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Topic 2.1
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ENTHALPY CHANGES
1. Exothermic and endothermic reactions
When a chemical reaction takes place, the products and
reactants have different stabilities and thus there is a change in potential
energy. However since total energy is always conserved, any change in potential
energy must be balanced by an equal and opposite change in kinetic energy.
The lower the chemical
potential energy of a given chemical species, the more stable it is. This means that stable species have a lower
potential energy than unstable species.
In some reactions, the products are more stable than the
reactants. The products therefore have less potential energy than the
reactants, and the potential energy of the reacting species decreases.
Since the total energy is always conserved, it follows that
the kinetic energy of the species must increase. The particles thus move faster
and the temperature increases. Reactions in which the products are more stable
than the reactants thus involve a transfer of energy from potential to kinetic
and an increase in temperature.
PE à KE
Such reactions give out heat and are thus said to be EXOTHERMIC.
In other reactions, the reactants are more stable than the
products. The products therefore have more potential energy than the reactants,
and the potential energy of the reacting species increases.
Since the total energy is always conserved, it follows that
the kinetic energy of the species must decrease. The temperature of the system
thus decreases. Reactions in which the reactants are more stable than the
products thus involve a transfer of energy from kinetic to potential and a
decrease in temperature.
KE à PE
Such reactions absorb heat and are said to be ENDOTHERMIC.
2. Standard enthalpy changes
The change in chemical potential energy during a chemical
reaction is known as the enthalpy change
for that reaction. It is given the symbol DH.
By convention, if the reaction is exothermic (ie heat is
given out) the enthalpy change is said to be negative: DH = -ve. If the reaction is endothermic (ie heat is absorbed)
the enthalpy change is said to be positive: DH
= +ve.
The enthalpy change of a reaction depends on the reaction
conditions. It is therefore necessary to specify standard conditions for the
measurement of enthalpy changes. These are taken to be atmospheric pressure (1
atm) and room temperature (298K). Enthalpy changes measured under standard
conditions are known as standard
enthalpy changes and are given the symbol DHo. During
these chemical changes, the pressure should be kept constant.
The enthalpy change
for a reaction is the heat energy change measured under conditions of constant
pressure.
The standard enthalpy
change for a reaction is the heat energy change measured under standard
conditions: 100 kPa and a stated temperature (usually 298K).
The enthalpy change also depends on the amount of substance
used. It is therefore necessary to specify the amount of reactants used.
Enthalpy changes are conventionally measured in kJmol-1.
So DH = heat energy
change/no. of moles
Given a reaction: A + 3B à
2C + 4D
The standard enthalpy change for this reaction is taken to
be the enthalpy change under standard conditions when one mole of A reacts with
three moles of B to give two moles of C and four moles of D.
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Eg If 0.2 moles of A react with 0.6 moles of B and 200 kJ
of energy are released, the enthalpy change is - 200/0.2 = -2000 kJmol-1
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3. Standard enthalpies of formation and combustion
The term "standard enthalpy of reaction" can be used
to describe any chemical reaction, but there are some important reactions which
have special names:
The standard enthalpy
of formation of a substance is the enthalpy change when one mole of that
substance is formed from the most stable allotropes of its elements in their
standard states under standard conditions.
It is given the symbol DHof.
Eg C(s) + 2H2(g) à
CH4(g), DH = -74.8 kJmol-1.
The standard enthalpy of formation of methane is -74.8 kJmol-1.
Eg H2(g)
+ 1/2O2(g) à H2O(l), DH = -285.8 kJmol-1
The standard enthalpy of formation of water is -285.8 kJmol-1.
Eg N2(g)
+ 2H2(g) + 3/2O2(g) à NH4NO3(s), DH = -365.6 kJmol-1
The standard enthalpy of formation of ammonium nitrate is
-365.6 kJmol-1.
Eg 1/2N2(g)
+ 1/2O2(g) à NO(g),
DH = +82.0 kJmol-1
The standard enthalpy of formation of nitrogen monoxide is +
82.0 kJmol-1.
The standard enthalpy
of formation of all elements in their standard states is zero.
The standard enthalpy
of combustion of a substance is the enthalpy change when one mole of that
substance is burned in an excess of oxygen under standard conditions.
It is given the symbol DHoc.
Eg H2(g)
+ 1/2O2(g) à H2O(l), DH = -285.8 kJmol-1
The standard enthalpy of combustion of hydrogen is -285.8
kJmol-1.
Eg CH4(g)
+ 2O2(g) à CO2(g) + 2H2O(l), DH = -890.3 kJmol-1
The standard enthalpy of combustion of methane is -890.3
kJmol-1.
Eg C2H5OH(l)
+ 3O2(g) à 2CO2(g) + 3H2O(l), DH = -1367.3 kJmol-1
The standard enthalpy of combustion of ethanol is -1367.3
kJmol-1.
Burning a substance in oxygen is almost always exothermic,
so standard enthalpies of combustion almost always have negative values.
Substances which do
not support combustion, like water, carbon dioxide and most other oxides, have
zero enthalpy of combustion.
4. Calculating enthalpy changes from temperature changes
Enthalpy changes are generally measured by carrying out a
reaction under controlled conditions in a laboratory and measuring the
temperature change.
The amount of heat required to change the temperature of a
system by 1K is known as the heat capacity of a system (Hc). It is
measured in JK-1.
The heat energy change for a given reaction can therefore be
calculated from the equation:
q = DT x Hc.
The specific heat
capacity (c) is the amount of heat required to heat 1g of a substance by
1K. So: heat capacity = specific heat capacity x mass
Mass = volume x density
So: heat capacity =
specific heat capacity x volume x
density
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q = VρcDT
or q = mcDT
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If a reaction is taking place in solution (and therefore
water is the main species present) it is reasonable to assume that the solution
behaves as if it were pure water.
The density of water is 1.0 gcm-3 and the
specific heat capacity of water is 4.18 Jg-1K-1.
So the calculation is q = (volume of solution) x 4.18 x DT. The enthalpy change can then be
calculated by dividing the energy change by the number of moles of reactants.
If the temperature goes up the enthalpy change is negative.
If the temperature goes down the enthalpy change is positive.
MEAN BOND ENTHALPIES
During a chemical reaction, the bonds in the reactants are
broken. This is an endothermic process; energy is required to do this. After
the bonds have been broken, however, the bonds in the products are formed. This
is an exothermic process; energy is released when this happens.
The enthalpy change for a chemical reaction can be deduced
from consideration of the energy required to break bonds in the reactants and
the energy released when the bonds in the products are formed. It can be
calculated from the following equation:
DH = Energy required
to break bonds in reactants - Energy required to break bonds in products.
This method can be used to calculate the enthalpy changes
for any reaction which does not involve ionic bonds. The breaking and making of
ionic bonds involves a more complicated sequence of energetic processes and
thus cannot be considered in this way.
a) Molecules in
the gaseous state
In gas phase reactions, the bonds which are broken and
formed are covalent bonds. The energy required to separate completely the atoms
in one mole of covalent bonds is known as the bond dissociation enthalpy of that bond (DHb).
A-B(g) à A(g) + B(g)
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Bond
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DH /kJmol-1
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C-H
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+413
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O-H
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+464
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C-C
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+347
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C=C
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+612
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C=O
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+805
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H-F
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+568
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H-Cl
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+432
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Cl-Cl
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+243
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Br-Br
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+193
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O=O
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+498
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These bond enthalpies are mean values; the exact strength of
a bond depends on its environment. Even the strengths of the same type of bond
in the same molecule may vary:
Eg in
water: H-O-H(g) à H-O(g) + H(g); H = +502 kJmol-1
H-O(g)
à H(g) + O(g); H = +427 kJmol-1
It is necessary therefore to take an average = ie (502 + 427)/2 = +465 kJmol-1.
Thus bond enthalpies calculated from different reactions may
vary slightly. They also do not take intermolecular forces into account. Mean bond enthalpies thus only give you
approximate values for enthalpy changes.
b) Solids and
liquids
In giant covalent substances, all the covalent bonds have to
be broken before free gaseous atoms can be formed.
In either case, the energy required to separate completely
the atoms in one mole of the substance is known as the atomisation energy (DHat).
M(s) à M(g), or AxBy(s)
à xA(g) + yB(g)
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Substance
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DHat/kJmol-1
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C(s) à C(g)
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+717
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Si(s) à Si(g)
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+377
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SiO(s) à Si(g) + 2O(g)
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+1864
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Using mean bond
enthalpies to calculate approximate enthalpy changes:
The approximate enthalpy change can be calculated by
considering the mean bond enthalpies of the bonds broken and bonds formed
Eg Consider the combustion of methane:
CH4(g)
+ 2O2(g) à CO2(g)
+ 2H2O(l)
Broken: 4 C-H bonds, 2 O=O bonds: total energy = (4 x 413) + (2 x 498)
= +2648 kJmol-1
Formed: 2 C=O bonds, 4 O-H bonds: total energy = (2 x 805) + (4 x 464) =
+3466 kJmol-1
So approximate ∆H = 2648 – 3466 = -818 kJmol-1
This answer is only approximate because it is using mean
bond enthalpies, not molecule-specific ones, and because it is ignoring changes
in intermolecular forces. The correct value of ∆H for the reaction is -890
kJmol-1.
Determining mean bond
enthalpies from given data:
If the enthalpy change for a reaction is known, and most of
the bond enthalpies are known, it is possible to calculate the mean bond
enthalpy of a particular bond:
Eg The enthalpy of formation of methane is known to be -76
kJmol-1. The bond enthalpy of a H-H bond is +436 kJmol-1,
and the enthalpy of atomization of carbon (C(s) à C(g)) is +713 kJmol-1.
Using this information, the mean bond enthalpy of a C-H bond
can be calculated.
C(s) + 2H2(g) à CH4(g)
The bonds broken are the bonds in carbon (C(s) à
C(g)) and 2 H-H bonds:
Total
energy = 713 + 2(436) = 1585
The bonds formed are four C-H bonds:
Total
energy = 4x (x = mean C-H bond enthalpy)
So 1585 – 4x = -76
So 4x = 1585 + 76 = 1661 kJmol-1
So x = 415 kJmol-1 = mean bond enthalpy of C-H
HESS' LAW
Hess' Law states
that "the enthalpy change for a
chemical reaction depends only on the initial and final states and is
independent of the path followed".
In other words whichever route, however direct or indirect,
by which the reaction proceeds, the overall enthalpy change for the reaction
will be same. It is an application of the principle of conservation of energy.
Hess' Law can be used to calculate many enthalpy changes
which cannot be measured directly.
Many enthalpy changes can be calculated simply by using
enthalpies of formation and combustion and applying Hess' Law.
1. Calculations
using enthalpies of formation
Consider the reaction CH4(g) + Br2(l) à CH3Br(g) + HBr(g)
One way by which this reaction could proceed would be to
convert all the reactants into their constituent elements in their standard
states and then to convert the elements into products.
Since the enthalpy of formation of CH4 is -74.8
kJmol-1 then the enthalpy change when one mole of CH4 is
converted into its elements will be +74.8 kJmol-1. Given that the
enthalpies of formation of CH3Br and HBr are -37.2 kJmol-1
and -36.4 kJmol-1 respectively, then the enthalpy change for the
reaction by this route can be calculated:
An energetic cycle such as this is known as a Hess' Law
cycle.
DH = -(-74.8) +
(-37.2) + (-36.4) = +1.2 kJmol-1
The enthalpy change of any reaction can be calculated if the
enthalpies of formation of all the reactants and products are known. The enthalpy
of formation of elements in their standard states is always zero.
The calculation can be condensed into a simple formula:
∆H = Σ[∆Hf(products)] – Σ[∆Hf(reactants)]
2. Calculations
using enthalpies of combustion
All organic compounds burn in air to give carbon dioxide and
water.
Consider the reaction C(s) + 2H2(g) à CH4(g)
One way by which this reaction could proceed is to convert
the reactants to carbon dioxide and water by combustion, and to convert carbon
dioxide and water to methane and oxygen (ie the reverse of combustion).
If the enthalpy of combustion of methane is -890 kJmol-1,
then the reverse must be +890 kJmol-1. The enthalpy of combustion of
hydrogen needs to be doubled because there are two moles of hydrogen in the
equation.
If all the relevant enthalpies of combustion are known, the
enthalpy change of the reaction by this route can be deduced.
DH = -394 + (2 x -286) - (-890) = -76 kJmol-1
The enthalpy change for a reaction can always be calculated
if the enthalpy of combustion of reactants and products are known.
The calculation can be condensed into a simple formula:
∆H = Σ[∆Hc(reactants)] – Σ[∆Hc(products)]
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