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TOPIC 5: Energetics (SL): Why do reactions happen?
Back ground information
Energetics is the study of the energy changes that happen during chemical reactions.
Energy = the ability or capacity of a system to do work or transfer heat.
(work = transfer of energy that is the result of a force moving a body) (heat = an energy transfer that is the
result of a difference in energy between a system and its surroundings) .
Important terms:
In energetics, we divide the Universe into 2 parts.


system = all the chemicals that are involved (reactants and products) in a chemical reaction (=reaction
mixture).
surroundings = anything else in the Universe which is not the reacting mixture, this includes the
apparatus (eg test tube) and the medium in which the reaction goes on eg water.
Every chemical reaction involves a transfer of energy between the reacting system (reactants and
products) and the surroundings the system is in. The surroundings include the water the reactants are
dissolved in, the test tubes/beakers, you, the universe…
This is the case because during any reaction the internal energy of the reactants (or system at the
start) is either smaller or greater than the internal energy of the products (system at the end).
If after a chemical reaction the reacting system has either:


more internal energy (in that case it must have taken energy from the surroundings) or
it has less internal energy ( in that case it must have given energy to the surroundings) after
the change.
The internal energy of a system at any time is made up by:
Kinetic energy (or thermal energy of the particles)
Refers to the energy a particle has because of its motion; it is the total of the vibrational, rotational and
translational energy of all particles in the system: the atoms, molecules or ions as well as the electrons and
nuclear particles. The greater the freedom of movement of a particle, the more kinetic energy it has. So a
free atom or electron has a large amount of kinetic energy, Ek. An electron in an atomic orbital has less
kinetic energy than an electron in a molecular orbital as the one in a molecular orbital (eg when a covalent
bond is formed between two atoms) has a larger region it can move in i.e. it can move around both nuclei
so it has more freedom.
Potential energy:
 The potential energy of a particle is the stored energy. It is the energy it has by virtue of its position
relative to other particles. Some of the stored energy is stored in chemical bonds; other in the nucleus.
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 Potential energy exists as a result of attractive or repulsive force acting on the particles within a system.
For instance, an electron has potential energy as a result of its position in relation to a nearby positive
nucleus which attracts it. At the same time, the nucleus also has potential energy as a result of it being
repelled by another nucleus of an atom in the same molecule.
 Potential energy also exists as the gravitational energy of the system
 During a reaction, some of this potential energy (or chemical energy) which atoms and molecules have
can be released as heat or work or light or sound or electrical energy.
Measurement of internal energy
Internal energy cannot be easily measured. However, we can measure the change in amount of internal
energy of a system; we can measure the amount that is transferred (released or absorbed), E, between
the system and the surroundings.
Einitial
Efinal
= internal energy of the system before the reaction
= internal energy of the system after the reaction
E = the difference between the internal energy of the system at the end and that at the start of the
reaction:
E = Efinal
-
Einitial
Stability and energetics
The more potential energy a system has, the less energetically stable it is!! For instance, snow at
the top of a mountain has more potential energy and is therefore less stable and more likely to be
converted in more kinetic energy when it slides down than the snow in a valley. The snow in a
valley is more stable as it has less potential energy.
As a result, within our study, we will also be concerned with how the stability of one system compares with
the stability of another system or how the stability of a system changes during a chemical reaction.
An example to explain the above postulate:
C + ½O2
+ ½O2
CO
- 393 kJ
+ ½O2
- 283 kJ
CO2
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The above sample shows that carbon dioxide is more stable than carbon monoxide as it has a lower
potential energy content which is why the formation of carbon dioxide (which is what happens in sufficient
oxygen) is favoured as opposed to the formation of carbon monoxide.
Systems with a high potential energy are less stable and more likely to undergo changes by converting
potential energy into kinetic energy which can do work on the surroundings or heat up the surroundings.
Higher Level: a practical problem
Most of the internal energy exchanged during a reaction between a system and it surroundings is
exchanged as either heat or work done.
Work is either done:
 by the surroundings on the system (e.g. contraction when the surroundings push on the system)
 or by the system on the surroundings (e.g. expansion: when the system produces a gas which
pushes back the surrounding air to allow the gas to escape).
The change in total internal energy, E, within the system is given by the relationship:
E = q + w
q = amount of heat transferred to/from the surroundings
w = amount of work done on/by the surroundings
By convention:


q has a + value if heat is given to the system while it is - value when heat is given out;
w is + if work is done on the system by the surroundings and - if work is done by the system on the
surroundings.

The total amount of internal energy of a system changes:
 when heat is added or removed and/or
when work is done by it or on it ( or when it produces electrical energy or even light but these are
two less common forms of energy).
The problem: getting rid of work
Work, w, is difficult to measure whilst q is a lot easier as you will see later; so in order to measure E
accurately it is advisable to either keep w to a minimum or get rid of it altogether by ensuring that the
system cannot do any work i.e. conduct the reaction in a closed system like a bomb calorimeter.
When a reaction is carried out at constant pressure (which means in the open in a laboratory where the
atmospheric pressure usually stays the same during an experiment) and there is no change in volume (e.g.
reactions involving solids, liquids or solutions in a bomb calorimeter in which expansion is impossible) than
w = 0.
In that case the above expression becomes E = q; (q can be calculated by using q = temperature
difference x mass x specific heat capacity; as you will see later) .
However, usually w is very small and can be ignored.
First Law of Thermodynamics
When studying energy changes, we must always accept the First Law of Thermodynamics which states
that energy may be exchanged between a system and its surroundings but the total energy of the system
and its surroundings (= the Universe) stays the same.
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5. 1. EXOTHERMIC AND ENDOTHERMIC REACTIONS
5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction ( _H Ö ) .
5.1.2 State that combustion and neutralization are exothermic processes.
5.1.3 Apply the relationship between temperature change, enthalpy change and the classification of a reaction as
endothermic or exothermic.
5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of the enthalpy
change for the reaction.
Enthalpy and enthalpy change
The energy changes during chemical reactions we are interested in are those energy changes as a result
of a change in the potential energy of the system. The study of the changes in potential energy increase
our understanding of the bonding in chemicals.
If we want to measure the changes in potential energy only, we need to:

eliminate the kinetic energy changes. We do this by considering both reactants and products at the
same level of kinetic energy. As Ek = ½mv2 , the kinetic energy will be the same before and after the
reaction if both temperature and mass are the same. As any chemical reaction conserves mass, all we
need to do is ensure that temperature at the start of the reaction is the same as at the end and
measure the energy change between these two temperatures.

ensure that no energy is transferred as work; this means that there should be no volume change during
the reaction and no change in atmospheric pressure. In reactions where there is no volume change and
which are carried out at constant atmospheric pressure, potential energy is always released or
absorbed as heat only.
In the conditions described above, we call a heat transfer or heat change an enthalpy change, or H.
Enthalpy change is the change in the heat content (or enthalpy) of a system during a reaction carried out at
constant pressure (e.g. in the lab, no volume change caused by change in pressure)(or if a gas is produced
it should be carried out in a bomb calorimeter in which the outside pressure does not have an impact and
so no work is done by the system), with the reactants and products at the same temperature.
If there is to be a change in enthalpy, then there must be an enthalpy. Enthalpy (H) = energy or heat
content of a system, it is the total amount of kinetic and potential energy of the substances in it. The
absolute enthalpy content, H, just like internal energy cannot be measured as stated earlier.
Enthalpy change, H, can be also expressed mathematically:
H = (enthalpy of the system at the end of the process) - (enthalpy of system at the start)
or
H = Hproducts -
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Hreactants
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Using enthalpy change to distinguish between endothermic and exothermic reaction
Whether a reaction is endothermic or exothermic depends on the sign of H.
 if using the above expressions, H = negative: the reaction is exothermic; the heat content or enthalpy
of the system has decreased; the final state (products) is more stable as it has a lower energy content;
the lower the enthalpy content the more stable so energy (which was stored in reactants as potential
energy) must have been released so that the system went to a more stable level. The system has a
lower H value; the temperature of the surroundings (e.g. water) increases as it receives the released
energy as kinetic energy. Energy must be released in going to a more stable system. The products are
energetically more stable. Neutralization and combustion reactions are exothermic - a temperature
increase is observed.
 if H = positive: the reaction is endothermic; the products are less stable relative to the reactants;
kinetic energy will have been removed from the surroundings (and converted into potential energy in the
system) which why it has a lower temperature than the system. The products are energetically less
stable. A temperature decrease is observed.
When considering the sign you should always consider it from the system’s point of view. A negative sign
indicates the system’s energy has decreased.
Standard enthalpy change: a few rules!!!!

There is a need to measure, calculate or state H in standard conditions because:
 this allows us to compare meaningfully the difference in enthalpy between reactants and products
as a result of the difference in potential energy; this means that if products are at a higher
temperature than the reactants they need to be cooled down. By doing so they heat up the
surroundings which is the rise that we measure (occurs during combustion);
 this also allows us to compare H of different reactions, the H are measured and calculated for the
same conditions as H changes with pressure and temperature (although to a lesser extent with
temperature).
These standard conditions are:
 298 K
 1 atmosphere pressure (101 kPa)
 1 mol dm-3 for solutions

We also indicate the standard conditions by using the symbol ; so H  means that the enthalpy
change has been measured in the standard conditions above.

The best way of showing the H of a reaction is by writing a balanced chemical equation with the H
written alongside it. You have to make sure you include the state symbols as different states have
different enthalpy changes.

As the amount of enthalpy change is directly proportional to the quantities of materials used, enthalpies
changes are usually expressed per mole with reference to the substance that has reacted.
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Why does (and must) the potential energy of a system change during a chemical reaction?
In any chemical reaction particles are re-arranged and this can only be done if existing bonds are broken
and new ones are made.
The following idea is important:
breaking of chemical bonds needs energy/is an endothermic process (as attraction needs to be overcome)
making new bonds releases energy/is an exothermic process (as kinetic energy is lost)
A chemical reaction results in a change in enthalpy because during any chemical reaction old bonds are
broken and new bonds of different strengths are being made.
When we change the bonds within a system, we are changing the way atoms and electrons interact with
each other, we change the way in which they are attracted and repelled to each other. This is why each
bond has its own strength and energy content because it involves different forces of attraction and
repulsion. The energy content of each bond is part of the total heat content of the system so as a result we
must also be changing the energy content of the system.
Whether a reaction is endothermic or exothermic now depends on how the amounts of energy needed for
the breaking of bonds and the amount released for the making of new bonds compare:
Exothermic reaction :
Energy needed for breaking old bonds

energy released when new bonds are made.
More energy is released than needed. This excess energy goes to the surroundings. This happens
because the bonds within the reactant particles are weaker (at a higher energy level i.e at a less negative
potential energy level) than the bonds within the products. Less energy is needed to make free atoms or
ions than what is released making the new bonds (=new attractions between these atoms or ions);
Endothermic reaction:
Energy needed for breaking old bonds

energy released when new bonds are made.
The difference in energy which the system needs is absorbed from the surroundings. The bonds within the
reactant particles were stronger than those within the products.
State functions
Enthalpy or Enthalpy change are thermodynamic (= refers to transfer of heat during chemical reactions)
state functions. A state function is a quantity which has a fixed value at a given state or set of conditions.
Its value depends on the current state of the system and does not depend on the detailed previous history
of the system. The change in function does not depend on the path followed in the change.
Other state functions are temperature, volume, potential and kinetic energy.
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5. 2. CALCULATION OF ENTHALPY CHANGES
5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed.
5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions.
5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants
and mass of water.
5.2.4 Evaluate the results of experiments to determine enthalpy changes.
Amount of heat transferred can be calculated using the expression below:
amount of heat transferred (q) = specific heat capacity (c) x mass (m) x T (in Kelvin)
T (in Kelvin) in the above expression = the difference between the highest (or lowest) and room
temperature.
Experimental procedures: how can we obtain T (in Kelvin)?
1. Coffee-cup calorimeter: Used for reactions in aqueous solutions, for example copper sulphate and
ammonium chloride reactions (constant pressure); heat evolved/absorbed is used to raise/lower the
temperature of the calorimeter which is the cup, the aqueous solution, the lid, the thermometer. For our
calculations, we only measure the heat transferred to the aqueous solution.
2. Bomb calorimeter: used to measure the enthalpy change for reaction between a substance (e.g. foods
and fuels) and oxygen i.e. combustion reactions; within the “bomb’ the system does not do any work as
the volume remains constant!
Problem in bomb calorimeters is the heat capacity of the bomb calorimeter itself. To use the
temperature rise of the water only to calculate the enthalpy change is inaccurate as some of the
heat released is also used to heat up the calorimeter which would not be considered when
calculating the enthalpy change. To solve this problem, the experiment is repeated with an electric
heater connected to a joule meter. The heater is switched on until it achieves the same temperature
rise of the water. The number of joules needed to achieve this (and the equivalent increase in
temperature of the calorimeter) is read from the joule meter and used to calculate the enthalpy
change.
Working out of T from raw data
The graphical method used to work out the difference in temperature is shown below. This method allows
us to compensate for heat loss to some degree.
How? A plot of temperature against time is drawn and by extrapolating the graph the temperature rise that
would have taken place had the reaction been instantaneous can be calculated as shown below. (image
obtained from http://www.creative-chemistry.org.uk/alevel/module2/documents/N-ch2-11.pdf on 28 Feb
2010)
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point X
The above graph has a positive slope up to point X followed by a negative slope. The positive slope is
caused by the reaction in the system giving out energy at a faster rate than the rate at which the system is
losing it to the surroundings.
The negative slope shows the reaction giving out less energy than what the system is releasing to the
surroundings.
At point X, the energy given out by the reaction is the same as that being released to the surroundings.
T is found by extrapolating the curve to the time of mixing (=t m); by doing so we allow for heat loss and
assume the heat loss is constant.
Data processing:
It is important to note the following when reading about enthalpies and looking at their values:

The “standard enthalpy of reaction” refers to the heat change which occurs when a reaction happens
according to its stoichiometric chemical equation i.e. involving the number moles as stated in the
balanced equation, and is expressed in kJ. So this could be for 0.025 moles, 1 mole or for 10 moles of
reactant.

If the standard enthalpy change is expressed in kJ mol-1 than the enthalpy change is referred to as the
molar enthalpy of reaction as it indicates the amount of energy transferred per mole of either reactant
(the limiting reagent!!!!) or product. Some commonly used enthalpy changes are given special names
and these are usually expressed in kJ mol-1 as they are for 1 mole. So look out for this.
However, at all times the values should be for standard conditions if the term standard is used!!!
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5.3 HESS’ LAW (1840) : CALCULATION OF ENTHALPY CHANGE USING AN INDIRECT METHOD
5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy
changes.
For some reactions the enthalpy change of reaction can not be measured directly using any type of
calorimeter. This can be because maybe the compound cannot be directly synthesised from its elements
or the reaction is too difficult to carry out or to isolate e.g. carbon monoxide formation.
We can still find the enthalpy change but we need to calculate it using enthalpy changes from reactions for
which we have enthalpy values. In these calculations we rely on the First Law of Thermodynamics which is
what Hess did.
If the reaction is the sum of two or more reactions, then H for the overall reaction is the sum of the H’s of
the constituent reactions.
A consequence of the First Law of Thermodynamics and is also sometimes referred to as the law of
constant heat summation.
Another way of stating Hess’s Law is that the enthalpy change of chemical change of reactants A and B
into X and Y is the same regardless of the reaction pathway provided the initial and final states are the
same.
Hess’s Law allows us to calculate enthalpy changes which cannot be found experimentally by using
enthalpy changes from reactions which can be found experimentally.
Hess’ Law can be applied in 3 ways:
1. by the vertical addition of equations for which the following rules need to be obeyed:
 substances on either side of the overall equation or sum are cancelled; this is the aim of the
addition: all species that appear on both sides are eliminated so that they do not appear in the
overall equation which might require changing some of the equations in the following way;
 reversing an equation; but then you also need to reverse the enthalpy sign;
 halving or doubling the coefficients but then you also need to half or double the enthalpy value;
 dividing or multiplying your final equation; again do the same to the enthalpy value.
2. drawing an energy cycle; there are two main types of energy cycles that can be drawn.
Example 1: using enthalpies of formation, Hf (HIGHER LEVEL)
Hreaction
reactants
products
Hf (reactants)
Hf (products)
elements
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Example 2: using enthalpies of combustion, Hc
Hreaction
reactants
products
Hc(reactants)
Hc (products)
combustion products
When using these cycles it is important that:



the equation for which the enthalpy change is unknown goes on the top;
all equations should be balanced;
the chemical alongside the arrows going up and down are the same and have the same
coefficients;
3. Using enthalpy diagrams: from the energy cycles enthalpy diagrams can be drawn:
C + ½O2
-110 kJ
+ ½O2
CO
- 393 kJ
- 283 kJ
CO2
The last two methods give a clear relationship between the equations used. However, the drawing of
cycles becomes very difficult when more than 2 equations are involved. The vertical addition of equations
is the best method.
Exercises
1. From the information below, calculate the enthalpy change, H , for the reaction
S (s) + O2 (g)  SO2 (g)
S (s) + 3/2 O2 (g)  SO3 (g)
SO2 (g) + ½O2 (g)  SO3 (g)
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H = -395 kJ
H = -98.3 kJ
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2. From the information below, calculate the enthalpy change, H , for the reaction
Fe2 O3 (s) + 3CO (g)  2Fe (s) + 3CO2 (g)
Fe2 O3 (s) + CO (g)  2FeO (s) + CO2 (g)
Fe (s) + CO2 (g)  FeO (s) + CO (g)
H = - 2.9 kJ
H = + 11.3 kJ
3. Sulfur dioxide, SO2 , may be catalytically oxidised to sulfur trioxide, SO3, as shown in reaction I below.
I. SO2 (g) + ½O2 (g)  SO3 (g)
II S (s) + O2 (g)  SO2 (g)
III S (s) + 1½O2 (g)  SO3 (g)
H =
H = -298.2 kJ
H = -399.4 kJ
Calculate the enthalpy change for reaction I.
4. Given the data:
2 H2S (g) + 3 O2 (g)  2 SO2 + 2 H2O (g) (g)
S (s) + O2 (g)  SO2 (g)
H2 (g) + ½O2 (g)  H2O (l)
H = - 1124.6 kJ
H = - 298.2 kJ
H = - 266.9 kJ
Calculate H in kJ for the reaction: H2 (g) + S (s)  H2S (g)
5. 4. BOND ENTHALPIES
5.4.1 Define the term average bond enthalpy.
5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic.
Alternatively, we could also determine the H of a reaction if we knew how much energy is needed to
break the bonds in the reactant and how much energy is released making new bonds in the products i.e:
H = amount of energy needed to break the bonds – amount of energy released when making bonds
We can find these amounts of energy by using values called bond enthalpies.
Bond enthalpy is the average amount of energy needed to break 1 mole of covalent bonds
(=endothermic process) to give individual gaseous atoms from a gaseous molecule.
Bond enthalpy is always a positive value as it is an endo process.
The equations below represents the bond enthalpy for H – Cl bond and I – I bond, all species should be in
the gaseous state:

HCl (g)
 Cl (g) +
 I2 (g)  2I (g)
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H (g)
(I2 should be (g) and not (s)
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Applying the First Law of thermodynamics, bond enthalpy also refers to the amount of energy released
when 1 mole of the same bond is formed.
*It is important to appreciate that the bond enthalpy is an average value e.g. the bond enthalpy for C-H is
the average bond enthalpy for all C-H bonds in any compound; it is the average of all bond dissociation
enthalpies which refer to the specific bond energy of a bond in a specific compound.
The actual amount of energy needed to break a particular bond in a particular compound depends on the
surrounding atoms and differs in different compounds eg the amount of energy needed to break a C-H in
an alkane molecule differs from the amount needed to break a C-H bond in an alcohol molecule.
Bond enthalpies can only be used in reactions involving molecular substances; they are an indication of the
strength of the forces holding together atoms in a covalently bonded molecule.
Example 1: Using bond enthalpies and Hess Law to calculate an enthalpy change of combustion of
methane:
H1
CH4 (g) + 2O2 (g)
H2 breaking bonds
CO2 (g) + 2H2O (g)
H3 making bonds
C(g) + 4H (g) + 4O (g)
We can use the above energy cycle to work out enthalpy of combustion, H1, like this:
H2 = enthalpy change when bonds are broken = 4 x BE(C-H) + 2 x BE(O=O) = + 2640 kJ mol-1 .
H3 = enthalpy change when bonds are made = 2 x BE(C=O) + 4 x BE(H-O) = - 3338 kJ mol-1 .
Using Hess Law: H1 = H2 + H3
= + 2640 kJ mol-1
- 3338 kJ mol-1 = -698 kJ mol-1
Example 2: Using bond enthalpies to calculate same enthalpy change of combustion of methane.
H of reaction =  bond enthalpies of reactants -  bond enthalpies of products
H of reaction = (4 x BE(C-H) + 2 x BE(O=O)) - (2 x BE(C=O) + 4 x BE(H-O))
= [4 mol (413 kJ mol-1) + 2 mol (498 kJ mol-1)] -[2 mol (746 kJ mol-1) + 4 mol (464kJ mol-1 )]
= 2648 kJ - 3348 kJ
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= - 700 kJ
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Bond enthalpy values are stated for 298K as they are different at other temperatures. See table 10 in your
data booklet. The H is different from the Hcombustion as this calculation uses average value; Hcombustion is
the more accurate value.
Uses of bond enthalpy:
 comparing covalent bond strengths between different elements;
 understanding structure and bonding e.g. benzene (see later) and resonance structures (HL);
 estimating heat of reactions e.g. see example above involving methane.
Which equation represents the bond enthalpy for the H-F bond?
A.
B.
C.
D.
HF (g)  H (g) + F (g)
HF (g)  ½ H2 (g) + ½ F2 (g)
HF (aq)  H+ (aq) + F- (aq)
HF (g)  H+ (g) + F- (g)
IB questions
PAPER 1
1.
2.
What energy changes occur when chemical bonds are formed and broken?
A.
Energy is absorbed when bonds are formed and when they are broken.
B.
Energy is released when bonds are formed and when they are broken.
C.
Energy is absorbed when bonds are formed and released when they are broken.
D.
Energy is released when bonds are formed and absorbed when they are broken.
The temperature of a 2.0 g sample of aluminium increases from 25°C to 30°C. How many
joules of heat energy were added? (Specific heat of Al = 0.90 J g−1K−1)
A.
3.
0.36
B. 2.3
C.
9.0
D.
11
Using the equations below:
C(s) + O2(g) → CO2(g)
Mn(s) + O2(g) → MnO2(s)
∆H = −390 kJ
∆H = −520 kJ
what is ∆H (in kJ) for the following reaction?
MnO2(s) + C(s)  Mn(s) + CO2(g)
A.
910
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B.
130
C.
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−130
D.
−910
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4.
Which statements about exothermic reactions are correct?
I.
A.
5.
II.
The products have a lower enthalpy than the reactants.
III.
The products are more energetically stable than the reactants.
I and II only
B.
I and III only
I.
The mass of the sample
II.
The density of the sample
III.
The specific heat capacity of the sample
I and II only
B.
A.
8.
D. I, II and III
I and III only
C. II and III only
D. I, II and III
The average bond enthalpies for O—O and O==O are 146 and 496 kJ mol−1 respectively.
What is the enthalpy change, in kJ, for the reaction below?
—O—H(g) + ½O==O(g)
H—O—O—
7.
C. II and III only
A sample of a metal is heated. Which of the following are needed to calculate the heat
absorbed by the sample?
A.
6.
H values.
– 102
B.
+ 102
C.
+ 350
D.
+ 394
What is the energy change (in kJ) when the temperature of 20 g of water increases by 10°C?
A.
20 × 10 × 4.18
C.
20  10  4.18
1000
B.
20 × 283 × 4.18
D.
20  283  4.18
1000
When the solids Ba(OH)2 and NH4SCN are mixed, a solution is produced and the
temperature drops.
Ba(OH)2(s) + 2NH4SCN(s) → Ba(SCN)2(aq) + 2NH (g) + 2H2O(l)
Which statement about the energetics of this reaction is correct?
A.
H is negative.
B.
H is positive.
C.
H is negative.
D.
H is positive.
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9.
Using the equations below
Cu(s) +
1
2
2Cu(s) +
O2(g) → CuO(s) ∆Hο = −156 kJ
1
2
O2(g) → Cu2O(s) ∆Hο = −170 kJ
what is the value of ∆Hο (in kJ) for the following reaction?
2CuO(s) → Cu2O(s) +
A.
10.
142
B.
1
2
O2(g)
15
C.
–15
D.
–142
D.
+844
Consider the following equations.
Mg(s) +
H2(g) +
1
2
1
2
O2(g) → MgO(s)
∆Hο = −602 kJ
∆Hο = −242 kJ
O2(g) → H2O(g)
What is the ∆H° value (in kJ) for the following reaction?
MgO(s) + H2(g) → Mg(s) + H2O(g)
A.
11.
B.
−360
C.
+360
Separate solutions of HCl(aq) and H2SO4(aq) of the same concentration and same volume
were completely neutralized by NaOH(aq). X kJ and Y kJ of heat were evolved respectively.
Which statement is correct?
A.
12.
−844
X=Y
B. Y = 2X
C.
X = 2Y
D. Y = 3X
Which statements are correct for an endothermic reaction?
A.
I.
The system absorbs heat.
II.
The enthalpy change is positive.
III.
The bond enthalpy total for the reactants is greater than for the products.
I and II only
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B.
I and III only
8 hours
C. II and III only
D. I, II and III
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The mass m (in g) of a substance of specific heat capacity c (in J g−1 K−1 ) increases by t °C .
What is the heat change in J?
13.
A.
mct (
B.
mc(t + 273)
C.
mct
1000
D.
mc (t  273)
1000
The average bond enthalpy for the C–H bond is 412 kJ mol−1 . Which process has an
enthalpy change closest to this value?
14.
A.
CH4(g) → C(s) + 2H2(g)
B.
CH4(g) → C(g) + 2H2(g)
C.
CH4(g) → C(s) + 4H(g)
D.
CH4(g) → CH3(g) + H(g)
15. Which bond energy combination is most likely in an exothermic reaction?
A.
B.
C.
D.
Small bond energies in the reactants and large bond energies in the products
Large bond energies in the reactants and small bond energies in the products
Large bond energies in both reactants and products
Small bond energies in both reactants and products
16. Which one of the statement below is correct for the following equation?
C (s) + 2H2 (g) 
A.
B.
C.
D.
CH4 (g)
H = -74.8 kJ
The energy absorbed during the process is 74.8 kJ
The energy released when one mole of CH4 (g) is formed is 74.8 kJ.
The energy required to break the bonds in one mole of CH4 is 74.8 kJ
The energy required to break one mole of H-H bonds is 74.8 kJ.
17. Using the information below:
H2 (g) + O2 (g)  H2O2 (l)
2H2 (g) + O2 (g)  2H2O (g)
H = -187.6 kJ
H = -571.6 kJ
what is the value of H (in kJ) for the following reaction
2H2O2 (l)  2H2O (l) + O2 (g)
A. –196.4
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B. –384.0
C. –759.2
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D. –946.8
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18. The enthalpy changes for two different hydrogenation reactions of C2H2 are ;
C2H2 (g) + H2 (g)  C2H4 (g)
C2H2 (g) + 2H2 (g)  C2H6 (g)
H1
H2
Which expression represents the enthalpy change for the reaction below?
C2H4 (g) + H2 (g)  C2H6 (g)
A. H1 + H2
B. H1 - H2
H = ?
C. H2 - H1
D. -H1 - H2
19. Use the bond energies for H-H (436 kJ mol-1) . Br-Br (193 kJ mol-1) and H-Br (366 kJ mol-1) to
calculate H (in kJ mol-1) for the reaction:
H2(g) + Br2 (g)  2HBr (g)
A. 263
B. 103
C. –103
D. -263
20. For the reaction
2C (s) + 2H2 (g)  C2H4 (g)
H1 = + 52.3 kJ
If H2 = - 174.4 kJ for the reaction: C2H2 (g) + H2 (g)  C2H4 (g)
what can be said about the value of H3 for the reaction below?
2C (s) + H2 (g)  C2H2 (g)
A.
B.
C.
D.
H3 must be negative.
H3 must be a positive number smaller than 52.3 kJ.
H3 must be a positive number larger than 52.3 kJ.
No conclusion can be drawn about H3 without the value of H for H2 (g).
PAPER 2
1. (N04)
(a) Define average bond enthalpy, illustrating your answer with an equation for methane, CH4. [3]
(b) Using the following average bond enthalpies (kJ mol-1):
[3]
C – H 412; O=O 496; C=O 743 ; O – H 463
Calculate the enthalpy change for the following reaction:
CH4 (g) + 2 O2 (g)
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 CO2 (g) + 2H2O (g)
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2. (M01)
(a) Hess law states that, whether a reaction occurs in one or several steps, the total enthalpy change is
the same. Illustrate your understanding of this law using the data below to calculate the
enthalpy change (H) when one mole of solid carbon is converted into carbon monoxide.
[3]
C (s) + O2 (g)  CO2 (g)
CO (g) + ½ O2 (g)  CO2 (g)
H = -393.5 kJ
H = -283.0 5 kJ
(b) State what is meant by the term endothermic reaction
[1]
(c) Enthalpy changes may also be calculated by using bond enthalpies, some values of which
(kJ mol-1) are provided below:
C =C 612; C –H 412; O – H 463; C=O 743; O = O 496
The balanced equation for the complete combustion of one mole of ethane, C2H4, in oxygen is
shown below
C2H4 (g) + 3O2 (g)  2CO2 (g) + 2H2O (g)
(i) Use the equation and the bond enthalpy data above to calculate the enthalpy change for the
complete combustion of one mole of ethane.
[4]
(ii) State, with a brief explanation, whether the reaction is endothermic or exothermic.
[1]
3 (M00)
(a) (i) Explain what is meant by the standard enthalphy change of reaction.
[3]
(ii) Describe an experiment to determine the enthalpy change of the reaction between dilute
hydrochloric acid and aqueous sodium hydroxide. Show how the value of H would be
calculated from the data obtained.
[9]
(iii) Draw an enthalphy level diagram for the neutralization reaction above. Indicate on your
diagram the enthalpy change of the reaction and hence compare the relative stabilities of
reactants and products.
[4]
(b) Explain, giving one example, the usefulness of Hess’s Law in determining H values.
[4]
4. (M98) Enthalpy changes of reaction can be calculated from average bond enthalpies.
(a) For the combustion of propane, C3H8:
(i) Describe the changes in bonding that take place during the reaction.
[4]
(ii) Write a balanced chemical equation for this reaction and calculate the enthalpy
change using bond enthalpies in the data book.
[6]
(b) It is possible, using Hess’s Law, to determine the enthalpy change, H, for a reaction such as the
combustion of propane by combining reactions such as
C (s) + O2 (g)

CO2 (g)
H2 (g) + ½ O2 (g)  H2O(l)
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H = x kJ mol-1
H = y kJ mol-1
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3C (s) + 4H2 (g)

C3H8 (g)
H = z kJ mol-1
Show how these enthalpy changes may be arranged to give the enthalpy change for the
combustion of propane.
[3]
(c) (i) Draw a labeled relative energy diagram for the combustion of propane, showing
clearly the enthalpy of reaction, H, and including its algebraic sign.
[3]
(ii) State the conditions needed when calculating a standard enthalpy change for a
reaction.
[1]
5. (N03) Given the following data:
C(s) + 2F2(g) → CF4(g);
F2(g) → 2F(g);
C(s) → C(g);
∆H1 = −680 kJ mol−1
∆H2 = +158 kJ mol−1
∆H3 = +715 kJ mol−1
calculate the average bond enthalpy (in kJ mol−1) for the C––F bond.
[4]
6. In aqueous solution, potassium hydroxide and hydrochloric acid react as follows.
KOH(aq) + HCl(aq)→ KCl(aq)+ H2O(l)
The data below is from an experiment to determine the enthalpy change of this reaction.
50.0 cm3 of a 0.500 mol dm−3 solution of KOH was mixed rapidly in a glass beaker with
50.0 cm3 of a 0.500 mol dm−3 solution of HCl.
Initial temperature of each solution = 19.6°C
Final temperature of the mixture = 23.1°C
(a)
State, with a reason, whether the reaction is exothermic or endothermic.
[1]
(b)
Explain why the solutions were mixed rapidly.
[1]
(c)
Calculate the enthalpy change of this reaction in kJ mol−1. Assume that the specific heat
capacity of the solution is the same as that of water.
[4]
(d)
Identify the major source of error in the experimental procedure described above.
Explain how it could be minimized.
(e)
[2]
The experiment was repeated but with an HCl concentration of 0.510 mol dm−3 instead
of 0.500 mol dm−3. State and explain what the temperature change would be.
[2]
7. (N99) In order to determine the enthalpy change of reaction between zinc and copper (II) sulfate,
a student placed 50.0 cm3 of 0.200 mol dm-3 copper (II) sulphate in a polystyrene beaker. The
temperature was recorded every 30 seconds. After two minutes 1.20 g of powdered zinc was added.
The solution was stirred and the temperature recorded every half minute for the next 14 minutes. The
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results obtained were then plotted to give the following graph
(a) Write the equation for the reaction taken place.
[1]
(b) Determine which of the two reagents was present in excess.
[2]
(c) The highest temperature is reached at point A. Explain what is happening in the
system at this point.
[1]
(d) By drawing a suitable line on the graph estimate what rise in temperature would have
been if the reaction had taken place instantaneously.
[2]
(e) Calculate how much heat was evolved during the reaction. Give your answer to three
significant figures.
[2]
(f) What is the enthalpy change of reaction in kJ mol-1.
[1]
(g) The accepted value for the enthalpy change of reaction is – 218 kJ mol-1. What is the
percentage error for the value obtained in this experiment?
[1]
(h) Suggest one reason why there is disagreement between the experimental value and
the accepted value.
[1]
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Marks scheme
SECTION A
1
2
3
4
5
6
7
D
C
B
D
B
A
C
8
9
10
11
12
13
14
B
A
C
B
A
A
D
15
16
17
18
19
20
A
B
A
C
C
C
SECTION B
1. (N04)
2.
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3. (M00)
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4. (M98)
(a) (i) bonds are broken between C-H (8 bonds) and C-C (2 bonds) which requires energy
bonds are made between C=O (6 bonds) and O-H (8 bonds) which releases energy
(ii)
C3H8 (g) + 5O2 (g)
 3CO2 (g) + 4H2O (g)
(b) H = B.E. reactants – B.E. products
(c)
H = 3x + 4y - z
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5. (N03)
6. (M05)
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7
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