Thermodynamics is the study of energy and its transformations. Thermochemistry is the subdiscipline involving chemical reactions and energy changes. The first law of thermodynamics states that energy is conserved as it transforms between forms. Energy exists in kinetic and potential forms. Kinetic energy is associated with motion, while potential energy is stored energy due to position or composition. Heat is the transfer of kinetic energy between objects of different temperatures until they reach equilibrium. Exothermic processes release heat from a system, while endothermic processes absorb heat into a system. Enthalpy changes (ΔH) quantify the heat absorbed or released by chemical reactions. Spontaneous processes are those that proceed without outside intervention, as dictated by a negative change

1. Chemical
Energetics
AS
Chemistry

2. Thermodynamics: the study of energy and
its transformations
Thermochemistry: the sub discipline involving
chemical reactions and
energy changes

3. Energy
•Energy is defined as the capacity to do work or to produce
heat.
•The 1st Law of Thermodynamics …. Energy is conserved
as it is converted between one form and another, it will be
neither created or destroyed, but simply change form…thus
making the energy of the universe constant!
•Energy can be classified as either Kinetic or Potential

4. Kinetic energy: energy associated with the motion of
atoms and molecules in a system.
KE = ½ mv2
Temperature is a measure of the average KE of a
collection of particles in a system.
RECALL…….

5. Heat vs Energy
•Thermal energy is the energy of the
object and is not in the process of being
transferred or moved.
•Heat is kinetic energy being transferred
–It moves from a hotter object towards cooler
object until the temperatures are the same.
–At this point the KE’s are at equilibrium.
–It is not a property of the substance.

6. System: the part of the universe we are
studying
Surroundings: everything else
Usually, energy is transferred to...
(1)
(2)
Change an object’s state of motion
like fuel in a vehicle.
Cause a temperature change like
a furnace warming a house.
In chemistry the system is the reaction that
we are interested in and the surroundings
could be the container that the reaction
takes place in.

7. Units of energy are
either
joules (J)
kilojoules (kJ)
CONVERSIONS:
Divide by 1000 to convert from J to KJ
Multiply by 1000 to convert from KJ to J
James
Prescott
Joule
(1818-1889)
4184 J = 4.184 kJ

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

9. Water Phase Change Diagram

10. • Exothermic process is any process that gives off heat –
The energy will be listed as a product.
2H2 (g) + O2 (g) 2H2O (l) + energy
H2O (g) H2O (l) + energy
Endothermic process is any process in which heat is
required by the system. The energy is listed as a reactant.
energy + 2HgO (s) 2Hg (l) + O2 (g)
energy + H2O (s) H2O (l)

12. ΔH = Hproducts – Hreactants
When ΔH is +, the system... has gained heat.
When ΔH is –, the system... has lost heat.
(ENDO)
(EXO)
Enthalpy is used to measure the heat that is
either gained or lost by a system that is at
constant pressure.
• Enthalpy is an extensive property, meaning that…the
amount of material affects its value

13. Enthalpy (H)

14. H2O (s) H2O (l) ΔH = 6.01 kJ/mol ΔH = 6.01 kJ
Thermochemical Equations
If you reverse a reaction, the sign of ΔH changes
H2O (l) H2O (s) ΔH =- 6.01 kJ
If you multiply both sides of the equation by a factor n,
then ΔH must change by the same factor n.
2H2O (s) 2H2O (l)
ΔH = 2 mol x 6.01 kJ/mol = 12.0 kJ

15. H2O (s) H2O (l) ΔH = 6.01KJ
The physical states of all reactants and products must be
specified in thermochemical equations.
Thermochemical Equations
H2O (l) H2O (g) ΔH = 44.0 KJ
Practice Question:
How much heat is evolved when 266 g of white
phosphorus (P4) burns in air? ΔHreaction = -3013 kJ

16. 2 H2(g) + O2(g) → 2 H2O(g) ΔH = – 483.6 kJ
What is the enthalpy change when 178 g of H2O(g)
are produced?
The space shuttle was powered
by the reaction above.

17. Calorimetry: the measurement of heat flow
A Calorimeter is used to measure the heat changes
molar heat (capacity): amt. of heat needed to raise
temp. of 1 mol of a substance
J/ C *mol or J/ K*mol
specific heat (capacity): amt. of heat needed to raise
temp. of 1 g of a substance
J/ C *g or J/ K*g

18. cX = heat of fusion (s/l)
or heat of vaporization (l/g)
We calculate the heat a substance loses or gains using:
q = heat
m = amount of substance
c = substance’s heat capacity
ΔT = temperature change
q = m c ΔT
(for within a given
state of matter)
AND q = m cX
(for between two states
of matter when temp is
constant)

19. Heat capacities of metals are very low when
compared to water or other substances.

20. 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.
q = m c ΔT

21. HEAT
Temp
.
s
s/l
l
l/g
g
Typical Heating Curve

22. Constant Pressure Calorimetry
• Commonly called
“COFFEE CUP” calorimetry
• It’s used to determine any
changes in enthalpy for
reactions occurring in
solution.
• Atmospheric pressure
remains constant during the
reaction .
Calorimetry:
The measurement of heat flow

23. Practice Problem
A lead (Pb) pellet having a mass of 26.47 g at 89.98°C
was placed in a constant-pressure calorimeter
containing 100.0 mL of water. The water temperature
rose from 22.50°C to 23.17°C.
What is the specific heat of the lead pellet?

24. 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.

25. Because the heat lost by the lead pellet is equal to the heat
gained by the water,
qPb = −280.3 J.

26. Combustion reactions
are studied using constant
volume calorimetry.
It requires a BOMB
CALORIMETER.

27. 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.

28. Hess’ Law 1840
•The change of enthalpy in a chemical
reaction is independent of the route by which
the chemical change occurs.
•This is true because enthalpy is a state
function , which is a value that does not
depend on the path taken

29. ● The ΔHrxns have been calculated and tabulated
for many basic reactions.
● Hess’s law allows us to put these simple
reactions together like puzzle piecesso that they
can add up to a more complicated reaction.
● By adding orsubtracting the ΔHrxns, we can
determine the ΔHrxn of the more complicated
reaction.
How Hess’s Law works

30. Hess's Law is saying:
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.

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

32. 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° = -1299.5 kJ
C(s) + O2(g) ---> CO2(g) ΔH° = -393.5 kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ

33. 1) Determine what must be done to the given equations to get the 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° = +1299.5 kJ
2C(s) + 2O2(g) ---> 2CO2(g) ΔH° = -787.0 kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ
Notice that the ΔH values changed as well
Add up ΔH values for our answer:
+1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ

34. Standard enthalpy of formation (ΔHf
0) is the heat change
that results when one mole of a compound is formed from
its elements at STP.
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.
ΔHf
0 (O2) = 0
ΔH0 (O3) = 142kJ/mol
f
ΔHf
0 (C, graphite) = 0
ΔHf
0 (C, diamond) = 1.90 kJ/mol

35. Some other important types of
enthalpy changes
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 at STP .
The enthalpy change of solution (ΔH soln) is the
heat generated or absorbed when a certain amount
of solute dissolves in a certain amount of solvent at
STP.

36. Standard enthalpy of a reaction (ΔHo
rxn):
Using Hess’s law, we can easily calculate
ΔHo
rxn from the ΔHf
o of all reactants and products by
using the following equation:
ΔHo
rxn = Σ (ΔH fproducts) – Σ (ΔH f reactants)

37. –238.6 kJ/mol
Approximate the enthalpy change for the
combustion of 246 g of liquid methanol.
CH3OH(l) O2(g) CO2(g) H2O(g)
+ +
2
X 2
2 4
–393.5 kJ/mol –241.8 kJ/mol
0 kJ/mol
X 2 X 4
–477.2 kJ –1754.2 kJ
ΔHo
rxn
=
(–1754.2 kJ) – (–477.2)
kJ)
= –1277 kJ
for 2 mol
(i.e., 64 g)
of CH3OH
So…
X = ΔH = –4910 kJ
3
(Look these up.
See App. 4,
P A19.)

38. Practice problem #1
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 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (g)

39. Practice Problem # 2
What is the Δ Hrxn for the complete
combustion of Butane, C4H10 (g)?
2 C4H10 (g) + 13 O 2(g) 8CO2 (g) + 10 H2O (g)

40. Chapter 17 Second Law of Thermodynamics

41. Think of these commonplace
experiences:
● A Hot frying pans cool down when
taken off the stove.
● Air in a high-pressure tire shoots out
from even a small hole in its side to the
lower pressure atmosphere.
● Ice cubes melt in a warm room.
● Iron exposed to oxygen and water will
form rust.

42. What’s happening in each of those processes?
● Energy of some kind is changing from being localized,
concentrated, and contained to becoming more spread
out and dispersed.
● Entropy is the measurement of disorder of a system
and it is given the letter S, and it is temperature
dependent.

43. 43
Entropy – quantitative measure of disorder

44. Ssolid < Sliquid < Sgas
• Entropy increases with dispersal of particles so,
entropies of gases are larger than liquids and liquid
entropies are larger than solids.
Entropies are greater for :
- more complex molecules
- Increased temperatures (KE)
- When volume increases for gases
+ΔS ………. Entropy increases
- ΔS ………. Entropy decreases

45. Which substance has the greater entropy?
CO2 (s) or CO2(g)
H2 (g) at 1 atm or H2 (g) at 1.0 x 10 -2 atm
What will the overall change in entropy?
● Solid sugar is added to water to make a sugar solution.
● Water vapor condenses.
● Ice melts

46. Entropy Changes in the System (ΔSsys)
When gases are produced (or consumed)
• If a reaction produces more gas molecules than it
consumes, ΔS0+
• If the total number of gas molecules diminishes, ΔS0-
What is the sign of the entropy change for the following
reaction?
2Zn (s) + O2 (g) 2ZnO (s)
The total number of gas molecules goes down….. ΔS0-

47. Chemical
Thermodynamics
Entropy on the Molecular Scale
• Ludwig Boltzmann described the concept of
entropy on the molecular level.

48. Chemical
Thermodynamics
Entropy on the Molecular Scale
Molecules exhibit several types of motion:
Translational: Movement of the entire molecule from
one place to another.
Vibrational: Periodic motion of atoms within a molecule.
Rotational: Rotation of the molecule on about an axis or
rotation about bonds.

49. Chemical
Thermodynamics
Entropy on the Molecular Scale
• Boltzmann envisioned the motions of a sample of
molecules at a particular instant in time.
This would be akin to taking a snapshot of all the
molecules.
• He referred to this sampling as a microstate of the
thermodynamic system.

50. Chemical
Thermodynamics
Entropy on the Molecular Scale
Implications:
• More Particles
-> more states -> more entropy
• Higher Temp
-> more energy states -> more entropy
• Less Structure (gas vs solid)
-> more states -> more entropy

51. The entropy change for a system(reaction) is calculated
from the entropies of the products and the reactants
ΔSo
system = Σ[So(products)] - Σ [So( reactants)]
ΔSo
system is Positive, then entropy increases
ΔSo
system is Negative,then entropy decreases
Calculations 

52. Chemical
Thermodynamics

53. Gibbs Free Energy
Gibbs Free Energy (G)
The energy associated with a chemical reaction
that can be used to do work.

55. What are Spontaneous Processes ?
Spontaneous processes are
those that can proceed without
any outside intervention.
For example, the gas in vessel
B will spontaneously effuse
into vessel A, but once the
gas is in both vessels, it will
not spontaneously revert to
its original state.

56. Chemical
Thermodynamics
Spontaneous Processes
Processes that are
spontaneous in one direction
are nonspontaneous in the
reverse direction.

57. For a constant-temperature process:
Gibbs free energy (G)
ΔG - The reaction is favorable (spontaneous) in the forward direction.
No outside energy is needed
Product formation is favored
ΔG + The reaction is unfavorable (non-spontaneous) as written.
The reaction is favorable (spontaneous) in the reverse direction
Reactant formation is favored
ΔG = 0 The reaction is at equilibrium and reactant and product
formation are equally favored

58. Recap: Signs of Thermodynamic Values
Negative Positive
Enthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy
(ΔG)
Favored Not favored