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Section 1 – Thermochemistry
Section 2 – Driving Force of Reactions
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Define temperature and state the units in which it
is measured.
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Define heat and state its units.
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Perform specific-heat calculations.
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Explain enthalpy change, enthalpy of reaction,
enthalpy of formation, and enthalpy of combustion.

Solve problems involving enthalpies of reaction,
enthalpies of formation, and enthalpies of
combustion.
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Thermochemistry: study of the transfer of
energy as heat that accompanies chemical
reactions and physical changes.
Virtually every physical change & chemical
reaction is accompanied by a change in
energy.
◦ What two ways can energy be involved?
 Absorbed (endothermic) or released (exothermic)
Solid
Liquid
Gas
What is a phase change?
• change from one state of matter to another
• during a phase change – temp. remains constant
Plasma

Heat: the energy transferred between samples
of matter because of a difference in their
temperatures.
◦ SI unit of heat and energy: Joule (J)
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Energy transferred as heat moves spontaneously
 from matter at a higher temp  lower temp
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Specific Heat (C): amount of energy required
to raise the temperature of 1 gram by 1°C or
by 1 K.
◦ Units: J/(g•°C) or J/(g•K)
◦ The temperature difference as measured in either
Celsius degrees or kelvins is the same.
◦ Specific heat allows us to compare various materials
to one another
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A higher C value:

A lower C value:
◦ More energy required to raise/lower temp.; heats /cools
relatively slow
◦ Can hold more energy overall
◦ Less energy required to raise/lower temp.; heats/cools
relatively quickly
◦ Hold less energy overall (
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Can the specific heat of a substance change
with a phase change?
◦ Yep
◦ Ice vs. water vs. steam – all have different values
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measure of the average kinetic energy of the
particles in a sample of matter.
◦ The greater the kinetic energy of the particles in a
sample, the hotter it feels.
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Remember: How do we convert Celsius to
Kelvin?
 K = 0C + 273
Q = m x C x ∆T
Q = heat energy lost/gained (J)
C = specific heat (J/g oC)
m = mass of the sample (g)
∆T = difference between the initial and final
temperatures. (oC or K)

Energy transferred depends on:
◦ The material itself (all substances have unique C’s)
◦ The mass
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During a phase change, temperature remains
constant
Can’t use previous equation
Q = mol x Hfus / Hvap
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Heat of Fusion: Hfus
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Heat of Vaporation: Hvap
◦ Heat energy needed to melt/freeze 1 mol
◦ Heat energy need to boil/condense 1 mol of a substance
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Step 1 (_________________) :
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Step 2 (_________________):
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Step 3 (_________________):
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Step 4: (_________________):
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Step 5: (_________________):
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Calorimeter: used to measure heat
absorbed or released.
Energy released from
the reaction is measured
from the temp change of
the water
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A 4.0 g sample of glass was heated from 274
K to 314 K, and was found to have absorbed
32 J of energy as heat.
a. What is the specific heat of this type of glass?
b. How much energy will the same glass sample
gain when it is heated from 314 K to 344 K?
Given:
m = 4.0 g
∆T = 314 – 274 = 40. K
q = 32 J
Unknown: a. cp in J/(g•K)
b. q for ∆T of 314 K → 344 K
Solution:
a.
q
32 J
cp 

 0.20 J/(g  K)
m  T (4.0 g)(40. K)
b.
q  c p  m  T
0.20 J
q
(4.0 g)(344 K  314 K)
(g  K)
0.20 J
q
(4.0 g)(30 K)  24 J
(g  K)
**See Practice Worksheet for more examples
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Assume we have 85 grams of ice at -10 °C . How
much energy is needed to convert it to water vapor at
120oC?
Step 1: -10°C to O°C (solid)
◦ What is happening to the particles as heat is added?
◦ General formula: q=(m)(cp)(∆T)
◦ Answer = 1751 J  1.8kJ
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Step 2: melting
◦ Notice how there is no increase in temp.
◦ Where is this energy going?
◦ General formula: q=(∆H fusion)(mol)
◦ Answer = 28 kJ
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Step 3: – 0 °C to 100 °C (liquid)
◦ What is happening to the particles as we heat?
◦ General formula: q=(m)(cp)(∆T)
◦ Answer = 36,000  36 kJ
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Step 4: Boiling
◦ Notice how there is no increase in temp.
◦ Where is this energy going?
◦ General formula: q=(∆H vap)(mol)
◦ Answer = 190 kJ
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Step 5: 100 °C to 120 °C (gas)
◦ What is happening to the particles as we heat?
◦ General formula = (m)(cp)(∆T)
◦ Answer = 3,200 J  3.2 kJ
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Now, add up all the steps for total heat.
◦ Answer = 259 kJ
Summary:
 When the substance is heating
◦ q=(m)(cp)(∆T)
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During phase change
◦ q=(mol)(∆H fus) OR
◦ q=(mol)(∆H vap)
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Enthalpy (H): Heat
◦ H is never directly determined, rather ∆H
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Enthalpy(∆H): “change in enthalpy”
energy absorbed as heat during a chemical
reaction at constant pressure
Enthalpy of reaction (∆Hrxn): heat energy
absorbed or released during a chemical
reaction

The difference between the enthalpies of
products and reactants.
∆Hrxn = Hproducts – Hreactants
◦ -∆Hrxn : exothermic, reaction releases heat
◦ +∆Hrxn : endothermic, reaction absorbs heat
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Equation that includes the quantity of energy
released or absorbed as heat
2H2(g) + O2(g) → 2H2O(g)
∆H = -483.6 kJ
◦ States of matter are ALWAYS shown because it can influence
the energy of released or absorbed.
◦ Coefficients are always viewed as number of moles in the
reaction.
Energy released
-∆H
Energy is absorbed
+∆H
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Compounds with a large negative enthalpy of
formation are very stable.
Compounds with positive values of enthalpies
of formation are typically unstable.
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∆Hcomb : enthalpy change that occurs during
the complete combustion of 1 mole of a
substance
**Enthalpy of combustion is defined in terms of
one mole of reactant
**Enthalpy of formation is defined in terms of
one mole of product
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molar enthalpy of formation: enthalpy change
that occurs when 1mole of a compound is
formed from its elements in their standard
state: 25°C and 1 atm.
◦ Enthalpies of formation are given for a standard
temperature and pressure so that comparisons
between compounds are meaningful.
H
0
f
“Standard
State”
Heat (enthalpy)
of FORMATION
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In your textbook! (or handout)
The values are given as the enthalpy of
formation for one mole of the compound
from its elements in their standard states.
◦ **because ΔHf is reported for the formation of 1
mol of a substance, it may be necessary to write on
½ mol for one of the reactants
◦ Ex:
1/2N2
(g) +
1/2O2
(g)  NO (g)
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Determined by using Hess’s Law:
the overall enthalpy change in a reaction is
equal to the sum of enthalpy changes for the
individual steps in the process.
*from enthalpy packet
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1. The sign ∆Hrxn of depends on the direction
of the reaction
 (If you flip the reaction around, the a positive enthalpy
would become negative)
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2. The magnitude ∆Hrxn of is proportional to
the amount of the substance in a reaction.
 (If you change of the coefficients then multiply the
too)
∆H,
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If you know the reaction enthalpies of
individual steps in an overall reaction, you
can calculate the overall enthalpy without
having to measure it experimentally.
What is the heat of formation of methane?
C(s) + 2H2(g) → CH4(g)
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The component reactions in this case are the
combustion reactions of carbon, hydrogen,
and methane:
Goal Reaction: C(s) + 2H2(g) → CH4(g)
C(s) + O2(g) → CO2(g)
Hc0  393.5 kJ
0
kJkJ)
HcH
2(285.8
285.8
c 
H
(g)2→
2 2[(g
H)2(+g)½O
+ 2½O
(g)H→
Hl2)O(l) ]
2O(
0
CH
gO(
)→
(g ) +
2H O(l)
2(2
CO42((gg) )++2O
2H
l) CO
→ 2CH
4(g) +2 2O2(g)
This reaction must x2!
The ΔH is x2!
0
H
890.8 kJkJ
c 0=+890.8
∆H
This reaction must be reversed!
The ΔH is changed to positive!
Hc0  393.5 kJ
C(s) + O2(g) → CO2(g)
2H2(g) + O2(g) → 2H2O(l)
Hc0  2( 285.8 kJ)
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g)
C(s) + 2H2(g) → CH4(g)
H 0  890.8 kJ
H  74.3 kJ
0
f
∆H0 = (ΣΔH0f products) – (ΣΔH0f reactants)
*Need to use reference sheet for Hf values or
table in back of textbook
**Keep in mind: elements in their standard
state have an enthalpy of formation of zero
◦ Ex: Ag(s) = 0

; N2 (g) = 0
See Practice WKST (Calculating Enthalpy from
Heats of Formation)
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Explain the relationship between enthalpy change
and the tendency of a reaction to occur.
Explain the relationship between entropy change
and the tendency of a reaction to occur.
Discuss the concept of free energy, and explain
how the value of this quantity is calculated and
interpreted.
Describe the use of free energy change to
determine the tendency of a reaction to occur.
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Most chemical reactions are exothermic,
liberating heat
◦ So products have less energy than the reactants
◦ Nature’s tendency is toward a lower energy state
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But, some endothermic processes are also
spontaneous
◦ Example: melting
(particles have more energy in the liquid state)
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What conclusion can we come to then?
◦ there is another factor that drives a reaction
◦ 2 Factors: Enthalpy, but also Entropy
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Entropy (S) : the degree of randomness or
disorder of the particles in a system
Think about your bedroom, does it always stay neat
and organized? Or does it get messier and messier?
Increase in disorder  Increase in entropy
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Nature’s tendency is toward more disorder or
greater entropy.
◦ Also means, the entropy of the universe is ALWAYS
increasing.
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When a solid dissolves in a liquid, the entropy
in the system increases.
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As you go from a solid  liquid  gas, the
entropy in the system ______________.
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When you dilute a substance, the entropy in
the system _______________.
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When you increase the temperature of a
system, the entropy ________________ .
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Change in Entropy ΔS
◦ Difference between the entropy of the products and
the reactants

Increase in Entropy  +ΔS
Decrease in Entropy  -ΔS
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Unit: kJ/mol . K or J/mol . K
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(more disorder)
(less disorder)
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Calculate entropy like calculating enthalpy
from heats of formation table, but use S
values instead!
∆S = Sproducts - Sreactants
See practice WKST, use thermodynamic data
sheet for Sf values
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Gibb’s Free Energy (G):
◦ Combined enthalpy-entropy function that assesses
how these two factors will affect the spontaneity of
a reaction
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Spontaneous Reactions: the ideal situation
◦ Large negative enthalpy – very stable
◦ Large positive entropy – greater disorder
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Only the change in free energy can be measured (KJ/mol)
At a constant pressure and temperature…
∆G0 = ∆H0 – T∆S0
*T must be in Kelvin since ∆S will be in units of
KJ/mol k
*∆S should also be in KJ
If -∆G  the reaction is spontaneous
+∆G  the reaction is not spontaneous
∆H
∆S
∆G
+
+
+
+
- /spontaneous
+ /not spontaneous
- / Low temps spontaneous
- / High temps spontaneous
See “Free Energy” worksheet to practice calculation
See Practice Worksheet – “Pg. 78 Gibbs Free Energy” to review concepts
For the reaction
NH4Cl(s) → NH3(g) + HCl(g)
∆H0 = 176 kJ/mol and ∆S0 = 0.285 kJ/(mol•K)
at 298.15 K. Calculate ∆G0, and tell whether
this reaction is spontaneous in the forward
direction at 298.15 K.
∆H0 = 176 kJ/mol at 298.15 K
∆S0 = 0.285 kJ/(mol•K) at 298.15 K
Unknown: ∆G0 at 298.15 K
Solution: The value of ∆G0 can be calculated
according to the following equation:
∆G0 = ∆H0 – T∆S0
∆G0 = 176 kJ/mol – 298 K [0.285kJ/(mol•K)]
∆G0 = 176 kJ/mol – 84.9 kJ/mol
∆G0 = 91 kJ/mol
Given: