Energetics and Bonding
A.
Introduction
Chemical reactions are usually associated with transfer energy; if the amount of energy
transferred can be measured, then it is possible to obtain vital information about the
interactions which occur during a reaction. Most frequently, energy transfer in
chemical reactions takes place in form of heat.
Exothermic reactions transfer heat to surroundings, for example,
4Al(s)
+ 3O2(g)
−−>
2Al2O3(s)
In the above reaction, 1,676kJ of heat are evolved to the surroundings for every mole
of aluminium oxide formed at 298K(at constant pressure).
Endothermic reactions take place with transfer heat from the surroundings to
chemical system, for example,
N2(g) + O2(g)
−−> 2NO(g)
In the above reaction, 90.4kJ of heat are absorbed from the surroundings for every
mole of nitrogen oxide formed at 298K(at constant pressure).
There are two types of reaction with the following energy profiles.
reagents
products
products
reagents
Endothermic reaction has positive sign of ∆H(= +ve).
Exothermic reaction has negative sign of ∆H (= -ve).
"Energetics" is the study of the energy changes of chemical reactions and a study of
the changes in heat energy accompanying chemical reactions is known as
"Thermochemistry".
In fact, we are not only interested in the values of energy or heat change. On the other
hand, we are interested in the reason and the consequences behind the energy changes.
Before we can interpret energy transfer (energy change), in terms of molecular
interactions taking place during chemical reactions (and more specifically in terms of
the breaking and forming of chemical bonds), we must look closely at what it is we
are measuring and at the conditions under which we are taking the measurements.
B.
Internal energy (U)
2
It is useful to think of a chemical system is possessing an internal energy, U. It is not
possible to measure the absolute value of this energy of a system but it is possible to
measure its change, ∆U.
The internal energy of a system is the sum of the internal energies of its constituent
molecules and these are comprised of several contributions, some kinetic and some
potential.
U =
UT
+
UR
T = Translational energy
R = Rotational energy
∆U
=
Ufinal -
+
UV
+
UE
V = Vibrational energy
E = Electronic energy
Uinitial
If the system transfers heat to the surrounding (that is exothermic reaction), then ∆U
is negative. Conversely, for endothermic reactions, the system gains heat from the
surroundings and ∆U is positive.
C.
Enthalpy (H)
Consider the energy change associated with the reaction,
Zn(s)
+
2H+(aq)
−−>
Zn2+(aq)
+
H2(g)
Work is done by Hydrogen gas in expanding against the atmospheric pressure. Since
the performance of work involves the transfer of energy, so the value of ∆U we
measured will not be true value. We now define a new term, the enthalpy change, ∆H,
which takes into account the work done when a system expands or contracts during a
reaction at constant pressure.
∆H = ∆U + p∆V
If work is done by the system, p∆V is positive. If work is done on the system p∆V is
negative. (Why?)
When a system absorbs heat energy, its internal energy increased. If the volume of the
system is fixed (∆V = 0), all heat absorbed will be changed to internal energy. That is
∆H = ∆U.
If the reaction takes place at constant pressure, then the heat absorbed is equal to the
increase in internal energy plus the work done.
The values of ∆H and ∆U only depend on the final and initial states of the system,
they do not matter how the change is made. They are state functions.
For example, the change in potential energy of a block of mass M when taking from
3
3m above sea level to 7m above sea level is
7m
3m
sea level
change in potential energy =
Such change in potential change ___________________________________________,
so it is a state function.
D.
Measuring the heat change
Basic concepts
In discussing enthalpy changes earlier, we spoke of heat transfers to measure them. In
practice, it is much easier to measure the enthalpy change of a reaction using a
calorimeter in which the system is insulated from the surroundings.
If the maximum temperature change of the system is recorded and if the heat capacity
of the system is known, it is easy to calculate the quantity of heat which would have to
be taken from or into the system to restore it to its initial temperature. This quantity of
heat is the enthalpy change.
Example:
50.0cm3 of 1.25 M hydrochloric acid and 50.0cm3 of 1.25 M potassium hydroxide
solution at the same temperature are put into a polystyrene cup as shown below. The
temperature of the resulting solution rises by 8.40oC. Calculate the standard enthalpy
change of neutralization. (specific heat capacity of solution is 4.18 kJ/K/kg, assume no
heat lost to the cup and surroundings and the density of the resulting solution =
1.0g/cm3)
Note: Enthalpy change of neutralization is the enthalpy change per mole of water
formed during neutralization.
heat gained by the resulting solution = 0.100(kg) x 4.18 (kJ/kg/K) x 8.40(K) = 3.51
(kJ)
no. of mole of water produced = 1.25 x 0.0500 = 0.0625 (mole)
3.51
0.0625
= 56.2(kJ/mol)
amount of heat produced by one mole of water =
Therefore heat of neutralization = -56.2kJ/mol(H2O)
(The negative sign means that neutralization is an exothermic reaction.)
Class work:
An excess of zinc powder was added to 50.0cm3 of 0.100M AgNO3 in a
4
polystyrene cup. Initially the temperature was 21.10oC and it rose to 25.40oC,
calculate the enthalpy change for the reaction:
Zn(s) + 2Ag+(aq)
−−>
Zn2+(aq) + 2Ag(s)
Assume that the density of the solution is 1.00g/cm-3 and its specific heat
capacity is the same as water. Ignore the heat capacity of the metals and the
polystyrene cup.
Answer: ∆Ho(298K) = -180kJ/mol(Ag)
E.
Conventions, symbols and standard states
As we know, the magnitude of heat energy transfer in a chemical reaction depends
upon the changes that occur in different factors which contribute to the internal energy
of the system. We want to eliminate all of these changes. Only then will it be possible
to make valid comparisons between ∆H values for different reactions. This means
that certain standard conditions must operate and certain conventions must be obeyed.
a.
Amount of substance
Clearly, the amount of energy transferred will depend upon the amount of
substances which reacted. So if we write
H2(g) + Cl2(g)
−−>
2HCl(g)
∆H = -184.6kJ
means that when one mole of Hydrogen gas reacts with one mole of Chlorine
gas to produce two moles of Hydrogen chloride gas completely, the change is
accompanied by the output of 184.6kJ of heat energy to the surroundings. For
producing one mole of HCl, the heat change will be -92.3kJ.
b.
Temperature
For the purposes of valid comparison, values of ∆H are usually quoted at a
standard temperature of 298K (although the choice of temperature is quite
arbitrary).
c.
Pressure
∆H values are quoted for a pressure of 1 atmosphere (standard atmospheric
pressure). Solutions must have unit concentration (1 moledm-3).
d.
Standard enthalpy changes
5
Enthalpy changes that have been measured under the conditions of
temperature and pressure stated above are known as standard enthalpy changes
and the symbol is written thus:
o
∆H (298K)
The temperature should be specified in a blanket, but if it is not, it should be
298K.
e.
State symbols
Since an appreciable energy change is involved in changing a substance from
one state of matter to another, it is important that a state symbol is attached to
all the formulae given in an equation, for example,
1
H2(g) +
O2(g) −−> H2O(l)
∆Ho (298K) = -285.9kJ/mol
2
1
H2(g) +
O2(g) −−> H2O(g)
∆Ho (298K) = -241.8kJ/mol
2
If a substance involved in a reaction can exist in more than one allotropic form,
then this should be specified, for example,
C(graphite) + O2(g)
C(diamond) + O2(g)
F.
−−> CO2(g)
−−> CO2(g)
∆Ho (298K) = -393.5kJ/mol
∆Ho (298K) = -395.4kJ/mol
Some special standard enthalpies
a.
Enthalpies of formation (∆Hfo)
The enthalpy change which occurs when one mole of a compound formed,
under standard conditions, from its elements in their standard states. For
example, the formation of ethene,
2C(graphite)
+ 2H2(g)
−−> C2H4(g)
∆Hfo (298K) = +52.3kJ/mol
From the view point of molecular interactions, the enthalpy change (∆Hfo) of a
compound represents the energy transferred to or from the surroundings when
chemical bonds in the elements are broken and new bonds are formed in the
compound. It is therefore a measure of the stability of the compound relative to
its constituent elements.
It is important to note that:
i.
It is helpful to compare∆Hfo of HF(g), HCl(g), HBr(g), and HI(g)
compound
∆Hfo(298K)/kJmol-1
HF(g)
-271.1
HCl(g)
-92.3
HBr(g)
-36.2
HI(g)
+26.5
Since it indicates ___________________________________________
_________________________________________________________
_________________________________________________________
The standard enthalpies of formation of the oxides of the short period
from sodium to chlorine are as follow:
6
Cl2O7
compound
Na20 MgO Al2O3 SiO2
P4O10 SO3
∆Hfo(mole of cpd)
-416
-2984 -395
-601
-1676 -911
+265
In order to be able to make valid comparison, enthalpies of formation
per mole of oxygen consumed are quoted also.
compound
∆Hfo(mole of O2)
Na2O MgO Al2O3 SiO2
-832 -1202 -1117 -911
P4O10 SO3 Cl2O7
-597 -264 +76
The variation in stability is clearly indicated, _____________________
_________________________________________________________
Another interesting comparison between standard enthalpies of
formation of the group IA oxides and those of the group VIB oxides is
as follow:
Group I oxides
Li2O Na2O K2O
∆Hfo(298K)/kJmol-1 -596 -416 -362
Group VI oxides
CO2
o
-1
∆Hf (298K)/kJmol -393
SiO2
-911
Rb2O Cs2O
-330 -318
GeO2 SnO2 PbO2
-551 -581 -277
_________________________________________________________
_________________________________________________________
_________________________________________________________
ii.
The word "stability" is used in the context of energetic stability.
For example,
H2(g)
+
O2(g)
−−>
H2O2(l)
∆Hfo(298K) =
-187.6kJ/mol
the negative sign does not mean that the reaction will necessary take
place at an observable rate. Furthermore, hydrogen peroxide is itself
energetically unstable with respect to water and oxygen:
1
H2O2(l) −−> H2O(l) +
O2(g) ∆Hfo(298K) = -98.3kJ/mol
2
which underlines the fact that ∆Hfo values refer only the elements as
standards.
iii.
The enthalpy changes are only an approximate guide to energetic
stability.
iv.
The ∆Hfo of any elements in its natural, most stable form is set to be
equal to zero
∆H o (any element) = 0
f
b.
Enthalpies of combustion (∆Hco)
7
The enthalpy change that takes place when one mole of a substance is
completely burned in oxygen is known as standard enthalpy of combustion. The
values are often commercial importance, particularly for internal combustion
engineers. For example, the hydrocarbon 3,3-dimethylpentane is an importance
component of high grade petrol.
∆Hco(298K) =
C7H16(g) + 11O2(g) −−> 7CO2(g) + 8H2O(l)
-4,802.8kJ/mol
Some standard enthalpies of combustion are listed below:
compound
Methane
Ethane
Propane
Butane
Methanol
Ethanol
Propan-1-ol
Butan-1-ol
c.
∆Hco(298K)/kJmol-1
Formula
CH4
C2H6
C3H8
C4H10
CH3OH
C2H5OH
C3H7OH
C4H9OH
-890
-1560
-2220
-2877
-726
-1367
-2017
-2675
Enthalpy of neutralization (∆Hno)
The enthalpy of neutralization of an acid with an alkali is the enthalpy change
which takes place when one mole of water is formed during neutralization. The
reaction is carried out in dilute aqueous solutions.
Effectively, the neutralization is the reaction between hydrogen and hydroxide
ions:
H+(aq)
+
OH-(aq)
−−>
∆Hno(298K) =
H2O(l)
-57.3kJ/mol
The value of enthalpy of neutralization should be the same for any acid and an
alkali but it is wrong.(Why?) Some examples are shown below:
HCl(aq) + NaOH(aq) −−>
HNO3(aq) + NaOH(aq) −−>
NaCl(aq)
+
H2O(l)
NaNO3(aq) + H2O(l)
1
1
H2SO4(aq) + NaOH(aq) −−> Na2SO4(aq) + H2O(l)
2
2
∆Hno= -57.3kJ/mol
∆Hno= -57.2kJ/mol
∆Hno= -57.1kJ/mol
CH3COOH(aq) + NaOH(aq) −−> CH3COONa(aq) + H2O(l) ∆Hno= -55.2kJ/mol
HF(aq) + NaOH(aq) −−>
NaF(aq) + H2O(l)
∆Hno= -68.6kJ/mol
CH3COOH(aq) + NH3(aq) −−> CH3COONH4(aq) + H2O(l) ∆Hno= -50.4kJ/mol
d.
8
Enthalpy of solution (∆Hso)
Enthalpy of solution is the heat (enthalpy change) when one mole of a solute
dissolved in a solvent to form an infinitely dilute solution. When the symbol (aq)
is used in a thermochemical equation, it can be assumed that the solution is at
infinite dilution.
Some standard enthalpies of solution of the alkali metal chlorides are listed
below.
Account for the solubility of the chlorides.
G.
Hess's Law
The total energy change accompanying a chemical change is independent of the route
by which the chemical change takes place.
The great value of Hess's Law that it can be used to calculate enthalpy change that
cannot be determine directly by experiments.
If the law is not so, the law of conservation of energy will be broken. Let's consider the
following example,
It would be possible to change A + B into C + D by route 1, and then change C + D
back into exactly same quantity, same substance A + B by route 2. If the law is not so,
the change may result in create or destroy energy. Consider
route 1
xkJ/mol
y
route 2
z
9
i.
suppose
x
>
(y
_______________________________________________,
+
z),
ii.
and suppose x < (y + z), ___________________________________________,
Therefore _____________________________________________.
An example, when you start from 2H2(g) + C(graphite) + 2O2(g) to 2H2O(l) + CO2(g)
by route 1 and route 2.
Actually, the numerical values of x is –965.3, y is –75.0 and z is –890.3.
In order to illustrate the conservation of energy in chemical processes, we can employ
energy level diagrams.
H.
Uses of Hess's law
a.
Calculating standard enthalpies of formation:
Very few enthalpies of formation can be measured directly by experiments.
However, Hess's law enables us to calculate enthalpies of formation from other
data, and especially from enthalpies of combustion which can be measured very
accurately by "bomb calorimeter". For example, calculate the standard enthalpy
of formation of methane from the following standard enthalpies of combustion.
C(graphite)
H2(g)
CH4(g)
+
+
O2(g)
−−>
∆H1o = -393kJ/mol
CO2(g)
1
O2(g) −−> H2O(l)
2
+ 2O2(g) −−> CO2(g)
∆H2o = -286kJ/mol
+ 2H2O(l)
∆H3o = -890kJ/mol
In this example, we want to calculate the standard enthalpy change of:
C(graphite)
+
2H2(g) −−>
CH4(g)
enthalpy change = ∆H4o
We should set up an "enthalpy cycle" as follow:
∆H4o
C(graphite) + 2H2(g)
CH4(g)
∆H1o + 2∆H2o
∆H3o
+ 2O2(g)
+ 2O2(g)
CO2(g)
+
2H2O(l)
10
From the above, we have
∆H4o
∆H1o + 2∆H2o – (∆H3o)
=
*the addition of ∆H's just like the addition of vectors
Thus, ∆H4o
Class work:
=
-393 + 2(-286) – (-890) =
Calculate the enthalpy formation of HCl(g)
Given:
H2(g) −−> 2H(g)
Cl2(g) −−> 2Cl(g)
2H(g) + 2Cl(g) −−> 2HCl(g)
Answer: H2(g) + Cl2(g)
b.
-75(kJ/mol)
−−>
∆H1o = +436kJ
∆H2o = +242kJ
∆H3o = -862kJ
∆H4o = -92kJ/mol
2HCl(g)
calculating enthalpies of reactions
Example:
Calculate the enthalpy change for the reaction.
1
CO(g) +
O2(g) −−> CO2(g)
2
Given:
C(graphite) + O2(g) −−> CO2(g)
1
C(graphite) +
O2(g) −−> CO(g)
2
Same as the above method, we should set up an enthalpy cycle:
∆H3o
1
O2(g)
2
CO(g) +
CO2(g)
∆H2o
∆H1o
C(graphite)
+
O2(g)
from the enthalpy cycle
∆H3o
Thus, ∆H3o =
= ∆H1o - ∆H2o
(-393) - (-111)
=
-282kJ/mol(CO2)
∆H1o = -393kJ
∆H2o = -111kJ
11
Class work: 1.
Calculate the enthalpy of the reaction
2H2S(g) + SO2(g)
−−>
3S(s) + 2H2O(l)
given:
H2(g) + S(s) −−> H2S(g)
S(s) + O2(g) −−> SO2(g)
1
H2(g) + O2(g) −−> H2O(l)
2
Answer:
2.
∆H1o = -20.6kJ
∆H2o = -296.9kJ
∆H3o = -285.9kJ
∆Ho = -233.7kJ/mol(SO2)
Given the following thermochemical equations:
2Cu(s) + S(s) −−> Cu2S(s)
S(s) + O2(g) −−> SO2(g)
Cu2S(s) + 2O2(g) −−> 2CuO(s) + SO2(g)
∆H1o = -79.5kJ
∆H2o = -296.9kJ
∆H3o = -527.5kJ
calculate the value of the standard enthalpy of formation of CuO(s).
Answer: -155kJ/mol.
12
c.
Calculating enthalpies of reactions by enthalpies of formation
Consider the following reaction:
B2H6(g) + O2(g) −−> B2O3(s) + 3H2O(l)
The enthalpy of the reaction can be found by using only the enthalpies of formation of
each compound or element in the equation.
Try to set up the enthalpy cycle:
∆Hro = ∆Hfo(B2O3) + 3∆Hfo(H2O) - ∆Hfo(B2H6).
In general, ∆Hro = Σ ∆Hfo(Products)
Example:
-
Σ ∆Hfo(Reactants)
Calculate the enthaply of the reaction
B2H6(g) + O2(g) −−> B2O3(s) + 3H2O(l)
given
∆Hfo(B2H6) = +32.0kJ/mol
∆Hfo(B2O3) = -1225.0kJ/mol
∆Hfo(H2O) = -286kJ/mol
by ∆Hro = Σ∆Hfo(Products) - Σ∆Hfo(Reactants)
∆Hro
Class work: a.
= [-1225.0 + 3(-286)] - (+32)
= -2115kJ.
Calculate ∆Hro for the reaction:
Na2O2(s) + H2O(l) −−>
given
b.
NaOH(s) + O2(g)
∆Hfo(Na2O2) = -505.6kJ/mol
∆Hfo(H2O) = -286kJ/mol
∆Hfo(NaOH) = -426.8kJ/mol
How many kJ of heat are liberated when 25.0g of Na2O2(s) react
according to this equation? (ram of Na = 23.0, O = 16.0)
(Do the class work on next page.)
13
Answer:
I.
a.
-124kJ
b.
20.2kJ
Bond energies and Enthalpy change
As we know (in the section of "enthalpy of formation"), the enthalpy change during
the formation of a compound from its elements is actually the energy difference
between the absorbed energy in breaking bonds in elements and the released energy
in forming bonds in the compound.
∆Hfo = Ebond breaking - Ebond forming
Thus, if the energy released is in greater amount than energy absorbed, the sign of
∆Hfo should be negative, that is an exothermic. Conversely, ∆Hfo will be positively
sign.
We can extend this idea to enthalpy of reaction.
∆Hro = Ebond breaking - Ebond forming
For example,
CH4(g) + Cl2(g)
−−>
CH3Cl(g) + HCl(g)
The enthalpy change of the reaction can be calculated by bond energies
Bond
C-H
Cl-Cl
C-Cl
H-Cl
Bond energy(kJ/mol)
+411.5
+244.0
+326.0
+431.0
Thus, energy required in breaking bonds = 1xE(C-H) + 1xE(Cl-Cl)
= 411.5 + 244.0
= 655.5
energy released in forming bonds
= 1xE(C-Cl) + 1xE(H-Cl)
= 326.0 + 431.0
= 757.0
∆Hro = 655.5 - 757.0 = -101.5(kJ/mol)
The value is in quite good agreement with the experimental value which is
14
-99.5kJ/mol.
Class work: Calculate the mean bond energy of C-C bond and C-H bond.
Given:
CH3CH2CH2CH3(g) −−> 4C(g) + 10H(g)
∆Ho = +5165kJ
CH3(CH2)3CH3(g)
−−>
∆Ho = +6337kJ
5C(g)+12H(g)
Answer:
E(C-C) = +347 kJ/mol, E(C-H) = +412 kJ/mol
Why the value of bond energy must be positive in sign?
Below are some values of mean bond energies of some common bonds:
bond
H-H
C-C
O-O
F-F
Cl-Cl
Br-Br
I-I
∆Ho/kJmol-1
+436
+346
+146
+158
+242
+193
+151
∆Ho/kJmol-1
+412
+388
+463
+562
+431
+366
+299
bond
C-H
N-H
O-H
F-H
Cl-H
Br-H
I-H
M. Bond energies and Hess's law
By using the concept of Hess's law, energy level diagram (or enthalpy cycle), we can
calculate some bond energies.
For example: Calculate the bond energy of HI(g).
Given:
Bond energy of H-H
Bond energy of I-I
H2(g) + I2(g) −−> 2HI(g)
= +436kJ/mol
= +151kJ/mol
∆Ho = -10.4kJ
In this example, we want to calculate the enthalpy of
HI(g) −−> H(g)
+
I(g)
Firstly, we should set up an enthalpy cycle (or energy level diagram)
∆Ho
HI(g)
-
H(g) + I(g)
1
1
x436 + x151 (kJ)
2
2
10.4
kJ
2
1
1
H2(g) + I2(g)
2
2
By Hess's law,
∆Ho =
– (-
10.4
1
)+
x436 +
2
2
1
x151 (kJ) = +298.7 kJ/mol
2
therefore the bond energy of H-I is +298.7 kJ/mol
15
Class work: Estimate the bond energy of C=O.
CO2(g)
Hint:
−−>
C(g)
+
2O(g)
∆Ho = xkJ/mol
1. Think of the ways to produce CO2(g) and C(g) + 2O(g) from the same origin.
2. Set up an enthalpy cycle or energy level diagram
3. The following enthalpies may be useful:
C(graphite) −−> C(g)
O2(g) −−> 2O(g)
C(graphite) + O2(g) 
 CO2
∆Ho = +716kJ/mol
∆Ho = +498kJ/mol
∆Ho = -393.7kJ/mol
Answer: +804J/mol
Appendix A
Internal energy
There are four main contributions of internal energy:
a.
b.
c.
d.
Translation energy - associated with the translation motions of a molecule. This is
only for atoms and molecules in gases and liquids.
Rotational energy - associated with the rotation of a molecule
about its centre
of gravity. There is no rotational contribution for a monatomic
gas.
Vibration energy - associated with the vibration motions of the atoms within the
molecule in solid, liquid, and gaseous states.
Electronic energy - associated with the electrostatic interactions between the
various charged particles that make up the molecule.
The first two contributions are entirely kinetic, the third is partly kinetic and partly potential,
while the forth is completely potential.
Gravitational energy is usually neglected, since it is very small relative to other forms of
energy. It is obvious that the internal energy of a substance depends on its physical state.
Thus a substance in the gaseous state will have greater internal energy when in the liquid or
solid states since the gaseous molecules will have greater translation, vibration and rotational
motion. The large number of energy terms that can contribute to the total internal
energy of a substance makes it impossible to determine the absolute value of this
quantity of substance in a given state. However, we are only interested in the change in
this property that occurs during a reaction.