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PART 3: ENERGETICS
6.1 What is Energetics
Energy Changes In Chemical Reactions
I.
Thermochemistry
Thermochemistry is concerned with the heat changes that accompany chemical changes and
/or phase changes.
Notes:
1. Energy is a measure of capacity to do work. It exists in many different forms, e.g. chemical
energy, light energy, electrical energy and heat.
2. In most chemical and physical changes, there is an associated energy change. The study of
these energies is known as energetic.
3. The energy changes occurring in chemical reactions are usually reflected as heat changes.
II.
Conservation of Energy
The first law of thermodynamic ( or law of conservation of energy ) states that energy may be
converted from one form to another, but it is never created nor destroyed.
Note: The total energy of a system and its surroundings is constant.
The usual form in which energy exchanged in chemical reaction is as heat, which causes
temperature changes.
III.
Enthalpy Changes
In thermochemistry, all substances are said to possess a heat content or enthalpy, H.
For a reaction carried out at constant pressure (usually at atmospheric pressure in an open
container ), and if no other work is involved, the heat absorbed or evolved is called the enthalpy
change.
The enthalpy change of reaction is the heat exchange ( either absorption or release )
with the surroundings at constant pressure before and after the reaction.
 H = Hp (enthalpy of products ) - Hr (enthalpy of reactants)
IV.
Endothermic and Exothermic Reactions
With respect to enthalpy change, reactions can be classified into exothermic and endothermic
reactions.
- For an exothermic reaction ( heat given out to the surroundings ), H is negative.
- For an endothermic reaction ( heat absorbed from the surroundings ), H is positive.
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The following table compares exothermic and endothermic reactions:
Exothermic reaction
Endothermic reaction
Definition
A reaction in which heat is released
to the surroundings.
A reaction in which heat is absorbed
from the surroundings.
Enthalpy of products and
reactants
-
Hp < Hr
The energy stored in the reactants as chemical energy is converted
into heat.
Hp > Hr
Energy is being stored in the
products as chemical energy.
Enthalpy change H
-
Negative
-
Postive
Temperature of reaction
-
The temperature increases and
-
The temperature decreases.
-
-
this energy is transferred to the
surroundings.
Once the reaction is complete,
the temperature of reaction
mixture falls.
Since the temperature of the
surroundings is now higher than
that of the reaction mixture,
energy flows into the reaction
mixture.
Once the reaction is complete,
the temperature rises.
mixtures
Enthalpy level diagram
Selected examples
-
Combustion reactions
Precipitation
Displacement reactions
Acid- alkali reactions
-
Cracking
Melting of solid
Boiling and evaporation of liquid
In an exothermic reaction, the energy released from the reacting system flows into the
surroundings. According to the law of conservation of energy, this same amount of energy is gained by
the surroundings. This is because energy can be neither be created or destroyed but can be transferred.
The energy of the universe is constant.
6.2 Endothermic and Exothermic Reactions related to the Breaking and Forming of Bonds
Nearly all of the heat energy released or absorbed in a chemical reaction can be accounted in terms
of the forming and breaking of bonds.
Note:
1. All the important forces between atoms, including those that hold atoms together in
molecules, are electrical in nature. This electrical forces are mostly attractive.
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2. To separate the atoms, work must be done against this attractive force, i.e. energy must be
done against this attractive force, i.e. energy must be applied.
3. The energy required to separated atoms is released when they join together again.
Any chemical reaction involved a rearrangement of atoms. Chemical bonds in the reactants are
broken and new bonds are formed in the products.
-
Bond breaking is an energy absorbing process. It takes in energy from the surroundings.
Bond forming is an energy releasing process. It gives out energy to the surroundings.
The energy involved can be in the form of heat.
Note:
1. Chemical reaction results in an overall enthalpy change because
the energy required to break the bonds in reactants to form the free atoms is
NOT EQUAL TO
the energy given out when the free atoms recombine to form different bonds in the
products.
2. As a result, the heat content of the reactants is not equal to the heat content of the products in
a chemical reaction.
3. Exothermic and endothermic reactions are related to bond breaking and formation in the
following way:
Exothermic reactions are reactions in which
the energy required to break the bonds in the reactants is smaller than the energy liberated in
forming the bonds in the products.
They give out heat to the surroundings.
Endothermic reactions are reactions in which
the energy required to break the bonds in the reactants is greater than the energy liberated in
forming the bonds in the products.
They absorb heat to the surroundings.
Note:
Average Bond Enthalpy values can be used to calculate the enthalpy change in a reaction.
In general, Enthalpy Change in Reaction = Sum of Bond Enthalpies of Reactants Sum of Bond Enthalpies of Products
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6.3 STANDARD ENTHLAPY CHANGES
I.
Standard Enthalpy Change
Standard Enthalpy Change for a reaction, symbolized as H0298 , is defined as
The enthalpy change when the molar quantities of reactants shown in a balanced
chemical equation completely react to from products under standard conditions.
Standard conditions are as follows:
- a pressure of 1 atmosphere
- a temperature of 298 K
- if solution is used, 1M is specified.
-
Elements or compounds in their normal stable state
Example: For the following reaction,
H2(g) + 1/2 O2(g)  H2O(l)
H0 = -285.9 kJ mol-1
When one mole of hydrogen and half mole of oxygen react to form one mole water,
285.9 kJ of heat are evolved under standard conditions. This is an exothermic reaction, H being
negative.
Note:
1. All the substance involved must be in their normal physical states under standard conditions.
2. The enthalpy of all elements in their most stable form and standard state is conventionally
taken as zero. For example,
H0298 [O2(g)] = 0 , H0298 [Na(s)] = 0
Therefore, the standard enthalpy of formation of a compound represents the enthalpy content
of the compound.
3.
The precise physical state and allotropic form, if any, must clearly specified. This is because
changes in state and allotropic form, involve energy changes even at a fixed temperature. A
examples:
(a) Change of State:
H2(g) + 1/2 O2(g)  H2O(g)
H2(g) + 1/2 O2(g)  H2O(l)
i.e. H2O(g)  H2O(l)
H0 = -242.0 kJ mol-1
H0 = -285.9 kJ mol-1
H0 = -44 kJ mol-1
(b) Change in allotropic form:
C ( graphite ) + O2(g)  CO2(g)
C ( diamond ) + O2(g)  CO2(g)
C ( graphite )  C ( diamond )
H0 –393.5 kJ mol-1
H0 –395.4 kJ mol-1
H0 = +1.9 kJ mol-1
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II.
Ch.6 : p.5
Standard Enthalpy of Neutralization and its Experimental Determination
Standard Enthalpy of Neutralization , H0n,298 is the enthalpy change when an acid and a
base neutralize to form one mole of water under standard conditions.
Example: The neutralization of sodium hydroxide and hydrochloric acid.
NaOH(aq) + HCl(aq) 
NaCl(aq) + H2O(l)
H0n,298 = -57.3 kJ mol-1
When one mole of sodium hydroxide reacts with one mole of hydrochloric acid to give
one mole of water under standard conditions, 57.3 kJ of energy are released to the
surroundings.
Note:
1. Neutralization is carried out in infinitely dilute solution.
2. The enthalpy of neutralization is always negative because neutralization is exothermic.
3.
The value of H0n,298 involving strong acids and bases is quite constant ( -57.3 kJ mol-1)
The heat producing reaction is essentially
H+(aq) + OH-(aq)  H2O(l)
4.
H0n,298 = -57.3 kJ mol-1
If either the acid or alkali is weak, the enthalpy of neutralization is less negative. (i.e. less
exothermic ) Energy is required to complete the dissociation of the weak acid or base as
reaction proceeds. The ionization process consumes some heat produced during
neutralization, thereby reducing the magnitude of H0n,298 .
(A) Experimental determination of Enthalpy of Neutralization
- Enthalpy of neutralization can be measured by mixing solutions of acids and alkalis in a
calorimeter and measuring the rise in temperature.
- In experiments, an insulated polystyrene cup or vacuum flask may be employed as the
calorimeter.
- The heat change of a process is determined by multiplying heat capacity (C) of the
calorimeter and its contents by the temperature change:
H = C x T
where heat capacity means the amount of heat per degree temperature change.
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-
the heat capacity of the calorimeter and the contents can be found by measuring the electrical
energy needed to raise the temperature by a certain degrees.
-
If
H is measured in kJ and T is measured in Kelvin (K),
Heat change = heat capacity x temperature change
-
Specific heat capacity ( J g-1 K-1 ) is defined as the amount of energy needed to raise the
temperature of 1g (or 1 kg) of a substance by 1K .
Specific Heat capacity =
Heat Change (J) / (mass x temp. change )
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Name: ___________________ Class No.: _____ Date: ____________
Exercise:
Ch.6 : p.7
Marks: ____________
Two experiments are conducted to determine the standard enthalpy of neutralization of sodium
hydroxide with 2 acids, hydrochloric acid and a carboxylic acid, RCOOH.
Experiment I
Procedure
Maximum Temp.
rise 0C
Equation
a)
b)
Experiment II
3
50 cm of 2.0 M NaOH solution
were added to 50 cm3 of 2.0 M
hydrochloric acid solution in a
polystyrene cup.
50 cm3 of 2.0 M NaOH solution
were added to 50 cm3 of 2.0 M
RCOOH solution in a polystyrene
cup.
13.0
H+(aq) + OH-(aq)  H2O(l)
10.5
RCOOH(aq) + OH-(aq)
RCOO-(aq) + H2O(l)
Calculate the standard enthalpy of neutralization for the two experiments. Account for the
differences in these two values.
Calculate the standard enthalpy of dissociation of RCOOH.
(Given at 298 K and 1 atm : specific heat capacity of solution = 4.2 J g-1 K-1,
density of final solution = 1.0 g cm-3 )
( 8 marks )
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III.
Ch.6 : p.8
Standard Enthalpy of Solution and its Experimental Determination
Standard enthalpy of solution, H0soln,298 is the enthalpy change when 1 mole of a substance
is completely dissolved in an infinite amount of solvent (usually water) under standard
conditions.
Example: The dissolution of sodium chloride in water
NaCl(aq) + aq  NaCl(aq)
Note:
H0 soln, 298 = +5.0 kJ mol-1
The solution should be dissolved in such a large excess of solvent that further
addition of solvent produces no further enthalpy change.
(A) Experimental Determination of Enthalpy of Solution
The enthalpy of solution can be determined by adding a known quantity of the substance in a
known volume of solvent. The solvent is stirred and the maximum ( or minimum) temperature of
the solvent is then determined. Knowing the heat capacity of the solvent and the container, the
enthalpy of solution is calculated.
Example: An experiment determining the approximate value for the enthalpy of solution of copper
(II) sulphate.
8 g of anhydrous copper (II) sulphate are added to 100 cm3 water in the polystyrene cup.
There is a temperature rise of 8 0C in the solution.
(a)
Calculate the molar enthalpy of solution of anhydrous copper (II) sulphate.
(b) Assumptions which are made in calculating the molar enthalpy of solution (a)
1. the thermal capacities of polystyrene cup and the thermometer are negligible.
2. the specific heat capacity of copper (II) sulphate solution is similar to that of water.
3.
4.
anhydrous copper (II) sulphate dissolves quickly such that heat losses are negligible.
further addition of the solution would cause no further heat change.
(c) Sources of errors and the methods of minimizing errors.
1) Heat may be lost or gained from the surroundings. Insulate the polystyrene cup or use a
vaccum flask calorimeter.
2) The molar heat capacity of copper (II) sulphate solution is not the same as that of water.
Determine the actual molar heat capacity of copper (II) sulphate solution.
3) Anhydrous copper (II) sulphate may have absorbed moisture from air. Gently heat the
sample in an oven for several hours and keep it in a desiccator.
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4)
Ch.6 : p.9
Heat may be absorbed by the polystyrene cup, i.e. its heat capacity may not be
negligible. Determine the value of the heat capacity of the cup, or use a container of
known heat capacity and take it into account of calculation.
5)
IV.
The copper (II) sulphate solution may not be sufficiently dilute to eliminate further heat
change on addition of water. Use smaller quantities of copper (II) sulphate in powder
form.
Standard Enthalpy of Formation
Standard Enthalpy of Formation , H0f,298 is the enthalpy change when 1 mole of a
substance is formed from its elements in their standard states at standard conditions.
Example: The formation of ethanol (C2H5OH) from its elements.
2C(s) + 3H2(g) + 1/2 O2(g) 
C2H5OH(l) H0f,298 [C2H5OH(l)] = -277 kJ mol-1
When one mole of ethanol is formed from its constituent elements ( in their normal
physical state ). 277 kJ of energy are released.
Note:
1) Elements in the standard state have zero enthalpy of formation.
2)
V.
Standard enthalpies of formation can sometimes be measured directly by using
calorimeter experiments. However, in many cases, the elements do not combine under
the experimental conditions. Their enthalpies of formation are measured indirectly,
using Hess’s Law.
Standard Enthalpy of Combustion and its Experimental Determination
Standard enthalpy of Combustion, H0c,298 is the enthalpy change when one mole
substance undergoes complete combustion ( being oxidizing by oxygen ) in the
standard state.
Example: Combustion of Sucrose
C12H22O11(s) + 12O2(g)  12CO2(g) + 11H2O(l)
Note:
H0c,298[C12H22O11(s)] = -5647 kJ mol-1
All enthalpies of combustion have a negative sign since combustion is exothermic and
always involves the evolution of heat.
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6.4 Experimental Determination of Enthalpy of Combustion
The enthalpies of combustion of gases or volatile liquids can be measured using a flame
calorimeter.
Method: Oxygen is passed through the combustion chamber. The energy evolved by combustion
heat the surrounding water bath. The temperature produced by the combustion of a
known mass of the compound is measured.
Example 1:
Pure oxygen is pumped into the calorimeter and 0.7 g of pure ethanol is burnt. Once ethanol starts
burning , it is enclosed to reduced heat loss. The heat produced raises the temperature of 250 cm3
of water by 20K.
(a) Suggest why the heat exchanger is made of copper in the calorimeter.
(b) Assume that the heat capacity of calorimeter is negligible and there is no loss,
determine the enthalpy of combustion of ethanol.
(c) Apart from the neglecting of heat capacity of the calorimeter and heat losses,
suggest two sources of experimental error.
( specific heat capacity of water = 4.18 J g-1 K-1 , density of water = 1.0 g cm-3 )
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6.5 Hess’s Law
The law of conservation of energy states that energy can be neither be created or destroyed ,
though one form of energy can be converted to another form.
This law therefore implies that the total energy content ( in the system ) is constant. Based on it,
Hess’s law arises.
I.
Hess’s Law
The standard enthalpy change of reaction depends only on the difference between the standard
enthalpy of the reactants and the standard enthalpy of the products, and not no the route by which
the reaction occurs.
Hess’s Law
The total enthalpy change accompanying a chemical change is independent of the route by
which the chemical change takes place. Therefore, the overall change in enthalpy is the
same, whichever route is followed.
In other words, for processes involving several stages, the enthalpy change for the reaction is
equal to the algebraic sum of the enthalpy change for each intermediate stage.
H2
H3
H1
H4
H5
The overall H for the exothermic reaction shown above is
H5 = H1 + H2 + H3 + H4
Example:
The enthalpy change for the oxidation of carbon (graphite) to carbon dioxide is the same
whether it is carried out Route A ( 1 stage ) or Route B ( 2 stages)
Route A: C(s) + O2(g)  CO2(g)
Ha = -393.5 kJ mol-1
Route B: C(s) + 1/2 O2(g)  CO(g)
Hb1 = -110.5 kJ mol-1
CO(g) + 1/2 O2(g)  CO2(g) Hb2 = -283.0 kJ mol-1
Hb1 + Hb2 = (-110.5) + (-283.0) = -393.5 kJ mol-1
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Ch.6 : p.12
Enthalpy Level Diagram and Enthalpy Cycle
The enthalpy change of a reaction can be represented by an enthalpy level diagram or enthalpy
cylcle.
(A) Enthalpy Level Diagram
The enthalpy level diagram for the oxidation of carbon is shown below:
C(s) + O2(g)
H1 = -110.5 kJ mol-1
-110.5 kJ mol-1
CO(g) + 1/2 O2(g)
m
Hc[C(s)]
H2 = -283.0 kJ mol-1
-393.5 kJ mol-1
CO2(g)
m
Note:
On drawing an enthalpy diagram.
1. Draw a line at 0 to represent the enthalpy level of elements.
2.
3.
4.
Draw a line above or below this zero line to represent the enthalpy level of
reactants.
Draw another line to represent the enthalpy level of intermediate products.
Draw another line to represent the enthalpy level of final products.
(B) Enthalpy Cycle ( Born-Haber Cycle )
The enthalpy cycle for the oxidation of carbon is shown below:
Hc[C(s)]
C(s) + O2(g)
CO2(g
)
H1 = -110.5 kJ mol
H2 = -283.0 kJ mol-1
-1
CO(g)
Note: On drawing the enthalpy cycle (Born-Haber cycle)
1. Relate the various equations involved in a reaction.
2. Write the equations in the form of a cycle and check if the cycle can ‘flow’ in one
direction.
The required enthalpy change can be found by following the path
Reactants
intermediate products
Final products
and doing simple arithmetic calculation.
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III. The Use of Hess’s Law in determining Enthalpy Changes
Some reactions proceed very slowly or involve formation of side products. Hess’s Law can be
used to determine the enthalpy changes of such reactions which cannot be determined directly by
calorimetry experiments.
Example 1:
It is difficult to determine the enthalpy change for
C(s) + 1/2 O2(g)  CO(g)
H1
because some graphite may oxidized completely to CO2(g)
This enthalpy change should be determined indirectly by calorimetry experiments using Hess’s
Law. Since
CO(g) + 1/2 O2(g)  CO2(g)
C(s) + O2(g)  CO2(g)
H2 = - 283 kJ mol-1
Hcomb = -393.5 kJ mol-1
These two values can be determined relatively easily in calorimetry experiments. The enthalpy
change for
C(s) + 1/2 O2(g)  CO(g)
H1 can be determined indirectly from these two values, using Hess’s Law
Hcomb. [C(s)] = H1 + H2
H1 =  Hcomb. [C(s)] - H2
= (-393.5) – (-283)
= -110.5 kJ mol-1
Example 2: Indirect determination of the enthalpy change of formation of calcium carbonate
<1> Reagents: Calcium, Calcium carbonate, a strong acid (e.g. HCl) , water
<2> Data Given: the enthalpy of formation of carbon dioxide gas the enthalpy change of
formation of water
<3> Description:
(a) Equation for the formation of calcium carbonate from its constituents:
Ca(s) + C(s) + 3/2 O2(g)  CaCO3(s)
Experimental difficulty in the direct determination of the enthalpy of formation:
1)
2)
3)
4)
the extent of the reaction cannot be controlled,
the heat evolved cannot be separated into appropriate terms,
direct combustion of calcium can be violent
side reaction may arise, e.g.
2Ca(s) + O2(g)  2CaO(s)
C(s) + O2(g)  CO2(g)
(b) Relevant equations for the indirect determination:
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
Ca(s) + 2HCl(aq)  CaCl2(aq) + H2(g)
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(c) Apparatus:
1) Calorimeter : plastic cup / insulated beaker / vacuum flask
2) Thermometer
3) Measuring cylinder / pipette / burette
4) Electronic balance
IV. Calculation involving Enthalpy Changes
Enthalpy changes can be calculated
- by algebraic method, and
- through enthalpy level diagram or enthalpy cycle
By Algebraic Method: The procedure is described below:
1.
2.
3.
4.
5.
Write down the equation involved if they are not given.
Find out the equation (s) that contain the reactants.
Find out the equation (s) that contain the products.
Reverse these equations when necessary and reverse the sign of the given enthalpy change.
Multiply the equations and the corresponding enthalpy changes by a suitable factor, such that
any intermediate reactants / products not in the required equation are cancelled out.
Sum up the equations to give the required equations.
Sum up enthalpy change in involved equation to give the enthalpy change of reaction.
6.
7.
By Enthalpy Level Diagram / Cycle Method : Procedure
1.
Write down the equation involved if they are not given.
2.
3.
4.
5.
Note:
Find out the equation(s) that contain the reactants.
Find out the equation(s) that contain the products.
Relate these equations and add them into the diagram / cycle to complete it.
Apply Hess’s Law and solve the resultant equation for one unknown ( the required
change)
Always remember to include in the answer
-
Units: kJ mol-1
Sign : to indicate an exothermic or endothermic reaction
In any equations, include
State symbols to show the physical conditions of elements / reactions / products.
enthalpy
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(A) Calculating Standard Enthalpy of Reaction from Standard enthalpy of Formation
Since the standard enthalpy of any element is zero,
Standard enthalpy of formation = enthalpy content of products
From this relation, it can be deduced that
Standard enthalpy change of a reaction =
Sum of standard enthalpies of formation of products –
Sum of standard enthalpies of formation of reactants
i.e. H0 = H0f[products] - H0f[reactants]
Example 1:
Find the standard enthalpy of the following reaction
CO(g) + 1/2 O2(g)  CO2(g)
Given that H0f [CO(g)] = -110.5 kJ mol-1 , H0f [CO2(g)] = -393.5 kJ mol-1
Example 2
Find the standard enthalpy of the following reaction
2SO2(g) + O2(g)  2SO3(g)
Given that H0f[SO2(g)] = -297.0 kJ mol-1 and H0f[SO3(g)] = -395.0 kJ mol-1
Ch.6 : p.15
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(B) Calculating Standard Enthalpy of formation from enthalpy of Combustion
When a compound is completely burnt, the combustion products are the same as those produced
when its constituent elements undergo complete combustion.
From Hess’s Law
H0f, 298
Elements
Compounds
H1
H2
Combustion Products
Therefore, H0f,298 = H1 - H2
Example 1
The standard molar enthalpies of combustion at 298 K for graphite , hydrogen , ethanol are –393.1
kJ mol-1 , -285.8 kJ mol-1 , and –1367 kJ mol-1 respectively.
From the above thermochemical data, calculate the standard molar enthalpy of formation for
ethanol at 298 K .
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Example 2
Given the following thermochemical data:
Reaction
C (graphite) + 2H2(g)  CH4(g)
C(graphite) + O2(g)  CO2(g)
H2(g)+ 1/2 O2(g)  H2O(l)
H0298 / kJ mol-1
-75
-393.5
-285.9
a)
Calculate the standard enthalpy change for the reaction given below:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
b)
Calculate the enthalpy change of vaporization of water at 298 K, given that the enthalpy
change is –801.7 kJ mol-1 for the reaction given below:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
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(C) Calculating Standard Enthalpy of formation from Standard Enthalpy of Solution
In practice, enthalpies of solution are easily measured and may be used to determine the standard
enthalpy of formation.
Note that on dilution of a solution which is not infinitely dilute, a further enthalpy change results.
This is referred to as the enthalpy of dilution.
Example 1
Given the following thermochemical data:
Standard enthalpy of solution of potassium hydroxide = -55 kJ mol-1
Standard enthalpy of formation of water = -286 kJ mol-1
Standard enthalpy change for the reaction between potassium and water = -195 kJ mol-1
Calculate the standard enthalpy of formation of potassium hydroxide.
Example 2:
Given the following thermochemical data:
H0soln[CuSO4 5H2O(s)] = + 8 kJ mol-1
H0soln[CuSO4(s)] = -66 kJ mol-1
H0f[CuSO4(s)] = -773 kJ mol-1
H0f[H2O(l)] = -286 kJ mol-1
Calculate H0f[CuSO4 5H2O(s)] from the above data.
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(D) Calculating Standard Enthalpy of Formation from Standard Enthalpy of Reaction
According to Hess’s Law, whether a compound is formed in one step or through many
intermediate steps, the enthalpy change is the same. This is because the enthalpy change depends
on the difference in the enthalpy contents of the reactants and the products.
From Hess’s Law,
H0 f,298 = H1 + H2
Example 1:
Calculate the enthalpy of formation of NaCl(s) from the following data
Reaction
H0298 / kJ mol-1
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
-57.3
HCl(g) + aq  HCl(aq)
-71.9
H2(g) + 1/2 O2(g)  H2O(l)
-285.9
1/2 H2(g) + 1/2 Cl2(g)  HCl(g)
Na(s) + 1/2 O2(g) + 1/2 H2(g)  NaOH(aq)
NaCl(s) + aq  NaCl(aq)
92.3
-425.6
+3.9
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
V.
Ch.6 : p.20
The use of Enthalpy Change as A measure of Feasibility of Reaction and Energetic Stability
(A) Feasibility of Reaction
The sign of the enthalpy change for a reaction is an indication of its feasibility.
As a chemical reaction proceeds, the change in enthalpy of the system can be illustrated by an
enthalpy profile. Enthalpy profiles for an exothermic and an endothermic reaction are shown
below:
Compared to the reaction products, the reactants are
1.
2.
energetically unstable if the reaction is exothermic (H is negative).
energetically stable if the reaction is endothermic (H is positive).
As regards the relationship between H value and the feasibility of a reaction,
The more negative the H value for a reaction, the more likely the reaction is
feasible.
Alternatively, reactions with high positive H values are not likely to be
feasible.
However, kinetic factors should also be considered in determining whether a reaction is feasible or
not.
1. Any reaction (exothermic or endothermic ) requires a certain minimum amount of energy
2.
3.
called activation energy to occur. It represents the energy barrier that the reactants
must overcome before they can combine to form products.
In the absence of this activation energy, a reaction does not occur, despite the magnitude
or sign of the enthalpy change evolved.
Therefore,
if the activation energy is very high, the reactants will be kinetically stable. It is
unlikely that a reaction occurs.
if the activation energy is very low, the reactants will be kinetically unstable. It is likely
that a reaction occurs.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Ch.6 : p.21
Example: Both allotropes of carbon, graphite and diamond have highly negative enthalpies of
combustion:
C(graphite) + O2(g)  CO2(g)
H = -393.5 kJ mol-1
C(diamond) + O2(g)  CO2(g) H = -395.4 kJ mol-1
1.
2.
Both graphite and diamond are energetically unstable with respect to the carbon dioxide.
Based on enthalpy of combustion only, it is likely that carbon would be oxidized readily, i.e.
combustion is highly feasible.
However, such a high activation energy is required to start combustion that diamond and
graphite appear inert. They are said to be kinetically stable.
In conclusion,
A reaction is more likely to occur if the reactants are both energetically and kinetically
unstable relative to the products.
(B) Energetically Stability
The value of the standard enthalpy of formation of a compound gives a good indication of its
energetically stability.
Compounds with high negative enthalpy of formation is very stable, so far as decomposition
into its elements is concerned.
This is because the enthalpy change for its decomposition would be highly positive, and therefore
unlikely to occur.
Compounds with high positive enthalpy of formation is very unstable and tend to
decompose into its elements.