IB Core Objective 5.1.1Define the terms exothermic reaction,
endothermic reaction and standard enthalpy change of reaction ( H o
) Define: Give the precise meaning of a word, phrase or physical
quantity.
Slide 3
5.1.1Define the terms exothermic reaction, endothermic reaction
and standard enthalpy change of reaction ( H o ) Thermochemistry =
study of energy changes in chemical reactions Most chemical
reactions absorb or evolve energy (usually as heat, sometimes as
light and mechanical energy) Energy is measured in Joules (J)
Slide 4
5.1.1Define the terms exothermic reaction, endothermic reaction
and standard enthalpy change of reaction ( H o ) Enthalpy ( H, heat
content) = the total energy of a system Some is stored as chemical
potential energy in the chemical bonds Energy is absorbed to BREAK
bonds Energy is released when bonds are MADE The potential energy
of the bonds changes in chemical reactions this is ENTHALPY
CHANGE
Slide 5
This quantity, H, is called the enthalpy of reaction, or the
heat of reaction. 5.1.1Define the terms exothermic reaction,
endothermic reaction and standard enthalpy change of reaction ( H o
)
Slide 6
1st Law of Thermodynamics: Energy can not be created nor
destroyed, but it can change form Energy lost = Energy gained
System + Surroundings = Universe ( Which is constant ) Reactions
Endothermic: Energy is used by the system Exothermic: Energy is
produced by the system The System represents the chemical
reaction
Slide 7
The system includes the molecules we want to study (here, the
hydrogen and oxygen molecules). The surroundings are everything
else (here, the cylinder and piston). 5.1.1Define the terms
exothermic reaction, endothermic reaction and standard enthalpy
change of reaction ( H o )
Slide 8
The change in enthalpy, H, is the enthalpy of the products
minus the enthalpy of the reactants: H = H products H reactants
5.1.1Define the terms exothermic reaction, endothermic reaction and
standard enthalpy change of reaction ( H o )
Slide 9
The change in enthalpy H = the net energy after bonds are
broken and remade H = ( Negative ) More energy was created than
used. EXOTHERMIC H = ( Positive ) More energy was used than
created. ENDOTHERMIC
Slide 10
IB Core Objective 5.1.2 State that combustion and
neutralization are exothermic processes. Combustion: CH 4(g) + 2O
2(g) CO 2(g) + 2H 2 O (g) H = - 882 kJ mol -1 Neutralization: HCl
(aq) + NaOH (aq) NaCl (aq) + H 2 O (l) H= - 57.3 kJ mol -1
Slide 11
IB Core Objective 5.1.3 Apply the relationship between
temperature change, enthalpy change and the classification of a
reaction a endothermic or exothermic. What happens to the
temperature when the reaction is exothermic? A: Temperature goes
up. What happens to the enthalpy change? A: It is a negative value.
If the same moles of substance were reacted in a larger container
of water, would the temperature and enthalpy values stay the same?
A: Enthalpy value would stay the same, the temperature however
would not go up as much.
Slide 12
IB Core Objective 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.
Slide 13
Exothermic Endothermic Products are more stable Heat energy is
RELEASED into the SURROUNDINGS Reactants are more stable Heat
energy is ABSORBED into the SYSTEM H is Negative H is Positive
Enthalpy
Slide 14
Standard State Enthalpy is affected by multiple factors
(concentration, pressure, state of reactants, temperature)) H Under
SATP (Standard Ambient Temperature and Pressure) P = 101.3kPa T =
25 o C or 298 K C = 1mol/dm 3 (Concentration for aqueous solutions
aq ) Standard state (physical state of element under these
conditions) (Theta) Means under SATP Different from STP
(Temperature is not 0 o C)
Slide 15
IB Core Objective 5.2.1 Calculate the heat energy change when
the temperature of a pure substance is changed.
Slide 16
Specific Heat Capacity The energy required to heat 1g of
substance 1 degree Celsius Specific Heat Capacity of water 1 g of H
2 O to heat 1 degree C requires 4.18 J Therefore the specific heat
capacity of water is 4180 J kg -1 K -1 4.18 Jg -1 K -1 Typically we
use this for labs due to the small quantities used.
Slide 17
5.2.1 Calculate the heat energy change when the temperature of
a pure substance is changed. Calorimetry To measure the energy of a
system is hard To measure the energy of surroundings is easy q =
mcT q = Energy change in the surroundings m = mass of surroundings
(g) c = Specific Heat Capacity of surrounding substance (Sometimes
s is used instead of c) T = Temperature change Energy must have
come from the system For us, this will be the mass of water
used
Slide 18
5.2.1 Calculate the heat energy change when the temperature of
a pure substance is changed. Questions How much heat is needed to
warm 250 g of water from 22 C to near its boiling point 98 C? A:
7.9 x 10 4 J or 79kJ. Calculate the quantity of heat used when you
mix 50cm 3 of 1.0M HCl and 50cm 3 of NaOH in a coffee-cup
calorimeter, and the temperature increases from 21.0 C to 27.5 C.
Assume the density is 1.00 g/cm 3. A: 2.7 kJ Calculate the enthalpy
change for the above reaction in kJ/mol. A: -54 kJ/mol
Slide 19
Questions Excess solid Magnesium is added to a 100g of a 2M
solution of Copper(II) Sulphate. The temperature increased from
20.0 o C to 65.0 o C. (Since the solution is largely water we will
assume specific heat capacity as 4.18 Jg -1 K -1. What is the
enthalpy change for the reaction? Dont forget it must be calculated
as a ratio of a full mol. Push for answer H = -94.1 kJ/mol.
Slide 20
IB Core Objective 5.2.2 Design suitable experimental procedures
for measuring the heat energy changes of reactions. 5.2.4 Evaluate
the results of experiments to determine enthalpy changes.
Slide 21
5.2.2 Design suitable experimental procedures for measuring the
heat energy changes of reactions. 5.2.4 Evaluate the results of
experiments to determine enthalpy changes. For liquids, a
calorimeter should be well insulated and the heat capacity should
be low. Calorimetry: Technique used to measure the enthalpy
associated with a particular change. Calorimetry depends on the
assumption that no heat is gained from or lost to the surroundings.
Even with well insulated calorimeters, heat exchange with
surroundings is a major source of error.
Slide 22
5.2.2 Design suitable experimental procedures for measuring the
heat energy changes of reactions. 5.2.4 Evaluate the results of
experiments to determine enthalpy changes. In combustion
experiments, where burning gas is used to heat liquid in a
calorimeter, the error is quite large. Temperature rises are much
less than expected and thus H values are less than literature
values.
Slide 23
IB Core Objective 5.2.3 Calculate the enthalpy change for a
reaction using experimental data on temperature changes, quantities
of reactants and mass of water.
Slide 24
5.2.2 Design suitable experimental procedures for measuring the
heat energy changes of reactions. Question: 20.0 cm3 of 2 mol dm-3
aqueous sodium hydroxide is added to 30.0 cm3 of hydrochloric acid
of the same concentration, the temperature increases by 12.0 C. For
dilute aqueous solutions, we can assume the density is 1.00 g cm-3.
What is H? A: -62.7 kJ mol -1
Slide 25
IB Core Objective 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 Hesss law.
Slide 26
5.3.1 Determine the enthalpy change of a reaction that is the
sum of two or three reactions with known enthalpy changes. Hesss
Law Heat of the whole = the sum of the parts Reaction that take a
direct route or multiple step route make no difference with
Enthalpy. ANALOGY I just bought a $1200 TV, I could either pay 12
equal instalments of $100, pay 2 instalments of $600 or pay it all
$1200 at once. How I do it doesnt matter, in the end I still pay
$1200. In enthalpy terms: H 1 = H 2 + H 3
Slide 27
5.3.1 Determine the enthalpy change of a reaction that is the
sum of two or three reactions with known enthalpy changes.
Calculate H f for C 2 H 2(g) C 2 H 2(g) 2O 2(g) 2CO 2(g) + H 2 O
(l) 2C (s) + H 2(g) H f = H c = H f = HH H c = 2O 2(g) CO 2 -395 kJ
mol -1 H2O -287 kJ mol -1 C 2 H 2 -1301 kJ mol -1 A: 224 kJ mol
-1
Slide 28
IB Core Objective 5.4.1 Define the term average bond enthalpy.
Enthalpies are a measure of the strength of a covalent bond: the
stronger the bond, the more tightly the atoms are joined together.
Breaking of a chemical bond requires energy, and is thus and
endothermic process. Average bond enthalpy: The bond enthalpy for a
compound will be affected by surrounding bonds, therefore we use
the average bond enthalpy.
Slide 29
IB Core Objective 5.4.2 Explain, in terms of average bond
enthalpies, why some reactions are exothermic and others are
endothermic.
Slide 30
Energy 5.4.2 Explain, in terms of average bond enthalpies, why
some reactions are exothermic and others are endothermic. It takes
energy to break bonds Energy is produced upon bond formation H =
Energy IN (Bonds broken) Energy OUT(Bonds formed) H = 1(H-H 436) +
(O=O 496) 2(O-H 463) H = 684 926 H = -242 (Energy is left over,
Exothermic) OO H H 496 kJ/mol 436 kJ/mol 463 kJ/mol
Slide 31
5.4.2 Explain, in terms of average bond enthalpies, why some
reactions are exothermic and others are endothermic. Sometimes Bond
energy is effected by surrounding bonds so an average must be used.
Calculate the Enthalpy of change for the following: CH 4(g) + O
2(g) CO 2(g) + H 2 O (g) Compare to enthalpy of combustion value
(Data book) Bond Bond Enthalpy kJ/mol H-H (g) 436 Cl-Cl (g) 242 F-F
(g) 158 H-Cl (g) 431 H-F (g) 562 Bond Avg. Bond Enthalpy kJ/mol C-C
(g) 348 C=C (g) 612 C=C (g) (in benzene) 518 C-H (g) 412 C=O (g)
743 O-H (g) 463 N-H (g) 388 O=O (g) 496 H = {Bonds broken Bonds
formed} Why the difference?
Slide 32
5.4.2 Explain, in terms of average bond enthalpies, why some
reactions are exothermic and others are endothermic. Given that the
enthalpy change for the reaction N 2 (g) + 3Cl 2 (g) 2NCl 3 (g) is
+688kJ/mol Calculate the enthalpy of the N-Cl bond, given that the
bond enthalpies in the nitrogen molecule and the chlorine molecule
are 944kJ/mol and 242kJ/mol. A: 164 kJ/mol
Slide 33
IB HL Objective 15.1.1 Define and apply the terms standard
state, standard enthalpy change of formation (H f ) and standard
enthalpy change of combustion (H c ) You have already learned about
standard state. What are the conditions? A: Form normally found at
298K, 101.3 kPa, 1 mol dm -3
Slide 34
15.1.1 Define and apply the terms standard state, standard
enthalpy change of formation (H f ) and standard enthalpy change of
combustion (H c ) H c = Standard enthalpy change of combustion
Enthalpy change when one mole of compound is burned in excess
oxygen under standard conditions. (Always exothermic). H f =
Standard enthalpy change of formation Amount of energy released or
absorbed in the formation of one mol of compound from elements in
their normal states. By using the definition of standard state, the
enthalpy of formation of an element in the standard state is
zero.
Slide 35
IB HL Objective 15.1.2 Determine the enthalpy change of a
reaction using standard enthalpy changes of formation and
combustion. H = H f (products) - H f (reactants)
Slide 36
15.1.2 Determine the enthalpy change of a reaction using
standard enthalpy changes of formation and combustion. For the
following reaction, find the enthalpy of formation: 2 C(graphite) +
3H 2 (g) + O 2 (g) C 2 H 5 OH(l) A: H f = -277 kJ mol -1 (All the
reactants are in their elemental, standard state, so they would be
zero!) Find the enthalpy of formation for the following
reaction
Slide 37
15.1.2 Determine the enthalpy change of a reaction using
standard enthalpy changes of formation and combustion. Find the
enthalpy of formation for the following reaction : NH 4 NO 3 (s) N
2 O(g) +2H 2 O(l) The enthalpies for formation for the above
compounds are: NH 4 NO 3 (s) :-366 kJ mol -1, N 2 O(g) +82 kJ mol
-1, and H 2 O(l) -285 kJ mol -1 A: -122 kJ mol -1 When we have
learned more about Hesss Law, we will return to this later
Slide 38
15.1.2 Determine the enthalpy change of a reaction using
standard enthalpy changes of formation and combustion. Find the
formation of combustion of benzene using enthalpies of formation
values. First, balance the equation: C 6 H 6 (l) + O 2 (g) CO 2 (g)
+ H 2 O(g) A: C 6 H 6 (l) + O 2 (g) 6CO 2 (g) + 3H 2 O(l) The
enthalpies of formation are: CO 2 (g): -393.5 kJ mol- 1, H 2 O(g):
-285.8 kJ, benzene (find in your data booklet), O 2 ??? A: -3267 kJ
mol -1 (Compare this answer with the data booklet).
Slide 39
IB HL Objective 15.2.1 Define and apply the terms lattice
enthalpy and electron affinity. Lattice enthalpy: Enthalpy change
to convert one mole of a solid ionic compound into gaseous ions or
vice versa. Electron affinity: Enthalpy change when one mole of
gaseous atoms or anions gain electrons to form a mole of negatively
charged gaseous ions.
Slide 40
15.2.1 Define and apply the terms lattice enthalpy and electron
affinity. Standard enthalpy change of atomization (also known as
standard enthalpy of vaporization). This is the enthalpy required
to change one mole of atoms from the standard state to the gaseous
state. Na(s) Na(g) H +103kJ mol -1 Exothermic or endothermic?
Slide 41
15.2.1 Define and apply the terms lattice enthalpy and electron
affinity. First ionization energy (remember this?) Na(g) Na+(g) +
e- H = +494 kJ mol-1 Enthalpy atomization of Cl Cl 2 (g) Cl(g) H =
+121 kJ mol-1 First electron affinity of Cl Cl(g) + e- Cl - (g) H =
-364 kJ mol-1
Slide 42
IB HL Objective 15.2.3 Construct a Born-Haber cycle for Group 1
and 2 oxides and chlorides, and use it to calculate an enthalpy
change.
Slide 43
Na (s) + Cl 2(g) NaCl (S) Na + (g) + Cl - (g) LATTICE ENTHALPY
H o f = Na (g) + Cl (g) Cl - (g) Na + (g)
Slide 44
15.2.3 Construct a Born-Haber cycle for Group 1 and 2 oxides
and chlorides, and use it to calculate an enthalpy change. Affinity
Formation of NaCl (s) from its gaseous elements. 1) Na (s) Na (g) H
o at = 103 kJ/mol 2) Cl 2(g) Cl (g) H o at = (242) kJ/mol 3) Cl (g)
+ e - Cl (g) - H o = -364 kJ/mol 4) Na (g) Na + (g) H o at = 500
kJ/mol Find the enthalpy of formation. Literature value for lattice
enthalpy can be found in the data booklet. A: -430 kJ/mol
Slide 45
15.2.3 Construct a Born-Haber cycle for Group 1 and 2 oxides
and chlorides, and use it to calculate an enthalpy change. Draw a
Born-Haber cycle for the formation of magnesium oxide, and
calculate the enthalpy of formation. The enthalpy of atomization of
magnesium is +150 kJ/mol, and for oxygen it is +249 kJ/mol. The
second ionization energy of magnesium is +1450 kJ/mol. Use the data
booklet to find other relevant information. A: -547 kJ/mol
Slide 46
IB HL Objective 15.2.2 Explain how the relative sizes and the
charges of ions affect the lattice enthalpies of different ionic
compounds. The relative value of the theoretical lattice enthalpy
increases with higher ionic charge and smaller ionic radius due to
increased attractive forces.
Slide 47
15.2.2 Explain how the relative sizes and the charges of ions
affect the lattice enthalpies of different ionic compounds. The
greater the charge on the ions, the greater the electrostatic
attraction and thus greater the lattice enthalpy, and vice versa.
The larger the ions, then the greater the separation of the charges
and the lower the lattice enthalpy, and vice versa. Lattice
enthalpy of MgO > NaCl. Why? A: Increased ionic charge. Lattice
enthalpy of KBr
15.3.1 State and explain the factors that increase the entropy
in a system. Entropy values for gas>liquid>solid. Gas
pressure increases, then entropy decreases (reduces volume for gas
particles to move in). When a solid or liquid dissolves in a
solvent, entropy increases. Increase the number of moles increases
the entropy. Heating a substance increases the entropy, since this
increases the movement of particles.
Slide 54
IB HL Objective 15.3.2 Predict whether the entropy change (S)
for a given reaction or process is positive or negative. S =
S(Products) S(Reactants) So if the products have more disorder than
the reactants, would the entropy change be positive or negative? A:
Positive
Slide 55
15.3.2 Predict whether the entropy change (S) for a given
reaction or process is positive or negative. Consider the following
equation: NH 4 Cl(s) NH 3 (g) + HCl(g) State S as positive or
negative, and explain why. A: Positive. (S=+285 J K -1 mol -1 )
More moles on product side, so more disorder. Also, solid has
become a gas, so more disorder.
Slide 56
IB HL Objective 15.3.3 Calculate the standard entropy change
for a reaction (S ) using standard entropy values (S ). S =
S(Products) S(Reactants)
Slide 57
15.3.3 Calculate the standard entropy change for a reaction (S
) using standard entropy values (S ). Calculate the entropy change
for the combustion of methane (CH 4 ). First, write the equation.
Make a prediction if it will be positive or negative, and explain
why. Next, plug in the values and calculate. You can find the
entropy value for methane in your data booklet. O 2 is 205, CO 2 is
214, and H 2 O is 70. A: -242 J K -1 mol -1
Slide 58
IB HL Objective 15.4.1 Predict whether a reaction or process
will be spontaneous by using the sign of G . A reaction will be
spontaneous when G has a negative value. Spontaneous: Once started,
a reaction will continue without any extra energy having to be
added. Remember redox? An electrochemical cell can create energy
through a spontaneous reactions. So if E cell is positive, what
would G be? A: Negative
Slide 59
15.4.1 Predict whether a reaction or process will be
spontaneous by using the sign of G . Is the combustion of coal
spontaneous at room temperature? Yes. The activation energy is
different than spontaneous reactions. You have to get the coal
going, but once you do, it continues to react. What about the
combustion of diamond at room temperature? Yes, although it has a
very high activation energy. Once this is reached, it will be a
spontaneous reaction.
Slide 60
IB HL Objective 15.4.2 Calculate G for a reaction using the
equation G = H TS and by using values of the standard free energy
change of formation, G f .
Slide 61
15.4.2 Calculate G for a reaction using the equation G = H TS
and by using values of the standard free energy change of
formation, G f . Lets go back to the combustion of diamond at room
temperature (298K): C (s) + O 2 (g) CO 2 (g) H = -395.4 kJ mol -1.
S = +6.6 J K -1 mol -1. What is G ? A: -397.4 kJ mol -1 Would this
reaction be spontaneous at all temperatures? Why or why not? Refer
to the equation! A: If H is negative and S is positive, then G will
always be negative.
Slide 62
IB HL Objective 15.4.3 Predict the effect of a change in
temperature on the spontaneity of a reaction, using standard
entropy and enthalpy changes and the equation: G = H TS If both S
and H are positive, then spontaneity depends on T. (Spontaneous at
higher temperatures). If both S and H are negative, then
spontaneity depends on T. (Spontaneous at lower temperatures).