1.4
Some enthalpy data patterns continued
1.4c(i) Enthalpies of
Neutralisation and Enthalpies of Formation of oxides and hydrides
|
Neutralisation reaction: base + acid ==> salt +
water |
ΔHneutralisation
kJ mol–1 |
|
1.
NaOH(aq) + HCl(aq) ==> NaCl(aq)
+ H2O(l) |
–57.1 |
|
2.
KOH(aq) + HCl(aq) ==> KCl(aq)
+ H2O(l) |
–57.2 |
|
3.
NaOH(aq) + HNO3(aq) ==> NaNO3(aq)
+ H2O(l) |
–57.3 |
|
4.
KOH(aq) + HNO3(aq) ==> KNO3(aq)
+ H2O(l) |
–57.3 |
|
5.
Ba(OH)2(aq) + 2HNO3(aq) ==>
Ba(NO3)2(aq) + 2H2O(l) |
–116.4 (–58.4 per H2O) |
|
6.
NaOH(aq) + HF(aq) ==> NaF(aq)
+ H2O(l) |
–68.6 |
|
7.
NH3(aq) + HCl(aq) ==> NH4Cl(aq) |
–52.2 |
|
8.
NaOH(aq) + CH3COOH(aq)
==> CH3COONaaq) + H2O(l) |
–55.2 |
|
9.
NH3(aq) + CH3COOH(aq)
==> CH3COONH4aq) |
–50.4 |
|
10.
KOH(aq) + HCN(aq) ==> KCN(aq)
+ H2O(l) |
–11.7 |
|
11.
NH3(aq) + HCN(aq) ==> NH4CN(aq) |
–5.4 |
Notes on the above examples
from the data table above.
Beware on comparing values!
(i)
NaOH(aq)
+ HCl(aq) ==> NaCl(aq) + H2O(l) ΔHθneutralisation
= –57.1 kJ mol–1
(ii)
Ba(OH)2(aq) + 2HNO3(aq)
==> Ba(NO3)2(aq) + 2H2O(l) ΔHθneutralisation
= –116.4 kJ mol–1
(iii)
1/2Ba(OH)2(aq) + HNO3(aq)
==> 1/2Ba(NO3)2(aq) +
H2O(l) ΔHθneutralisation
= –58.2 kJ mol–1
Note! It looks as if the
enthalpy of neutralisation of barium hydroxide is approximately
double that of sodium hydroxide ie ~ twice as exothermic! Well
yes it is! and no it isn't!
Yes – ~twice as
much energy is released per mole of soluble base/alkali.
No – however, on the basis of
heat released per mole of water formed, they are actually
very similar. In other words, which value you
quote, depends on which point you want to make.
Yet another example of carefully
qualifying enthalpy values with respect to the context.
Comparing Weak & strong
acids or bases (soluble–alkali)
Neutralisation
reactions 1., 2., 3., 4. and 5. are reactions between ~fully
ionised strong acids and strong bases and the only real chemical
change is formation of a water molecule from hydrogen/oxonium ions
and hydroxide ions.
H3O+(aq)
+ OH–(aq) ==> 2H2O(l)
or more simply H+(aq) + OH–(aq)
==> H2O(l) ΔH = ~–57 kJ mol–1
All the
other ions e.g. Na+, K+, Ba2+,
Cl–, NO3–, SO42–
etc. are spectator ions and take no part in the reaction.
The
resulting salt solution has a pH of ~7.
Wherever one
of the acids or bases is 'weak' the energy change is less exothermic
compared to a pair of 'strong' reactants.
The reason for
this relates to the fact that if an acid or base is not fully
ionised, then energy must be expended facilitate the ionisation.
Weak acids or weak bases are on partially ionised (e.g. 2%) in
aqueous solution. Or, you can argue theoretically another way in
terms of salt hydrolysis i.e. the ions of the salt react with water
to 'reform' some of the acid or the base. In other words, you don't
really get complete neutralisation of the acid and base on a
strictly stoichiometric molar basis. (for more on acid–base theory
see
Equilibria Part 5.)
Reaction 7. is
the neutralisation of the weak base ammonia and the strong
hydrochloric acid.
NH4+(aq)
+ H2O(l)
 H3O+(aq) + NH3(aq)
ammonium
chloride solution has a pH of ~4
Reaction 8. is
the neutralisation of the strong base sodium hydroxide and the weak
organic carboxylic acid, ethanoic acid.
CH3COO–(aq)
+ H2O(l)
CH3COOH(aq) + OH–(aq)
sodium
ethanoate solution (CH3COONa(aq)) solution
has a pH of ~9
Reaction 9. is
even less exothermic because this is the neutralisation of a weak
base (ammonia) and a weak acid (ethanoic acid).
See the
equations for reactions 7. and 8. above
Reaction 10. is
the neutralisation of an extremely weak acid (HCN) and, despite KOH
being a strong base, only about 1/5th of the
energy of a strong acid – strong base neutralisation is released.
CN–(aq)
+ H2O(l)
 HCN(aq) + OH–(aq)
potassium
cyanide solution (KCN(aq)) has a pH of over 10.
Reaction 11. is
the neutralisation of an extremely weak acid (HCN) and a weak base
(NH3) in which only a 1/10th of the maximum possible
energy is released.
See
equations for reaction 7 and 10.
Reactions 6 is
unusually high despite the fact that hydrofluoric acid is a weak
acid.
TOP OF PAGE
1.4c(ii) Examples of series
of enthalpies of formation of similar compounds
Most of the time
there is little point in comparing the enthalpies of formation of
dissimilar compounds, but for a series of like compounds useful
patterns emerge e.g.
|
Series and comments: |
Members of the series: |
|
Group VII Halogen
Halides – the value becomes less exothermic from F ==>
I, a reflection of the increasing size of the halogen atom,
but emphasises the particularly high stability of hydrogen
fluoride. |
HF(g) |
HCl(g) |
HBr(g) |
HI(g) |
|
Enthalpy of formation ΔHf,298(HX(g))
for 1/2H2(g) + 1/2X2(g/l/s)
==> HX(g) in kJ mol–1 |
–271.1 |
–92.3 |
–36.2 |
+26.5 |
|
The organic homologous
series of hydrocarbons the Alkanes – where the
enthalpy of formation becomes progressively more exothermic
as the series is ascended. |
CH4(g) |
C2H6(g) |
C3H8(g) |
C4H10(g) |
|
Enthalpy of formation ΔHf,298(CnH2n+2)
for nC(s) + (n+1)H2(g) ==> CnH2n+2(g)
in kJ mol–1 |
–74.9 |
–84.7 |
–104.0 |
–125.0 |
Sometimes for a
series of compounds different ways of viewing the enthalpy values
help in comparing values e.g. the enthalpy of formation of the
Period 3 oxides with the element in its highest oxidation state.
|
Period 3 oxide |
Na2O(s) |
MgO(s) |
Al2O3(s) |
SiO2(s) |
P4O10(s) |
SO3(g) |
Cl2O7(g) |
|
Enthalpy of formation ΔHf,298(per
mole of oxide) in kJ mol–1 |
–416 |
–601 |
–1676 |
–911 |
–2984 |
–395 |
+265 |
|
Enthalpy of formation ΔHf,298(per
mole of oxygen) in kJ mol–1 |
–832 |
–1202 |
–1117 |
–911 |
–597 |
–264 |
+76 |
|
Oxidation state of the
Period 3 element Na to Cl in its highest valency oxide |
+1 |
+2 |
+3 |
+4 |
+5 |
+6 |
+7 |
|
Type of structure |
ionic lattice |
ionic lattice |
ionic lattice |
giant covalent |
small molecule |
small molecule |
small molecule |
-
By comparing the
values per mole of oxygen you get a better comparison of the
energetics of the reaction of the element with oxygen.
-
For example
it looks as if phosphorus(V) oxide is the most energetically formed
and seems way above the other oxides but when compared on a 'per
mole of oxygen basis' it fits in with a more 'gentler' pattern of
initially rising and then gradually falling enthalpy values of oxide
formation.
-
As a very rough rule of thumb, very negative values
suggest high thermodynamic stability e.g. the high energy needed via
electrolysis to extract reactive metals like sodium, magnesium and
aluminium or the very stable giant covalent structure of silicon
dioxide.
-
The positive value of the enthalpy of formation of
chlorine(VIII) oxide suggests it might be unstable and it is! It
decomposes readily on heating and is a highly reactive oxidising
agent.
-
Other variations in the data for a series of compounds can be
due to differences in structure.
-
So, PLEASE do not always expect
nice consistent patterns in enthalpy data that you see in enthalpies
of combustion of alkanes/alcohols or enthalpies of formation of
alkanes – here you are dealing with similar molecules with
~identical C–H, C–C or C–O covalent bonds and constant 'incremental'
changes in the length of the molecule.
TOP OF PAGE
1.4d Enthalpies of
Hydrogenation of unsaturated hydrocarbons and benzene structure
-
The
enthalpy of hydrogenation can be defined as the energy released
when one mole of an unsaturated hydrocarbon molecule becomes
saturated on reaction with hydrogen.
-
For
comparative purposes, the values of ΔHhydrogenation
quoted here are for gaseous species only at 355K and 101.3
kPa.
-
Since ΔHθf(hydrogen)
= 0 by definition for an element, at 298K and 101.3 kPa
-
ΔHθhydrogenation,298(compound)
= ΔHθf(saturated compound) = ΔHθf(unsaturated
compound)
-
Examples:
-
CH2=CH2
+ H2 ==> CH3CH3
ΔHhydrogenation(ethene) = –157 kJ mol–1
-
CH3CH=CH2
+ H2 ==> CH3CH2CH3
ΔHhydrogenation(propene) = –126 kJ mol–1
-
CH3CH2CH=CH2
+ H2 ==> CH3CH2CH2CH3
ΔHhydrogenation(but–1–ene) = –127 kJ mol–1
-
CH3CH=CHCH3
+ H2 ==> CH3CH2CH2CH3
ΔHhydrogenation(but–2–ene) = –116/120 kJ
mol–1
-
CH2=CH–CH=CH3
+ 2H2 ==> CHCH2CH2CH3
ΔHhydrogenation(but–1,3–diene/1,3–butadiene) =
–239 kJ mol–1
-
+ H2 ==>
or
+ H2 ==>
-
+ 2H2 ==>
or
+ 2H2 ==>
-
C6H6
+ 3H2 ==> C6H12 or
+ 3H2 ==>
-
hydrogenation of
benzene to cyclohexane
-
ΔHhydrogenation(benzene)
= –208 kJ mol–1
-
Compare with the fictitious equation
10.
-
C6H5CH3
+ 3H2 ==> C6H11CH3
or
+ 3H2 ==>
-
+ 3H2 ==>
-
Discussion
of the above enthalpy of hydrogenation values:
-
Apart
from the first and simplest alkene, ethene, most enthalpies
of hydrogenation are about –120±6
kJ mol–1 (for 1 mole H2 per mole
alkene) and in the case of dienes about double that for two
moles of hydrogen per diene, which is what you might
reasonably expect.
-
The special case of
benzene – the aromatic ring structure of arenes –
aromaticity
-
However,
on the basis of these trends, the expected value for benzene
and other aromatic compounds with a single benzene ring
would be around 3 x –120 = –360 kJ mol–1, but not so!
-
In
fact the energy released on hydrogenating benzene (208
kJ/mole) is even less than hydrogenating a diene!
-
So,
something must be different about benzene but it can be
explained with the enthalpy level diagram shown below and an
examination of possible molecular structures.
-
-
The
'top' molecule in the diagram shows the theoretical
structure of a triene with the same molecular formula of
benzene (C6H6) and, if it existed in
this form it would be called cyclohexa–1,3,5–triene or
1,3,5–cyclohexatriene.
-
This
molecular structure assumes there are simple alternate
single (C–C) and double (C=C) carbon-carbon bonds.
-
BUT,
according to the actual thermochemical data calculated and
derived from e.g. enthalpies of hydrogenation, benzene is
already more stable by ~152kJ mol–1, so, whatever
its structure, it cannot have this 'triene' structure.
-
 What
happens in reality, is that the equivalent of 3 double bonds
(C=C) and 3 single bonds (C–C) 'merge' to form a six equal
bonds each involve an average of 3 shared electrons.
-
Two of
the electrons for each bond are concentrated between the two
carbon atoms OR between a carbon atom and hydrogen atom,
both equivalent to a single bond known as a sigma
(σ) bond.
-
The 4th electron per
carbon atom is located in a ring
orbital, of two sections, above and below the plane of the
hexagonal ring of carbon atoms.
-
These are known as pi (π)
orbitals and each contains 3 pi (π) electrons which are
delocalised around the ring.
-
Whenever charge is delocalised
or 'spread out' the potential energy of the system is
lowered and in the case of benzene about 160 kJ per mole.
-
This is indicated by the ring in the centre of the ring
either in skeletal formula or structural formula
-
Apart
from the thermochemical evidence argued above, actual bond
length measurements from X–ray crystallography back up the
aromatic ring theory and show that there are six carbon
carbon bonds all of equal length and intermediate between
single and double bonds – this gives the 'benzene or
aromatic ring' a symmetrical hexagonal shape.
-
Typical
bond lengths: single bond C–C is 0.154 nm (bond order 1)
e.g. in alkanes, aromatic ring bond
is
0.139
nm (bond order 1.5) e.g. in benzene or methylbenzene etc.
and a double bond C=C is 0.134 nm (bond order 2)
e.g. in alkenes. Incidentally the triple bond in alkynes has a
typical length of 0.120 nm (bond order 3). There term bond
order refers to the 'electronsworth' of the bond i.e. the
average number of shared electrons involved in that
particular bond.
-
For a wider discussion on
bond lengths.
See also
A set of
enthalpy calculation problems with worked out answers – based on
enthalpies of reaction,
formation, combustion and bond
enthalpies
[SEARCH
BOX]
TOP OF PAGE
|
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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