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3.2 Table of
entropy data and comments
The examples below
illustrates some of the ideas described in section 3.1 above.
S denotes the entropy content
of a substance and it always has a positive value. It increases with
increase in temperature as more energy levels (rotational, vibrational,
electronic etc.) become more available as energy is absorbed and
redistributed as the temperature increases.
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Examples
– and brief explanation of entropy trend |
Entropy values
below are the Sθ298K,1atm
in J K–1 mol–1 |
|
Group 0 Noble Gases:
A series of monatomic molecules – single atoms – an increasing
complexity of electron arrangement – more electrons and energy levels used to
distribute the electronic energy. |
He(g) |
Ne(g) |
Ar(g) |
Kr(g) |
Xe(g) |
|
126 |
146 |
155 |
164 |
170 |
|
Linear alkanes
– increasingly complex molecules, more electrons, more bond
vibrations, rotations of groups of atoms. Sθpentane
quoted for the gas only, even though it is a liquid at 25oC
and 1atm pressure. |
CH4(g) |
C2H6(g) |
C3H8(g) |
C4H10(g) |
C5H12(g) |
|
186 |
230 |
270 |
310 |
348 |
|
Water and
benzene –
illustrating the difference between the entropy of a gas,
liquid and solid. The more condensed liquid form offers less ways of arranging
the molecules. The entropy decrease for highly polar water from
gas ==> liquid is particularly marked because even in the liquid
there are clumps of molecules held together by hydrogen bonds – a bit
of 'pseudo solid' structure decreasing the entropy more! This
'clumping' is far less likely in benzene in which only Van der
Waals forces are operating so the entropy is greater. The
entropy for a solid is even less than for that of the liquid –
more ordered arrangement and less variation of possible states,
though the entropy decrease from liquid to solid is not as
marked as the entropy decrease from a liquid to gas. In an
interactive sense i.e. intermolecular forces AND the possible
ways of 'molecular arrangement', a liquid is closer to a solid
than to a gas. |
H2O(g) |
H2O(l) |
H2O(s) |
C6H6(g) |
C6H6(l) |
|
189 |
70 |
44.8 |
269 |
173 |
3.3 ΔS, Entropy changes
and the feasibility of a change – physical or chemical
3.3a The Entropy
Change of a state changes
The entropy change associated with a
physical state change:
The direction of
change of a system which is not at equilibrium (physical or chemical) is
determined by the overall entropy change. Up to now we haven't considered
the whole situation where there is potential for change. The whole
situation meaning system and surroundings.
From data tables of
entropy (e.g. as in table 3.2 above) you can readily calculate the
entropy change of the system.
ΔSθsys
= ΣSθfinal
state – ΣSθinitial
state
The entropy change
for the surroundings is minus the enthalpy change of the system change
divided by the absolute temperature in Kelvin.
ΔSθsurr
= –ΔHθsys/T(K)
You do not need
to know where this comes from.
The total entropy
change is then given by the 'simple' equation i.e. the sum of all the
entropy changes.
ΔSθtot
= ΔSθsys + ΔSθsurr
For a change to
be feasible the total entropy change must be at least greater or equal
to zero.
i.e. ΔSθtot
>= 0
but note that
when ΔSθtot
= 0, an equilibrium formed, in fact this is the criteria for
equilibrium formation!
In this section
3.3a we will apply this rule to two physical state changes.
3.3a1 The melting of a pure
substance – water
ΔSH2O
melting = +22.0 J mol–1 K–1, ΔHmelt/fusion(solid
H2O) = +6.01 kJ mol–1, T must be in K
For the change:
H2O(s)
==> H2O(l)
ΔSθsys
= Sθfinal
state – Sθinitial
state
ΔSθsys
= SθH2O(l)
– SθH2O(s) = +22.0 J
mol–1 K–1
(I couldn't find
the individual S values for water at 0oC, just the delta
S value)
ΔSθsurr
= –ΔHθfusion ice/T = –(+6010/T)
therefore
for ice melting (endothermic) ...
ΔSθtot
= ΔSθsys + ΔSθsurr
= 22.0 – 6010/T
watch the
units! convert kJ to J here and T in Kelvin
So what we can
do now is to try three temperatures in the above expression (i) 268
K (–5oC), (ii) 273 K(0oC) and (iii) 278 K(5oC)
and see what the result is i.e. what is predicted ...
For (i) ΔSθtot
= 22.0 – 6010/268 = –0.425 J mol–1 K–1
negative
total entropy change, therefore melting not possible at 268 K
(–5oC)
For (ii) ΔSθtot
= 22.0 – 6010/273 = –0.014 J mol–1 K–1
(theoretically it is zero with sufficiently accurate data)
total
entropy change close to zero at 273 K (0oC), in fact
at 273.15 K ice and water would coexist in equilibrium and delta
S would zero. Whether melting or freezing would take place at
273.15 K depends on the direction of heat transfer to or from
the surroundings – remember there is no temperature change in
this state change but melting occurs if heat is absorbed by the
system or freezing happens if heat is lost to the surroundings
in this case.
For (iii) ΔSθtot
= 22.0 – 6010/278 = +0.38 J mol–1 K–1
positive
total entropy change, therefore melting possible at 278 K (5oC)
This simple
example of a physical state change exemplifies feasibility
calculations i.e. what changes are possible theoretically?
TOP OF PAGE
3.3a2 The freezing of a mixture
e.g. salt solution
Approach (i)
The spreading of
salt (sodium chloride) on roads is a tried and tested method of de–icing
roads or preventing ice formation. So, why does it work? why is the
melting/freezing point of water lowered? – well its all down to the
entropy change! Unfortunately I can't find accurate enough data to do an
illustrative calculation. However, using the entropy equations, we can
do a 'thought experiment' and reason out why adding salt lowers the
freezing point of water, or to state it another way, why salt solution
has a lower freezing point than pure water.
The most important
point to grasp is that salt solution has a greater entropy than pure
water. There are now three different particles in the liquid instead of one i.e.
sodium and chloride ions as well as water molecules. When salt solution
freezes pure water will crystallise out of the liquid.
For the change:
H2O(l) ==> H2O(s)
(but from salt solution)
ΔSθsys
= Sθfinal state – Sθinitial
state
ΔSθsys
= SθH2O(s) – SθNaCl(aq)
= –(22.0 + x) J mol–1 K–1
Note (i) The
entropy change is negative as we are going from liquid to the
more ordered solid state – decreased possible particle
arrangements.
(ii) x
represents the extra entropy of the salt solution compared to
pure water and the more salt in the water, the higher the
entropy of the mixture, so x will be even bigger.
ΔSθtot
= ΔSθsys + ΔSθsurr
= –(22.0 + x) + (+6010/T)
Note that ΔSθsurr
= –ΔHθfusion ice/T = –(–6010/T) = +6010/T
Its
reasonable here to assume ΔHmelt/freeze is ~same for
the salt solution i.e. the crystallisation of pure water to form
ice, but the sign is changed, freezing is an exothermic process.
ΔSθtot
= –(22.0 + x) + 6010/T
Now think! the
greater x, the more salt, the greater the entropy change, then the smaller T must be to give
a positive total entropy change. When x is zero, T is 273,
when x is over zero then T must be less than 273 to give a
positive ΔSθtot. So lowering of the
freezing point of salt solution is all down to the higher entropy
state of salt solution.
Two other
approaches to the salt solution–ice situation
(ii) Alternative
calculation approach but using the data from above:
When water
and ice co–exist in equilibrium ΔSθtot
= 0 = ΔSθsys + ΔSθsurr
= ΔSθsys + (–ΔHθsurr/T)
so if x
represents the extra entropy change for ice <=> salt solution
compared to just ice <=> pure water ...
for melting
(endothermic ΔH +ve, increase ΔS +ve), so 0 = +(22 +
x) –
(+6010/T)
for freezing
(exothermic ΔH –ve, decrease ΔS –ve), so 0 = –(22 + x)
–(–6010/T)
therefore
either way ΔSθsys = ΔSθsurr
at equilibrium and if x increases then T decreases numerically
in order for the criteria for equilibrium to be satisfied.
(iii) A simple
'limited' conceptual
approach:
In terms of
entropy: ice + salt < ice + pure water < ice + salt solution
(<== increasing order)
Therefore
the highest entropy state will be attained at the lowest
possible temperature.
However,
unless you put data into the equations it doesn't exactly prove
a lot. However using the total entropy equation plus a
conceptual picture is a most satisfying situation – the ideal in
chemistry!
A simple experiment to determine
the enthalpy of vaporistion of water and the accompanying entropy change
Procedure
You need an electric kettle of
known power that can be accurately weighed to the nearest 0.1g.
Add about a litre of water to the
kettle and get the water to boiling point.
Turn off the power and weigh the
kettle and water.
Quickly, reconnect with the power
and boil for two minutes, stop boiling and quickly reweigh the
kettle.
Typical results
Mass of water evaporated/boiled
off = 119.0 g.
Time of boiling = 60 x 2 = 120 s
at 100oC (373 K)
Calculation
(i) Enthalpy of vaporisation of
water
Assuming the kettle's power
rating is 2.4 kW = 2400 W = 2400 Js-1
Energy transferred = power x time
= 2400 x 120 = 288,000 J
Ar's: H = 1, O = 16, Mr(H2O)
= 18, Moles of water = 119/18 = 6.611
Enthalpy of vapourisation =
288,000/6.611 = 43563 Jmol-1
Therefore ΔHvap(H2O(l))
= 43.6 kJmol-1
The quoted book value of the
enthalpy of vaporization of water is 41.1 kJmol-1
(ii) Entropy change of the
vaporisation of water
Since heat is transferred to
liquid water converting it to gaseous water (steam!) there is a
large positive increase in entropy.
ΔS = ΔH/T, ΔS =
entropy change, ΔH = enthalpy change,
T = temperature K at which
the state change takes place
ΔSvap(H2O(l))
= ΔHvap(H2O(l))/Tbpt =
43563/373 kJmol-1 = +117 Jmol-1K-1
I couldn't find the entropy data
for 373 K, but at 298 K, the entropy change can be calculated as
follows
ΔSθsys
= ΣSθfinal
state – ΣSθinitial
state
ΔSθsys
= SθH2O(g)
– SθH2O(l) = 189 - 70 = +119 J
mol–1 K–1
Its a simple experiment to do,
not that accurate, but very instructive!
Energetics-Thermochemistry-Thermodynamics Notes INDEX
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QUICK INDEX for
Energetics:
GCSE Notes on the basics of chemical energy changes
– important to study and know before tackling any of the three Advanced Level
Chemistry pages Parts 1–3 here
* Part 1a–b
ΔH Enthalpy Changes
1.1 Advanced Introduction to enthalpy changes
of reaction,
formation, combustion etc. : 1.2a & 1.2b(i)–(iii)
Thermochemistry – Hess's Law and Enthalpy
Calculations – reaction, combustion, formation etc. : 1.2b(iv)
Enthalpy of reaction from bond enthalpy
calculations : 1.3a–b
Experimental methods
for determining enthalpy changes and treatment of results and
calculations :
1.4
Some enthalpy data patterns : 1.4a
The combustion of linear alkanes and linear
aliphatic alcohols
:
1.4b Some patterns in Bond
Enthalpies and Bond Length : 1.4c
Enthalpies of
Neutralisation : 1.4d Enthalpies of
Hydrogenation of unsaturated hydrocarbons and evidence of aromatic
ring structure in benzene
:
Extra Q page
A set of practice enthalpy
calculations with worked out answers **
Part 2 ΔH Enthalpies of
ion hydration, solution, atomisation, lattice energy, electron affinity
and the Born–Haber cycle : 2.1a–c What happens when a
salt dissolves in water and why? :
2.1d–e Enthalpy
cycles involving a salt dissolving : 2.2a–c
The
Born–Haber Cycle *** Part 3
ΔS Entropy and ΔG Free Energy Changes
: 3.1a–g Introduction to Entropy
: 3.2
Examples of
entropy values and comments * 3.3a ΔS, Entropy
and change of state : 3.3b ΔS, Entropy changes and the
feasibility of a chemical change : 3.4a–d
More on ΔG,
free energy changes, feasibility and
applications : 3.5
Calculating Equilibrium
Constants from ΔG the free energy change : 3.6
Kinetic stability versus thermodynamic
feasibility - can a chemical reaction happen? and will it happen?
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