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TOPIC 15: ENERGETICS (HL)
In this section we will study why some chemical reactions are spontaneous (or feasible in terms of energetics)
and other reactions are not. The spontaneity or feasibility of a change (physical or chemical) depends on the
interaction of two concepts, one of which was introduced in the SL course:
 Enthalpy: during a spontaneous change most systems show a tendency to release energy to take on a
more stable lower energy state; however, this is not always the case e.g. dissolving of ammonia chloride in
water that is endothermic. This shows the limitations of using enthalpy change only to explain spontaneity
and proves that reducing the enthalpy of a system is not the only factor that determines spontaneity.
 Entropy which refers to the state of disorder; most spontaneous changes involve an increase in disorder but
not all e.g. condensation. Such examples also show the limitations of using entropy to explain spontaneity
of changes.
To indicate the combined effect of these two concepts (or how they interact) a third concept has been
introduced which is called “free energy” which relates the enthalpy and entropy changes to the system itself
(as opposed to also considering the surroundings) which gives us the advantage that we can measure both
the enthalpy and entropy changes.
15. 1 STANDARD ENTHALPY CHANGES OF REACTION
15.1.1 Define and apply the terms standard state, standard enthalpy change of formation and standard
enthalpy change of combustion
15.1.2 Determine the enthalpy change of a reaction using standard enthalpy changes of formation and
combustion.
Standard enthalpy changes, indicated by the symbol H, are enthalpy changes measured or calculated with
both reactants and products in standard
conditions i.e. 101 kPa (=1 atm) and 298 K and
in the case of solutions a concentration of 1.0 mol dm-3.
The state of substances in which they exist in these standard conditions is called the standard (physical) state.
Therefore in an equation defining a standard enthalpy change, for whichever type of enthalpy change, we
must make sure reactants and products are in their standard physical states and if that is the case the
enthalpy change is indicated by the symbol H.
When we refer to an enthalpy of reaction we are referring to the energy change when the reaction is carried
according to the stoichiometric equation (balanced equation as given) and not necessarily for 1 mole of
reactant or product.
Standard enthalpy of formation (for compounds only)= Hf 
The standard enthalpy of formation = the enthalpy change of the reaction in which one mole of a substance is
formed from its elements with all chemicals in their standard states. As it is molar it is in kJ mol -1.
Examples:
H2 (g)
¼P4 (s)
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+
+
½ O2 (g) 
3/2 Cl2 (g) 
H2O (l)
PCl3 (l)
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Hf  of water = - 285.8 kJ mol -1
Hf  of PCl3 = - 319.7 kJ mol -1
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The above 2 equations tell us that in both reactions the products, i.e. the compounds, are more stable than the
reactants, i.e. the elements as they are at a lower enthalpy level.
Reference point:
As the absolute enthalpies of chemicals cannot be measured and a reference point on the vertical axis
of a energy level diagram is needed, it has been decided to give the absolute enthalpy content of an
element in its standard state (which is their most stable state) the relative value of zero enthalpy.
It is the standard against which the other enthalpies and energetic stability are measured.
As a result of this agreement, the standard enthalpy of formation of elements in their standard states is also
zero as no formation reaction is needed. The formation of an element from its elements is not considered a
reaction as no new substance is formed so therefore it cannot have an en!
Examples of standard enthalpy changes of formation for some compounds
reaction
Ca (s) + ½O2 (g) CaO (s)
Hf  in kJ mol -1
-635.5
Na (s) + ½Cl2 (g) NaCl (s)
-411.0
C (s) + O2 (g)  CO2 (g)
-393.5
H2 (g) + ½O2 (g) H2O(l)
-285.9
H2 (g) + ½O2 (g) H2O(g)
-241.8
C (s) + 3H2 (g) + ½O2 (g)  C2H5OH (l)
-277.7
½H2 (g) + ½F2 (g) HF (g)
-271.1
8C (s) + 4H2 (g)  C8H8 (l)
-224.4
2H2 (g) + N2 (g)  N2H4 (l)
+ 50.6
3/2 H2 (g) + ½N2 (g)  NH3 (l)
- 46.0
Using standard enthalpy changes of formation to find enthalpy change of reaction
In any reaction, the reactants and the products are made form the same elements in their standard states.
Hreaction = a Hf (products)
-
b Hf (reactants)
The above mathematical expression means that to get the enthalpy change of a reaction simply add up the
enthalpies of formation of the products (multiplied by their respective coefficients symbolised by a) and
subtract from this the sum of the enthalpies of formation of the reactants (multiplied by their respective
coefficients, b).
The sum of the enthalpies of formation of the reactants is the energy released or absorbed when the reactants
are decomposed; by subtracting this from the sum of the products we are reversing the sign of the individual
enthalpies of formation of the reactants that makes sense as during the decomposition of the reactants the
formation reaction is reversed.
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So if the above expression is used you do not need to reverse any signs of any enthalpies of formation which
is why the expression is constructed as it is. It is also logically that if we are going to calculate a difference we
subtract what we have after the reaction with what we had before the reaction.
Worked example 1:
Using enthalpies of formation, calculate the enthalpy of combustion of benzene as shown by the equation
below. (find the values in table 11 in your IB data booklet or in the table on the previous page)
C6H6 (l) + 7½O2 (g)  6CO2 (g)
Hreaction  = a Hf (products)
-
+
3H2O (l)
b Hf (reactants)
= [6 Hf (CO2 (g)) + 3Hf (H2O (l))] - [7½ Hf (O2 (g)) + 1Hf (C6H6 (l)) ]
= [6 mol (-393.5 kJ mol-1) +3 mol (-285.8 kJ mol-1)] - [7½mol (+0 kJ mol-1) +1mol (+49 kJ mol-1)]
= - 3267 kJ
Worked example 2:
Calculate the standard enthalpy for the combustion of 1 mole of propane.
C3H8 (l) + 5 O2 (g)  3 CO2 (g)
+
4H2O (l)
Hr  = [3 Hf (CO2 (g)) + 4 Hf (H2O (l))] - [7 ½ Hf (O2 (g)) + 1 Hf (C3H8 (l)) ]
= [ 3 mol (-393.5 kJ mol-1) + 4 mol (-285.9 kJ mol-1)] - [ 7½ mol (+0 kJ mol-1) +1 mol (-105 kJ mol-1)]
= - 2219.1 kJ
Exercises
Use enthalpies of formation values in your IB data booklet table 11 and the values below to complete the
following calculations.
Hf in kJ mol-1 for
CH4 (g)
-74.8
NaHCO3 (s)
-948
O3 (g)
+143
Na2CO3 (s)
-1131
CO2 (g)
-394
N2O (g)
+81.6
H2O (l)
-286
SO2 (g)
-296.9
C6H6 (l
+ 49
H2O (g)
- 242
CaBr2 (s)
-682.8
Ca2+ (g)
+1925.9
Br- (g)
-233.9
CuO (s)
-157.3
1. Using standards heats or enthalpies of formation, calculate H in kJ for the reaction below
3CH4 (g) + 4O3 (g) 
3CO2 (g)
+ 6H2O (l)
2. The combustion of benzene may be written as
2C6H6 (l) + 15O2 (g) 
12CO2 (g)
+ 6H2O (l)
Using Hf calculate the standard enthalpy change of combustion for benzene, kJ mol-1.
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3. Using the enthalpies of formation and the equation below,
N2O (g) + 3H2 (g)  N2H4 (l)
+ H2O (l)
H = -317.0 kJ
calculate the standard enthalpy of formation of hydrazine, N2H4 (l).
4. Given the following information
2CH3OH (l) + 3O2 (g)  2CO2 (g) + 4H2O (l)
H = -727.0 kJ mol-1
calculate the standard enthalpy of formation of methanol, CH3OH(l).
5. Given the enthalpies of formation, what is the value of H in kJ for the reaction below?
CaBr2 (s)  Ca2+ (g) + 2Br- (g)
2Cu2O (s) + O2 (g) 
6. Given the thermochemical equation:
4CuO (s)
H = -292.0 kJ
calculate Hf in kJ mol-1 for Cu2O (s)
7. Nitroglycerine (Hf (2C3H5 (NO3 )3 ) = – 364 kJ mol-1 ) decomposes violently when it is detonated
according to the equation:
2C3H5 (NO3 )3 (l)
 ½O2 (g) + 6CO2 (g) + 3N2 (g) + 5H2O (l)
What is the enthalpy change for the decomposition of 2 moles of nitroglycerine?
8. Use the standard enthalpies of formation, calculate the enthalpy for the reaction
2NaHCO3 (s)
 Na2CO3 (s) + CO2 (g) + H2O (l)
Enthalpies of formation also allow us to comment on state the stability of a compound relative to its constituent
elements.
Example: H2 (g) + ½ O2 (g) 
H2O (l)
Hf  = - 285.8 kJ mol –1
Using the equation above, 1 mole of water molecules has an absolute enthalpy which is 285.8 kJ lower than
the absolute enthalpy of oxygen and hydrogen combined and is therefore more stable than these two
elements.
Using standard enthalpy of combustion
Hc =
standard enthalpy change of combustion for 1 mole of the substance.
This is the amount of energy released when 1 mole of a substance is combusted in sufficient supply of oxygen
in standard conditions.
The equation below is the equation describing the standard enthalpy of formation of water but it is also the
standard enthalpy of combustion of hydrogen!!
H2 (g)
+
½ O2 (g) 
H2O (l)

Hc (H2(g)) = - 285.8 kJ mol –1
The same relationship applies to carbon and carbon dioxide.
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Let’s now look at a few more examples of how we can use enthalpies of formation or combustion in enthalpy
cycles. On the previous page we have already calculated the standard enthalpy for the combustion of 1 mole
of propane using a formulae.
C3H8 (l) + 5 O2 (g)  3 CO2 (g) + 4H2O (l)
HC  = - 2219.1
kJ
We can also solve this using energy cycles or enthalpy level diagrams.
Enthalpy diagram
(energy in kJ mol-1)
3C (s) + 4H 2 (g) + 5 O2 (g)
0 Hf (reactants) = - 105 kJ = H1
C3 H8 (g) + 5 O2 (g)
Hf (products) = -2324.1 kJ = H2
Hreaction = - 2219.1 kJ
3CO2 (g) +
Hreaction =
H2
-
4H2O (g)
H1 = Hf (products) - Hf (reactants) =
- 2324.1 kJ - (- 105 kJ )
= - 2219.1 kJ
Enthalpy cycle: for H2 we can use both the enthalpy of combustion of hydrogen and carbon or the enthalpy of
formation of carbon dioxide and water.
H1
3C (s) + 4H 2 (g)
C3 H8 (g)
H2
Hreaction
+ 5 O2 (g)
+ 5 O 2 (g)
3CO2 (g) +
Hreaction =
H2
-
4H2O (l)
H1 = Hf (products) - Hf (reactants)
=
-2324.1 kJ - (- 105 kJ )
= - 2219.1 kJ
General cycle
Hreaction
elements/reactants
compound
H2
H1
combustion products of both elements and compound
We can draw the above enthalpy cycle differently
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Hreaction
C3 H8 (g) + 5O2 (g)
3CO2 (g) +
H2
4H2O (l)
H1
3C (s) + 4H 2 (g) + 5O2 (g)
Hreaction =
H1
H2 =
-
-2324.1 kJ - (- 105 kJ )
= - 2219.1 kJ
General cycle
Hreaction
reactants
products
H2
H1
elements in standard states of both reactants and products
Worked example: Using both enthalpies of formation and combustion

Construct a simple enthalpy cycle and calculate the value of Hf (C2 H5OH(l)) given the following data
Hf in kJ mol-1
-286
-394
compound
H2O (l)
CO2 (g)
C2 H5OH (l)
Hcombustion in kJ mol-1
-1371
Answer: The equation for the enthalpy of formation for ethanol is:
3C (s) + 4H 2 (g) + ½ O2 (g)  C2 H5OH (l)
Hf
3C (s) + 4H 2 (g) + ½ O2 (g)
(reactants)
C2 H5OH (l)
Hf (CO2 and H20)
Hcombustion
+ 3 O2 (g)
+ 3 O 2 (g)
3CO2 (g) +
Hf (C2 H5OH ) = Hf (products) - Hcombustion=
4H2O (l)
- 1645 kJ - (- 1371 kJ )
= - 274 kJ
Worked example: Using standard enthalpy of combustion only to calculate an enthalpy change
Calculate the enthalpy change for hydrogenation of propene using enthalpy of combustion only. You will find
enthalpy of combustions in your data booklet and propene = -2058 kJ mol-1. This only works if all species are
flammable.
Equation for hydrogenation:
C 3H6 (g) + H 2 (g)  C3 H8 (g)
Answer: use the enthalpy cycle below
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Hhydrogenation
C3H6 (g) + H 2 (g)
Hcombustion
+ 5 O2 (g)
C3 H8 (g)
Hcombustion
+ 5 O 2 (g)
3CO2 (g) +
Hhydrogenation = Hc (reactants) - Hc (product) =
4H2O (l)
(-2058 + -286 kJ) - (- 2219 kJ )
= - 125 kJ
Exercise
Calculate the enthalpy of hydrogenation of ethane first using enthalpies of formation and then using enthalpy
of combustion. In each case draw enthalpy cycles and add equations.
15. 2. Born-Haber cycle (FOR IONIC SUBSTANCES ONLY!)
15.2.1 Define and apply the terms lattice enthalpy and electron affinity.
15. 2.2. Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds.
15.2.3 Construct a Born–Haber cycle for group 1 and 2 oxides and chlorides, and use it to calculate an
enthalpy change.
15.2.4 Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds
in terms of their covalent character.
Lattice enthalpy is the ionic equivalent to bond enthalpies. Bond enthalpies in molecular substances indicate
the strength of covalent bonds. Lattice enthalpy is an indicator of the strength of an ionic bond within an ionic
lattice.
Lattice enthalpy = is the enthalpy change when one mole of a solid ionic crystal is separated into each of its
component ions in the gaseous state (this is to indicate that the ions should be completely separated and free
to move independently), at standard temperature and pressure. This an endothermic process and therefore it
has a positive sign. A negative lattice enthalpy value refers to the formation of an ionic lattice which is
shown by reversing the equation shown below.
separation:
NaCl (s)
formation:
Na+ (g) +

Na+ (g) +
Cl- (g)
Cl- (g)  NaCl (s)
Hlattice  = + 776 kJ mol -1
Hlattice  = - 776 kJ mol -1
The lattice energy is an also indicator of the stability of an ionic compound: the larger the value the more
stable the compound, the more tightly held are the ions within the lattice.
Calculation of lattice energy using Born-Haber cycles
Lattice enthalpies cannot be measured directly but can be calculated using Born-Haber cycles which are
thermodynamic cycles based on Hess’s Law.
Born-Haber cycles do this by relating lattice energy to ionisation energies, electron affinities and other atomic
and molecular properties.
Each Born-Haber cycle involves, starting from the elements in their standard state, all the changes which
occur to form an ionic compound from the elements:
 atomisation of both metal (from solid to gas = sublimation) and non-metal
 ionisation of metal
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 electron affinity of non-metal
 lattice energy
Electron affinity
 the first electron affinity of an element is the enthalpy change that occurs when one electron is gained by
each atom in a mole of gaseous atoms of the element to give one mole of ions, each with a single negative
charge (at standard temperature and pressure); most common first enthalpies are negative.
 the second electron affinity is the enthalpy change when a second electron is gained by one mole of
gaseous ions with one negative charge to give one mole of gaseous ions with a double negative charge;
most second electron enthalpies are positive as it takes energy to add an electron to an already negative
ion; however, in most cases that is compensated by the gain in lattice energy as a result of the stronger
attraction caused by an ion with a negative charge of –2 as opposed to –1.
Standard enthalpy of atomisation of an element
Standard enthalpy of atomisation of an element is the enthalpy increase that takes place when one mole of
gaseous atoms is made from the element in its standard physical state; in the case of hydrogen this is equal to
half the value of the H-H bond enthalpy so the same value is known under two different terms. In the case of
solid elements it is the same value as the heat of sublimation.
Examples:
 carbon: enthalpy of atomisation = Hsublimation (carbon) =
 Na (s)
 Na (g) = + 108.4 kJ mol-1
C (s)  C (g) = 715 kJ mol-1
Factors affecting lattice enthalpy
Study the chart below.
larger anions
larger
cations
larger anions
F-
Cl-
Br-
I-
O2-
S2-
Li+
+1031
+848
+803
+759
Be2+
+4443
+3832
Na+
+918
+780
+742
+705
Mg2+
+3791
+3299
K+
+817
+711
+679
+651
Ca2+
+3401
+3013
Rb+
+783
+685
+656
+628
Sr2+
+3223
+2843
Cs+
+747
+661
+635
+613
Ba2+
+3054
+2725
larger
cations
The lattice enthalpy depends on:


size of ion or inter-ionic distance: the smaller the ion or the inter-ionic distance, the larger the attraction
charge of the ion: the larger the charge, the greater the attraction
Uses of energy cycles: to test ionic and bonding models
See table 13 in your IB data booklet which compares experimental values (obtained using Born-Haber) and
theoretical values. If the experimental value is greater than the theoretical value calculated using electrostatic
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principles, this means that the actual bonding is stronger than what the electrostatic model suggest i.e.
bonding has a stronger covalent character as opposed to an ionic one.
Conclusions from the data in the table:
 Na-compounds: theoretical (predicted from physical calculations) and experimental (from Born-Haber
which uses experimental values) enthalpies show little discrepancy which means that the ionic model (ions
are discrete spherical ions) is correct.
 Ag-compounds: large discrepancy: actual lattice energy are higher than predicted by ionic model. Silver is a
transition metal in which the 3d electrons are less effective at shielding the attractive effect of the nucleus
which can attract the electron cloud from the anion. This polarises the anion and making it return partly the
electron that silver had donated to it. This gives the bond a more covalent character which at the same
time makes it stronger than an ionic bond.
15. 3. ENTROPY
15. 3. 1 State and explain the factors that increase the entropy in a system.
15. 3. 2 Predict whether the entropy change (ΔS) for a given reaction or process is positive or negative.
15. 3. 3 Calculate the standard entropy change for a reaction ( S°) using standard entropy values (S°) .
Entropy
Energetics or Thermodynamics allows us to determine the feasibility or spontaneity of a reaction; it establishes
whether a reaction is possible in principle (kinetics establishes whether it is possible in practice). To determine
the spontaneity we need to find out what is the driving force of spontaneous changes.
Reactions which are highly exothermic are generally spontaneous but lowering of the enthalpy is not always
the driving force e.g. evaporation and the dissolving of ammonium nitrate both of which occur spontaneously
although they are endothermic. There are other factors which also affect spontaneity and they are included in
the concept of entropy which is =
 a measure of the randomness or disorder of a system; the less regularly arranged, the greater the entropy
or disorder
 a measure of the random dispersal of energy of a system.
Entropy increase
We are interested in the entropy change, S, of a reaction.
an increase in entropy: S = +
a decrease in entropy: S = -
A proper workable definition of entropy is that it is a measure of the different ways in which molecules can be
arranged and their quanta of energy be distributed; as such the entropy of any system can be increased in the
following different ways:
 number of molecules/moles increased i.e. when there are more molecules/moles on the product side than
on the reactant side of a reaction (e.g. decomposition of a metal carbonate);
 the production of a greater number of gaseous particles (greatest increase in entropy);
 when molecules are more spread out (greater distance between them) e.g. formation of a gas, increase in
volume of a gas, evaporation, melting (more disorder over a greater area);
 increase in complexity of molecule:
 larger number of atoms/electrons; more ways of distributing energy/more energy levels;
 non-linear molecules instead of linear which fit together nicely, more orderly
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 increase in temperature (more units of energy or quanta available for distribution amongst the particles)
also resulting in greater movement of particles;
Simple rules for recognising entropy increases:






change in state: entropy always increases when a substance changes from solid to liquid to gas;
greatest increase = increase in number of particles in the gaseous state;
when a pure solid/liquid dissolves in a solvent;
when heating a substance; increased movement of particles;
when the number of moles is greater on the product side;
when a gas is produced during a reaction.
Predicting the sign of entropy S
Predict whether the entropy change is positive, negative or close to zero for each of the following reactions,
give reasons for your answer:
(a) S (g) +
O2 (g)  SO2 (g)
(b) O (g) +
O (g)  O2 (g)
(c) NH4NO3 (s)  2H2O(g) + N2O (g)
(d) Zn (s) + 2HCl (aq)  ZnCl2 (aq) + H2 (g)
(e) 2H2O2 (l)  2H2O (l) + O2 (g)
(f) Ca2+ (aq) + CO32- (aq)  CaCO3 (s)
(g) 2LiOH (aq) + CO2 (g)  Li2CO3 (aq) + H2O (l)
(h) N2(g)

(i)
N2(g) +
(j)
HCl (g) +
2N (g)
O2 (g) 
NH3 (g)  NH4Cl (s)
O2 (g)  2SO3 (g)
(k) 2SO2 (g) +
(l) 2HBr (g) +
F2 (g)  2HF (g) + Br2 (g)
3O2 (g)  2CO2 (g) + 4 H2O (g)
(m) 2CH3OH (l) +
(n) 4FeO (s) +
2NO (g)
O2 (g)  2Fe2O3 (s)
Standard entropy
The standard entropy is the entropy at 1 atm and 298 K which is the temperature and pressure at which most
reactions occur and is therefore measured in J K-1 mol –l . As entropy is a much smaller quantity than
enthalpy it is measured in joules and not kilojoules which means either enthalpy or entropy will need to be
converted when both quantities are involved in the same calculations.
Unlike enthalpy, absolute entropy can be measured and is expressed on a positive scale.
The scale starts with zero enthalpy which is the entropy of a crystal at O K; it has been given a value of 0 J K1
mol-1 because it has the most perfect order (this is also known as the 3rd Law of Thermodynamics).
As this is now an absolute value – it is the most orderly system and therefore has the lowest entropy possible
– standard entropies of any other substance, elements and compounds, are therefore always positive.
Examples of some values of standard entropies: these are absolute values as entropy can be measured.
Substance
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S J K-1 mol-1
substance
4 hours
S J K-1 mol-1
substance
S J K-1 mol-1
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H2 (g)
N2 (g)
O2 (g)
Cl2 (g)
I2 (s)
H2O(l)
H2O(g)
NH3 (g)
SO2 (g)
SO3 (g)
CO (g)
Br2(g)
F2 (g)
+ 130.6
+ 191.4
+ 204.9
+ 223.0
+ 116.1
+ 70.0
+ 188.7
+ 193
+ 248
+ 257
+ 198
+ 245.3
+202.7
CO2 (g)
NO (g)
C6H6 (l)
CH4 (g)
C2H4 (g)
C2H5OH(l)
CHCl3 (l)
HCl (g)
HBr (g)
CH3OH (l)
HF (g)
NH4Cl (s)
NO2 (g)
+ 213.6
+ 210.5
+ 172.8
+ 186.2
+ 219.5
+ 160.7
+ 201.8
+ 187
+ 199
+ 127
+ 173.51
+94.6
+240.45
C (s) (graphite)
S (s)
Na (s)
Zn (s)
Cu (s)
CaCO3 (s)
CuSO4.H2O (s)
CaO (s)
FeO (s)
Fe2O3 (s)
+ 5.7
+ 31.9
+ 51.0
+ 41.4
+ 33.3
+ 92.9
+ 305.4
+ 39.8
+ 61
+ 90
Calculating the standard entropy change for a reaction
The entropy change can be calculated using the following equation:
Ssystem = a S (products)
-
b S (reactants)
You need to multiply by the coefficients because the greater the number of particles the greater the entropy.
Worked examples:
Using absolute entropy values given, calculate the standard entropy change for the following reactions at
25C. The product is in J K-1 but could also be left in J K-1 mol-1for 1 mole of reactant as written in the
equation.
(a) CaCO3 (s) 
CaO (s)
+
CO2 (g)
Using the above expression:
Ssystem = [S (CaO) + S (CO2 )] – [S (CaCO3)]
= [ (1 mol) (39.8 J K-1 mol-1) + (1 mol)(213.6 J K-1 mol-1)] - [(1 mol)(92.9 J K-1 mol-1)]
= 160.5 J K-1
(increase in entropy for this reaction)
3H2 (g) 
(b) N2(g) +
2NH3 (g)
(unit of result left in J K-1 mol-1)
Ssystem = [2S (NH3)] – [ S (N2) + 3S (H2)]
= [( 2 x (193 J K-1 mol-1) - (192 J K-1 mol-1)] + 2 x (131 J K-1 mol-1)
= -199 J K-1
(decrease in entropy)
(c) Cl2(g) +
H2 (g) 
2HCl (g)
Ssystem = [2S (HCl)] – [ S (Cl2) + S (H2)]
= [ (2 mol) (187 J K-1 mol-1)] - [(1 mol)(223 J K-1 mol-1) + [(1 mol)(131 J K-1 mol-1)]
= 20 J K-1
(increase in entropy)
Exercises
For the following reactions, first predict whether the entropy of the system will increase (= + sign), decrease
(= - sign) or be close to zero and then calculate the standard entropy change:
(a) N2(g) +
(b) HCl (g) +
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O2 (g) 
2NO (g)
NH3 (g)  NH4Cl (s)
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(c) 2SO2 (g) +
O2 (g)  2SO3 (g)
(d) 2CO (g) +
O2 (g)  2CO2 (g)
(e) 2HBr (g) +
F2 (g)  2HF (g) + Br2 (g)
(f) 2NO (g) +
O2 (g)  2NO2 (g)
3O2 (g)  2CO2 (g) + 4 H2O (g)
(g) 2CH3OH (l) +
O2 (g)  2Fe2O3 (s)
(h) 4FeO (s) +
(i) CH4 (g) +
2O2 (g)  CO2 (g) + 2H2O (g)
15. 4. SPONTANEITY OF A REACTION
15.4.1 Predict whether a reaction or process will be spontaneous by using the sign of G°.
15.4.2 Calculate G°for a reaction using the equation G° = H° − T S° and by using values of the standard free
energy change of formation, ΔGf°
15.4.3 Predict the effect of a change in temperature on the spontaneity of a reaction using standard entropy and
enthalpy changes and the equation G°= H°− T S°.
According to the second law of thermodynamics, when a spontaneous process occurs, there must be an
increase in the entropy of the universe.
The entropy of the universe can be divided into the entropy of the system and the entropy of the surroundings.
For a reaction to be spontaneous, the entropy change of the system and the entropy change of the
surroundings that it caused by the reaction in the system when added together must have a positive value.
S universe
=
S surroundings
+
S system
During a reaction, the entropy change of the system causes a change in entropy of the surroundings in either
of the following two ways:
1.
In an exothermic reaction heat energy flows from the system to the surroundings; this causes a
decrease in the entropy of the system (S system = negative ) but causes the particles of the
surroundings to move faster and so they become more disordered. This entropy change of the
surroundings, S surroundings , will be positive.
2.
In an endothermic reaction, the surroundings lose heat to the system. This has the opposite effect: the
entropy of the surroundings will decrease as the particles become more ordered and S surroundings will
be negative. The entropy of the system however increases so S system = positive.
The spontaneity of a reaction depends on how these entropy changes interact as shown by the table below.
Predicting whether a reaction will be spontaneous or not using, S uni , S surr and S sys
S sys
if
if
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+
-
S sur
+
-
spontaneous?
S uni
+
4 hours
S universe = S surroundings+ S system
Yes
No
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+
-
if
if
+
Yes if S sys  S surr
+/+/-
Yes if S system   S surr
If we can calculate S universe we can have an idea of the spontaneity of a reaction. The problem is we can not
measure S surroundings.
But we know S surroundings depends on the enthalpy change of the system and the temperature of the
surroundings at the time of the reaction which are two quantities we can measure.
S surr
- Hsystem
T
=
We can now substitute S surroundings in the expression below.
=
+
S universe
-H system/T
S system
Rearranging the above expression produces a new expression and quantity called free energy. In our
calculations we are interested in the change in free energy in a reaction. Change in free energy is expressed
in kJ mol-1.
G = Hsystem - TSsystem
For a reaction to be spontaneous its free energy change must have a negative value: G  0.
We can use the above expression to predict the spontaneity of a reaction as shown in the table below
H (KJmol-1)
if
positive
S universe
(JK-1mol-1)
positive
if
positive
if
if
G (KJmol-1)
negative
depends on
temperature
positive
Spontaneity
Spontaneous but only at high temperatures when T
x S system  H
Reaction will not occur (reaction will occur in the
opposite direction)
negative
positive
negative
Spontaneous at all temperatures
negative
negative
depends on
temperature
Spontaneous but only at low temperatures when T x
S system  H
Calculations of G to find out if a reaction is spontaneous or not
There are 3 different ways in which we cam calculate the G of a reaction.
Using standard free energies of formation

Standard free energy of formation is the free energy change when 1 mole of the compound is formed from
its elements in their standard states;

Standard free energy of formation of elements in their standard states is
spontaneously as they already exist in that state.
Greaction = a Gf (products)
-
zero as they are not formed
b Gf (reactants)
Worked examples: Calculate the standard free-energy changes for the following:
(a)
CH4 (g) +
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2O2 (g)  CO2 (g) + 2H2O (l)
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Greaction = [Gf (CO2) + 2Gf (H2O)] - [Gf (CH4) + 2Gf (O2)]
= [(1 mol) (-394.4 kJ mol-1) + (2 mol) (-237.2 kJ mol-1)]
-
[(1 mol) (-50.8 kJ mol-1) +(2 mol) (0 kJ mol-1)]
= - 818.0 kJ this means reaction is very spontaneous!! (the answer could also be stated in kJ mol-1)
(b) 2MgO (s)
 O2 (g) + 2Mg (s)
(unit left in kJ mol-1)
Greaction = [2Gf (Mg) + Gf (O2)] - [2Gf (MgO)]
= [2 x (0 kJ mol-1) + (0 kJ mol-1)]
= 1139 kJ mol-1
[2 x (-569.6 kJ mol-1)]
-
which means reaction is not spontaneous at all.
Using Hess Law
We can also add up equations of which we know the free energy values to find the free energy value of
another equation; again the same rules apply.
Using G = Hsystem - TSsystem
The main problem in this type of calculation is that entropy and enthalpy are measured in different units; either
you change the entropy into kJ or do it the other way around.
Worked example:
Calculate G for the following reaction is:
3H2 (g) + N2 (g) 
2NH3(g) at 500 C
G = Hsystem - TSsystem
= - 92.38 kJ - (773 K) (0.198.3 kJ K-1) = - 92.38 kJ + 153.29 kJ = 60.91 kJ
This positive value means that at 500C the forward reaction is not spontaneous.
Exercises
1. Calculate the free energy for the following reaction using the free energies of formation in the table
below:
substance
NH3 (g)
HCl (g)
H2O (l)
MgCl2(s)
MgO (s)
N2H4 (g)
H2O (g)
Gf (in kJ mol-1)
- 26.5
-95.27
-236.81
-592.1
-569.6
159.4
-105.98
(a) N2(g) +
3H2 (g) 
(b) H2(g) +
Cl2 (g) 
(c) H2O (l) +
MgCl2 (s)  MgO (s) + 2HCl (g)
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substance
N2H4 (g)
CH3OH (l)
CO2 (g)
NO (g)
SO2 (g)
SO3 (g)
Gf (in kJ mol-1)
159.4
-166.23
-394.4
86.71
-300.4
-370.4
2NH3 (g)
2HCl (g)
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(d) 2NH3(g)  N2H4 (g) + H2 (g)
(e) 2CH3OH (l) +
(f)
N2(g) +
3O2 (g)  2CO2 (g) + 4 H2O (g)
O2 (g) 
(g) 2SO2 (g) +
2NO (g)
O2 (g)  2SO3 (g)
2. Calculate the free energy of the following reactions when given H298 and S298.
(a) 2SO2 (g) + O2 (g)  2SO3 (g) at 400 K.
H298 = -196.6 kJ
(b) H2O2 (g)  H2O (g) + ½O2 (g) at 298K
H298 = -106 kJ
(c) N2(g) +
O2 (g) 
2NO (g) at 298K
(d) Ca2+ (aq) + CO32- (aq)  CaCO3 (s) at 298K
S = - 189.6 J K-1.
S298 = + 58 J K-1.
H298 = +180 kJ mol-1
S298 = + 58 J K-1mol-1.
H298 = +13 kJ mol-1
S298 = -205 J K-1mol-1.
3. From the information below, calculate the standard free energy change, G , for the reaction
Fe3 O4 (s) + CO (g)  Fe (s) + Fe2 O3 (s) + CO2 (g)
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Fe2 O3 (s) + 3CO (g)  2Fe (s) + 3CO2 (g)
G = - 29.4 kJ
3Fe2 O3 (s) + CO (g)  2 Fe3 O4 (s) + CO2 (g)
G = - 61.6 kJ
4 hours
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IB questions
PAPER 1
1. (M08) Which change does not lead to an increase in entropy?
A.
B.
C.
D.
Mixing nitrogen and oxygen gases at room temperature
Cooling steam so that it condenses to water
Heating hexane to its boiling point
Dissolving sugar in water
2. (M08)
3.
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(M08)
4 hours
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4.
(M08)
5. (M08) Under what circumstances is a reaction spontaneous at all temperatures?
A.
B.
C.
D.
∆Hο
+
+
–
–
∆Sο
+
–
–
+
6. (M08)
A.
 x  y
B. x y
C.  x y
D. x y
7. (M08)
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4 hours
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8. (M08) Which compound has the highest lattice enthalpy?
A. CaO
B. CaS
C. LiF
D. LiI
9. (M08)
10. (N07)
A. 2.5 g
B. 5.0 g
C. 10 g
D.
160 g
11. (N07)
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4 hours
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–105 B.
A.
–48
C. +66
D. +123


12. (N07) The H and S values for a reaction are both negative. What will happen to the spontaneity of
this reaction as the temperature is increased?
A. The reaction will become more spontaneous as the temperature is increased.
B. The reaction will become less spontaneous as the temperature is increased.
C. The reaction will remain spontaneous at all temperatures.
D. The reaction will remain non-spontaneous at any temperature.
13. (N07) Which combination of ion charge and ion size produces the greatest lattice enthalpy?
A. High charge, large size
C. Low charge, small size
B. High charge, small size
D. Low charge, large size
14. (M07)
A. –297
C. –493
B. +297
D. +493
15. (M07) Which statement is correct for an endothermic reaction?
A.
B.
C.
D.
Bonds in the products are stronger than the bonds in the reactants.
Bonds in the reactants are stronger than the bonds in the products.
The enthalpy of the products is less than that of the reactants.
The reaction is spontaneous at low temperatures but becomes non-spontaneous at high
temperatures.
16. (M07)
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4 hours
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17. (M07)
18.
What is ∆H for the reaction below in kJ?
CS2(g) + 3O2(g)  CO2(g) + 2SO2(g)
[∆Hf / kJ mol–1: CS2(g) 110,
A. −570
19.
Topic15
CO2(g) – 390,
SO2(g) – 290]
B. −790
C. −860
D. −1080
Which reaction occurs with the largest increase in entropy?
A.
Pb(NO3)2(s) + 2KI(s) → PbI2(s) + 2KNO3(s)
B.
CaCO3(s) → CaO(s) + CO2(g)
4 hours
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22.
23.
3H2(g) + N2(g) → 2NH3(g)
D.
H2(g) + I2(g) → 2HI(g)
The ∆Hο and ∆Sο values for a certain reaction are both positive. Which statement is correct about
the spontaneity of this reaction at different temperatures?
20.
21.
C.
A.
It will be spontaneous at all temperatures.
B.
It will be spontaneous at high temperatures but not at low temperatures.
C.
It will be spontaneous at low temperatures but not at high temperatures.
D.
It will not be spontaneous at any temperature.
For which of the following is the sign of the enthalpy change different from the other three?
A.
CaCO3(s) → CaO(s)+ CO2(g)
B.
Na(g) → Na+(g) + e−
C.
CO2(s) → CO2(g)
D.
2Cl(g) → Cl2(g)
Which reaction has a positive entropy change, ∆Sο?
A.
H2O(g) → H2O(l)
B.
2SO2(g) + O2(g) → 2SO3(g)
C.
CaCO3(s) → CaO(s)+ CO2(g)
D.
N2(g) + 3H2(g) → 2NH3(g)
For a certain reaction at 298 K the values of both ∆Hο and ∆Sο are negative. Which statement
about the sign of ∆Gο for this reaction must be correct?
A.
It is negative at all temperatures.
B.
It is positive at all temperatures.
C.
It is negative at high temperatures and positive at low temperatures.
D.
It cannot be determined without knowing the temperature.
24. (N03) For which of the following compounds is the theoretical lattice energy likely to show the greatest
deviation from that calculated with a Born-Haber cycle?
A. NaCl
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B. CaF2
C. ZnF2
4 hours
D. ZnS
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25. In which of the following reactions is the entropy change the most positive?
2C (s) + O2 (g)  2CO (g)
CaO (s) + CO2 (g)  CaCO3 (s)
H2 (g) + Cl2 (g)  2HCl (g)
2SO2 (g) + O2 (g)  2SO3 (g)
A.
B.
C.
D.


26. For which one of the following is H of the reaction NOT equal to H f of the product?
A.
B.
C.
D.
½ H2 (g) + ½ I2 (g)  HI (g)
H2 (g) + ½ O2 (g)  H2O (l)
Ca (s) + C (s) + ½ O2 (g)  CaCO3 (s)
CO (g) + O (g)  CO2 (g)
27. The Born-Haber cycle for the formation of potassium chloride includes the steps below:
K (g)  K+ (g) + e½Cl2 (g)  Cl (g)
Cl (g) + e-  Cl- (g)
K+ (g) + Cl- (g)  KCl (s)
I.
II.
III.
IV.
Which of these steps are exothermic?
A
I and II only
B III and IV only
C
I, II and III only
D I, III and IV only
28. For the Born-Haber cycle shown below, all data are cited in kJ mol-1.
H = 1696.9
2Li (g) + O (g)  2 Li+ (g) + O2- (g)
Hat (Li) = 159.4
Hat (O) = 249.2
H = 2814
Hf
2Li (s) + ½O2 (g)  Li2O (s)
What is the Hf value for lithium oxide ( in kJ mol-1 )?
A. + 708. 5
B -673.7
C
-708.5
D -549.1
29. Which equation corresponds to the lattice enthalpy for AgI (s)?
A. AgI (s)
 Ag (s) + I (g)
B. AgI (s)  Ag (s) + ½ I2 (g)
C. AgI (s)  Ag+ (aq) + I- (aq)
D. AgI (s)  Ag+ (g) + I- (g)
30. In which of the following are the compounds CaF2, CaCl2, CsF and LiF arranged in increasing order of
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lattice enthalpy
A. CaCl2, CaF2, CsF, LiF
B. CsF, LiF, CaCl2, CaF2,
C. CaCl2, CaF2, LiF, CsF
D. LiF, CaF2, CsF, CaCl2
31. When the Hreaction is equal to Hformation then the
I.
II.
III.
reactants must be elements
reactants must be gases
reaction must be carried out at 25C and 1 atmosphere pressure.
A. I only
B. I and II only
C. I and III only
D. I, II and III
32. All of the following have a standard heat of formation value of zero at 25 C and 1.00 atm except
A.
Cl2 (g)
B. N (g)
C.
Fe (s)
D. Hg (l)
33. The standard heat of formation of C2H6 (g) at 25 C and 1 atm is represented by
A. 2C (g) + 6H (g)  C2H6 (g)
C. 2C (s) + 6H (g)  C2H6 (g)
B.
D.
2C (g) + 3H2 (g)  C2H6 (g)
2C (s) + 3H2 (g)  C2H6 (g)
34. Of the species listed below, the only one for which the standard enthalpy of formation is zero is
A. Na (l)
B. Hg (l)
35. For the reaction;
C. I 2 (l)
3HC=CH (g)  C6H6 (g)
D. H2O (l)
(there should be a triple bond)
H = -597.3 kJ and S = - 0.33 kJ K-1. This reaction
A. is spontaneous at 300K and becomes non-spontaneous at higher temperatures.
B. is spontaneous at 300K and becomes non-spontaneous at lower temperatures.
C. is non-spontaneous at 300K and becomes spontaneous at higher temperatures.
D. is non-spontaneous at 300K and becomes spontaneous at lower temperatures.
36. (N04) Using the equations below:
Mg (s) + ½ O2 (g)  MgO (s)
H2 (g) + ½ O2 (g)  H2O (g)
H = - 602 kJ
H = - 242 kJ
what is H value (in kJ) for the following reaction?
MgO (s) + H2(g)  Mg (s) + H2O (g)
A. -844
B. -360
C. +360
D. +844
Paper 2
1. (M09) Two students were asked to use information from the Data Booklet to calculate a value for the
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enthalpy of hydrogenation of ethene to form ethane.
C2H4(g) + H2(g)→C2H6(g)
John used the average bond enthalpies from Table 10. Marit used the values of enthalpies of combustion
from Table 12.
(a)
Calculate the value for the enthalpy of hydrogenation of ethene obtained using the average bond
enthalpies given in Table 10.
[2]
(b)
Determine the value for the enthalpy of hydrogenation of ethene using the values for the
enthalpies of combustion of ethene, hydrogen and ethane given in Table 12.
[2]
(c)
Suggest one reason why John’s answer is slightly less accurate than Marit’s answer and calculate
the percentage difference.
[2]
(d)
John then decided to determine the enthalpy of hydrogenation of cyclohexene to produce
cyclohexane.
C6H10 (l) + H2 (g) → C6H12 (l)
(i) Use the average bond enthalpies to deduce a value for the enthalpy of hydrogenation
of cyclohexene.
[1]
(ii) The percentage difference between these two methods (average bond enthalpies and
enthalpies of combustion) is greater for cyclohexene than it was for ethene. John’s hypothesis
was that it would be the same. Determine why the use of average bond enthalpies is less
accurate for the cyclohexene equation shown above, than it was for ethene. Deduce what
extra information is needed to provide a more accurate answer.
[2]
2. (M09)
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3. (N08)
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4. (M07)
5. (N03) For the process:
C6H6(l)  C6H6(s)
the standard entropy and enthalpy changes are:
∆Hο = −9.83kJ mol−1 and ∆Sο = −35.2J K mol−1.
Predict and explain the effect of an increase in temperature on the spontaneity of the process.
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[3]
6. (M04)
(a) The standard enthalpy change of formation of Al2O3(s) is −1669 kJ mol−1 and the standard
enthalpy change of formation of Fe2O3(s) is −822 kJ mol−1.
(i)
Use these values to calculate ∆Hο for the following reaction.
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
(ii)
(iii)
State whether the reaction is exothermic or endothermic.
[3]
Draw an enthalpy level diagram to represent this reaction. State the conditions under which
standard enthalpy changes are measured.
[3]
Estimate, without doing a calculation, the magnitude of the entropy change for this reaction.
Explain your answer.
[3]
(b) Explain in terms of Gο, why a reaction for which both Hο
sometimes
be spontaneous and sometimes not.
Sο values are positive can
[4]
(c ) Consider the following reaction.
N2(g) +
3H2(g) → 2NH3(g)
Hο, for this
(i)
reaction.
[3]
(ii)
The magnitude of the entropy change, S, at 27 °C for the reaction is 62.7 J K−1 mol–1. State,
with a reason, the sign of S.
[2]
(iii)
Calculate G for the reaction at 27 °C and determine whether this reaction is spontaneous at this
temperature.
[3]
Topic15
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Mark scheme
Paper 1
1
2
3
4
5
6
B
C
B
B
D
A
7
8
9
10
11
12
C
A
C
A
D
B
13
14
15
16
17
18
B
A
B
C
D
D
19
20
21
22
23
24
B
B
D
C
D
D
25
26
27
28
29
30
A
D
B
C
D
B
31
32
33
34
35
36
Paper 2
1.
2.
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C
B
D
B
A
C
3.
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4.
5.
T increases, −T∆Sο becomes larger / more positive;
∆G increases / becomes more positive / less negative;
process becomes less spontaneous / reverse reaction favoured;
3
6.
(a)
∆H = ∆Hf (products) −∆Hf (reactants) /= (−1669) − (–822)
= −847 kJ
Ignore units;
exothermic (ECF from sign of ∆H) ;
3
(ii)
2
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R
e
a
c
ta
n
ts
e
n
th
a
lp
y/e
n
e
rg
y
ignore lines connecting reactants and products
P
ro
d
u
c
ts
[1] for the diagram
ECF from sign of ∆H in (i)
298 K / 25°C AND 1 atm / 101(.3) kPa;
Both needed for the mark.
(iii)
entropy change will be zero / very small;
(structure of / bonding in) reactants and products similar / only solids involved;
the disorder / randomness will not change;
3
[8]
(b) a reaction is spontaneous when ∆Gο is negative / non-spontaneous when ∆Gο is positive;
at high T, ∆Gο is negative;
(because) T∆Sο is greater than ∆Hο;
at low T, ∆Gο is positive because T∆Sο is smaller than ∆Hο / OWTTE;
4
[4]
(c)
selection of all the correct bonds or values from Data Booklet;
∆H = (N≡≡N) + 3(H—H) − 6(N—H)/944 + 3(436) − 6(388);
= −76 (kJ);
3
Allow ECF for one error (wrong bond energy / wrong coefficient / reverse
reaction) but not for two errors
(so –611, –857, +76, +1088 all score 2 out of 3).
(ii)
(iii)
negative;
decrease in the number of gas molecules / OWTTE;
∆G = ∆H − T∆S
∆G = −76.0 − 300 (−0.0627);
Award [1] for 300 K.
Award [1] for conversion of units J to kJ or vice versa.
Allow ECF from c(i) from ∆H.
Allow ECF from c(ii) for sign of ∆S.
2
3
= −57.2 (kJ mol−1) is spontaneous / or non-spontaneous if positive value
obtained;
[3 max]
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