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Entropy
What is Entropy and why do we use it?
Entropy is described as “the number of ways atoms can share quanta of energy”
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Entropy is essentially a measure of disorder
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This is a driving force behind many reactions
and causes them to be spontaneous (occur without any external influence)
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System and Surroundings:
When considering entropy, you have to consider system and surroundings
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It doesn’t change temperature or pressure
and mass cannot be transferred to the surroundings
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What factors affect Entropy of System? (Sosystem)
Physical State affects entropy:
Solid particles will just vibrate about a fixed point, there’s hardly any disorder and
therefore solids will have the lowest entropy
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Dissolving a solid also increases its entropy
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For example:
The standard entropy of water vapour (H2O(g)) is 188
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91 clearly showing that the
physical state affects the entropy of a substance
Temperature affects entropy:
Temperature also affects entropy
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The higher the temperature, the more
energy particles will have and therefore the more disorder, as a result the greater the
entropy
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For example; liquid benzene (C6H6) has an entropy of 173
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98
The complexity of a substance affects entropy:
The more complex a substance, the more ways the particles can arrange themselves in,
the more disorder as a result and therefore the greater the entropy

Predicting change in entropy of system:
Based on a chemical equation, you can actually predict if the entropy of system will
increase or decrease in a given reaction
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In general; a significant increase in the entropy will occur if;
There is a change of state from solid or liquid to gas
There is a significant increase in the number of molecules between products and
reactants

Total Entropy Change:
During a reaction, there’s an entropy change between the reactants and the products –
the entropy change of system (as mentioned before)
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The total entropy change is
the sum of both the entropy change of system and surroundings
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This particular equation is useless unless you know the values
of the entropy change of system and the entropy change of surroundings
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The individual
entropies of the reactants and products will be given in the exam question; you just
need to calculate the difference
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This involves
dividing the enthalpy change (in J mol-1) by the temperature and turning this value
negative
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Never kJ K-1 mol-1 which explains why
the units for the enthalpy change (when calculating the entropy change of
surroundings) must always be J mol-1
Example:
Calculate the total entropy change for the reaction of ammonia and hydrogen chloride
under standard conditions
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3 J K-1 mol-1
S[HCl(g)] = 186
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6 J K-1 mol-1
First off, find the entropy change of the system;
We know that from the previous equation to calculate the entropy change of system, we
need to know the total entropy of the reactants and the products
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Sreactants = 192
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8 = 379
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6
Now we work out the difference between the values:

Ssystem = 94
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1 = -284
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This needs to
be converted so we multiply it by 1000 to change the units from kJ mol-1 to J mol-1
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So the value of T is 298

So now, here’s the equation:

∆Ssurroundings = -

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= -(-1057) = +1057 J K-1 mol-1

The minus sign is really important here
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To calculate the total entropy change, you just need to work out the sum of the entropy
change of system and that of surroundings like such;

∆Stotal = ∆Ssystem + ∆Ssurroundings
∆Stotal = -284
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5 J K-1 mol-1
Effect of Temperature on Total Entropy:
The equation ∆Stotal = ∆Ssystem + ∆Ssurrounding is essentially:

∆Stotal = ∆Ssystem + -

∆

If a reaction is exothermic and the temperature is high, the total entropy change
reduces
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If a reaction is exothermic and the temperature is low, the total entropy change will
increase
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If however a reaction is endothermic and the temperature is high, the total entropy
change will increase rather than decrease
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The entropy change of surroundings will increase and therefore have
more of an effect of the total entropy change
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The enthalpy change value is positive as the reaction is endothermic
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A reaction becomes thermodynamically feasible when the total
entropy change equals 0 or more
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Therefore make ∆Stotal 0
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Kinetic Favourability and Thermodynamic Feasibility:
Now, a reaction which is thermodynamically feasible isn’t a reaction which can always
happen
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You’ve got to consider the activation energy
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Sometimes in the exam, you’ll be asked to explain why a certain reaction cannot occur;
you need to remember to recall both ideas of thermodynamic feasibility and kinetic
favourability
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If a reaction is thermodynamically feasible, the reactants will be thermodynamically
unstable relative to the products
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If a reaction has a positive entropy (so is thermodynamically feasible) however doesn’t
happen, the reactants will be kinetically stable relative to the products as the activation
energy is very high
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Yes, there is a
Hess’ Law that you need to be aware of, but this one isn’t as bad as the ones you’ve
encountered at AS

Enthalpy of Lattice Formation:
By definition the enthalpy of lattice formation is the following:
The standard enthalpy change when 1 mole of an ionic crystal lattice is formed from its
constituent ions in gaseous form
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For instance in this case, the value would be +787kJ mol-1
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As Lattice Formation is always exothermic, the
value will be negative so if the value is negative, the value is representative of Lattice
Formation
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Now you might ask why is that always exothermic/endothermic? Lattice Formation is a
process where bonds are being formed correct? As bonds are made, energy is released,
therefore the process is exothermic (bond making is always exothermic)
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Energy is
required to break down bonds and therefore the process is endothermic
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Enthalpy of Hydration:
By definition the enthalpy of hydration is the following:
The standard enthalpy change when 1 mole of gaseous ions becomes hydrated such that
further dilution causes no further heat change
For example:
Li+(g) + aq Li+(aq)
F-(g) + aq F-(aq)

∆Hhyd = -519kJ mol-1
∆Hhyd = -506kJ mol-1

This is always an exothermic process because bonds are made between the ions and
water molecules realising energy in the process
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This is essentially what
hydration is; the formation of
bonds between ions and water
molecules

Negative ions are
attracted to the
δ+ hydrogen
atoms on the
water molecule
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You will be expected to calculate this during a Hess’ Law in the actual exam
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This value will determine whether or not, a particular solid can dissolve in solution or
not
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However if the value is negative (and therefore the
process is exothermic), the solid is SOLUBLE
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The strength of an enthalpy of lattice formation depends on two factors:
The size of the ions:
The larger the ions, the less negative the enthalpies of lattice formation (i
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a
weaker lattice)
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g
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For example; if you compared NaCl to MgCl2, you’ll realise that MgCl2 has a more
negative lattice dissociation enthalpy value
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This isn’t as bad as you encountered at AS but nonetheless
still evil
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This is energy required to break the lattice down to its constituent ions and
then the energy released from the bonds made between the ions and water molecules
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MgCl2
The ionic equation for dissolving this salt is;
MgCl2(s) + aq

Mg2+(aq) +2Cl-(aq)

∆Hsolution
Mg2+(aq) +2Cl-(aq)

MgCl2(s)

∆Hlattice dissociation

∆Hhydration
Mg2+(g) +2Cl-(g)

The Hess’ Law diagram is for the hydration of Magnesium Chloride
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The arrow for the enthalpy change of lattice dissociation can actually be reversed so
that both are going up, however this can only be done if the value becomes the enthalpy

change of lattice formation
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Based on this Hess’ Law, the equation to calculate the value for the enthalpy change of
solution is:

∆Hlattice dissociation + ∆Hhydration

∆Hsolution

If one may choose to use the enthalpy change of lattice formation rather than the
enthalpy change of dissociation, the equation will be:

-∆Hlattice formation + ∆Hhydration

∆Hsolution

Do note that the value for the enthalpy change of lattice formation is the same as the
negative value for the enthalpy change of lattice dissociation
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However
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Entropy ties in with the topic of solubility and
dissolving
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A solid salt will have a lower entropy compared to a
dissolved salt because a dissolved salt is more disordered; it’s more “spread out” in the
solution rather than clumped together
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The entropy of
surroundings looks at the enthalpy change and temperature correct? Here the enthalpy
change will be that of solution
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However when the enthalpy change of solution is positive (endothermic), the equation
will look like this;

∆Ssurroundings = -

∆
or alternatively

-∆

So the overall entropy of surrounding will be negative
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When looking at the total entropy change, you’ve got to factor in both surroundings and
system
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