IB Chemistry HL1
1
GRADE 11: UNIT 6- ENERGETICS
IB TOPIC 5
2/2013
System and Surroundings
2
 The system is the name we give the sample or
reaction vessel of interest.
 The surroundings are everything else in the universe.
 When a chemical change happens in an open system
matter and energy can be exchanged between the
system and the surroundings.
 In a closed system only energy can be exchanged
with the surroundings.
2/2013
System and Surroundings
3
2/2013
Energy
4
 Energy is defined as the ability to do work. Energy is
often converted from one form to another during
physical and chemical changes.
 Thermochemistry is the study of energy changes
associated with chemical reactions.
 Chemical energy is the energy stored in chemical
bonds. It is a type of potential energy.
2/2013
Thermochemistry
5
 Most reactions absorb or evolve energy usually in the
form of heat but chemical reactions can also produce
light, electricity and mechanical energy – used to do
work.
 Energy is measured in joules, J.

1000 J = 1 kJ.
 Physical changes like change of state/phase also have
heat energy changes.
2/2013
Exothermic and Endothermic Reactions
6
 5.1.1 Define the terms exothermic reaction,
endothermic reaction and standard enthalpy
change of reaction (ΔH°).
 Standard enthalpy change is the heat energy
transferred under standard conditions—pressure
101.3 kPa, temperature 298 K. Only ΔH can be
measured, not H for the initial or final state of a
system.
2/2013
Exothermic and Endothermic Reactions
7
 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 reactant and products, and the
sign of the enthalpy change for the reaction.
2/2013
Enthalpy
8
 Enthalpy is the total energy of a system, some of
which is stored as chemical potential energy in the
chemical bonds.
 Enthalpy is given the symbol H.
 Enthalpy is also known as the heat content of a
system.
 We cannot measure the enthalpy content of a system
but we can measure changes in it.
2/2013
Enthalpy Change
9
 In chemical reactions, bonds are broken and made, but the
energy absorbed breaking bonds is almost never exactly
equal to that released in making new bonds.
 All reactions are accompanied by a change in the potential
energy of the bonds and hence an ENTHALPY CHANGE.
 There is no “absolute zero” for enthalpy so absolute
enthalpies cannot be measured only the change in enthalpy
that occurs during a reaction.
2/2013
Enthalpy Change
10
 The enthalpy change of a reaction is given the
symbol ∆H.
 ∆H is the difference in the enthalpy between the
products and the reactants.
 ∆H = H(products) - H(reactants) when at constant
pressure.
 Enthalpy level diagrams are used to show the change
in enthalpy of a system during a change.
2/2013
Enthalpy Level Diagram
11
2/2013
Enthalpy Level Diagram
12
2/2013
Endothermic Reaction
13
 An endothermic reaction is one where energy is transferred
from the surroundings to the system.
 If energy is absorbed during a reaction then the enthalpy of
the products will be higher than that of the reactants.
 This means the enthalpy change will have a positive sign.
 This reaction will either get cooler or heat will need to be
supplied - temperature decreases.
2/2013
Exothermic Reaction
14
 An exothermic reaction is one where energy is transferred
from the system to the surroundings.
 If energy is released during a reaction then the enthalpy of
the products will be lower than that of the reactants.
 This means the enthalpy change will have a negative sign.
 This reaction will “feel” hotter - temperature increases.
2/2013
Stability
15
 Reactions in chemistry tend towards products that
are more stable.
 Stability increases as energy decreases so exothermic
reactions increase the stability of the substances.
 This is why most chemical reactions that occur in
nature are exothermic.
 Endothermic reactions usually need “help” in the
form of energy to allow them to occur.
2/2013
Stability
16
 Remember that lower energy is more stable. We can
compare this to standing on top of a high building
where you have more potential energy than someone
on the ground.
 If you fall to the ground you lose some of that
potential energy you had on the roof but you are now
more stable.
2/2013
Exothermic and Endothermic Reactions
17
 5.1.2 State that combustion and neutralization are
exothermic processes.
2/2013
Combustion
18
 Combustion is the scientific word for burning.
 Most hydrocarbons burn easily in excess oxygen.
 When they burn they produce carbon dioxide and
water.
 Ex. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
 Combustion produces lots of heat which is
transferred to the surroundings so it is very
EXOthermic.
2/2013
Combustion
19
 When 1 mole of a substance is burned the energy
released is called the Enthalpy Change of
Combustion, ΔHc°.
 The ° sign indicates this was measured under
standard conditions in a controlled environment.
 Standard conditions for thermochemistry
experiments are T = 298K and P = 101.3kPa (= 1
atm).
2/2013
Neutralization
20
 Neutralization reactions involved acids and bases.
 If an acid and a base react completely the resulting solution




will be pH neutral.
The products are a salt (ionic compound) + water.
Ex. NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
When weak acids and bases are involved such as NaHCO3
then CO2(g) may be another product.
These reactions release energy to the surroundings so they
are EXOthermic.
2/2013
Negative ΔH
21
 As combustion and neutralization reactions are
always exothermic the enthalpy changes will always
have negative values.
 You can find many ΔH values in scientific literature
ex. your textbook, the books in the classroom and
online.
2/2013
Spontaneous Reactions
22
 A spontaneous reaction is one that occurs when the
reactants are mixed without the need to be heated or
have some other outside influence.
 Most spontaneous reactions are EXOthermic but
there are some spontaneous endothermic reactions
ex. Dissolving NH4Cl in water.
2/2013
Summary of Enthalpy Changes
23
2/2013
Enthalpy Changes
24
 In an exothermic reaction the products are more
stable than the reactants so the bonds made are
stronger than the bonds broken.
 In an endothermic reaction the products are less
stable than the reactants so the bonds made are
weaker than the bonds broken.
2/2013
Enthalpy Changes
25
 Enthalpy changes are usually written alongside
the chemical equation for the process with a
positive or negative sign.
 State symbols are VERY IMPORTANT as
changes of state have their own enthalpy
change values.
 Enthalpy changes are usually reported per mole
so the units are kJ mol-1.
 If this is not the case then just kJ is used.
2/2013
Enthalpy Changes
26
 For example:
 2NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)
 H = +91.6 kJ mol-1
 The + sign indicates it’s an endothermic reaction.
2/2013
Standard Conditions
27
 To compare enthalpy changes conditions must be the same.
 The thermochemical standard conditions are
 Temperature = 25°C = 298K (this is room temp)
 Pressure = 1 atm = 101.3kPa
 Solutions have a concentration of 1 mol dm-3
 Standard conditions are sometimes indicated by the symbol  or °:
H orΔH°
 Sometimes the temperature is included too: H 298.
 The values in bold are the SI units.
2/2013
Learning Check: Do NOW
28
1.
2.
3.
4.
5.
6.
2/2013
Write the equation for the formation of chlorine
oxide, Cl2O from its elements
What bonds are broken and what bonds are made
in this process?
Do the processes in 2. absorb or release energy?
What is an enthalpy change?
In this reaction the bonds made are less strong
than those broken. Will the enthalpy change be
positive or negative?
Will this be an exothermic or endothermic
reaction?
Calculation of Enthalpy
Changes
29
 5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
 Students should be able to calculate the heat energy change
for a substance given the mass, specific heat capacity and
temperature change using q = mcΔT.
 5.2.2 Design suitable experimental procedures for
measuring the heat energy changes of reactions. Students
should consider reactions in aqueous solution and
combustion reactions.
2/2013
Calculation of enthalpy changes
30
 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. Students should be
aware of the assumptions made and errors due to
heat loss.
2/2013
Temperature
31
 Temperature is a measure of the average kinetic





energy of the particles in a system.
Units are K or °C.
Note: there is no ° sign used with K!
Heat is a measure of the total energy in a substance.
T°C + 273 = TK
TK – 273 = T°C
2/2013
Specific Heat Capcity
32
 Some substances will conduct heat and therefore
change temperature more easily than others.
 Ex. A metal pan on a stove will become very hot
before the water inside it does.
 A measure of how easily something changes
temperature is called SPECIFIC HEAT CAPACITY.
2/2013
Specific Heat Capacity
33
 Specific heat capacity is the amount of heat energy
required to increase 1 g of a substance by 1K or 1°C.
 It is used in the following equation:
 q = m x c x T where q is heat energy, m is mass, c is
specific heat capacity and T is change in
temperature.
2/2013
Specific Heat Capacity
34
 Units of specific heat capacity (c) are J g-1K-1
 Units of heat energy (q) are J or kJ
 Units of mass (m) are g or kg
 Units of temperature (T) are K or °C
2/2013
Calorimetry
35
 Calorimetry is a method used to measure the
enthalpy associated with a particular change.
 The temperature change of a liquid is measured
inside a well insulated container called a calorimeter.
 Often a styrofoam cup is used as it has a very low
heat capacity and is a good insulator.
2/2013
Measuring Energy Changes
36
 To measure the enthalpy
change of a reaction that
occurs in solution you can
carry it out in a styrofoam
(polystyrene) cup and
monitor the temperature
during the reaction.
 Styrofoam is a good
insulator so the amount of
heat lost to the
surroundings will be
reduced.
2/2013
Measuring Energy Changes
37
 Burning substances in a
“bomb calorimeter”
measures the
temperature change in
the water surrounding
the burning item.
 This system is also well
insulated to try and
reduce heat loss.
 There may be some
losses from incomplete
combustion.
2/2013
Calorimetry
38
 If calorimeters made of other materials are used then the
heat absorbed by the calorimeter must be added to that
absorbed by the liquid:
 Heat absorbed = (mcT)liquid + (mcT)calorimeter
 Calorimetry assumes no heat is transferred to or from the
surroundings so they must be well insulated.
 However this is hard to achieve and is a major source of
error in high school labs.
2/2013
Calorimetry
39
 Once you have calculated the energy released from a
process then you can calculate how many kJ of
energy were released per mole of the reactant. This is
usually referred to as ΔH or molar enthalpy in a test
question.
 ΔH = q/n
Where n represents the number of moles of the
reactant that is reacting/burning.
2/2013
Sample Problems
40
Specific heat, c of liquid water = 4.18 kJ dm-3K-1
= 4.18J g-1K-1
1. How much heat energy is required to increase the
temperature of 10 g of nickel (c = 440 J g-1K-1)
from 50°C to 70°C?
1.
2/2013
The enthalpy of combustion of ethanol (C2H5OH)
is 1370 kJ mol-1. How much heat is released when
0.200 mol undergo complete combustion?
Sample Problems
41
3. H2(g) + 1/2 O2(g)  H2O(l)
∆H for the reaction above is -286 kJ mol-1. What
mass of oxygen must be consumed to produce 1144
kJ of energy?
4. Calculate the molar enthalpy change when excess
zinc is added to 50 cm3 of a 1 mol dm-3 solution of
CuSO4. The temperature increases from 20°C to
70°C when the zinc is added. Assume the solution
has the same density as water = 1.00 g cm-3
2/2013
Bond Enthalpies
42
 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.
2/2013
Bond Enthalpies
43
 All chemical reactions involve the making and
breaking of bonds.
 The bonds in the reactants are broken which absorbs
energy so this is an endothermic process.
 The bonds in the products form which releases
energy so this is an exothermic process.
2/2013
Bond Enthalpy
44
 Bond enthalpy is defined as the energy needed to
break one mole of bonds in gaseous molecules
under standard conditions.

Ex. ½ H2(g) + ½ Cl2(g)  H(g) + Cl(g)
 Breaking a bond is endothermic so these values are
always positive.
 Bond enthalpies depend on the rest of the molecule
so values are usually averages.
2/2013
Bond Enthalpy
45
 The energy released when a bond is made is the same
value as the bond enthalpy but with a negative sign.
 A higher bond enthalpy indicates a stronger bond.
2/2013
Bond Enthalpy
46
 If the bonds being broken (bonds in the reactants)
are weaker than those being made (bonds in
products) then the reaction will be exothermic.
 If the bonds being broken are stronger than those
being made then the reaction will be endothermic.
2/2013
Bond Enthalpies
47
 As bond enthalpies are averages and they are only for
gases they are not the most accurate way to calculate
an enthalpy change but they are usually within about
10% of more accurate values and are a useful tool.
 ∆Hºreaction= ∑BEbonds broken- ∑BEbonds made
Where BE stands for bond enthalpy
2/2013
Bond Enthalpy
48
 Bond enthalpy values are given in the data booklet. A
sample is shown here. When using these values be
careful to check if bonds are single or double etc.
2/2013
Bond
E / kJ mol-1
H-H
436
C-H
413
C-C
347
C=C
612
O=O
498
Bond Enthalpy Calculations
49
 When water is formed from its elements, what bonds




are broken and formed? What is the enthalpy change
predicted by bond enthalpies?
2H2(g) + O2(g)  2H2O(l)
Bonds broken: 2(H-H), 1 (O=O)
Bonds formed: 4 (O-H)
ΔH = Σbonds broken – Σbonds formed
2/2013
Bond Enthalpy Calculations
50
 = (2(436) + 498 ) – (4(464))
 = (872 + 498) – (1856)
 =1370 – 1856 = -486 kJ
 What does this sign tell you about the reaction?
 Does this make sense?
2/2013
Practice Problem
51
 Find the enthalpy change when CO2(g) is formed
from its elements using bond enthalpy values in the
data booklet.
2/2013
Hess’s Law
52
 5.3.1 Determine the enthalpy change of a reaction
that is the sum of two or three reactions with known
enthalpy changes.
 Students should be able to use simple enthalpy cycles
and enthalpy level diagrams and to manipulate
equations. Students will not be required to state
Hess’s Law.
2/2013
Hess’s Law
53
 Hess’s Law is a special case of the law of
conservation of energy.
 It states that the enthalpy change for a reaction will
have the same value no matter how many steps were
taken to go from reactants to products.
 Another way we can use it is to say the enthalpy
change for the reaction is equal to the enthalpy of the
products - enthalpy of reactants.
2/2013
Hess’s Law
54
 Hess’s Law enables us to calculate an enthalpy
change for a reaction without carrying out the actual
reaction.
 We do this by measuring the enthalpy change for
other related reactions.
 Hess’s Law is very useful for reactions that are
difficult to carry out in a lab or that do not occur.
2/2013
Hess’s Law
55
 For example sodium hydrogen carbonate can be
reacted with hydrochloric acid as follows
NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l) ∆H1
 To give the same products we could carry out 2
other reactions:
2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(l)
∆H2
Na2CO3(s) + 2HCl (aq)  2NaCl(aq) + CO2(g) + H2O(l) ∆H3
2/2013
Hess’s Law
56
 The sodium hydrogencarbonate could be heated to
form sodium carbonate which is then reacted with
hydrochloric acid.
2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(l) ∆H2
Na2CO3(s) + 2HCl (aq)  2NaCl(aq) + CO2(g) + H2O(l) ∆H3
2/2013
Hess’s Law
57
 If these two equations are added together the Na2CO3
cancel out and the result is twice the overall equation.
 2NaHCO3(s) + 2HCl(aq) --> 2NaCl(aq) + 2CO2(g) + 2H2O(l)
 Hess’s Law says that the enthalpy change for the two stage
reaction must be equal to the single stage process:
 2∆H1 = ∆H2 + ∆H3
The number 2 is present in front of ΔH1 as the equations must
all be balanced.
2/2013
Hess’s Law
58
 This can also be shown using an enthalpy cycle:
2/2013
Hess’s Law
59
 To read an energy cycle you must identify the
reactants and products that you are being asked
about and make sure all 3 sides are balanced.
 Then you must identify the alternative route you will
take to get from reactants to products.
2/2013
Hess’s Law
60
 If you are following an arrow then you add all those
ΔH values. You then subtract any ΔH values when
you go in the opposite direction of the arrow.
2/2013
Hess’s Law
61
 You can also use a series of equations and compare
them to find the unknown enthalpy change.
 For example:
S(s) + 1½O2(g)  SO3(g) ΔH°=-395 kJ ΔH1
SO2(g) + ½O2(g)  SO3(g) ΔH°=-98 kJ ΔH2
 Calculate the standard enthalpy change, ΔH° for
the reaction:
 S(s) + O2  SO2(g)
2/2013
Hess’s Law
62
 ΔH1 has the same reactants as the reaction in
question.
 ΔH2 relates SO3(g) to SO2(g) but the SO2 is on the
opposite side of the equation from the reaction in
question. This means we reverse the chemical change
and the sign of ΔH2.
 The equations can now be combined and the values
added to find the ΔH°:
ΔH°= -395 + 98 = -297 kJ
2/2013
Hess’s Law Activity Worksheet
63
 1. Calculate ΔH for the reaction
C2H4 (g) + H2 (g)  C2H6 (g),
from the following data.
 C2H4 (g) + 3 O2 (g)  2 CO2 (g) + 2 H2O (l)
ΔH = -1411. kJ/mole
 C2H6 (g) + 7/2 O2 (g) 2 CO2 (g) + 3 H2O (l) ΔH = -1560. kJ/mole
 H2 (g) + 1/2 O2 (g) H2O (l)
ΔH = -285.8 kJ/mole
2/2013
Solution to #1 problem
64
 Reactions that were reversed or multiplied by a
constant are shown in italics.
 1. ΔH = -137. kJ
 C2H4 (g) + 3 O2 (g)  2 CO2 (g) + 2 H2O (l)
 2 CO2 (g) + 3 H2O (l)  C2H6 (g) + 7/2 O2 (g)
 H2 (g) + 1/2 O2 (g)  H2O (l)
2/2013
ΔH = -1411. kJ
ΔH = +1560. kJ
ΔH = -285.8 kJ
Hess’s Law Worksheet
65
 2. Calculate ΔH for the reaction
 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g),
 from the following data.
 N2 (g) + O2 (g) 2 NO (g)
 N2 (g) + 3 H2 (g) 2 NH3 (g)
 2 H2 (g) + O2 (g)  2 H2O (g)
 Answer:
 ΔH = -1628. kJ
2/2013
ΔH = -180.5 kJ
ΔH = -91.8 kJ
ΔH = -483.6 kJ