Energy and
Thermochemistry
1
Energy
The ability to do work
 2 types

Potential: stored energy
 Kinetic: energy in motion

2
Thermochemistry
Changes of heat content and heat transfer
 Follow Law of Conservation of Energy
 Or, 1st Law of Thermodynamics


Energy can neither be created nor destroyed
 Thus,
total energy of universe constant
3
Temperature & Heat

Heat not same as temperature


Heat = energy transferred to one system by another
due to temperature difference
Temperature = measure of heat energy content &
ability to transfer heat



Thermometer
Higher thermal energy, greater motion of
constituents
Sum of individual energies of constituents = total
thermal energy
4
Systems and Surroundings




System = the object in question
Surrounding(s) = everything outside the system
When both system and surrounding at same
temperature  thermal equilibrium
When not




Heat transfer to surrounding = exothermic
(you feel the heat)  hot metal!
Heat transfer to system = endothermic
(you feel cold)  cold metal!
5
Math!

Joules (J) used for
energy quantities


But usually kJ (1000 J)
used
Ye Royal Olde School
used calorie (cal)

cal = amt of heat required
to raise the temperature of
1.00 g of water by 1C


1 kg  m
Joule (J) =
2
s
1 cal = 4.184 J (SI-unit)
But…Calorie (Cal) = 1000
cal

Used in nutrition science
and on food labels
6
2
Heat Capacity

Specific heat capacity

Quantity of heat
required to raise the
temp of 1 gram of any
substance by 1 K
J
C=
gK

Molar heat capacity

Quantity of heat
required to raise the
temp of 1 mole of any
substance by 1 K
J
c=
mol  K
4.184 J
specific heat capacity of water =
75.4 J
g  K molar heat capacity of water =
mol  K
7
Calculating heat transfer
Q = C  m  T
Q = transferred heat, m = mass of substance, T = temperature change

FYI

Specific heat capacity
of metals is very low


 < 1.000 J/(gK)
What does this tell us
about heat transfer in
metals?
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Let’s do an example

In your backyard, you have a swimming
pool that contains 5.19 x 103 kg of water.
How many kJ are required to raise the
temperature of this water from 7.2 °C to
25.0 °C?
9
Example solved
Q = C  m  T = (4.184

J
)  (5.19 x 106 g)  (298.2 K - 280.4 K) = 3.87  108 J = 3.87  105 kJ
gK
Trick:  T in K = T in °C
10
Practice

How many kJ are required to raise the
temperature of 25.8 g of quicksilver from
22.5 °C to 28.0 °C? CHg = 0.1395 J/(gK)
11
But what if there’s a change of
state?

Temperature constant throughout
change of state

Added energy overcomes inter-molecular
forces
12
Change of state

What do the flat areas represent?
13

qtot = qs + qsl + ql + qlg + qg

qsl = heat of fusion

Heat required to convert solid at melting pt. to liq


Ice = 333 J/g
qlg = heat of vaporization

Heat required to convert liq. at boiling pt. to gas

Water = 2256 J/g
14
Practice
How much heat is required to vaporize
250.0 g of ice at -25.0 °C to 110.0 °C?
 Given:

Specific heat capacity of ice = 2.06 J/gK
 Specific heat capacity of water = 4.184 J/gK
 Specific heat capacity of steam = 1.92 J/gK


Let’s do this
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Calorimetry






The process of measuring heat transfer in
chemical/physical process
qrxn + qsoln = 0
qrxn = -qsoln
Rxn = system
Soln = surrounding
What you’ll do in lab



Heat given off by rxn
Measured by thermometer
Figure out qrxn indirectly
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Enthalpy
= heat content at constant pressure
 If H = “+”, process endothermic
 If H = “-”, process exothermic
 Enthalpy change dependent on states of
matter and molar quantities
 For example:

Is vaporizing ice an exothermic or
endothermic process?
 Thus, will H be “+” or “-”?

17
Hess’s Law
If a rxn is the sum of 2 or more other
reactions, H = sum of H’s for those rxns
 So, Htot = H1 + H2 + H3 + … + Hn

18
Let’s solve a problem




C(s) + 2S(s)  CS2(l); H = ?
Given:
C(s) + O2(g)  CO2(g); H = -393.5 kJ/mol
S(s) + O2(g)  SO2(g); H = -296.8 kJ/mol

CS2(l) + 3O2(g)  CO2(g) + 2SO2(g); H = -1103.9 kJ/mol

How do we manipulate the 3 rxns to achieve the
necessary net rxn?
Does H change if the rxns are reversed and/or
their mole ratios are changed?
Let’s talk about this on the next slide


19
Let’s work it out
1. Switch this rxn: CS2(l) + 3O2(g)  CO2(g) + 2SO2(g); H = -1103.9 kJ
Thus, CO2(g) + 2SO2(g)  CS2(l) + 3O2(g) ; H = + 1103.9 kJ
Thus, -Hfwd = +Hrev
2. Double this rxn: S(s) + O2(g)  SO2(g); H = -296.8 kJ
Thus, 2S(s) + 2O2(g)  2SO2(g); H = (-296.8 kJ) x 2 = -593.6 kJ
Since H is per mole, changing the stoichiometric ratios entails an equivalent change in H
3. Keep this rxn: C(s) + O2(g)  CO2(g); H = -393.5 kJ
4. Add those on same side of rxns/eliminate those on opposite sides of rxn:
CO2, 2SO2, 3O2
5. Net rxn: C(s) + 2S(s)  CS2(l)
H = 1103.9 kJ - 593.6 kJ – 393.5 kJ = 116.8 kJ
Is it an exo- or endothermic rxn?
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Practice
Given:
 CH4(g)  C(s) + 2H2(g); H = 74.6 kJ/mol
 C(s) + O2(g)  CO2(g); H = -393.5 kJ/mol
 H2(g) + O2(g)  H2O(g); = -241.8 kJ/mol
 CH4(g) + 2O2(g)  CO2(g) + 2H2O(g); Hrxn = ?

21
Standard Energies of Formation

Standard molar enthalpies of formation = Hf = enthalpy
change for formation of 1 mol of cmpd directly from
component elements in standard states

Standard state = most stable form of substance in physical state
that exists @ 1 bar pressure & a specific temp., usually 25C
(298K)

1 bar = 100kPa




101.325 kPa = 1 atm
 So 1 bar  1 atm (an SI unit)
Example
C(s) + O2(g)  CO2(g); H = Hf = -393.5 kJ
Hf = 0 for elements in standard state
22
Enthalpy Change for a Rxn
Must know all std molar enthalpies
 Hrxn = Hfprods - Hfreactants
 Given to you in a table in the back of the
book
 Again, keep in mind the mole ratios for
each species involved!

23
Example

Hrxn for 10.0 g of nitroglycerin?

2C3H5(NO3)3(l)  3N2(g) + ½O2(g) + 6CO2(g) + 5H2O(g)
C3H5(NO3)3(l) = -364 kJ/mol
 CO2(g) = -393.5 kJ/mol
 H2O(g) = -241.8 kJ/mol

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Solution
1)H rxn = [3 mol  0
kJ
kJ
kJ
kJ
+ 1 mol  0
+ 6 mol  -393.5
+ 5 mol  -241.8
]
2
mol
mol
mol
mol
kJ
] = -2842 kJ
mol
1mol -2842kJ
2)10.0 g 

 62.6kJ
227.2g
2mol
- [2 mol  -364
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Practice
Determine H°rxn for:
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
 Given:
 NH3(g) = -45.9 kJ/mol
 NO(g) = 91.3
 H2O(g) = -241.8

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