1. Thermochemistry
Chapter 6
AP Chemistry

2. thermodynamics: the study of energy and
its transformations
-- thermochemistry: the subdiscipline involving
chemical reactions and
energy changes

3. + + Energy kinetic energy: energy of motion; KE = ½ mv2
-- all particles have KE
-- Thermal energy is due to the KE of particles.
We measure the average KE of a
collection of particles as...
temperature.
potential energy: stored energy
Chemical potential energy
is due to electrostatic forces
between charged particles. + +
-- related to the specific
arrangement of atoms in the substance

4. nutritional calories (Cal or kcal).
SI unit
Units of energy are
joules (J),
kilojoules (kJ),
calories (cal), or
nutritional calories (Cal or kcal).
James
Prescott
Joule
( )
-- conversions:
4184 J = kJ = 1000 cal = 1 Cal = 1 kcal

5. system: the part of the universe we are studying
surroundings: everything else
-- In chemistry, a
closed system
can exchange energy
but not matter with
its surroundings.
-- Usually, energy is transferred to...
…(1)
change an object’s state of motion
...or... (2)
cause a temperature change

6. Work (w) is done when a force moves
through a distance.
W = F d
Heat (q) is an amount of energy transferred from
a hotter object to a colder one.

7. Find the kinetic energy of a
single dinitrogen monoxide
molecule moving at 650 m/s.
N2O
(laughing gas)
KE = ½ mv2 ?
m = 44 g
KE =

8. First Law of
Thermodynamics
Law of Conservation
of Energy =
-- Energy morphs between its various forms,
but the total amount remains the same.

9. In endothermic processes, heat is _________ by
the system.
absorbed
e.g.,
melting
boiling
sublimation
In exothermic processes, heat is ________ by
the system.
released
e.g.,
freezing
condensation
deposition

10. 2H2 (g) + O2 (g) 2H2O (l) + energy
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g) H2O (l) + energy
H2O (g) H2O (l) + energy
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
energy + 2HgO (s) Hg (l) + O2 (g)
energy + H2O (s) H2O (l)

11. internal energy (E) of a system: the sum of all the
KE and PE of the components of a system
The change in the internal energy
of a system would be found by:
ΔE = Efinal – Einitial
And for chemistry, this equation would become:
ΔE = Eproducts – Ereactants
ENDOTHERMIC gains energy
ΔE is
ΔE is –
EXOTHERMIC loses energy

12. But we ARE able to find ΔE by measuring two
types of “energy” quantities:
ΔE = q + w
q = heat:
+/– q = system absorbs/releases heat
w = work:
+/– w = work done on/by system
** KEY: Based on the system’s point of view.

13. Heike Kamerlingh Onnes
To go further, we must introduce
the concept of enthalpy (H).
-- Enthalpy (H) is defined as... H = E + PV
where E = system’s internal energy
P = pressure of the system
Heike Kamerlingh Onnes
1853–1926
The Dutch physicist and Nobel laureate H.K. Onnes coined the term enthalpy, basing it on the Greek term enthalpein, which means “to warm.”
V = volume of the system

14. Enthalpy is used to quantify the heat that
is either gained or lost by a system that
Is at constant pressure.
ΔH = Hfinal – Hinitial = qP
P indicates constant pressure conditions.
When ΔH is +, the system...
has gained heat.
(ENDO)
When ΔH is –, the system...
has lost heat.
(EXO)
Enthalpy is an extensive property, meaning that…
the amount of material affects its value.

15. In the burning of firewood at constant pressure, the
enthalpy change equals the heat released. ΔH is (–)
and depends on the quantity of wood burned.

16. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance
H2O (s) H2O (l)
ΔH = 6.01 kJ/mol ΔH = 6.01 kJ
If you reverse a reaction, the sign of ΔH changes
H2O (l) H2O (s)
ΔH = kJ
If you multiply both sides of the equation by a factor n, then ΔH must change by the same factor n.
2H2O (s) H2O (l)
ΔH = 2 mol x 6.01 kJ/mol = 12.0 kJ

17. Thermochemical Equations
The physical states of all reactants and products must be specified in thermochemical equations.
H2O (s) H2O (l)
ΔH = 6.01 kJ
H2O (l) H2O (g)
ΔH = 44.0 kJ
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) ΔHreaction = kJ
1 mol P4
123.9 g P4 x
3013 kJ
1 mol P4 x
266 g P4
= 6470 kJ

18. What is the enthalpy change when
2 H2(g) + O2(g) → 2 H2O(g) ΔH = –483.6 kJ
What is the enthalpy change when
178 g of H2O are produced?
178 g H2O
ΔH = –2390 kJ
The space shuttle was powered
by the reaction above.

19. Calorimetry: the measurement of heat flow
-- device used is called a...
calorimeter
molar heat capacity: amt. of heat needed to raise
temp. of 1 mol of a substance 1 K = 1oC
specific heat (capacity): amt. of heat needed to raise
temp. of 1 g of a substance 1 K = 1oC
molar heat capacity = molar mass X specific heat

20. We calculate the heat a substance loses or gains using:
q = m cP ΔT
AND
q = +/– m cX
(for within a given
state of matter)
(for between two
states of matter)
where q = heat
m = amount of substance
cP = substance’s heat capacity
ΔT = temperature change
cX = heat of fusion (s/l)
or heat of vaporization (l/g)

21. In an experiment it was determined that 59
In an experiment it was determined that 59.8 J was required to change the temperature of 25.0 g of ethylene glycol (a compound used as antifreeze in automobile engines) by 10.0 C. Calculate the specific heat capacity of ethylene glycol from these data.
q = m cP ΔT

22. Typical Heating Curve g
← heat removed (–q)
l/g l
Temp.
s/l
heat added (+q) → s
HEAT

23. qwater = m x s x t

24. Important Relationships in Calorimetric equations
Heat lost by chemicals = Heat gained by Solution
Heat gained by water = Heat lost by metal object
Heat lost by hot water = Heat gained by cold water
m x s x t = m x s x t

25. A lead (Pb) pellet having a mass of 26. 47 g at 89
A lead (Pb) pellet having a mass of g at °C was placed in a constant-pressure calorimeter containing mL of water
The water temperature rose from 22.50°C to °C.
What is the specific heat of the lead pellet?

26. A sketch of the initial and final situation is as follows:
We know the masses of water and the lead pellet as well as the initial and final temperatures. Assuming no heat is lost to the surroundings, we can equate the heat lost by the lead pellet to the heat gained by the water. Knowing the specific heat of water, we can then calculate the specific heat of lead.

27. The heat gained by the water is written as:

28. Because the heat lost by the lead pellet is equal to the heat gained by the water,
qPb = −280.3 J.
Solving for the specific heat of Pb

29. Combustion reactions are studied using constant-
volume calorimetry.
bomb calorimeter
This technique requires
a bomb calorimeter.
-- The heat capacity of
the bomb calorimeter
(Ccal) must be known.
unit is J/K
(or the equivalent)

30. Again, we assume that no energy escapes into the
surroundings, so that the heat absorbed by the bomb
calorimeter equals the heat given off by the reaction.

31. Henri Hess, in 1840, discovered a very useful principle which is named for him:
Hess's law states that the change of enthalpy in a chemical reaction (i.e. the heat of reaction at constant pressure) is independent of the route by which the chemical change occurs.
This is true because it is a state function,which is a value that does not depend on the path taken

32. Hess’s Law
The ΔHrxns have been calculated and tabulated for many basic reactions.
Hess’s law allows us to put these simple reactions together like puzzle pieces such that they add up to a more complicated reaction that we are interested in.
By adding or subtracting the ΔHrxns as appropriate, we can determine the ΔHrxn of the more complicated reaction

33. Hess's Law is saying that if you convert reactants A into products B, the overall enthalpy change will be exactly the same whether you do it in one step or two steps or however many steps.
The overall enthalpy change must be the same, because it is controlled by the relative positions of the reactants and products on the enthalpy diagram.

34. However many stages the reaction is done in, ultimately the overall enthalpy change will be the same, because the positions of the reactants and products on an enthalpy diagram will always be the same.

35. Important things to remember when using Hess’s Law to calculate Enthalpy changes.
If a reaction is reversed, the sign of the H is also reversed.
The size of H is directly related to the quantities of reactants and products in the reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of H is multiplied by the same integer.

36. Calculate the enthalpy for this reaction 2C(s) + H2(g) ---> C2H2(g)
ΔH° = ??? kJ
Given the following thermochemical equations:
C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ)
ΔH° = kJ
C(s) + O2(g) ---> CO2(g)
ΔH° = kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ)
ΔH° = kJ

37. 2CO2(g) + H2O(ℓ) ---> C2H2(g) + (5/2)O2(g) ΔH° = +1299.5 kJ
1) Determine what we must do to the three given equations to get our target equation:
a) first eq: flip it so as to put C2H2 on the product side
b) second eq: multiply it by two to get 2C
c) third eq: do nothing. We need one H2 on the reactant side and that's what we have.
2CO2(g) + H2O(ℓ) ---> C2H2(g) + (5/2)O2(g) ΔH° = kJ
2C(s) + 2O2(g) ---> 2CO2(g) ΔH° = -787 kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = kJ
Notice that the ΔH values changed as well
Add up ΔH values for our answer:
kJ + (-787 kJ) + ( kJ) = kJ

38. ΔHf0 (C, diamond) = 1.90 kJ/mol
Standard enthalpy of formation (ΔHf0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm and a temperature of 25oC (298oK)
Whenever a standard enthalpy change is quoted, standard conditions are assumed.
The standard enthalpy of formation of any element in its most stable form is zero.
ΔHf0 (O2) = 0
ΔHf0 (C, graphite) = 0
ΔH0 (O3) = 142 kJ/mol f
ΔHf0 (C, diamond) = 1.90 kJ/mol

39. Standard states
For a standard enthalpy change everything has to be present in its standard state. That is the physical and chemical state that you would expect to find it in under standard conditions.
standard state for water, is liquid water, H2O(l) - not steam or water vapour or ice.
Oxygen's standard state is the gas, O2(g) - not liquid oxygen or oxygen atoms.
Elements which have allotropes (two different forms of the element in the same physical state), the standard state is the most energetically stable of the allotropes.
Carbon exists in the solid state as both diamond and graphite. Graphite is more stable than diamond, and so graphite is taken as the standard state of carbon.

40. Some other important types of enthalpy change
Standard enthalpy change of combustion, ΔH°c
The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one mole of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.
Standard enthalpy change of reaction, ΔH°rxn
The standard enthalpy change of a reaction is the enthalpy change which occurs when equation quantities of materials react under standard conditions, and with everything in its standard state.
The enthalpy of solution (ΔHsoln) is the heat generated or absorbed
when a certain amount of solute dissolves in a certain amount of solvent.

41. Standard enthalpy of a reaction (ΔHorxn):
Using Hess’s law, we can easily calculate ΔHorxn from the ΔHfo of all R and P.
ΔHorxn = Σn ΔHfo(products) – Σm ΔHfo(reactants)
Where n and m are the coefficients in the balanced equation.

42. Approximate the enthalpy change for the
(Look these up.
See App. 4,
P A19.)
Approximate the enthalpy change for the
combustion of 246 g of liquid methanol. 2
CH3OH(l) + 3
O2(g) 2
CO2(g) + 4
H2O(g)
–238.6 kJ/mol
0 kJ/mol
–393.5 kJ/mol
–241.8 kJ/mol
X 2
X 2
X 4
–477.2 kJ
– kJ
ΔHorxn =
– kJ – (–477.2 kJ)
= –1277 kJ
for 2 mol
(i.e., 64 g)
of CH3OH
So…
X = ΔH = –4910 kJ

43. Benzene (C6H6) burns in air to produce carbon dioxide and liquid water
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)

44. Solution
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
ΔH0
rxn
ΔH0 (products) f = Σ
ΔH0 (reactants) - ΔH
rxn
6ΔH0 (H2O) f
12ΔH0 (CO2) = [ + ] -
2ΔH0 (C6H6) ΔH
rxn =
[ 12 × × ] – [ 2 × ] = kJ
-6535 kJ
2 mol
= kJ/mol C6H6