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Section 1 – Thermochemistry
Section 2 – Driving Force of Reactions

Define temperature and state the units in which it
is measured.

Define heat and state its units.

Perform specific-heat calculations.

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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Virtually every chemical reaction is
accompanied by a change in energy.
◦ What two ways can energy be involved?

Thermochemistry: study of the transfer of
energy as heat that accompanies chemical
reactions and physical changes.


Calorimeter: used to measure heat
absorbed or released.
Energy released from
the reaction is measured
from the temp change of
the water

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.

Remember: How do we convert Celsius to
Kelvin?
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Heat: the energy transferred between samples
of matter because of a difference in their
temperatures.
◦ Joule: SI unit of heat and energy

Energy transferred as heat moves spontaneously
 from matter at a higher temp  lower temp

Energy transferred depends on:
◦ The material
◦ The mass
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Specific Heat: specific heat of a substance is
the amount of energy required to raise the
temperature of one gram by one Celsius
degree (1°C) or one kelvin (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:
cp 
q
m  T
Heat released
or absorbed:
q  c p  m  T
cp = specific heat at a given pressure
q = energy lost or gained
m = mass of the sample
∆T = difference between the initial and final
temperatures.

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
How much energy does it take heat water from ice
to vapor??
(2 equations: 1 for change of temp./ 1 for phase
change)
Phase Change:
 Heat of Fusion: Hfus
◦ Heat energy needed to melt 1 mol (s to l)

Heat of Vaporation: Hvap
◦ Heat energy need to boil 1 mol of a substance ( l – g)

Enthalpy(∆H):
“change in enthalpy”
energy absorbed as heat during a chemical
reaction at constant pressure

The difference between the enthalpies of
products and reactants.
∆H = Hproducts – Hreactants

Equation that includes the quantity of energy
released or absorbed as heat
2H2(g) + O2(g) → 2H2O(g) + 483.6 kJ
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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.
If 2x as many reactants were provide, 2x as many
moles of water would be produced and 2x as many
kJ of energy would be released.
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chemical reaction that releases energy
the energy of the products is less
than the energy of the reactants
◦ example:
2H2(g) + O2(g) → 2H2O(g) + 483.6 kJ

chemical reaction that absorbs energy
the products have a larger enthalpy than the
reactants
2H2O(g) + 483.6 kJ → 2H2(g) + O2(g)

Enthalpy is often not included in the reaction
expression, instead as a ∆H value
◦ Example:
2H2(g) + O2(g) → 2H2O(g)
∆H = –483.6 kJ
◦ Negative ∆H is exothermic – releases heat
◦ Positive ∆H is endothermic – absorbs heat
Energy released, ∆H is negative.
Energy is absorbed, ∆H is positive.

molar enthalpy of formation: enthalpy change
that occurs when one mole of a compound is
formed from its elements in their standard
state at 25°C and 1 atm.
◦ Enthalpies of formation are given for a standard
temperature and pressure so that comparisons
between compounds are meaningful.
change in
enthalpy
H
0
f
standard
states
heat 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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
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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∆Hc : enthalpy change that occurs during the
complete combustion of one 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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

Review: ∆Hrxn energy absorbed or released
during a chemical reaction
Use 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.
This means that the energy difference
between reactants and products is
independent of the route taken to get from
one to the other.
*from enthalpy packet

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)

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)

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 = Σ[(ΔH0f of products) × (mol of products)]
– Σ[(ΔH0f of reactants) × (mol of reactants)]
*Need to use reference sheet for Hf values
**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.

Majority of spontaneous reactions are
exothermic
◦ Reactions proceed in a direction that leads to a
lower energy state

But, some spontaneous reactions are
endothermic
◦ Example: melting
(particles have more energy in the liquid state)
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So there are 2 factors that drive a reaction
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2 Factors: Enthalpy, but also Entropy
Entropy (S) : the degree of randomness or
disorder of the particles in a system
◦ Increases when a system goes from one state to
another (even if there is no enthalpy change)
2NH4NO3(s)  2N2(g) + 4H2O(l) + O2(g)

The arrangement of particles on the righthand side of the equation is more random
than the arrangement on the left side and
hence is less ordered.

There is a tendency of nature to go from
order to disorder
◦ Think about your bedroom, does it always stay neat
and organized? Or does it get messier and messier?
◦ Increase in disorder  Increase in entropy
◦ The entropy of the universe is ALWAYS increasing

How does entropy change with temperature?
As you go from a solid  liquid  gas
What happens to the entropy?
Entropy increases!
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As you go from a gas  liquid  solid
What happens to the entropy?
Entropy decreases
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What about when you dissolve a solute?
Or what about when you dilute a substance?
Entropy increases!
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Unit: kJ/(mol K) or J/mol K
Change in Entropy ΔS
◦ Difference between the entropy of the products and
the reactants
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
Increase in Entropy  +ΔS
Decrease in Entropy  -ΔS
(more disorder)
(less disorder)
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Calculate entropy like calculating enthalpy
from heats of formation, 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

Spontaneous Reactions:
◦ Least enthalpy (very negative #)
◦ Greatest entropy (large positive #)
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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: