Download 3 CO 2 (g) + 4 H 2 O(l)

CHAPTER FIVE: THERMOCHEMISTRY
Every chemical species has what chemists call enthalpy. The enthalpy of a substance is the ability of that species to
produce Heat. Almost every chemical reaction produces or absorbs heat and we will learn how to calculate the
amount of heat associated with any particular chemical change.
When we heat a chemical in isolation (i.e. when no reaction occurs) two things can happen. Either the chemical will
simply get hotter without changing its phase or at a certain temperature the species will undergo a phase change.
In chapter 1, we quantified the amount of products that could be expected from the stoichiometry of the chemical
reaction. However it is not only new chemical species which are produced in a chemical reaction. Energy, in the
form of heat, can also be produced. In fact some chemical reactions can absorb heat. Therefore when dealing with
chemical reactions it is not only the stoichiometry between the reactants and the products which is important (the
mass balance) but also how much heat is absorbed or released during the reaction (the energy balance). The mass
and energy balance are linked of course. The amount of heat absorbed or released will depend upon how much of
the reactants we use in the reaction – the greater the amount of reactants the greater the heat liberated or absorbed.
This aspect of stoichiometry is known as Thermochemistry.
5.1
Heat
Let’s start at the very beginning on this one just to make sure we know what we’re talking about. Heat is not a
component of a system. We cannot say, for instance, that hot objects have a lot of heat or that colder objects have
less heat. Rather, heat is the mechanism for the transfer of energy between one system and another.
Heat is energy transferred from one system to another only because of a difference in temperature.
What do we mean by the word “system”? A system is “officially” a selected part of the universe that we are
interested in – sounds very complicated! However, for the purposes of the chemist, a system is simply our reaction
vessel and its contents. It is necessary to define a specific system when discussing heat as it must be remembered
that the overall energy in the universe remains constant i.e. energy is conserved. This is a statement of the First Law
of Thermodynamics which we will return to later in the course. However if we look at a specific system then we
can measure the changes in energy that occur when heat is transferred from our system to the surroundings or vice
versa. The “surroundings” is simply another system that encompasses our system. Therefore our system is the
chemical reaction we are interested in. If our system is at a higher temperature than the surroundings heat will be
transferred from our system to the surroundings.
High Temperature
Energy transfer via
heat
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Low Temperature
If energy is released during a chemical reaction, the temperature of our system will increase and heat will be
transferred to the surroundings. This is called an exothermic process. If instead energy is absorbed during a
chemical reaction then the temperature of our system will decrease.
Heat will then be transferred from the
surroundings to the system. This is called an endothermic process.
Most chemical reactions take place at constant pressure. For instance if we add a solution of NaOH to a solution of
HCl in a test tube then the reaction will take place at the prevailing atmospheric pressure. The heat that is produced
or absorbed by a chemical reaction at constant pressure is termed the enthalpy change and is given the symbol,
ΔH.
For an endothermic reaction ΔH is positive
For an exothermic reaction ΔH is negative
Thus every chemical reaction has a value of ΔH associated with it. ΔH values are normally quoted in kilojoules
(kJ). For instance the equation
2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(s)
is an exothermic reaction and has a ΔH value of – 851.5 kJ. We can thus write
2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(s)
ΔH = - 851.5 kJ
Note it is vitally important that the physical state of the reactants and products is specified in thermochemistry (we
will see why later on). The above equation can be described in words as :
When 2 moles of solid aluminium react with 1 mole of solid iron (III) oxide the heat produced is 851.5 kJ.
Question 5.1: How much heat is produced if 4 moles of solid aluminium are allowed to react with 2 moles of iron
(III) oxide?
Question 5.2: Is heat transferred from the system to the surroundings or from the surroundings to the system for an
exothermic reaction?
Question 5.3: Another unit used to measure heat is the calorie (cal). This is a non-SI unit but is still commonly used
in North America. A calorie is defined as the heat required to raise 1 g of water through 1°C. (1 cal = 4.184 J)
(i) Convert –851.5 kJ into kcal., (ii) Which is the bigger unit – the joule or the calorie? and (iii) How many grams of
water can have their temperature raised by 1°C by 1 J of energy?
5.2
Calorimetry
How do we measure the amount of heat that is produced or absorbed by a chemical reaction? The simplest way is to
do the chemical reaction in an insulated reaction vessel and to measure the temperature change that occurs. If for
instance the reaction takes place in water, we can calculate the heat change associated with the reaction by
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measuring the change in temperature of the water. The insulated reaction vessel is commonly known as a
calorimeter and the process of measuring the change in temperature is known as calorimetry.
Example 5.1: A chemical reaction is performed in a calorimeter containing 342 g of water. The
temperature of the water is raised by 5.42oC. How many kilojoules of heat did the reaction produce?
Solution:
Question 5.4: A chemical reaction releases 1 240 J of energy. This energy is all absorbed by the water (423 g)
containing the reactants. If the water was initially at 19.6 °C, what will be its final temperature?
Question 5.5: 1.264 g of water changes its temperature by 3.11 °C during an exothermic chemical reaction in a
calorimeter. The final temperature of the water was 20.4 °C. Calculate
(i) the amount of energy released by the reaction and (ii) the initial temperature of the water
5.2.1
Specific Heat Capacities
We have seen in the previous section that we can use the temperature change of the reaction medium (i.e. water) to
measure the energy produced or absorbed by a chemical reaction. This required us to know how much energy is
necessary to change the temperature of a known amount of water.
Thus
1 J of energy changes the temperature of 0.2390 g of water by 1 °C
or, instead we could write
4.18 J of energy changes the temperature of 1.000 g of water by 1 °C
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4.18 J g-1 °C-1 is known as the specific heat capacity of water.
It is of course not only water which has a specific heat capacity. All substances will have a specific heat capacity,
the energy required to change the temperature of 1.00 g of that substance by 1 °C. Some typical specific heat
capacities are listed below
Substance
Specific heat capacity (J g-1 °C-1)
Liquid water
4.18
Air
1.01
Copper
0.38
Ethanol
2.42
Stainless steel
0.51
The energy to change the temperature of a substance does not have to come from a chemical reaction directly. We
could for instance heat up a kettle of water using an electric hotplate. Irrespective of the source of the energy, the
energy required to heat up (or cool down) a given amount of a substance by a given temperature change is constant.
The energy that is used to change the temperature of a substance, without changing its state, is sometimes called
sensible heat.
Example 5.2: Calculate the quantity of heat required to change the temperature of a 455 g block of copper
sitting on a 1 450 g block of stainless steel from 22.5 °C to 104 °C.
Solution:
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Example 5.3: Water is to be used to cool down a stainless steel industrial chemical reactor from 550 °C to 120 °C.
The reactor has a mass of 1475 kg. The initial and final temperatures of the water are 15.6 °C and 37.2 °C
respectively. How much water is required?
Solution:
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Example 5.4: A 12.5 g block of copper at 75.4 °C is cooled in 100 g of water at 15.5 °C. What temperature will the
mixture reach assuming that there is no heat loss?
Solution:
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Question 5.6: Calculate the heat produced from the reaction of 15.0 g of O 2 and 8.20 g of SO2 given by the equation:
2SO2(g) + O2(g)  2SO3(g)
ΔH = - 198 kJ
[Hint: Determine the limiting reagent!]
Question 5.7: The specific heat capacity of benzene is 1.055 J g-1 oC-1. Calculate the heat required to raise 124 g of
benzene from 13.2 °C to 20.2 °C.
Question 5.8: A 24.6 g sample of a metal at 98.7 °C is placed in a calorimeter containing 42.5 g of water at 18.6 °C.
The final temperature of the mixture is 27.8 °C. Calculate the specific heat capacity of the metal.
Question 5.9: Calculate the final temperature of the mixture when 455 g of stainless steel at 79.8 °C is placed in a
calorimeter containing 101 g of H2O at 17.8 °C.
5.3
Enthalpy
We have already defined the change in enthalpy as the heat released or absorbed at constant pressure.
The change in enthalpy is:
ΔH = Hfinal – Hinitial
Again, we must reiterate that heat is not a component of a system. Enthalpy however is a component of a system.
Enthalpy is not stored as heat. Rather enthalpy is a measure of the capacity of the system to supply energy as heat
when a certain change occurs. The enthalpy of a substance will depend on its physical state (solid, liquid or vapour)
and its temperature (it will also depend upon the pressure but this is a small effect and we will largely ignore it in
this course). Therefore when discussing enthalpies of substances we must specify both the physical state and the
temperature of the system. We must also realise that the enthalpy, the store of energy, will also depend upon how
much of the substance we have, this should be obvious from the calculations in the previous section. As enthalpy
depends upon the amount of the substance it is said to be an extensive property.
An extensive property is a physical property which depends upon the size of the sample.
The opposite to an extensive property is an intensive property.
An intensive property is a physical property which is independent of the size of the sample.
Question 5.10: Divide the following physical properties into two groups, those that are intensive and those that are
extensive.
Mass, specific heat capacity, volume, temperature, density
Another very important property of enthalpy is that it is a state property. This means that the enthalpy of a system
does not depend upon the past history of the system, but only on the current conditions of temperature, pressure,
amount and state. This has very important consequences when dealing with changes in enthalpy. For example if
we change the temperature of 1.0 × 102 g of water from 15.6 °C to 27.4 °C then
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ΔH = 1.0 × 102 g × 4.18 J g-1 °C-1 × (27.4 – 15.6) °C = + 4.9 × 103 J
i.e., our system has gained 4.9 × 103 J of enthalpy
If we now take our 1.0 × 102 g of water at 15.6 °C and heat it to 94.6 °C and then cool it to 27.4 °C, then the overall
energy change can be calculated as
ΔH (15.6 °C  94.6 °C) = 1.0 × 102 g × 4.18 J g-1 °C-1 × 79.0 °C = + 3.30 × 104 J
ΔH (94.6 °C  24.4 °C) = 1.0 × 102 g × 4.18 J g-1 °C-1 × -67.2 °C = -2.81 × 104 J
 overall, ΔH = (3.30 × 104 – 2.81 × 104) J = 4.9 × 103 J (as above)
Therefore the overall enthalpy change is the same, independent of the route we took to get there.
Therefore, as a consequence of enthalpy being a state property, the difference in enthalpy between two systems
under a given set of conditions is constant.
As far as chemists are concerned there are three different types of enthalpy change of interest

The enthalpy change that occurs when we heat or cool a substance without changing its state. This is what
we termed sensible heat. The enthalpy change can be measured if we know the specific heat capacity of
each of the components in the system.

The enthalpy change associated with changing the state of the system. For instance if we heat liquid water
at a normal atmospheric pressure of 1 atm the temperature will continue to rise until 100oC is reached. At
this temperature liquid water will be converted to water in the vapour state. This requires energy to break
the intermolecular forces holding the water molecules together in the liquid state. We term this type of
enthalpy input latent heat. Latent heat changes occur at constant temperature.

The enthalpy change which occurs during a chemical reaction.
We have already mentioned sensible heat changes and you will certainly see more of this during the physics
modules. The latter two enthalpy changes are more important to the chemist and we will expand upon each of these
types of energy change.
5.3.1
Enthalpies of physical change
If we turn water into steam do we have to add energy? Of course we do. We could use an electric kettle to supply
the required energy or put a pot of water on the fire. The point is if we wish to turn a liquid into a vapour then we
need to add energy to our system. The enthalpy change associated with a liquid-vapour state change is known as the
enthalpy of vapourisation, ΔHvap.
Are ΔHvap values positive or negative?
For water at 100 °C,
ΔHvap = Hvapour – Hliquid = + 40.7 kJ mol-1
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Therefore if we take 18.02 g (1 mol) of liquid water at 100 °C we would have to add 40.7 kJ to turn the liquid into a
vapour. During this process the temperature of the system will remain at 100 °C. The energy we have added has
not been used to increase the temperature of the system. Instead it has been used to break the intermolecular
attractive forces between the molecules.
What we have done therefore is the following change
Liquid water at 100 °C  Water vapour at 100 °C
A similar enthalpy change will occur when converting a solid to a liquid. This enthalpy change will occur at the
melting point of the solid. For water at 1 atm this will be at 0 °C.
Ice at 0 °C  water at 0 °C
ΔHmelt = 6 kJ mol-1
The enthalpy change is termed ΔHmelt, the enthalpy of melting. Again the enthalpy change is endothermic as energy
is required to break the intermolecular forces. The ΔH melt value is generally smaller than ΔHvap for a given
substance (Why?). One other phase change of interest to chemists is when a solid is converted to a gas without
going through the liquid phase. This is known as sublimation and the associated enthalpy change is the enthalpy of
sublimation, ΔHsublimation.
Tables of these enthalpy changes can be found in most chemistry reference textbooks. However, we must make two
things clear here. Enthalpies are dependent upon pressure. Therefore enthalpy values are normally listed at a
standard pressure (1 atm). If ΔH values are given the superscript ΔH o this signifies the value is quoted at standard
pressure (1 atm). Also ΔH values will depend upon temperature. Latent heats are normally quoted at the phase
transformation temperature and 1 atm. Thus ΔH vap for H2O is normally quoted at 100 °C. However sometimes
ΔHvap values are quoted at 25 °C. If for instance we look up ΔH vap at 25 °C for H2O we will find a number of
44.0 kJ mol-1 as opposed to the number at 100 °C, 40.7 kJ mol-1.
Question 5.11: Consider the following enthalpy values*
Substance
ΔHomelt/kJ mol-1
ΔHovap/kJ mol-1
Water
6.01
40.7
Ethanol
4.60
43.5
Argon
1.2
6.5
Acetone
5.72
29.1
Methane
0.94
8.2
methanol
3.16
35.3
*the values are quoted at the transition temperature
(i) Why are the values for methane and argon so much lower than the other values?
(ii) Why does ethanol have a higher ΔHovap than methanol?
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Question 5.12: Sodium has the following enthalpy change values at 25 °C:
ΔHovap = 98 kJ mol-1, ΔHomelt = 2.6 kJ mol-1.
Calculate ΔHosub at 25 °C for sodium. Hint: remember enthalpy is a state property.
5.3.2
Enthalpies of chemical change
If we want to be exact about a chemical reaction then what we should write down is the thermochemical equation
for the reaction. This type of equation shows the stoichiometry of the reaction and the enthalpy change associated
with it. For example if we wanted to give a full description of the Ostwald Process to make nitric acid we would
write down the following:
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
ΔH = - 905 kJ*
2 NO(g) + O2(g)  2 NO2(g)
ΔH = - 113 kJ
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g)
ΔH = - 138 kJ
* ΔH values quoted at 25 °C
You might have thought it was bad enough having to remember the mass balance of the equations, now we’ve got
ΔH values as well! Don’t worry – there is no need to remember ΔH values. The standard enthalpy change of any
chemical reaction can be calculated from data readily available in tables. If we know the standard enthalpy change
for a given stoichiometry then we can calculate the enthalpy change for any given amount of reactant or product.
Example 5.5: Calculate the enthalpy change in making 25.4 g of gaseous nitric oxide from gaseous ammonia at
25 °C.
Solution:
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Question 5.13: (i) Calculate the enthalpy change involved in making 22.6 g of NO 2(g) from NO(g) and (ii) How
much heat will be produced if 543 g of NO2(g) is reacted with excess water to form HNO3(aq) and NO(g).
Make sure you get the vocabulary right. We talk of enthalpy changes not heat changes. We rather talk of heat
being produced or absorbed. This is what we mean when we say that enthalpy is a component of the system but
heat is not.
You’ll notice that we have been very careful in specifying the physical state of the reactants and products. That is
because the enthalpy change depends upon the physical state. Take the following reactions for instance:
 C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
ΔH = - 2 220 kJ
 C3H8(l) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
ΔH = - 2 204 kJ
 C3H8(l) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
ΔH = - 2 028 kJ
You will notice from reactions  and  that gaseous propane will produce more heat than liquid propane when
burnt in oxygen. Reactions  and  show that the physical state of the products must also be specified. If water in
the gaseous state is produced rather than liquid water then less heat will be produced. This of course makes perfect
sense in that if we need to transform water from a liquid to a vapour then energy will be consumed.
H2O(l)  H2O(g)
ΔHvap = + 44.0 kJ
As we have produced 4 mol of H2O in reaction  this will consume 176 kJ to convert the liquid into water vapour.
Those of you who are really alert will notice that the difference in the enthalpy change between reactions  and  is
exactly 176 kJ. This is of course a direct result of enthalpy being a state property.
Enthalpy changes are sometimes put into enthalpy diagrams to better illustrate the consequences of the ‘state
property’.
For reaction  above we could draw the following diagram:
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C3H8(g) + 5 O2(g)
ΔH = - 2 204 kJ
Enthalpy
3 CO2(g) + 4 H2O(l)
The information from reaction  allows us to add to the diagram
C3H8(g) + 5 O2(g)
Enthalpy
ΔH = - 2 028 kJ
ΔH = - 2 204 kJ
3 CO2(g) + 4 H2O(g)
176 kJ
3 CO2(g) + 4 H2O(l)
If you now calculate ΔH for the following reaction C3H8(l) → C3H8(g) you should get an answer of 16 kJ (Hint:
Look at equations  and ). Now we can add this to the diagram.
C3H8(g) 16
+ kJ
5 O2(g)
Enthalpy
C3H8(l) + 5 O2(g)
H = - 2028 kJ
H = - 2204 kJ
3 CO2(g) + 4 H2O(g)
176 kJ
3 CO2(g) + 4 H2O(l)
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H = - 2220 kJ
You will notice that all the enthalpy changes are self-consistent. It’s a bit like the energy diagrams you can draw in
spectroscopy problems isn’t it? Here, just as in spectroscopy, we are measuring changes not absolute values.
Enthalpy diagrams of the sort we have just been looking at are very useful in that they allow us to immediately
realise that a very important concept must be true. A reaction which is exothermic in one direction (i.e. moving
down the diagram) must be endothermic in the reverse direction. If for instance we want to calculate the enthalpy
change of the following reaction:
3 CO2(g) + 4 H2O(g) → C3H8(l) + 5 O2(g)
ΔH = ?
then the diagram gives a value of + 2 204 kJ
C3H8(l) + 5 O2(g)
- 2 204 kJ
Enthalpy
+ 2 204 kJ
3 CO2(g) + 4 H2O(g)
The enthalpy change of any reverse process is the negative of the enthalpy change of the forward process
Enthalpy diagrams are very useful in helping us to picture what is going on but we don’t really need to draw a
diagram everytime we want to do an enthalpy change calculation. We now know the important things to do with
enthalpy changes:

Enthalpy is a state property

The enthalpy change depends upon the amount of the reactants used

The sign of the enthalpy change for the reverse reaction is opposite to that of the forward reaction
An important statement of these facts is Hess’s Law.
A reaction enthalpy is the sum of the enthalpies of any sequence of reactions (under the same conditions of
amounts, temperature and pressure) into which the overall reaction can be divided
Hess’s Law allows us to calculate the enthalpy change for any chemical reaction, even those that don’t happen! For
instance we can write an equation for the formation of ethanoic acid from ethane.
2 C2H6(g) + 2 O2(g) → 2 CH3COOH(g) + 2 H2(g)
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This reaction does not occur in the laboratory. However, we can still work out the enthalpy change associated with
this reaction by employing Hess’s Law.
We can measure the following enthalpy changes
C2H6(g) → C2H4(g) + H2(g)
ΔH = 137 kJ
C2H4(g) + H2O(g) → C2H5OH(g)
ΔH = - 45.8 kJ
2 C2H5OH(g) + O2(g) → 2 CH3CHO(g) + 2 H2O(g)
ΔH = - 345 kJ
2 CH3CHO(g) + O2(g) → 2 CH3COOH(g)
ΔH = - 544 kJ
Can we use these values to calculate our unknown ΔH? The answer is yes. If we follow a careful procedure then
there should be no problem. Go through them slowly and carefully.

Choose an equation which contains at least one of the reactants of the desired equation. In this case our
desired equation is
2 C2H6(g) + 2 O2(g) → 2 CH3COOH(g) + 2 H2(g)
So let’s choose ethane. We have the following reaction
C2H6(g) → C2H4(g) + H2(g)
ΔH = 137 kJ
Our final equation has 2 C2H6(g) on the LHS so multiply this equation by 2
2 C2H6(g) → 2 C2H4(g) + 2 H2(g)
ΔH = + 274 kJ
Remember to multiply ΔH by 2 as well!

Next choose an equation which contains one of the desired products and add it to the previous equation
2 C2H6(g) → 2 C2H4(g) + 2 H2(g)
+ 2 CH3CHO(g) + O2(g) → 2 CH3COOH(g)
ΔH = + 274 kJ
ΔH = - 544 kJ
2 C2H6(g) + 2 CH3CHO + O2(g) → 2 C2H4(g) + 2 H2(g) + 2 CH3COOH(g)

ΔH =-270 kJ
Now we need to get rid of the unwanted reactants and products which do not appear in the final equation.
Let’s start with the ethanal CH3CHO. Remember we need an equation with the unwanted species on the
opposite side of the equation.
2 C2H6(g) + 2 CH3CHO(g) + O2(g) → 2 C2H4(g) + 2 H2(g) + 2 CH3COOH(g)
ΔH = -270 kJ
2 C2H5OH(g) + O2(g) → 2 CH3CHO(g) + 2 H2O(g)
ΔH = -345 kJ
2 C2H6(g) + 2 C2H5OH(g) + 2 O2(g) → 2 C2H4(g) + 2 H2(g) + 2 CH3COOH(g) + 2 H2O(g) ΔH = -615 kJ
Now the ethanol (C2H5OH), ethene (C2H4), and water
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2 C2H6(g) + 2 C2H5OH(g)+ 2 O2(g) → 2 C2H4(g) + 2 H2(g)+ 2 CH3COOH(g) + 2 H2O(g)
ΔH = -615 kJ
2 C2H4(g) + 2 H2O(g) → 2 C2H5OH(g)
ΔH = -91.6 kJ
2 C2H6(g) + 2 O2(g) →2 H2(g) + 2 CH3COOH(g)
ΔH = -707kJ
So our theoretical equation has a ΔH value of – 707 kJ.
Example 5.6: Calculate the enthalpy change for the synthesis of hydrogen chloride gas at 25°C
H2(g) + Cl2(g) → 2 HCl(g)
given the following data at 25°C
NH3(g) + HCl(g) → NH4Cl(s)
ΔH= -176.0 kJ
N2(g) + 3 H2(g) → 2 NH3(g)
ΔH= - 92.22 kJ
N2(g) + 4 H2(g) + Cl2(g) → 2 NH4Cl(s)
ΔH= - 628.86 kJ
Solution:
104
5.3.3
Enthalpies of formation
Throughout this tutorial we have been talking about enthalpy changes. That is what is important in chemistry- how
much heat will be absorbed or produced. However, to allow us to calculate these enthalpy changes from tabulated
values, we need to have some reference against which to measure the enthalpies of each species.
For
thermochemistry the elements in their standard state are defined as having zero enthalpy.
The standard state of an element is its pure form at 1 atm pressure
The standard enthalpy of formation of a compound is the standard reaction enthalpy, per mole of compound
produced, for the compound’s synthesis from its elements in their most stable form at 1 atm and the specified
temperature.
Thus the standard enthalpy of formation of MgCl2 at 25 °C is the enthalpy change associated with the following
reaction
Mg(s) + Cl2(g) → MgCl2(s)
ΔHof = - 641.8 kJ mol-1
The standard heats of formation are given the symbol ΔHof and are normally quoted at 25 °C. ΔHof (25 °C) for CO2
would be for
C(s) + O2(g) → CO2(g)
ΔHof = - 393.5 kJ mol-1
A consequence of the definition of standard enthalpies of formation is that ΔH of is equal to zero for the elements
themselves. For example
ΔHof (H2) = 0 kJ mol-1 and ΔHof (Ar) = 0 kJ mol-1
Standard enthalpies of formation are invaluable when calculating the amount of heat produced or absorbed in a
reaction. From now on all standard enthalpies of formation will be quoted at 25 °C.
Example 5.7: Calculate the enthalpy change at 25 °C for the following reaction
2 NO(g) + O2(g) → 2 NO2(g)
given the following data:
ΔHof / kJ mol-1
NO(g)
+ 90.37
NO2(g)
+ 33.8
Solution:
105
Example 5.8: Calculate the ΔHreaction for the following reaction at 25 °C.
C3H8(l) + 2 O2(g) → C2H2(g) + CO(g) + 3 H2O(l)
Solution:
106
Standard enthalpies of formation for common chemicals are given in the table below. You will need this ΔHoformation
data to answer the following questions.
ΔHoformation kJ mol-1
ΔHocombustion kJ mol-1
+ 82.93(g)
- 3 301.5(g)
+ 48.66(l)
- 3 267.6(l)
-123.1(g)
- 3953.0(g)
-156.2(l)
- 3 919.9(l)
- 124.7(g)
- 2 878.5(g)
- 147.0(l)
- 2 855.6(l)
but-1-ene
+ 1.17
- 2 718.6(g)
propanol
- 257.5(g)
Compound
benzene
cyclohexane
n-butane
- 304.6(l)
CO2
- 393.5
H2O (l)
- 285.84
H2O (g)
- 241.83
CH4
- 74.85
- 890.36
H 2S
- 19.96
- 562.59
SO2
- 296.90
Question 5.14: Calculate the enthalpy of reaction for the hydrogenation of 2 745 g of benzene to form cyclohexane
C6H6(l) + 3 H2(g) → C6H12(l)
Question 5.15: Calculate the enthalpy of reaction for the dehydrogenation of 74.5 g of n-butane to but-1-ene.
C4H10(g) → C4H8(g) + H2(g)
Question 5.16: Write equations that correspond to ΔHof for the following compounds:
(i) propanol, CH3CH2OH
(ii) 3-chlorobutanoic acid, CH3CHClCH2COOH
Question 5.17: Given ΔHof (CH4(g)) = - 74.85 kJ mol-1, ΔHof (HCl(g)) = -92.31 kJ mol-1 and
CH4(g) + 4 Cl2(g) → CCl4(g) + 4 HCl(g)
calculate ΔH f for CCl4(g).
o
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ΔH = - 402 kJ
5.3.4
Enthalpies of Combustion
Although combustion reactions can be treated in the same way as any other type of chemical reaction they are so
common in chemistry that standard enthalpies of combustion, ΔH°c, are listed separately (see the Table on the
previous page).
A combustion results in the oxidation of carbon and hydrogen to carbon dioxide and water respectively. For instance
the combustion of butane is represented by the equation
C4H10(l) + 13/2 O2(g) → 4 CO2(g) + 5 H2O(l)
Of particular note is that standard enthalpies of combustion, ΔH oc, are quoted for liquid water as a product. Again
these enthalpies are normally quoted at 25 °C. Remember however that combustion reactions normally take place at
much higher temperatures and it is likely that the actual product will be gaseous water. This will result in less
energy being produced (enthalpies of combustion are always exothermic) than the theoretical enthalpy of
combustion. For instance
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
ΔHoc = - 890 kJ mol-1
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
ΔHoc = - 802 kJ mol-1
The difference between the two enthalpies being due to
2 H2O(l) → 2 H2O(g)
ΔH = + 88 kJ
Note again ΔHoc values are quoted in kJ mol-1.
Example 5.9: Given the standard enthalpy of combustion of methane is -890.4 kJ mol-1, the standard enthalpy of
formation for CO2(g) is -393.5 kJ mol-1, the standard enthalpy of formation for H2O(g) is -241.83 kJ mol-1 and the
enthalpy of vapourisation of water is 44.0 kJ mol-1 (all values quoted at 25°C); calculate the standard enthalpy of
formation for CH4(g) at 25 °C.
Solution:
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Example 5.10: Calculate the ΔHoc for H2S(g) given its ΔHof value is - 19.96 kJ mol-1.
Solution:
The key to thermochemistry is to write balanced chemical equation
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Question 5.18: Given the following information calculate the standard enthalpy of combustion of liquid benzene.
Remember to quote the value in kJ mol-1.
Substance
ΔHof / kJ mol-1
C6H6(l)
49.0
CO2(g)
- 393.51
H2O(l)
- 285.4
Question 5.19: When 1.92 g of magnesium reacts with excess nitrogen to form magnesium nitride, 12.2 kJ of heat is
produced. Calculate ΔHof for Mg3N2. (Remember to quote in kJ mol-1).
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