) ΔS; no change in number of molecules but the heavier HCl(g) will be moving less quickly;
therefore less disorder.
9. The standard entropies of the gases F2, Cl2 and FCl are +203, + 223 and + 218 JK‾1 mol-1
respectively. Calculate the standard entropy of formation of FCl.
ΔS = Σs products – Σs reactants
The equation:
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F2(g) + Cl2(g) → 2FCl
ΔS = 2(+218) – (203 +223)
= +436 – 426
= +10 JK‾1 mol‾1
10. Consider the following standard entropies, SO, for the reactants and products in the
reaction to reduce CuO(s) to Cu(s) using H2(g) at STP:
CuO(s) = + 44 J K-1 mol-1
H2(g) = + 131 J K-1 mol-1
Cu(s) = + 34 J K-1 mol-1
H2O(l) = + 70 J K-1 mol-1
Does the entropy change favour this reaction? Give your reason.
The equation is: CuO(s) + H2(g) → Cu(s) + H2O(l)
ΔS = Σs products – Σs reactants
ΔS = +34 +131 –(+34 +70)
= +165 – 104
= +61 J K-1 mol‾1
Therefore increase in entropy so reaction is favoured by the entropy change.
In determining whether or not a reaction will occur, three factors must be considered:
temperature, enthalpy and entropy.
15
15.4 Exercises
1. Define, and give the symbol for the standard free energy change of a reaction.
It is the change in free energy that is produced during a chemical reaction under standard
conditions. [T=293 K, P = 1 atom]. Symbol, ∆G
2. Supply the missing words:
For a reaction to be spontaneous two forces are required. The first is towards more stable
products; products that have less heat content than do the reactants. ∆H for these
exothermic reactions has a negative sign. The second force is towards products, which are
more disordered than are the reactants. Changes in which ∆S has a positive sign are
favoured, while those with a negative sign are not favoured. We can predict that reactions
with a ∆H that has a negative sign and a ∆S that has a positive sign will be spontaneous. But
which of these will dominate if they are in opposition?
The quantity that links together ∆H and ∆S, and answers this question, is called the free
energy change, symbol ∆G, and for standard state conditions (which are temperature 293 K
and pressure 1 atmosphere). The equation is: ∆GO = ∆HO - T∆SO. For spontaneous reactions
to occur, ∆G is always negative.
3. (a) What is ∆G?
Gibbs free energy, determined by the equation: ∆GO = ∆H O – T∆S O.
(b) When the T∆S (system) is small compared to ∆H, which approximation applies:
A. ∆G > ∆H
B. ∆G < ∆H
C. ∆G = ∆H
D. None of these
If the value of T∆S is small compared to the value of ∆H it can be ignored hence the answer is C
(c) Explain why some exothermic reactions do not occur spontaneously.
Spontaneity is determined by ∆GO = ∆H O – T∆S O. If entropy decreases (∆S O has a – sign) this can
change the –T∆S into a positive factor and if the ∆H O is negative (reaction is exothermic and
small), overall ∆GO becomes +; therefore reaction not spontaneous.
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4. The T∆S (system) factor becomes significant when:
A. T is high
B. ∆S is low
C. T is low
D. Both T and ∆S must be high
Increasing temperature influences T∆S because it is a multiplication so the answer is A. A high T
can counter a low ΔS.
5. Hydrogen peroxide can be synthesized in two ways:
The first method involves reduction of oxygen by hydrogen:
H2(g) + O2(g)  H2O2(l)
The second method involves oxidation of water:
2H2O(l) + O2(g)  2H2O2(l)
Find the free energy of formation, ∆Go, for both processes and predict which process is
more efficient for the commercial preparation of hydrogen peroxide. The ∆Go values for
H2O2(l), H2O(l) and H2(g) are - 113.7, - 237.0 and 0 kJ mol-1 respectively.
Equation 1:
∆Gof = ∆Go products – ∆Go Reactants.
= –113.7 – (–0 –0) = –113.7 kJ mol‾1
Equation 2:
∆Gof = 2x –113.7 – (–237.0 – 0)
= –227.4 + 237.0 = +9.6 kJ mol‾1
The first reaction is spontaneous, hence more energy efficient.
Remember that the superscript o shows standard state and hence temperature is in K. Also kJ is
NOT the same as J.
6. Determine ∆So, ∆Ho, and ∆Go for this reaction, at 25 oC:
CO(g) + Cl2(g)  COCl2(g).
The information required is given in this table:
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SO in J mol-1 K-1
ΔHO(f) in kJ mol-
CO
+ 197.7
- 110.3
Cl2
+ 228.8
0
COCl2
+ 288.8
- 222.9
∆So for CO(g) + Cl2(g) → COCl2(g)
= Σ products – Σ reactants.
= +288.8 – (+197.7 + 228.8)
= +288.8 – 4265 – 177.7 JK‾1
= –0.1777 kJ
∆Ho = – 229.9 – (–110.3 –0)
= –119.6 kJ mol‾
Therefore ∆Go = ∆Ho – T∆S = –119.6 – 293 x –0.1777
= –119.6 + 52.1
= –67.5 kJ mol‾1
7. Is the following reaction, CCl4(l) +
H2(g)

HCl(g) +
CHCl3(l), spontaneous at 25oC
under standard state conditions? At 25 oC ∆Ho = - 91.1 kJ and ∆So = + 41.5 J K-1.
∆Go = ∆Ho – T∆So
= –91.1 – 293 x 0.0415
= –91.1 –12.3
= –103.4 kJ mol‾1
∆Go is negative and therefore the reaction is spontaneous.
8. What is the free energy change, ∆GO, for changing liquid water at 100 OC and 1
atmosphere to water vapour at the same conditions? The molar enthalpy of vaporization,
∆HO, for water is 40630 J and the standard entropies for H2O(l) = +69.9 JK‾1 mol‾1 and for H2O
(g)
= 188.7 JK‾1 mol‾1.
H2O(l) → H2O(g)
∆GO = ∆Ho– T∆So
= ∆Ho– T(Σs products – Σs reactants)
= 40630 – 373(188.7-69.9)
= -36.82
kJ K‾1 mol‾1
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9. What is the value for ∆G in a reaction at equilibrium?
∆G = zero
10. For the decomposition of water into its elements under standard conditions ΔG = + 237
kJ mol-1. What does this mean?
It means that water will not spontaneously decompose into its constituent elements as shown by
the equation: H2O(l) → H2(g) + ½O2(g)
11. Which of the following quantities is independent of pressure?
A.
ΔHo
B.
ΔSo
C.
ΔGo
D. None of these
A pressure change results in a change in concentration. Concentration changes affect ΔH, ΔS and
hence ΔG. Therefore the answer is D.
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Your Exam Questions and Answers
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