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Unit 3 Energy Changes in Chemical Reactions
1.
THE NATURE OF ENERGY:
The production of energy is one of the most important and pervasive aspects of chemistry. Think for a moment
about all of the chemical reactions that we use to produce energy during the course of a typical day in our lives:
(Obtain student examples.) We eat foods to produce the energy needed to maintain our biological function. We
burn fossil fuels to produce the electrical energy that is central to our modern society, to produce heat for our
homes, and to produce power for planes, trains, and automobiles. We use ice cubes to cool our drinks and we
use heat to convert raw dough into baked bread. Green plants absorb energy from the Sun to fuel the chemical
reactions that lead to their growth. All of these examples illustrate the general point that chemical reactions
involve energy. Some reactions, such as the burning of gasoline, release energy. Others, such as the splitting of
water into hydrogen and oxygen, require the addition of energy. Over 90% of the energy produced in our
society comes from chemical reactions, mostly from the combustion of coal, petroleum products, and natural
gas. And this will be of major concern to humanity over the next 25 years, as reserves of fossil fuels run out.
If you yelled for 8 years, 7 months and 6 days you would have produced enough sound energy to heat one cup
of coffee.
Banging your head against a wall uses 150 calories an hour. This is about 630 J/h
In understanding the concepts of energy change in a chemical context, we need to understand energy, work and
heat.
Energy is the ability to do work or to transfer heat. Work is performed when a force moves an object over a
distance. Work is done when you lift an object. Work is not being done when you hold it over your head, no
matter how heavy the object is.
Heat is the energy transferred form a hotter object to a cooler object when the two are placed in thermal contact
with one another. (Thermal contact, means that the heat has a way of moving from the warmer to the cooler
object)
The three primary sources of available energy are solar energy, chemical energy and nuclear energy. Obtain
from students examples of chemical energy. These should include: chemical batteries, food, fossil fuels.
The Law of Conservation Energy, is the First Law of Thermodynamics, and states that energy cannot be
created or destroyed. But as energy actually can be created during a nuclear reaction, there is the law of
conservation of mass-energy, which states that mass and energy can be transformed into each other but that
mass-energy can be neither created nor destroyed.
Einstein saw mass as concentrated energy, and related mass and energy in the equation E=mc2.
The change in mass during a nuclear reaction will equal the energy released divided by c, the speed of light.
There are two categories into which all energy can be classified: potential and kinetic.
Potential energy is stored energy that matter possesses because of its position, condition, composition or
electric charge.
Kinetic energy is the energy of motion. Ek=mv2. This is zero a zero Kelvin.
Chemical energy is a sum of the kinetic and potential energy of the molecule. The origin of chemical energy
lies in the position and motion of atoms, molecules and subatomic particles. The total energy possessed by a
molecule is the sum of all the forms of potential and kinetic energy associated with it.
If you consider a single molecule of say hydrogen chloride: you can look at a variety of different energy types.
Vibrational energy is one form of kinetic energy, the energy produced by the vibration of the atoms with
respect to each other. Use a model and stretch, contract the spring.
Translational energy is a type of kinetic energy associated with gas molecules as they move linearly from one
point in space to another. Et=  v2 and will be 0 at 0K.
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Unit 3 Energy Changes in Chemical Reactions
A molecule has rotational energy if it is rotating about an axis through its centre of mass. Within this hydrogen
chloride molecule, there are electrons moving about the atomic nuclei. The electron motion about the nuclei is
a form of kinetic energy referred to as electron energy, Ee Most of the energy of a substance relates to the
electron energy. In chemistry, we are mainly concerned with the electron energy involved in the making a
breaking of chemical bonds.
The potential energy of a molecule results form the interaction between electrons and nuclei both within and
between atoms. This interaction gives rise to chemical bonds, and the energy associated with this interaction
between positive and negative charges is called bond energy.
There is also nuclear energy related to the forces that hold the nucleus together and gravitation energy. It is
very difficult to determine the total energy of a molecule by the energy that we measure experimentally is that
associated with a system changing from an initial to a final state. We are interested in energy changes, rather
that absolute total energy content. And we are primarily concerned with energy in the form of heat that is
absorbed or released by a reaction at constant pressure. This quantity of energy is know as change in enthalpy
(heat content) of the system and has the symbol H.
When chemicals such as fuels react, energy is released. The chemical energy of these substances is due to the
potential energy stored in the arrangement of the atoms of the substance.
The SI unit of energy is the joule named in honor of James Joule (1818-1889). As joules are rather small,
chemists often use the unit kilojoule, kJ. The non-SI unit of energy is the calorie. When we speak of food
calories, we are actually talking about Kcal. 1 claorie = 4.184J exactly, i.e. by definition.
Work and Heat
Energy can be transferred in one of two ways. Energy can be used to do work, that is to cause an object to
move against a force. An example of work is pulling a sled across a snowy field. The force of friction is
resisting the smooth passage of the sled, and there is work done to overcome the friction. Energy can also be
transferred by heat (which is a form of kinetic energy). Heat is the energy transferred from a hotter object to a
colder one. A combustion reaction releases the chemical energy stored in the molecules of the fuel in the form
of heat. The heat raises the temperature of the surrounding (cooler) objects. Energy in the form of heat is
transferred from the system to the surroundings. In order for there to be “heat” there must be two substances
that are of different temperatures. Heat is a relative term. There is heat in the warm soup which is at 60°C. But
the soup is cool compared to a cup of boiling water. The water is cool compared to a fire. Some flames are
hotter than others, and hence classify as ‘cool’. Our sun is a cool star especially compared to a supernova, but
out star is 38,000 °C in the sunspots, the cool spots, and 58,000 °C at the surface otherwise.
Energy
Thus we have a definition for energy. Energy is the capacity to do work or to transfer heat. Now imagine a
clay ball being lifted to the top of a wall. This lifting is work because of the force of gravity. The energy we
transfer to the ball by doing work on it increases its potential energy because the ball is now at a greater height.
If the ball now rolls off the wall, its downward speed increases as its potential energy is converted to kinetic
energy. When the clay ball strikes the ground, it stops moving and its kinetic energy drops. Some of the kinetic
energy is used to do work in squashing the ball, and the rest is dissipated to the surrounding as heat during the
collision with the ground and as sound.
Extensive and Intensive properties.
Imagine a system that is uniform throughout. Now divide it into smaller ‘subsystems’ but do not alter it in any
other way.
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An extensive property is one that can be written as the sum of the corresponding properties of the subsystems.
e.g. the mass of a large system is the sum of the masses of the subsystems making up this large system. Volume
and energy are also extensive properties. (You can extend systems and you will affect the extensive property)
An intensive property is one this is the same as corresponding properties of each of the subsystem. e.g. the
pressure of the subsystem is the same as the larger system. So is the temperature. (The intensity of the
perceived temperature and pressure are not different if your system is big or little)
Heat and Temperature. Heat and temperature are not the same. Imagine a kg of water (1 L) and a kg of
aluminum are heated for identical lengths of time by identical heat sources. The final temperature of the two
will be quite different, although they both will have received the same amount of heat. Temperature is an
intensive property and heat is an extensive property. There is more heat (energy to be transferred) in a large
object than in a small one.
Heat (q) is the thermal energy transferred between two bodies in when the two are placed in thermal contact
with one another. Heat transfers from a body at warmer temperature to a body at a lower temperature.
Temperature (T) is the degree of hotness or coldness of an object as measured on some defined scale.
Temperature related to the absolute kinetic energy of the particles in an object.
When chemists do experiments and observe changes in energy, it is easiest to detect the change in heat energy.
this heat detected is in the surroundings, not the chemical system, because the chemical system is absorbing
heat and changing into chemical potential energy, or releasing chemical potential energy in the form of heat. So
chemists define the system as the chemicals that are reacting and the rest of the universe as the surroundings.
System and Surroundings.
An open system is one in which both materials and energy can flow in or out. An example is an open beaker
on a lab bench.
A closed system is one in which the chemicals are contained, but energy is not. Reactants and products do not
get added or removed, but energy does. An example is a sealed glass tube.
An Isolated system is not able to transfer any material or any energy. And isolated system is only a
hypothetical construct, and cannot exist in reality, unless you wish to consider the universe in its entirety as an
isolated system.
THE FIRST LAW OF THERMODYNAMICS
The first law of thermodynamics is not new to you. It states that energy is conserved. Any energy that is lost
by the system must be gained by the surroundings and vice versa. In mathematical terms this is:
E  E final  Einitial (the symbol  means change)
Calorimetry is based on the first law of thermodynamics.
CALORIMETRY
When heat flows into, or out of, a substance, the temperature of the substance changes. Experimentally, we can
determine the heat flow associated with a chemical reaction by measuring the temperature change it produces.
The measurement of heat flow is calorimetry and the apparatus that measures heat flow is a calorimeter.
(Refer students to the picture in text on page 309)
When an object sits in the sun and absorbs heat, its temperature changes. The temperature change experienced
by an object when it absorbs a certain amount of energy is determined by its heat capacity. Consider placing
100mL (100g) of water in the sun and then 100 g of copper beside it. Which will be untouchably hot first? Heat
capacity is defied as the amount of heat required to raise its temperature by 1K. The greater the heat
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capacity of a body, the greater the heat required to produce a given rise in temperature. The heat capacity of 1
mol of a substance is called it molar heat capacity. The heat capacity of 1 g of substance is called it specific
heat capacity. The heat capacity is the heat required to raise the temperature by 1 ºC. Sample size is not
variable. Units are J/ºC
quantity of heat transferred
The specific heat capacity (C) of a
( grams of substance)  (temperature change)
substance can be determined experimentally by measuring the temperature change, T, that a known mass, m of
the substance undergoes when it gains or loses a specific quantity of heat, q.
Specific heat capacity 
Example: It requires 209 J to increase the temperature of 50.0 g of water by 1.00 K. Thus the specific heat
209 J
J
capacity of water is: Specific heat capacity 
 4.18
(50.0 g )(1.00 K )
g.K
The temperature change in this example is positive as T=Tf-Ti, so if the temperature increases, T is positive.
Example: Calculate the molar heat capacity for water, given that the specific heat capacity is
4.18 J/g.K
H2O is 18.0 g/mol
4.18 J/g.K x 18.0g/mol = 75.2 J/mol.K
Example: How much heat is needed to warm 250 g of water (about 1 cup) from 22C to near its boiling point,
98C?
Solution:
Tf = 98C
Ti = 22C
T = 98C-22C = 76C=76K
q = (specific heat of H2O) x (mH2O) x T
= (4.18 J/g.K)(250 g )(76 K)
= 7.9 x 104 J
Thus 7.9 x 104 J of heat is required to heat this cup of water. As heat can get quite large in value, kilojoules
are often used. The answer can also be reported as 79 KJ of heat.
An aside about the heat capacity of water: It is one of the highest known for any substance. It is about five
times as great as that of say aluminum metal. The high specific heat of water serves the important role of
keeping the temperatures of the oceans relatively resistant to change. This is critical, as even a change of only
2C in the oceans' temperatures could cause a mini ice age like the one that occurred between the 12 th and 17th
centuries. The high heat capacity of water also helps humans to stay at a relatively constant temperature, even
in temperature extremes, which is critical as many of the bodies’ chemical reactions are very temperature
sensitive. For example, if the body heats to more than 40°C (104°F) then brain damage can occur due to
undesirable chemical reactions.
Calorimetry is the name given to the experimental technique of determining heats of reaction. It involves
carrying out a given reaction in an enclosed container (called a "calorimeter"). The heat produced by the
reaction is absorbed by the fluid (often water) in the calorimeter. Depending on the composition of the
calorimeter, some of the heat may be also be absorbed by the container itself (if it is not a good insulator). The
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Unit 3 Energy Changes in Chemical Reactions
mass and temperature change of the fluid can be measured. The specific heat capacity (C) of the fluid is usually
obtained from a reference text. The quantity of heat absorbed by the calorimeter is assumed to be equal to
the heat produced by the reaction (in practice some heat is always lost to the surroundings). The quantity of
heat (q) absorbed by the calorimeter can then be calculated according to the following formula.
Quantity of Heat q=mfluid x Tfluid x Cfluid or q=mCT
The molar heat of reaction in kJ/mol of the reactant may then be calculated using the formula
Molar Heat of Reaction H rx  q where n = # moles reactant
n
The -ve sign is introduced since the heat gained by the calorimeter (+ve ,H) is heat produced (lost) by the
reaction.
Example: When 2.00 g of magnesium is completely burned the heat given off causes the temperature of 500 mL
of water in an insulated calorimeter to change from 28.5C to 52.1 C. Calculate the heat of combustion
(Hcomb) of magnesium from this data.
Solution :
m(Mg) = 2.00 g
M =24.3 g/mol
2.00 g
n( Mg )  m 
 0.0823 mol
M 24.3g / mol
Vwater = 500 mL
 mwater = 500 g = 0.500 kg (recall density of water = 1.00 g/mL)
Ti=28.5C
Tf= 52.1C
T= Tf- Ti = 52.1C - 28.5C = 23.6C
C(H2O) = 4. 186kJ./(kg.C)
qfluid = mfluid x Tfluid x Cfluid = (0.500 kg) (23.6C ) (4. 186kJ./(kg.C ) = 49.4 kJ
q
49.4 kJ

 600 kJ / mol
 Hrx 
n 0.0823 mol
 the heat of combustion of magnesium is -600 kJ/mol.
Bomb Calorimetry: One of the most important types of reactions studied using calorimetry is combustion. A
compound, usually organic, is allowed to react completely with excess oxygen in a bomb calorimeter. The
substance to be studied is placed in a small cup within sealed vessel called a bomb. The bomb, which is
designed to withstand high pressures, has an inlet valve for adding oxygen and also has electrical contact to
initiate the combustion. After the sample has been placed in the bomb, the bomb is sealed and pressurized with
oxygen. It is then placed in the calorimeter, which is essentially an insulated container, and covered with an
accurately measured quantity of water. When all the components within the calorimeter have come to the same
temperature, the combustion reaction is initiated by passing an electrical current through a fine wire that is in
contact with the sample. When the wire gets sufficiently hot, the sample ignites. The energy of the reaction
which causes the temperature to rise is calculated by using the change in temperature of the water and the
calorimeter because it is assumed (with a great deal of certainty) that the heat lost by the system is absorbed by
the calorimeter only. Although the bomb calorimeter is not a constant pressure system, the difference between
E and H is very small.
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Unit 3 Energy Changes in Chemical Reactions
Sample problem:
1.435 g of naphthalene (C10H8) a.k.a. moth balls, was burned in a bomb calorimeter. The temperature of the
water rose from 20.17 ° C to 25.84 ° C. If the water surrounding the calorimeter was exactly 2000g and the heat
capacity of the bomb calorimeter was 1.80 kJ/° C, calculate the heat of combustion of naphthalene on a molar
basis; that is: find the molar heat of combustion. (We will assume that all the heat released is captured by the
H2O )
Solution:
First calculate the heat change for the bomb and for the water.
q=mCT
qwater = (2000g)(4.184 J/g.° C) (25.84° C-20.17° C)
= 4.74 x 104 J
qbomb = (1.80 x 103J/° C) (25.84° C-20.17° C)
= 1.02 x 104 J
Heat lost in the reaction is the sum of these heats. So heat of reaction is the negative sum of these heats.
qrxn = -(4.74 x 104 J + 1.02 x 104 J)
= - 5.76 x 104J
MC10H8 = 128.2 g/mol
molar heat of combustion =
-5.76 104 J 128.2 g C10 H 8

1.435 g C10 H 8 1 mol C10 H 8
 5.15 106 J / mol
 5.15 103 kJ / mol
Calorimetry Problems
4.1
3.00 g of potassium hydroxide (KOH) pellets are put into a Styrofoam calorimeter (i.e. a used
Styrofoam coffee cup) containing 250 mL of water. The initial temperature of the water is
22.3°C. The mixture is then stirred until all of the KOH is dissolved. The final temperature of
the
solution is measured and recorded as 29. 7C . Calculate the heat of solution for potassium hydroxide. The
specific heat capacity of the solution may be assumed to be the same as that of water since this is a dilute
solution (3 g in 250 mL is roughly a 1 %+ solution). Assume no heat loss from the calorimeter. Ignore the mass
of solute when calculating q. (-144J/mol)
4.2
12.72 g of ammonium chloride is dissolved in 100 mL of water in an insulated calorimeter so that the
temperature changes from 24. 7C to 7. 2C . Calculate the heat of solution of ammonium chloride assuming
that the specific heat capacity of the solution is the same as that of water. Ignore the mass of solute when
calculating q. (30.7kJ/mol)
4.3
A sample of sucrose (C12H22O11) with a mass of 1.32 g is burned in a bomb calorimeter. The overall heat
capacity (C x m) of the calorimeter has been previously determined to be 9.43 kJ.C-1. The temperature of the
calorimeter changed from 25.00°C to 27.31 C. Calculate the heat of combustion of sucrose. (-5.64 x
103kJ/mol)
4.4
A sample of carbon with a mass of 4.00 g is burned in a 500 g "bomb " calorimeter made of nickel metal
[CNi = 0.444 kJ/(kg.C)]. The bomb calorimeter is immersed in an insulated container containing 1.50 kg of
water. The temperature is measured before and after and found to go from 22.75°C to 42.98°C. Calculate the
molar heat of combustion of carbon. (-394 kJ/mol)
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Unit 3 Energy Changes in Chemical Reactions
Assign also the Calorimetry Worksheet
ENTHALPY
Absolute values of Energy cannot be measured by chemists, but relative amounts, changes in values can be
measured. Back to our definition of energy as the ability to do work or to transfer heat. If we set up our
experiment so that in the reaction the pressure does not change, then we can measure the change in energy as
the heat given off. This heat is called Enthalpy (from the Greek word enthalpein, meaning to warm).
Enthalpy is the heat content of a substance. The change in enthalpy in a reaction is the difference
between the heat contained in the products and the heat contained in the reactants. The enthalpy of reaction is
heat absorbed at constant pressure.
H = qp
Enthalpy was defined so that we can talk about the amount of heat released or absorbed by a reaction when any
work being done is PV work occurring at constant pressure.
H < 0 then qp is negative which means heat is leaving the system. A process which releases heat is exothermic.
H > 0 then qp is positive which means heat is being absorbed by the system. A process which absorbs heat is
endothermic.
Remember this does not represent the change in the total energy content (internal energy, E) it just tells us how
much heat energy is released during a change occurring under very specific conditions. The approximation is
good unless there is a lot of gas produced (i.e. work is done)
In general, a reaction either releases or absorbs heat. This heat called the enthalpy of reaction is defined as the
difference between the enthalpy of products and enthalpy of reactants.
H=Hf –Hi just as T=Tf-Ti , Hf is the enthalpy of the products and Hi is the enthalpy of the
reactants. So H= Hproducts-Hreactants. If the enthalpy of the products is greater than the enthalpy of the reactants,
then the H is positive. Also the products must have absorbed heat from the surroundings and the temperature
around the reaction has decreased. So when we see a temperature drop, it means that the enthalpy change of the
chemicals, the part we are interested in, the system, is positive. i.e. a reaction that appears to get colder, as its
surrounding drop in temperature, has a positive enthalpy change.
Enthalpy is a State Function meaning that its value is independent of the route taken to reach it. (A state
function is a property of a system that is uniquely determined by the present state of the system and not at all by
its history) An analogy is change in altitude. If you go from Kingston to Whistler BC, it does not matter what
route you take, the altitude change will be the same. You can fly or take a train or drive up from the south or
from Kelowna in the north, but the change in altitude will always be the same.
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Unit 3 Energy Changes in Chemical Reactions
A function which is not a state function is work. If I need to move 100 books up three flights of stairs, I can
carry them up one at a time, I can carry them up 20 at a time, or I can tie them together and pull them up using
a pulley system. In each different scenario, I am doing a different amount of work to get the same result.
Other properties which are state functions are temperature, volume, pressure, enthalpy and the internal energy.
When a reaction occurs in which the system (the reactants and products) absorb heat, we say that the process is
endothermic. During an endothermic process, heat flows into the system from its surroundings. Melting of ice
is endothermic, as is the dissolution of ammonium chloride, the salt used in athletic 'instant ice packs'.
A process which results in the evolution of heat is called exothermic. During an exothermic process, heat flows
out of the system and into its surroundings. The combustion of gasoline is an example of an exothermic
process. Another exothermic reaction is the dissolution of calcium chloride in water, a reaction exploited in
hand warmers. The convention is that if H> 0 then the process is endothermic. The system has absorbed heat,
taken heat from the surroundings. Perhaps the test tube got colder as the system took heat from it.
A positive H means the reaction is endothermic.
products
H is positive: H = Hfinal - Hinitial
Energy
reactants
time
Endothermic Reaction
Thus the convention also is that if H< 0 then the process is exothermic. If the system lost heat, it must give
that heat to the surroundings and the surrounding increased in temperature.
A negative H means the reaction is exothermic.
reactants
H is negative : H = Hfinal - Hinitial
Energy
products
time
Exothermic Reaction
Do the demo of the endothermic reaction of NH4NO3 dissolving in water. An exothermic dissolution could
be NaOH in water.
Although you know of reactions that are both exothermic and endothermic, there are many more reactions that
are exothermic than endothermic. This will make sense when we think of what happens to a brick as it is
dropped. It loses potential energy. This is natural. Chemical systems also tend to lose energy, and thus an
exothermic reaction is more likely than an endothermic reaction. There is another factor at work here (which
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Unit 3 Energy Changes in Chemical Reactions
we will discuss later) which allows some endothermic reactions to occur. Heat not only drives some reactions,
it retards others. Chemists can use the change in heat of reactants to make predictions. Thermodynamics’ goal
is to predict what types of chemical and physical processes are possible, and under what conditions, and to
calculate quantitatively the properties of the equilibrium state that ensues when a process is carried out.
From the law of conservation of energy, the enthalpy change for a reaction is equal in magnitude but opposite in
sign to H for the reverse reaction. Consider the thermochemical reaction below. The presence of the H at
the end of the equation, a thermodynamic property, makes this a thermochemical equation.
Example:
CH4 (g) + 2O2(g)  CO2(g) + 2H2O(l) + 890 kJ
Heat is given off and this is an exothermic reaction. The products have less heat in them than the
reactants. Hproducts - Hreactants is negative. Thus H for the reaction is -890 kJ/mol
CH4 (g) + 2O2(g) CO2(g) + 2H2O(l) H = -980 kJ
and
CO2(g) + 2H2O(l)  CH4 (g) + 2O2(g) H = 980 kJ
The reverse reaction must have an equal and opposite enthalpy of reaction in order for there to be conservation
of energy. This change of 980 kJ is per 1 mol of CO2 per 2 mol of H2O per 1 mol of CH4 etc.
If the equation were rewritten as
2CH4 (g) + 4O2(g)  CO2(g) + 4H2O(l) then H = -890 kJx2 = -1780kJ
Some Different Types of Enthalpy Change
While the enthalpy change for a chemical reaction can always be called H, there are several special forms of
this notation.
For a chemical reaction, as opposed to a physical change, H is often written as Hr. The subscript "r"
symbolizes "reaction".
. For reactions that occur at 25°C and 100 kPa, the enthalpy change is labeled H° -note the superscript "°" .
. When a compound is formed directly from its elements in their standard state, the enthalpy change can be
called Hf. The subscrpit "f" indicates "formation". (Standard state refers to physical state-solid, liquid, gas-at
25°C, 100 kPa.) For example C(graphite) + O2(g)  CO2(g) Hf= -393.5 kJ
This means that the formation of carbon dioxide gas, at 25°C and 100 kPa, directly from solid graphite and
oxygen gas (i.e. their standard state) liberates 393.5 kJ.
. The enthalpy change corresponding to a physical change such as vapourization or fusion (melting) is
symbolized Hvap or Hfus.
. The enthalpy change corresponding to a combustion reaction can be symbolized Hcomb. An example is the
combustion of propane, C3H5.
C3H5(g) + O2(g) 3 CO2 (g) + 4 H2O(g) H = H comb.
. The enthalpy change corresponding to a neutralization reaction can be symbolized Hneut
It is important to note that all of the above are simply more specific forms of H. The H notation encompasses
all of the above.
Thermochemical Equations.
Since chemical reactions involve the gain or loss of energy, the energy term may be included in the balanced
equation of a reaction. When this is done the equation is termed a thermochemical equation.
Example#1 The synthesis of one mole of carbon dioxide form its element s is accompanied by the release of
393.7 kJ of energy;; i.e. the 393.7 kJ is a product of the reaction just as is the carbon dioxide. The
thermochemical equation for this reaction may therefore be written as:
C(s) + O2(g)  CO2(g) + 393.7 kJ  Hrxn = -393.7 kJ
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Unit 3 Energy Changes in Chemical Reactions
Example #2 When one mole of HI(g) is produced by the reaction of hydrogen gas with iodine vapour 26.0 kJ
of energy is absorbed in the process. This may be represented as:
½H2(g) + ½O2(g) + 26.0 kJ  HI(g)  Hrxn = +26.0 kJ
Example #3 The combustion of one mole of ethane (C2H6) releases 1428 kJ of energy. The reaction my be
represented by
C2H6 + 7 2 O2(g)  2CO2 + 3 H2O(g) + 1428 kJ
 Hrxn = -1428 kJ
Hrxn = -1428 kJ/mol C2H6
Hrxn = -1428 kJ/ 7 2 mol O2(g) = -408 kJ/mol O2(g)
Hrxn = -1428 kJ/2 mol CO2(g) = -714 kJ/mol CO2(g)
Hrxn = -1428 kJ/3 mol H2O(g) = -476 kJ/mol H2O(g)
Note: the sign of Hrxn is negative in each case since the reaction as written is exothermic.
See also practice questions from pg 319 of text.
5.1
Rewrite each of the equations below as a thermochemical equation.
5.1.1 NO(g)  ½N2(g) + ½ O2(g)
Hrx = -90.4 kJ
5.1.2 CO2(g) +3/2 H2O(g)
1/2 C2H6(g) + 7/4 O2(g)
Hrx = +714.1 kJ
5.1.3 NH3(g) + 7/4 O2(g) NO 2(g) + 3/2 H2O(g)
Hrx = -283.0 kJ
For each of the equations below express the heat of reaction in Hrx notation per mole of the bolded
quantity. Classify each reaction as exothermic or endothermic.
Example.
Ca(s) + ½ O2(g)  CaO(s) + 636 kJ
rx is exothermic
Answer: Hrx = -636 kJ / ½ mol O2 = -1272 kJ/mol O2
= -1.27 x 103 kJ/mol O2 (max 3 SD!)
5.2
Hrxn = 33.9 kJ/mol NO2
5.2.1
½ N2(g) + O2 (g) + 33.9 kJNO2(g)
5.2.2
½ N2(g) +
5.2.3
C3H8(g) 3 C(s) + 4 H2(g) + 104kJ
5.3
A given mass of ammonium chloride dissolves in 50.0 mL of water causing a temperature change from
23.0°C to 14.7°C. Is the solubility of ammonium chloride exo- or endothermic?
5.4
2.50 g of sodium hydroxide dissolves in 100 mL of water causing a temperature change from 21.6°C to
27.4°C. Classify the solubility of sodium hydroxide as exothermic or endothermic.
Are combustion reactions in general exothermic or endothermic? Explain.
5.5
5.6
5.6.1
5.6.2
5.6.3
5.6.4
3
2
H2  NH3 + 46.2 kJ
Hrxn = -30.8 kJ/mol H2
Hrxn = -26 kJ/mol H2
Write thermochemical equations for the combustion of one mole of each of the given compounds.
Assume that the products are "the usual" gaseous products for burning hydrocarbons, at standard
conditions. The standard heats of combustion are given in brackets.
acetylene, C2H2(g) ............................................................. (-1299.6 kJ/mol)
methyl alcohol, CH3OH(l) ................................................ (- 726.51 kJ/mol)
diethyl ether, C4H10O(l) .................................................... (-2751.1 kJ/mol)
toluene, C7H8(l) ................................................................. (-3909 kJ/mol)
Do also Section 5.3 questions from pg 320 of text.
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SCH 4UK
Unit 3 Energy Changes in Chemical Reactions
6.
Hess' Law Not all reactions can be monitored and examined in a calorimeter. Some are too dangerous.
Some are too slow!! It is often possible to calculate the H for a reaction from the tabulated H values of other
reactions. Because enthalpy is a state function, the enthalpy change, depends only on the amount of matter that
undergoes change, and on the nature of the initial state of the reactants ant the final state of the product. This
means that if a particular reaction can be carried out in one step or in a series of steps, the sum of the enthalpy
changes associated with the individual steps must be the same as the enthalpy change associated with the onestep process.
It is important to specify the states of the reactants and products in thermochemical equations. It is assumed
that unless told otherwise the temperature of all products and reactants is 25C or 298K.
Example 1
A +B → C
 H1
C → D+E
 H2
A + B + C → D + E +C
 H =  H1 +  H2
Because C occurs on both sides, it can be cancelled like in an algebraic equation.
A + B→ D + E
 H =  H1 +  H2
Example 2
CH4(g) + 2O2(g)
 CO2(g) + 2H2O(g)
H = -802 kJ
(add)
2H2O(g)  2H2O(l)
H = -88KJ
CH4(g) + 2O2(g) + 2H2O(g)  CO2(g) + 2H2O(g) + 2H2O(l) H = -890 kJ
net equation
CH4(g) + 2O2(g)CO2(g) + 2H2O(l)
H = -890 kJ
Because 2 H2O(g) occurs on both sides, it can be canceled, like an algebraic quantity which appears on both
sides of an equal sign.
Hess' Law, states: if a reaction is carried out in a series of steps, the enthalpy change of the reaction will be
equal to the sum of the enthalpy changes for the individual steps. The overall enthalpy change for the process
is independent of the number of steps or the particular nature of the path by which the reaction is carried out.
We can calculate H for any process, as long as we find a route for which H values are known for each step.
The analogy is my bank balance. It is the sum of all of the credits and debits made. As long as I know the
nature of the transaction and the $ transferred, I know the final balance.
Example: The enthalpy of combustion of C to CO2 is -393.5 kJ/mol C, and the enthalpy of combustion of CO
to CO2 is -283.0 kJ/mol CO:
1. C(s) + O2(g)  CO2(g) H = -393.5 kJ
2. CO(g) + ½O2(g)  CO2(g)
H = -282.0 kJ
Using these data, calculate the enthalpy of combustion of C to CO:
3. C(s) + ½O2(g)  CO(g)
Solution: In order to use equations (1) and (2), we arrange them so that C(s) is on the reactant side and CO(g) is
on the product side of the arrow, as in the target reaction, equation (3). Because equation (1) has C(s) as a
reactant, we can use that equation just as it is. Notice that the target reaction has CO(g). Thus, we need to turn
equation (2) around so that CO(g) is a product. Remember that when reactions are turned around, the sign of
H is reversed. We arrange the two equations so that they can be added to give the desired equation.
11
SCH 4UK
Unit 3 Energy Changes in Chemical Reactions
C(s) + O2(g) CO2(g)
H = -393.5 kJ
CO2(g)  CO(g) + ½O2(g)
H = 283.0 kJ
___________________________________________________
C(s) + ½O2(g)  CO(g)
H = -110.5 kJ
7.
Calculating Heats Of Reaction - The Method Of Summing Equations
This method of calculating heats of reaction, as the name implies, involves expressing the reaction in question
(the "target equation") as the sum of two, or more, other equations (the "component equations"). The Table of
Standard Enthalpies of Formation, is a convenient source of component equations. The Handbook of Chemistry
and Physics has a more exhaustive list. Below is a summary of the steps involved in using the Table to calculate
heats of reaction. We will assume standard conditions throughout this unit unless information to the contrary is
specifically given.
1. For each compound in the target equation obtain from Table #1 a standard state formation equation for that
compound. Write it down along with its H value.
2. Rearrange component equation(s) as needed to show reactants in the target equation as reactants in the
component equations, products as products.
3. Change the sign of the enthalpy term, H, for each equation reversed!
4. Where necessary, multiply each component equation and corresponding H so that the component
equation shows the same number of moles of the compound as the target equation.
5. Number the target equation and component equations along with their corresponding H values.
6. Sum the component equations according to the rules of algebra. Treat "" as "=".
7. Simplify the resulting equation and compare with the target equation. If they are identical then Hess's Law
applies.
8. Calculate the heat of reaction for the target equation by summing the H values of the component
equations.
Example: Calculate the heat of reaction, Hrxn, for the reaction represented by the following equation.
Rx1 12 CO2(g) + 6 H2O(l) 2 C6H6(l) + 15 O2(g) Hrxn =?
Solution: From the Table we can obtain standard state formation equations for each of the compounds in the
target equation.
C(s) + O2(g)CO2(g)
H= -393.7 kJ
H2(g) + ½ O2(g) H2O(l)
H= -285.9 kJ
6 C(s) + 3 H2(g) C6H6(1)
H= +48.6 kJ
Since CO2(g) and H2O(l) appear on the left side of the target equation, they must appear on the left side of the
component equations. Therefore their formation equations are reversed. NOTE: H signs must be changed
accordingly.
CO2(g)  C(s) + O2(g)H = +393.7 kJ
H2O(l)  H2(g) + ½O2(g)
H = +285.9 kJ
12
SCH 4UK
Unit 3 Energy Changes in Chemical Reactions
Also the target equation shows 12 mol CO2, 6 mol H2O and 2 mol C6H6 . Therefore we must multiply the
component equations by 12, 6, and 2 respectively. Since energy changes are proportional to the amount of
matter involved therefore the corresponding H values must also be multiplied by these respective
coefficients. This will give us the component equations that should sum together to give us the target equation
as illustrated.
(x12) Rx2
(x6) Rx3
(x2) Rx4
12 CO2(g)  12 C(s) + 12 O2(g)
6 H2O(1) 6H2(g) + 3 O2(g)
12 C(s) + 6H2(g) 2 C6H6 (l)
H2 = 12(+393.7 kJ) = +4724.4 kJ
H3 = 6(+285.9 kJ) = + 1715.4 kJ
H4 = 2(+48.6 kJ) = +97.2 kJ
Rx(2+3+4) 12 CO2(g) + 6 H2O(l)  2 C6H6(l) + 15 O2(g)
A quick check of our target eqn shows that Rx2 + Rx3 + Rx4 = Rx1
Therefore, by Hess's Law ..... H1  = H2 + H3 + H4
= +4724.4 kJ + 1715.4 kJ + 97.2 kJ
= +6537 kJ
Therefore the heat of reaction for the given equation is +6537 kJ.
7.1
What is the connection between Hess' Law and the fact that H is a state function:
7.2
Consider the following hypothetical reactions:
AB
H = + 30 kJ
BC
H = + 60 kJ
Use Hess' Law to calculate the enthalpy change for the reaction A  C
7.3
Suppose you have the following hypothetical reactions:
XY
H = -50kJ
XZ
H = -70 kJ
Use Hess' Law to calculate the enthalpy change for the reaction Y  Z
7.4 Use the METHOD OF SUMMING EQUATIONS to calculate the heat of reaction (enthalpy change) for
each of the reactions represented by the equations below:
7.4.1
7.4.2
7.4.3
7.4.4
C2H6(g) + 7/2 O2(g)  2 CO2(g) + 3 H2O(l)
C6H6(g) + 6 H2(g)  3 C2H6(g)
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
H2O(g) + C(s)  CO(g) + H2(g)
7.5
Calculate the entha1py change for the reaction P4(s) + 10 C12(g)  4 PC15(s) given the following
equations and corresponding enthalpy changes:
P 4(s) + 6 Cl2(g)  4 PCl3 (1)
PCl3(1) + Cl2(g)  PCl5(s)
7.6
H = -1272.5 kJ
H = -137.3 kJ
Calculate the heat of formation of methanol, CH3OH(l), given the following:
13
SCH 4UK
Unit 3 Energy Changes in Chemical Reactions
CH3OH(l) + 3/2 O2(g)  CO2(g) + 2 H2O(l) H = -715.4 kJ
C(s) + O2(g)  CO2(g)
H= -393.7 kJ
H2(g) + ½ O2(g)  H2O(l)
H = -285.9 kJ
The Standard State of any reactant or product is its most stable form at 101.3 kPa and the specified
temperature. Unless indicated otherwise, always assume that a solid or liquid is pure. The standard state of a
dissolved substance is the substance at a concentration of 1 mol/L. There is no standard temperature, so the
temperature should be specified. If no temperature is specified, then 25 C should be assumed. This is different
from the STP conditions! Don't confuse them!
The reason that the state of the reactants is important is that there is energy converted in the change from one
state to another.
e.g H2O(l)  H2O(g)
H = +44 kJ
Producing H2O(g) is 44kJ more expensive in energy terms that producing H2O(l). It is like buying the food
already prepared.
8
Standard Enthalpy of Formation of a compound, Hf, is the change in enthalpy for the reaction that
forms 1 mol of the compound from its elements, with all substances in their standard states. The temperature
for Hf is usually 298K. If an element exists in more than one form under standard conditions, the most stable
form of the element is used for the formation reaction. For example, the standard enthalpy of formation for
ethanol, C2H5OH, is the enthalpy change for the following reaction:
2C (graphite) + 3H2 + ½ O C2H5OH(l) Hf = -277.7 kJ
The elemental source of oxygen is O2 not O3 or O, because O2 is the stable form of oxygen at 298 K and
standard atmospheric pressure. Similarly, the elemental source of carbon is graphite and not diamond, because
the former is the more stable (lower energy) form at 298 K and standard atmospheric pressure. Likewise the
most stable form of hydrogen under standard conditions is H2(g) so this is used as the source of hydrogen in the
equation.
The stoichiometry of formation reactions always indicates that 1 mol of the desired substance is produced. As a
result, enthalpies of formation are reported in kJ/mol of the substance. As well, in order to make the equations
accurate, it is sometimes necessary for fractional molar coefficients to be used. While it is never possible to
have ½ molecule of O2 and have it still be O2, it is quite possible to have ½ mol of O2. This is simply 3.01 x
1023 molecules instead of the usual 6.03 x 1023 molecules.
By definition, the standard enthalpy of formation of the most stable form of the element is zero because there
is no formation reaction needed when the element is already in its standard state. Thus the values of Hf for C,
H2(g) O2 and the standard states of all other elements are zero by definition.
Using Enthalpies of Formation to calculate Enthalpies of Reaction. We can use Hess' Law to calculate the
standard enthalpy change for any reaction for which we know the Hf values for all the reactants and products.
Using the formula method to calculate the heats in a reaction:
Hrn
products
reactants
-Hf reactants
Hfproducts
Thus we haveelements
Hrn = H
in elemental
form
f products - H
f reactants
14
SCH 4UK
Unit 3 Energy Changes in Chemical Reactions
A general statement is that the standard enthalpy of the reaction is equal to the sum of the standard enthalpies of
formation for the products (multiplied by their appropriate molar coefficients) minus the sum of the standard
enthalpies of formation of the reactants (also multiplied by their appropriate coefficients.)
Example: the combustion of propane gas C3H8(g) with oxygen to form CO2(g) and H2O(l)
The balanced equation is C3H8(g) + 5O2  3CO2(g) 4 H2O(l)
We can write this equation as the sum of three formation reactions:
C3H8(g)  3C(s) + 4H2
3C(s) + 3O2  3CO2(g)
4H2 + 2O2  4H2O(l)
C3H8(g) + 5O2 3CO2(g) 4 H2O(l)
+
+
H1 = -Hf [C3H2]
H2 = 3 Hf [CO2(g)]
H3 = 4Hf [H2O(l)]
Hrx= H1 + H2 + H3
eq'n (1)
eq'n (2)
eq'n (3)
eq'n (4)
Hrx= H1 + H2 + H3 = -Hf [C3H8(g)]+ 3 Hf [CO2(g)] + 4Hf [H2O(l)]
= -(-103.85 kJ) + 3(-393.5 kJ) + 4(-285.8 kJ) = -2220 kJ
Points to note:
1. The first equation is the reverse of the formation reaction for C3H8(g). As a result, the enthalpy
change for this reaction is -Hf [C3H8(g)] .
2. The second equation is the formation reaction for 3 mol of CO2(g). Because enthalpy is
extensive, that is it is dependent upon quantity of material present, the enthalpy change for this
step is 3 Hf [CO2(g)].
3. Similarly for equation three, we multiplied the Hf [H2O(l)] by 4. Note that we were careful to
specify the formation of H2O (l) not H2O (g).
In mathematical terms: for the reaction wA + xB + …  yC + z D + ….
Hrxn = [yHf (C) + zHf (D) + …] -[wHf (A) + zHf (B) + …]
Note on Standard enthalpy of Formation of Elements. The standard enthalpy of formation of any element is
zero by definition. For this reason the Hf of any element may be omitted from the formula when calculating
heats of reaction
e.g. The Hrxn for the reaction O2O2 is zero since not change has occurred.
Exercises.
8.1
Use the formula method to calculate the heat of reaction for each of the following equations.
(i.e. Hrn = Hf products - Hf reactants.)
8.1.1 2C6H6(g) + 15 O2(g)  12 CO2(g) + 6H2O(l)
8.1.2 C6H6(g) + 6 H2  3 C2H6(g)
8.1.3 C3H8(g) + 5O2(g) 3 CO2(g) + 4 H2O(l)
8.1.4 H2O(l) + C(s)  CO(g) + H2(g)
8.2
8.3
Calculate the heat of combustion of methane assuming that the products are gaseous.
Calculate the heat of combustion of ethylene (C2H4(g), assuming that the products are gaseous.
Optional
Chemical Bonds
Chemical bonds form when the attractive forces between two atoms are greater than the
repulsive forces. The electrons of one atom have an attraction for the nucleus of another atom. The electrons of
one atom will repel the electrons of another atom. Initially, attractions are greater than repulsions causing
atoms to continue to come ever closer together. As the atoms get even closer both attractions and repulsions
increase. However at some point, as the atoms get still closer together, repulsions increase faster than
15
SCH 4UK
Unit 3 Energy Changes in Chemical Reactions
attractions and eventually catch up to and become equal to attractions. The point at which attractions equal
repulsions is the point at which the molecule is formed and becomes a stable particle. The distance between
the nuclei of two atoms at the point when attractions equal repulsions is called the Bond length of the molecule.
The bonded condition is usually a more energetically stable one than an unbonded one. The energy
required to overcome attractive forces (and break bonds) between bonded particles raises the energy of the
system. Thus, bonded atoms are in a lower energy, more stable state than unbonded ones.
Relating bond strengths and Enthalpies of Reactions.
Bond enthalpy (aka bond dissociation energy or bond energy) is the energy required to break a chemical
bond. Bond energies are expressed in kJ/mol. Breaking bonds is always an endothermic process. i.e. it always
takes energy to break a bond.
Cl
Example:
Cl
2
Cl
H = 242 kJ = D(Cl-Cl)
The designation D(bond type) represents bond enthalpies. It take 242 kJ to break 1 mol of
Cl-Cl bonds.
Example:
H
H
C
H
(g)
C
(g)
4
H (g)

1660 kJ
H
Because there are four equivalent C-H bonds in methane, the heat of atomization is equal to the total bond
enthalpies of the four C-H bonds
.
16
SCH 4UK
Unit 3 Energy Changes in Chemical Reactions
Therefore, the average C-H bond enthalpy is D(C-H) = (1660/4) kJ/mol =415 kJ/mol.
The bond enthalpy for a given set of atoms, say C-H, depends on the rest of the molecule of which it is a part.
However, the variation form one molecule to another is generally small this supports the idea that the bonding
electron pairs are localized between atoms. If we consider C-H bond enthalpies in many different compounds,
we find that the average bond enthalpy is 413 kJ/mol which compares closely with the 415 k/Jmol value
calculated from CH4
Calculating Enthalpy changes from Bond Energies. If we were to make something of Lego and recorded
each type of connection that was made, and the cost of each connection we could come up with a price for
creating any Lego structure. If we first had to take apart some previously made structures we could add that
energy into our total calculations too. Similarly we can assign the energy cost of making each type of bond and
come up with an energy cost of making any compound if we know the cost of each type of bond. This is a
straightforward application of Hess' Law. We imagine that the reaction occurs in two steps; (1) We supply
enough energy to break those bonds in the reactant that are not present in the products. In this step the enthalpy
of the system in increased by the sum of the bond enthalpies of the bonds that are broken. (2) We make the
bonds in the products that were not present in the reactants. This step will release energy, and it lowers the
enthalpy of the system by the sum of the bond enthalpies of the bonds that are formed. The enthalpy of the
reaction, Hrxn, is estimated as the sum of the bond enthalpies of the bonds broken, minus the sum of the bond
enthalpies of the new bonds formed.
Hrxn = (bond enthalpies of bonds broken) - (bond enthalpies of bonds formed)
Example: What is the energy f the reaction between methane, CH4 and chlorine to produce methyl chloride,
CH3Cl?
H-CH3(g) + Cl-Cl  Cl-CH3 + H-Cl
Hrxn = ?
Bonds broken:
1 mole C-H, 1 mol Cl-Cl
Bonds made:
1 mol C-Cl, 1 mol H-Cl
We first supply enough energy to break the C-H bond and Cl-Cl bond, which will raise the enthalpy of the
system. We then form the C-Cl and H-Cl bonds, which will release energy and lower the enthalpy of the
system. But using data from bond energy tables, we estimate the enthalpy of the reaction as
Hrxn =(bond enthalpies of bonds broken) - (bond enthalpies of bonds formed)
= [D(C-H) + D(Cl-Cl)] - [D(C-Cl) + D(C-Cl)]
= (413) kJ + 242 kJ) - (328 kJ + 431 kJ) = -104 kJ
The reaction is exothermic because the bonds in the products (especially the H-Cl bond) are stronger
than the bonds in the reactants (especially the Cl-Cl bond). This method is less accurate than using actually Hf
 values because the bond energies are only averages. Diagram below shows the energy changes in the reaction
of CH4 with Cl2 to form HCl and CH3Cl.
If you look at the table of bond enthalpies, the N  N , C  O , and C=O bonds are among the strongest. So not
surprisingly, explosives are usually designed to produce the gaseous products N2(g), CO(g) and CO2(g).
e.g. nitroglycerine decomposes: 4C3H5N3O9(l) 6N2(g) + 12CO2(g) + 10H2O + O2(g)
This is almost a perfect explosive because it does not need an exterior source of O2.
Calculate the molar bond enthalpy of nitroglycerine decomposition from its bond enthalpies.
17
SCH 4UK
H H H
H C
N
C
Unit 3 Energy Changes in Chemical Reactions
C H
O O O
N O
O O
O O
Nitroglycerine, a pale yellow, oily liquid that is highly shock-sensitive: Merely shaking
the liquid can cause its explosive decomposition into nitrogen, carbon dioxide water and oxygen gases.
(Nitroglycerine is absorbed into cotton or clay to stabilize it. Alfred Noble, after whom the Nobel Prizes are
named, absorbed nitro onto silica, which stabilized it, and named this dynamite. Nobel focused on the
development of explosives technology as well as other chemical inventions, including such materials as
synthetic rubber and leather, artificial silk etc. By the time of his death in 1896 he had 355 patents. Another use
of 'nitro' is as a muscle relaxant. People with angina inhale a capsule of nitroglycerine and it releases NO into
the blood stream which dilates the blood vessels and increases the blood flow, thus reducing the strain on the
heart.)
O
N
From Bonds formed, H
H
Bonds Broken:
=[6 D(NN) + 24 D(C=O) + 20 D(H-O) +D(O=O)]÷4
= [6(946) + 24(724) + 20(465) + (502)] ÷4
= 32 854 kJ / 4 mol
= 8 213.5/mol
=5D(C-H) + 3D(C-O) + 6D(N-O) + 3D(N=O) + 2D(C-C)
=5(414) + 3(352) + 6(201) + 3(607) + 2(347)
=6 847 kJ/mol
Hrxn = (bond enthalpies of bonds broken) - (bond enthalpies of bonds formed)
= 6 847/mol -8 213 kJ/mol
=1366 kJ/mol
The main ingredient of C4 plastic is similar. It is a 6 membered ring with alternating C and N atoms in the ring.
Each N is bonded to a nitro group.
NO2
N
O2N
N
N
NO2
9.1
The combustion of propane is represented by the balanced chemical equation
C3H8 + 5O2  3CO2 + H2O. Calculate the heat of reaction for the reaction using bond energy
data.
9.2
If the heat of combustion of methanol, Hcomb {CH3OH(l)} = -715 kJ, calculate the standard heat
of formation of methanol using bond energy data.
9.3
Calculate the heat of formation ( Hf) of ammonia using bond energy data.
3H2 + N2  2NH3
18