Download 5 Energetics/thermochemistry

5
Energetics/thermochemistry
ESSENTIAL IDEAS
■
The enthalpy changes from chemical reactions can be calculated from their effect on the
temperature of their surroundings.
■ In chemical transformations energy can neither be created nor destroyed (the first law of
thermodynamics).
■ Energy is absorbed when bonds are broken and is released when bonds are formed.
5.1 Measuring energy changes – the enthalpy changes from
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chemical reactions can be calculated from their effect on the temperature of their
surroundings
■■ Heat
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If you touch a hot surface, thermal energy will enter your hand because the surface is warmer
than your hand. But if you touch ice, thermal energy will leave your hand and pass into the colder
ice. The direction of spontaneous energy transfer is always from a warmer substance to a cooler
substance. The thermal energy that is transferred from an object to another because of a temperature
difference between them is called heat. The term ‘heat’ is best regarded as a description of a process
rather than the name of a form of energy. We say heat is transferred between a system and its
surroundings if the transfer of energy occurs as a result of a temperature difference between them.
When heat flows from one object to another object, they are said to be in thermal contact.
During thermal contact, heat will flow from the hotter object to the cooler object until they are
both at the same temperature. This is known as thermal equilibrium and the rates of heat flow
between the two bodies are the same. Heat can never flow on its own from a cooler object to a
hotter object. This is against the laws of thermodynamics.
Physicists use the concept of internal energy, which is the sum of all types of energies inside
a substance, including potential energy due to forces between molecules. A substance does not
contain heat but it does contain internal energy.
■■ Temperature
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All three states of matter – solids, liquids and gases – are made of atoms, ions or molecules. These
particles are vibrating in solids and in liquids, and moving around in gases (translational motion).
Because of these random motions at all temperatures above absolute zero (0 K or −273 °C), the
particles in matter have kinetic energy. The average kinetic energy of these individual particles
causes what we perceive through sense perception as warmth. Whenever a substance becomes
warmer, the average kinetic energy of its particles has increased.
The random kinetic energy of particles in matter can be increased in a number of ways. For
example, solar energy from the Sun being absorbed by sea water, striking a piece of metal with a
hammer many times, compressing the air in a tyre pump for a bicycle or simply using a flame.
Temperature is directly related to the random motion of particles in substances at temperatures
above absolute zero. In the case of an ideal gas, the absolute temperature (in kelvin) is directly
proportional to the average kinetic energy of translational motion (from one place to another). In
liquids and solids there is still a similar relationship. So the warmth you feel when you touch a hot
surface is the kinetic energy transferred by molecules in the hot surface to your colder fingers.
Be aware that temperature is not a measure of the total kinetic energy of all the molecules in
a substance. There is twice as much molecular kinetic energy (in joules) in 2 dm3 of boiling water
as in 1 dm3 of boiling water at the same temperature. However, the temperatures of both cubic
decimetres (litres) of water are the same because the average kinetic molecular energy is the same.
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The Fahrenheit scale
C°
100°
90°
180°
80°
160°
70°
140°
60°
120°
50°
100°
40°
80°
60°
32°
40°
20°
0°
30°
20°
10°
0°
–10°
–17.8°
1 Find out how to interconvert temperatures between Fahrenheit and Celsius.
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■■ Conservation of energy
Energy is conserved during chemical reactions. The amount of energy in the Universe at
the end of the chemical reaction is the same as at the beginning. This is known as the law
of conservation of energy. This implies that the amount of energy in the Universe remains
constant and energy is neither created nor destroyed. Energy can be converted from one form
to another, from potential or chemical energy to internal energy and heat energy, but the total
energy within the Universe remains fixed.
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■■ Figure 5.1
Fahrenheit and
Celsius scales on
a thermometer
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F°
212°
200°
The physical quantity that tells us how hot a sample of a substance is, is temperature.
Temperature is expressed by a number that corresponds to a degree mark on a chosen scale.
The Celsius and Fahrenheit scales are based on the physical properties of water but the absolute
or thermodynamic scale of temperature is not based on the physical properties of any substance.
The kelvin is the SI unit of temperature and is based on absolute zero (approximately −273 °C)
– the temperature at which the particles of a substance have no random kinetic energy. The
kelvin has the same incremental scaling as the degree Celsius. So a change of 1 K is the same as
a change of 1 °C.
Nearly all matter expands when its temperature increases and contracts when its temperature
decreases. A mercury or alcohol thermometer measures temperature by showing the expansions
and contractions of a liquid in a sealed glass tube using a calibrated scale.
The Fahrenheit scale (Figure 5.1) of temperature is widely used in the United States. The
number 32 indicates the temperature at which water freezes and the number 212 indicates the
temperature at which water boils. The Fahrenheit scale is not part of the metric system. It is
named after the German physicist Gabriel Fahrenheit (1686–1736).
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2 Find out about perpetual motion machines.
Fundamental principle – conservation of energy is a fundamental
principle of science
The law of conservation of energy states that energy cannot be created or destroyed and that it
is simply changed from one form to another. Hence, whenever one form of energy disappears,
an equal amount of energy in some other form appears. When the Universe is taken into
consideration, there is one quantity that does not change: energy.
The law of conservation of energy is also represented in the first law of thermodynamics,
the study of energy changes during chemical reactions and physical changes. The first law
of thermodynamics includes the principle that the amount of energy in an isolated system is
constant: when one form of energy disappears, an equal amount of energy in another form is
produced.
It is now known that energy can be produced by the loss of mass during a nuclear reaction.
Energy and mass are related by Einstein’s mass-energy equivalence relationship E = mc2, where
c is the velocity of light. The modified law, therefore, states that the total mass and energy of an
isolated system remain constant.
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Nature of Science
■■ Exothermic and endothermic reactions
Enthalpy changes
Chemical reactions involve a transfer of energy. Chemical substances contain chemical
energy, a form of potential energy. Many chemical reactions involve a transfer of chemical
energy into heat.
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5.1 Measuring energy changes 167
For example, when methane (the major component of natural gas) burns in excess oxygen,
chemical energy is transferred to the surroundings as heat. The products of this combustion
reaction are water and carbon dioxide:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
CaCO3(s) → CaO(s) + CO2(g)
When an exothermic reaction transfers heat energy to the surroundings the chemical reactants
lose potential energy. The products have less potential energy than the reactants. This potential
energy stored in the chemical bonds is known as enthalpy and is given the symbol H. The
transfer of heat energy that occurs (at constant pressure) during a chemical reaction from the
reaction mixture (known as the system) to the surroundings is known as the enthalpy change,
ΔH, where the Greek letter delta means ‘change in’.
The negative and positive signs in enthalpy
changes do not represent ‘positive’ and ‘negative’
energy. They simply indicate the direction of the
flow of heat energy (Figure 5.4). Enthalpy changes
are usually measured in units of kilojoules per
HOT
COLD
mole (kJ mol−1). The value of the enthalpy
change for a particular reaction will vary with the
conditions, especially concentration of chemicals.
Hence, standard enthalpy changes, ΔH , are
measured under standard conditions:
Exothermic reactions
Endothermic reactions
■ a pressure of 1 atmosphere or 100 kPa
give out heat. This
take in heat. This cools
■ a temperature of 25 °C (298 K) (though in
warms the mixture
the mixture at first and
and then heat is lost
then heat is gained from
theory any temperature can be specified)
to the surroundings
the surroundings
−3
■ concentrations of 1 mol dm . If carbon
involved, then it is assumed to be in the form ■■ Figure 5.4 The directions of heat flow
during exothermic and endothermic reactions
of graphite (unless diamond is specified).
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■■ Figure 5.3
Burning Camping Gaz
(compressed butane)
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■■ Figure 5.2
A burning magnesium
sparkler (an
exothermic reaction)
The majority of chemical reactions release heat energy to their surroundings (Figures 5.2
and 5.3). This type of reaction is known as an exothermic reaction. A few chemical reactions
absorb heat energy from their surroundings: these reactions are known as endothermic
reactions. An example of an endothermic reaction is the thermal decomposition of calcium
carbonate to form calcium oxide and carbon dioxide:
Further points about enthalpy changes
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In a reversible reaction (Chapter 7), if the forward reaction is exothermic, then the reverse
reaction is endothermic, for example:
N2(g) + 3H2(g)
2NH3(g)
2NH3(g)
N2(g) + 3H2(g)
ΔH = −92 kJ mol−1
ΔH = +92 kJ mol−1
Similarly, if the forward reaction is endothermic, the reverse reaction is exothermic.
The enthalpy change depends on the amounts of reactants used. If the coefficients of
the thermochemical equation are multiplied or divided by a common factor, the value of the
enthalpy change is changed by the same factor. For example:
if
CO(g) +
1
2
O2(g) → CO2(g)
ΔH = −283 kJ mol−1
then 2CO(g) + O2(g) → 2CO2(g) ΔH = 2 × −283 = −566 kJ mol−1
Thermochemical equations are often manipulated according to these rules when solving
problems using Hess’s law.
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■■ Calculation of enthalpy changes
Specific heat capacity
thermometer
heater
solid
insulation
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When a substance is heated, the temperature
of the substance increases. The size of the
increase depends on the heat capacity of the
substance. The heat capacity of the substance is
the amount of heat energy required to raise the
temperature of a substance by one degree Celsius
or one kelvin. Heat capacity has units of joules
per degree Celsius (J °C−1) or joules per kelvin
(J K−1).
The specific heat capacity (Figure 5.5)
is the amount of heat required to raise the
temperature of a unit mass of the substance by
one degree Celsius or one kelvin. Specific heat
capacity, c, often has units of joules per gram
per degree Celsius (J g−1 °C−1). The lower the
specific heat capacity of a substance, the greater
its temperature rise for the same amount of heat
absorbed:
glycerine to improve thermal
contact between solid and thermometer
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■■ Figure 5.5 Measuring the specific heat
capacity of a solid
heat energy (q) = mass of object (m) × specific heat capacity (c) × temperature rise (ΔT) i.e.
q = mcΔT
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Worked examples
Calculate the heat capacity of 80.0 grams of water. The specific heat capacity of water is 4.18 J g−1 °C−1.
Heat capacity = 80.0 g × 4.18 J g−1 °C−1 = 334.4 J °C−1
How much heat energy is required to increase the temperature of 20 grams of nickel (specific heat capacity
440 J kg−1 °C−1) from 50 °C to 70 °C?
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q = mcΔT
q = 0.02 kg × 440 J kg−1 °C−1 × 20 °C = 176 J
Using q = mcΔT to calculate the heat change when the temperature
of a pure substance is changed
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A 10.0 g piece of pure iron has its temperature changed from 25 °C to 500 °C. The specific heat
capacity of iron is 0.45 J g−1 °C−1. Calculate the amount of heat energy in joules required to bring
about this change.
q = 10.0 g × 0.45 J g−1 °C−1 × 475 C = 2137.7 J
Calculate the mass of gold (specific heat capacity 0.13 J g−1 °C−1) that can be heated through the
same temperature difference when supplied with same amount of thermal energy.
q
m = c ΔT
m=
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2137.7 J
0.13 J g−1 °C−1 × 475 °C = 34.61 g
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5.1 Measuring energy changes 169
■■ Evaluating the results of a laboratory calorimetry experiment
to determine an enthalpy change
The amount of thermal energy (heat) required to convert a solid to a liquid at constant
temperature and pressure depends on the amount of substance present. However, the amount
of heat needed to melt a specific amount, known as the heat of fusion, is always the same. It is
often useful to refer to the amount of heat needed per gram or per mole of substance (molar heat
of fusion). The molar heat of fusion is therefore the energy needed to change one mole of a solid
into liquid at the melting point and it is specific to that substance. The involvement of thermal
energy can be shown in equation form as follows:
H2O(s) → H2O(l)
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An excess amount of ice is placed in a calorimeter (polystyrene cup) with a known amount of
hot water. Enough time is allowed for the ice to melt to bring the temperature of the ice-water to
the melting point of ice, approximately 0 °C. At this time the extra ice is removed. The volume
of the melted ice and water in the cup will be used with temperature data for the calorimeter
to calculate the molar heat of fusion of ice. The density of water (1.00 g cm−3) will be used to
convert volume to mass where needed.
Results from the experiment
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Sample calculation (with no random errors or uncertainties,
to simplify the calculation)
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Volume of hot water = 90.4 cm3
Final minimum temperature (water and ice) = 0.70 °C
Temperature of hot water = 53.6 °C
Volume (hot water and melted ice) = 151.9 cm3
Steps in calculating the enthalpy change
1 Calculate the change in temperature of the original hot water in the calorimeter.
The water cooled from 53.6 °C to 0.7 °C.
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ΔT = 53.6 °C − 0.70 °C = 52.9 °C ≈ 53 °C
2 Calculate the heat lost by this water (qw) in joules.
Since the density of water is 1.00 g cm−3, the mass of 90.4 cm3 of hot water is 90.4 g.
qw= mcΔT = (90.4 g) × (4.18 J g−1 °C−1) × (53 °C) = 20 027 J
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3 Calculate the heat used to melt the ice (qi). Assume that all of the heat lost by the water
went into melting the ice.
Therefore, qi = qw = 19 989 J
4 Calculate the volume of water that is from melted ice. The difference between the final
volume and the initial volume is due to melted ice.
Vi = 151.9 cm3 − 90.4 cm3 = 61.5 cm3
5 Calculate the mass of melted ice. The melted ice is now liquid water. Use the volume
you just calculated. With a density of 1.00 g cm−3, 61.5 cm3 of water has a mass of 61.5 g.
6 Calculate the amount (mol) of melted ice. Use the molar mass to convert. Ice has the
formula H2O.
61.5 g 1 mole ice
×
= 3.41 mol
1
18.02 g
7 Calculate the enthalpy of fusion.
Enthalpy of fusion =
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qi
19 989 J
heat to melt ice
=
= 5862 J mol−1
=
moles
3.41 mol
amount of ice
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170 5 Energetics/thermochemistry
The literature value for the enthalpy of fusion of ice is 5660 J mol−1.
An evaluation of this experiment would include a calculation of the percentage error and
a comparison to the total percentage error due to the all measurements. Likely systematic errors
should be identified, discussed and suggestions made for ways of reducing them and the random
errors. Any assumptions and limitations should be discussed.
■■ Measuring enthalpy changes
Enthalpy changes are usually measured by their effect on a known volume of water in
a container known as a calorimeter. A chemical reaction involving known amounts of chemicals
dissolved in the water may be performed and the temperature increase or decrease measured.
Alternatively, a combustion reaction may be performed and the temperature increase in the
water bath recorded.
The heat produced or absorbed may be calculated from the following expression:
total mass of water
specific heat capacity temperature
× × change
or solution
of water
In symbols:
q (J) = m (g) × c (J g−1 °C−1) × ΔT (°C)
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heat change =
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The calculations of heat transferred to the water are based on the following assumptions:
■ The reaction is assumed to occur sufficiently rapidly for the maximum temperature to be
reached before the reaction mixture begins to cool to room temperature. This occurs if the
next condition is completely fulfilled.
■ There is no heat transfer between the solution, thermometer, the surrounding air and the
calorimeter.
(In practice neither of these conditions is completely fulfilled, but the rate of heat transfer in or
out of the calorimeter is tracked and extrapolated back to the moment the reaction began.)
■ The solution is sufficiently dilute that its density and specific heat capacity are taken to be
equal to those of water, namely, 1 g cm−3 and 4.18 J g−1 °C−1.
The heat change is for the specific amount of chemicals used in the reaction. This is usually less
than one mole, so by simple proportion or ‘scaling up’ the heat change is then calculated for the
amounts of chemicals shown in the chemical equation.
Worked example
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50.00 cm3 of 0.100 mol dm−3 silver nitrate solution was put in a calorimeter and 0.200 g of zinc powder
added. The temperature of the solution rose by 4.3 °C. Deduce which reagent was in excess and then
calculate the enthalpy change for the reaction (per mole of zinc that reacts). Assume that the density of
the solution is 1.00 g cm−3 and the specific heat capacity of the solution is 4.18 J g−1 °C−1. Ignore the heat
capacity of the metals and dissolved ions.
q = 50.00 g × 4.18 J g−1 °C−1 × 4.3 °C = 898.7 J
Amount of silver nitrate =
Amount of zinc =
50.0
dm3 × 0.100 mol dm−3 = 0.00500 mol
1000
0.200 g
65.37 g mol−1
= 0.0031 mol
2AgNO3(aq) + Zn(s) → Zn(NO3)2(aq) + 2Ag(s)
Zinc is the excess reactant (Chapter 1) and hence the temperature change and the enthalpy change are
determined by the limiting reactant, the silver nitrate. Therefore:
ΔH = −0.8987 k J = −180 kJ mol−1
0.00500 mol
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5.1 Measuring energy changes 171
The enthalpy changes of reactions in solution can be easily measured with the
simple apparatus shown in Figure 5.6. A lid may be fitted to minimize heat
transfer. More accurate measurements could be performed in a calorimeter based
around a Thermos or vacuum flask.
thermometer
lid
Problems with calorimetry
There are three problems associated with the use of calorimeters:
■■ T
■ he desired reaction does not (fully) occur. This is relevant to enthalpies of
combustion where incomplete combustion occurs.
■■ Loss of heat to the surroundings (exothermic reactions); absorption of heat
from the surroundings (endothermic reactions). This unwanted flow of heat
can be reduced by ‘lagging’ the calorimeter to ensure it is well insulated.
■■ U
■ sing an incorrect specific heat capacity in the calculation of the heat
change. If a copper can is used as a calorimeter (Figure 5.7) during an
enthalpy change of combustion investigation, then its specific heat capacity
needs to be taken into account in the calculation. For example:
heat transferred = mass of water × specific heat capacity of water
× temperature change + mass of copper × specific heat capacity of copper
× temperature change
solution
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polystyrene
cup
■■ Figure 5.6 A simple calorimeter:
polystyrene cup, lid and thermometer
thermometer
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Temperature corrections
Accurate results can be obtained by using simple calorimeters (e.g. a polystyrene
cup fitted with a lid) for fast reactions, such as neutralizations or precipitations.
However, for slower reactions such as metal ion replacement, the results will be
less accurate with the same apparatus. This is because there is heat loss to the
surroundings; this will increase if the reaction is slow because the heat will be
lost over a longer period of time. Consequently, the temperature rise observed
in the calorimeter is not as great as it should be. However, an allowance can
be made for this by plotting a temperature–time graph (or cooling curve). The
method is described below.
One reagent is placed in the polystyrene cup and its temperature recorded at,
say, 1 minute intervals for, say, 4 minutes, stirring continuously. At a known time,
say 4.5 minutes from the start, the second reagent is added, stirring continuously,
and the temperature recorded until the maximum temperature is reached. As the
reaction mixture starts to cool, temperature recording and stirring are continued
for at least 5 minutes. A graph of temperature against time is then plotted. Dataloggers can be used to record temperature changes in solutions.
The lines are extrapolated to the time of mixing to determine the temperature
change that would have occurred had mixing of the reagents been instantaneous
with no heat loss to the surroundings. Graphs are given for an exothermic
(Figure 5.8) and an endothermic reaction (Figure 5.9).
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copper
calorimeter
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water
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spirit
burner
■■ Figure 5.7 Simple apparatus used
to measure enthalpy changes of
combustion of liquids
A
C
∆Tc
D
C
0
1
2
3
4
5
6
Time/min
7
8
9
10
■■ Figure 5.8 A temperature correction curve for an exothermic
reaction
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Temperature/°C
Temperature/°C
B
D
∆Tc
A
B
0
1
2
3
4
6
5
Time/min
7
8
9
10
■■ Figure 5.9 A temperature correction curve for an
endothermic reaction
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172 5 Energetics/thermochemistry
ToK Link
What criteria do we use in judging discrepancies between experimental and theoretical values?
It is important to recognize the distinction between a theoretical value and a literature value: discrepancies
between an experimentally determined value and a literature value are due to random and systematic
errors, but a discrepancy between an experimental value and a theoretical value may be indicative of
inappropriate assumptions in the theoretical model. For example, calculated lattice enthalpies usually
assume a purely ionic model and therefore a discrepancy with experiment may indicate a covalent character
in the bonding.
Which ways of knowing do we use when assessing experimental limitations and theoretical assumptions?
Sense perception, reason and emotion are certainly all important here. Sometimes reason and emotion can
conflict with one another. Classic examples of the influence of emotion are cold fusion and the recently
reported ‘faster than light’ neutrinos. It is an emotional response to rush to publish a scientific paper with a
result that runs counter to a well-tested prevailing theory, whereas it would have been preferable to ensure
that all variables had been accounted for, before making the results public.
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■■ Enthalpy change of combustion
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The standard enthalpy change of combustion for a substance is the heat energy released
when one mole of the pure substance is completely burnt in excess oxygen under standard
conditions.
An example of the enthalpy change of combustion is the combustion of methane. The
reaction can be described by the following thermochemical equation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH c = −698 kJ mol−1
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Enthalpy changes of combustion are always negative as heat is released during combustion
processes.
Measuring enthalpy changes of combustion of fuels
Enthalpy changes for the combustion of liquids can be measured in a flame combustion
calorimeter (Figure 5.10). The investigation is carried out as follows:
Place a known volume and hence known mass of water into the calorimeter.
Stir and record the temperature of the water.
Record the mass of the spirit burner.
Turn on the pump so that there is steady flow of air and hence oxygen through the
copper coil.
Use the electrically operated heating coil to light the wick.
Slowly stir the water throughout the experiment.
Allow the spirit burner to heat up the water.
Record the maximum temperature of the water.
Re-weigh the spirit burner to determine the mass of liquid fuel combusted.
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1
2
3
4
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5
6
7
8
9
■■ Figure 5.10
Measuring the enthalpy
change of combustion
of a liquid fuel using
a flame combustion
calorimeter
thermometer
to pump
stirrer
copper coil
water
heating coil
wick
fuel
air
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5.1 Measuring energy changes 173
Worked example
Some example results are given below for the combustion of methanol.
Volume of water = 100 cm3
Temperature rise = 34.5 °C
Mass of methanol burned = 0.75 g
Specific heat capacity of water = 4.18 J g−1 °C−1
Use these results to calculate the molar enthalpy change of combustion. Compare the value with the
literature value.
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Heat energy transferred = 100 g × 4.18 J °C−1 g−1 × 34.5 °C = 14 421 J
Amount of methanol burnt = 0.75 g = 0.023 mol
32 g mol−1
14 421 J
Amount of energy released per mole of methanol =
= 627 000 J mol−1
0.023 mol
Hence the enthalpy change of combustion of methanol is −627 kJ mol−1.
The experimental literature value for the standard enthalpy change of combustion of methanol is
−726 J mol−1. The absolute error is (726 − 627), that is, 99 kJ mol−1. The percentage error is (726 − 630)/726 ×
100, that is 14% (Chapter 11).
The bomb calorimeter
The experimental literature values for standard enthalpy changes of combustion are obtained
using a more accurate bomb calorimeter. These can be obtained for fuels as well as energy
contents of dried foodstuffs.
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Additional
Perspectives
ed
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The error is due to the large heat losses that occur during the use of the flame combustion
calorimeter. Heat losses to the surrounding air are relatively large, despite the use of heat shields.
Also, heat energy from the flame heats up the material of the calorimeter itself, as well as the water.
A correction can be made for the heat losses to the calorimeter if the specific heat capacity of
copper is known or if the apparatus is standardized using a substance of known enthalpy change of
combustion.
This technique involves combusting a known mass of dried food in the presence of
excess oxygen and recording the change in temperature of water. This apparatus gives
accurate readings because heat losses to the surroundings are minimized. Strictly speaking
the bomb calorimeter measures internal energy changes, but these can be converted to
enthalpies.
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A sample of dried bread with a mass of 1.48 g was completely combusted (in a bomb calorimeter)
and was found to raise the temperature of 100 g of water by 71 °C. The quantity of heat
transferred, q, is given by the expression, q = mcΔT; q = 100 g × 4.18 J g−1 °C−1 × 71 °C = 29678 J.
Hence the energy content of the dried bread is 29678 J/1.48 g = 20052 J/g.
The energy values of food
Although the cells of our body do not burn digested food (nutrients) in the same way fossil
fuels are burnt, the outcome is the same. Molecular oxygen is still required and the energy
obtained from compounds is similar as if they were combusted in a calorimeter. When glucose
is respired (‘burnt’) in the cells of our body there are a series of enzyme-controlled steps each
catalysed by an enzyme. The enthalpy change of combustion is its energy value and for ease of
comparison, values are usually given for 100 g of dried food. Some typical values of food from
a database are given in Table 5.1.
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174 5 Energetics/thermochemistry
■■ Table 5.1
The energy values of
selected food
Food
Apple
Milk
Potatoes
White bread
Bacon
Cheddar cheese
Butter
Energy content (kJ/100 g)
200
270
370
900
1470
1700
3040
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When food is consumed some of its energy context is lost through our faeces in the form of
fibre. If we consume more energy than we expend, the surplus is stored as fat. Most of the energy
we consume is used up by our bodies. The exact amount we use depends on our age, sex, height,
weight and the amount of activity we do. For men, the range is usually between 9200 kJ and
12100 kJ per day although for manual workers it can be more. For women, the range is 6700 to
8800 kJ per day. Energy is needed for the enzyme-controlled chemical changes that occur inside
our body cells. This termed metabolism. For example, energy is needed to metabolize the food
we consume. Energy is needed to keep us warm in cold weather.
■■ Enthalpy change of neutralization
Pr
The standard enthalpy change of neutralization is the enthalpy change that takes place when
one mole of hydrogen ions is completely neutralized by an alkali under standard conditions.
An example of the enthalpy change of neutralization is the reaction between sodium
hydroxide solution and hydrochloric acid. The reaction can be described by the following
thermochemical equation:
ed
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
ΔH = −57 kJ mol−1
The enthalpy change of neutralization of a strong acid with a strong alkali is almost the same
for all strong acids and strong alkalis. This is because strong acids and strong alkalis undergo
complete ionization or dissociation in water:
ev
is
NaOH(s) + (aq) → Na+(aq) + OH−(aq)
HCl(g) + (aq) → H+(aq) + Cl−(aq)
The reaction between a strong base and a strong acid is the combination of hydrogen and
hydroxide ions to form water molecules. The other ions are spectator ions (they take no part in
the reaction). The reaction can be described by the following ionic equation:
R
H+(aq) + OH−(aq) → H2O(l)
ΔH = −57 kJ mol−1
For sulfuric acid, a dibasic acid, the enthalpy of neutralization equation is
1
H SO4(aq)
2 2
1
2
+ KOH(aq) → K2SO4(aq) + H2O(l)
ΔH = −57 kJ mol−
and not
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
This is because during the neutralization process each sulfuric acid molecule releases two
hydrogen ions:
H2SO4(aq) → 2H+(aq) + SO42−(aq)
However, where neutralizations involve a weak acid, a weak base, or both, then the enthalpy
of neutralization will be smaller in magnitude than −57 kJ mol−1, that is, slightly less
exothermic.
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5.1 Measuring energy changes 175
For example, the enthalpy of neutralization for ethanoic acid and sodium hydroxide
is −55.2 kJ mol−1 because some of the energy released on neutralization is used to ionize or
dissociate the acid:
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
CH3COOH(aq) → CH3COO−(aq) + H+(aq)
This can be shown in the form of an enthalpy level cycle (Figure 5.11).
∆H1
∆H = –55.2 kJ mol–1
CH3COO– + H2O
∆H = –57 kJ mol–1
CH3COO– + H+ + OH–
oo
f
A simple method for determining the enthalpy
change of neutralization involves mixing equal
volumes of dilute solutions of a strong acid
and a strong base of known concentration and
measuring the temperature rise. A thick plastic
cup fitted with a lid makes a cheap and effective
calorimeter.
The maximum temperature can be deduced
from the graph (see Figure 5.12) by extrapolating
both lines back to find out where they intersect.
Pr
■■ Figure 5.12 A
graph of temperature
versus volume of acid
for an investigation
to determine the
enthalpy change of
neutralization for an
acid
CH3COOH + OH–
Temperature/°C
■■ Figure 5.11 Enthalpy
level cycle for the
neutralization of
ethanoic acid
ed
Volume of acid added/cm3
Worked example
ev
is
50.00 cm3 of 1.0 mol dm−3 hydrochloric acid was added to 50.00 cm3 of 1.0 mol dm−3 sodium hydroxide solution.
The temperature rose by 6.8 °C. Calculate the enthalpy change of neutralization for this reaction. Assume that
the density of the solution is 1.00 g cm−3 and the specific heat capacity of the solution is 4.18 J g 1 °C−1.
q = mcΔT
q = 100 g × 4.18 J g 1 °C−1 × 6.8 °C = 2842 J
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
R
Amount of hydrochloric acid =
Amount of sodium hydroxide =
50.00
dm3 × 1.0 mol dm−3 = 0.050 mol
1000
50.00
dm3 × 1.0 mol dm−3 = 0.050 mol
1000
Enthalpy change of neutralization =
−2.842 kJ
= 56.8 kJ mol−1 ≈ 57 kJ mol−1
0.050 mol
(The answer is expressed to two significant figures because 6.8 °C has only two significant figures.)
Additional
Perspectives
Enthalpy change of solution
The standard enthalpy change of solution is the enthalpy change that occurs when one mole
of a solute dissolves in a large excess of water, so that no further heat change occurs when
more water is added to the solution.
For example, the enthalpy change of solution when dissolving sodium chloride in water is the
enthalpy change of the following reaction:
NaCl(s) + (aq) → Na+(aq) + Cl−(aq)
829055_05_IB_Chemistry_165-198.indd 175
ΔH = +3.9 kJ mol−1
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176 5 Energetics/thermochemistry
This enthalpy change of solution is positive, that is, the reaction is endothermic. This reaction
proceeds because a large positive entropy change occurs during the dissolving process
(Chapter 15). Many ionic compounds have enthalpies of solution that are negative, that is, the
reactions are exothermic.
Worked example
0.848 grams of anhydrous lithium chloride, LiCl, are added to 36.0 grams of water at 25 °C in a
polystyrene cup acting as a calorimeter. The final temperature of the solution was 29.8 °C. Calculate the
enthalpy change of solution for one mole of lithium chloride.
Amount of lithium chloride =
Therefore,
36.0 g
= 2.00 mol
18.02g mol−1
oo
f
Amount of water =
0.848 g
= 0.0200 mol
42.4g mol−1
amount of LiCl
1
=
amount of H2O 100
ΔH
H = 36.0 g × 4.18 J g−1 °C−1 × 4.8 °C = −0.72 kJ
ΔH
H = −0.72 kJ ×
Pr
‘Scaling up’ to molar quantities:
1 mol
= −36 kJ
0.020 mol
Therefore:
ed
LiCl(s) + (aq) → LiCl(aq) ΔHsol = −36 kJ mol−1 (to two significant figures)
■■ Enthalpy change of formation
ev
is
The enthalpy change of formation, ΔH f , of a substance is the heat change (at constant
pressure) on production of one mole of the pure substance from its elements in their standard
states under standard thermodynamic conditions (298 K and 1 atm pressure).
The standard state is generally the most thermodynamically stable form of the pure element
that exists under standard thermodynamic conditions. For carbon it is graphite and for
phosphorus it is white phosphorus, P4(s). (However, red phosphorus is more stable than white
phosphorus.)
The enthalpy change of formation of silver bromide, AgBr, is the enthalpy change for the
reaction:
1
2
R
Ag(s) + Br2(l) → AgBr(s)
ΔH f = −99.5 kJ mol−1
The following balanced equations do not represent enthalpy changes of formation:
2Ag(s) + Br2(l) → 2AgBr(s)
Ag(s) +
1
Br (g)
2 2
→ AgBr(s)
two moles of silver bromide are formed
bromine is not in its standard state
Enthalpy changes of formation are often difficult to measure in practice due to competing side
reactions and slow rates of reaction. For example, methane and potassium manganate(vii) cannot
be prepared from their elements via the following thermochemical equations:
C(s) + 2H2(g) → CH4(g)
ΔH f = −75 kJ mol−1
K(s) + Mn(s) + 2O2(g) → KMnO4(s)
ΔH f = −813 kJ mol−1
The IB Chemistry data booklet tabulates enthalpies of formation for selected organic compounds
on page 12. Note that enthalpy changes of formation for elements (in their standard states)
are zero since the thermochemical equation representing the formation of an element is a null
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5.1 Measuring energy changes 177
reaction: no reaction is involved in their formation. For example, the standard enthalpy change
of formation of oxygen is represented by:
O2(g) → O2(g)ΔH f = 0 kJ mol−1
However, the enthalpy changes of formation for ozone (O3(g)) and diamond (C(s, diamond)) are
not zero since these are not the standard states of the elements oxygen and carbon.
A number of enthalpies of formation can be directly determined, for example the enthalpy of
formation of water:
H2(g) +
1
2
O2(g) → H2O(l) ΔH f = −286 kJ mol−1
H2(g) +
1
2
oo
f
This enthalpy change is also equivalent to the enthalpy of combustion of hydrogen. The
enthalpy of combustion is the enthalpy change (at constant pressure) when one mole of a pure
substance undergoes complete combustion under standard thermodynamic conditions.
For example, the standard enthalpies of combustion of hydrogen, methane and ethanol are
represented by:
O2(g) → H2O(l) ΔH c = −286 kJ mol−1
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH c = −890 kJ mol−1
CH3CH2OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
ΔH c = −1370 kJ mol−1
ed
Pr
Enthalpies of formation are usually calculated indirectly from other enthalpy changes of reaction
including bond enthalpies. Enthalpy changes of formation are commonly used to calculate
enthalpy changes of reaction, using Hess’s law.
Enthalpy changes of formation are usually negative, that is, the corresponding reactions
are exothermic. However, some compounds have positive enthalpies of formation, for example
benzene and nitrogen monoxide:
6C(s) + 3H2(g) → C6H6(l) ΔH f = +49 kJ mol−1
1
N (g)
2 2
1
2
+ O2(g) → NO(g)
ΔH f = +90 kJ mol−1
R
ev
is
These compounds are energetically unstable relative to their elements. The Gibbs free energy
change of formation, ΔG f , is the criterion that determines the thermodynamic stability of
a compound relative to its elements (Chapter 15). Both benzene and nitrogen monoxide
are kinetically stable – there is a large activation energy barrier to decomposition at room
temperature. However, nitrogen monoxide does undergo significant decomposition in the
presence of a platinum catalyst, as the catalyst lowers the activation energy barrier to the
reaction (Chapter 6).
The normal ‘rules’ for manipulating enthalpy changes apply to enthalpies of formation. If a
thermochemical equation is reversed, then the sign of the enthalpy change is reversed. If the
balanced thermochemical equation is multiplied (or divided) by a constant, the enthalpy change
is multiplied (or divided) by the same constant.
Nature of Science
829055_05_IB_Chemistry_165-198.indd 177
Making careful observations – measuring energy transfers between system
and surroundings
The measurement of accurate enthalpy changes relies upon the accurate measurement of
temperature changes due to an exchange of heat between the system and surroundings.
When analysing energy changes, chemists need to focus on a limited and well-defined part
of the Universe to keep track of the heat energy changes that occur. The part chosen for study is
the system; everything else is termed the surroundings. When you study the heat energy changes
that accompany a chemical reaction in the laboratory, the reactants and products form the
system. The container or calorimeter and everything beyond it are the surroundings.
Systems may be open, closed or isolated (Figure 5.13). An open system is one where energy
and matter can be interchanged with the surroundings. A beaker of boiling water over a Bunsen
burner is an open system. Heat energy can enter the system from the flame, and water is slowly
released into the surroundings as steam.
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178 5 Energetics/thermochemistry
■■ Figure 5.13 Types
of thermodynamic
systems
surroundings
matter
energy
a open system
surroundings:
cylinder, piston
and everything
beyond
ev
is
■■ Figure 5.14
A closed system containing a
mixture of hydrogen and oxygen
gases
oo
f
system:
H2(g) and O2(g)
Most chemical systems studied in energetics or thermochemistry are closed systems that
can exchange energy but not matter with their surroundings. Consider a mixture of
hydrogen and oxygen gas in a cylinder fitted with a friction-less piston (Figure 5.14). The
system is the oxygen and hydrogen gases; the cylinder, piston and everything else in the
Universe are the surroundings.
If the gases react to form water, heat energy is released. There is no change
in mass but the system can exchange energy with its surroundings in the form
of work and heat. If the piston is moved up then the system has done work. This
is orderly motion. If the surrounding air outside the piston is heated, then the
random movement of particles in the air increases. This is due to the absorption
of heat.
An isolated system is one in which neither energy nor matter can be exchanged
with the surroundings. An insulated vacuum flask containing hot coffee
approximates to an isolated system. However, the hot coffee eventually cools and
enters into state of thermal equilibrium with the surroundings.
A characteristic of a system is called a property. Intensive properties do not
directly depend on the amount of matter in the system (for example temperature
and pressure). Extensive properties, for example mass and volume, do directly
depend on the number of particles. The ratio of two extensive properties becomes
intensive in nature.
A system in equilibrium experiences no changes when it is isolated from
its surroundings. A system is in thermal equilibrium if the temperature is the
same throughout the system. It is in mechanical equilibrium if the pressure stays
constant.
Pr
matter cannot
enter or leave
the system
c isolated system
b closed system
ed
energy can enter or
leave the system as
heat or as work is
done on the piston
energy
5.2 Hess’s law
– in chemical transformations energy can neither be created
or destroyed (the first law of thermodynamics)
■■ Hess’s law and enthalpy change
R
Hess’s law states that if a reaction consists of a number of steps, the overall enthalpy change is
equal to the sum of the enthalpy changes for all the individual steps. Hess’s law (Figure 5.15)
states that the overall enthalpy change in a
Route 1
reaction is constant and not dependent on the
ΔH this way ...
pathway taken.
When reactant A is converted directly into
ΔH1
product
D by route 1, or indirectly by route 2
D
(via intermediates B and C), then according to
Hess’s law the enthalpy change in route 1 will
Route 2
... is the same as
ΔH4
equal the enthalpy changes of the reactions in
ΔH this way
route 2.
In symbols:
A
ΔH2
B
ΔH3
C
■■ Figure 5.15 An illustration of the principle of Hess’s law
829055_05_IB_Chemistry_165-198.indd 178
ΔH1 = ΔH2 + ΔH3 + ΔH4
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