Thermochemistry

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Thermochemistry
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THERMOCHEMISTRY
The study of heat released or required by
chemical reactions
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• Energy is the capacity to do work there are
  many types of energy!
• Thermal energy is the energy associated with the
  movement of molecules in a substance it is the
  same as kinetic energy!
• Chemical energy is the energy stored within the
  bonds of chemical substances
• Nuclear energy is the energy stored within the
  collection of neutrons and protons in the atom
• Electrical energy is the energy associated with
  the flow of electrons
• Potential energy is the energy available by
  virtue of an objects position

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Two main general forms of energy

• Energy is measured in the standard unit of
  Joules
• 1 J 1 kg m2/s2
• mass must be in kg
• velocity must be in m/s
• height must be in meters
• g acceleration due to gravity must be in m/s2

Potential energy (EP) mgh
Kinetic energy (EK) ½ mv2
Energy due to position (stored energy)
Energy due to motion
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Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between
two bodies that are at different temperatures.
Temperature is an indirect measurement of the
thermal or heat energy.
Temperature is NOT heat Energy
400C
greater temperature But less heat energy
There is a greater amount of Heat energy in a
bathub at 40 degrees Than in a coffee cup at 90
degrees!
900C
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UNITS OF ENERGY
S.I. unit of energy is the joule (J) Heat and
work ( energy in transit) also measured in
joules The calorie (cal) is another metric unit
for energy 1 cal 4.184 J A Food calorie,
with a capital C is equal to a 1000 chemistry
calories 1 Food Calorie (1 Calorie) 1000
calories A Candy bar with 480 Calories actually
contains 480,000 chemistry calories!
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The specific heat (C) of a substance is the
amount of heat (q) required to raise the
temperature of one gram of the substance by one
degree Celsius.
To measure the Heat (q) absorbed or released by
a substance
q m C Dt
Q heat absorbed or released m mass of
substance C specific heat of substance Dt
tfinal - tinitial
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How much heat is given off when an 869 g iron bar
cools from 940C to 50C?
C of Fe 0.444 J/g 0C
Dt tfinal tinitial 50C 940C -890C
q mCDt
869 g x 0.444 J/g 0C x 890C
-34,000 J
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Exothermic process is any process that gives off
heat transfers thermal energy from the system
to the surroundings.
Endothermic process is any process in which heat
has to be supplied to the system from the
surroundings.
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Fireworks exploding is an exothermic reaction
Photosynthesis is an endothermic reaction
(requires energy input from sun)
Ice melting is an endothermic process!
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Enthalpy (H) is used to quantify the heat flow
into or out of a system in a process that occurs
at constant pressure. (comes from Greek for heat
inside)
DH H (products) H (reactants)
DH heat given off or absorbed during a reaction
at constant pressure
Hproducts lt Hreactants
Hproducts gt Hreactants
DH -
DH
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Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH is positive!
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
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Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH is negative!
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
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Heat, or DH, can be written as part of a chemical
reaction!
If the reaction has a positive DH, then the
reaction needs heat, and heat is written on the
left side of the arrow as a reactant If the
reaction has a negative DH, then the reaction
releases heat, and heat is written on the left
side of the arrow as a product
DH -2043 kJ
or
C3H8 (g) 5 O2 (g) 3 CO2 (g) 4H2O (g)
2043 kJ
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We can treat heat, then, like a reactant or
product. And perform stoichiometry problems
How many kJ of heat are released when 355 grams
of propane are burned with excess oxygen?
355 grams C3H8 x 1 mole C3H8 8.07 moles
C3H8
44 grams 8.07 moles C3H8 x
2043 kJ heat 16,483.3 kJ heat released
1 mole
C3H8
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How do we measure the heat of a reaction in an
experiment?

• There are three ways
• Using a calorimeter
• Using Hess Law
• Using a table of heats of formation
• Lets look at each method.

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1. Using a calorimeter

• A calorimeter is an insulated device used to
  capture all of the heat either absorbed or
  released by a reaction!
• The reaction is usually surrounded by
  water.why?
• Water is stable, and has a high specific heat
• It changes temperature slowly!
• q reaction - q surroundings
• By measuring the heat that the water absorbs or
  releases, we can calculate the heat of the
  reaction!

No heat enters or leaves!
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A 0.1964-g sample of solid quinone (C6H4O2) is
burned in a bomb calorimeter that contains 373
grams of water. The temperature of the water
inside the calorimeter increases by 3.2C.
Calculate the energy of combustion of quinone per
mole.

• First write a balanced chemical equation!
• 1 C6H4O2 (s) 6 O2 (g) ? 6 CO2 (g) 2 H2O
  (g)
• The heat released by the reaction is absorbed by
  the calorimeter
• q reaction - q calorimeter
• q mcDT
• q (373 grams H2O)(4.184 J/g0C)(3.2C)
  4,994.02 J gained by calorimeter
• q reaction -4,994.02 J (released by reaction)

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This is not the DH, though!

• 1 C6H4O2 (s) 6 O2 (g) ? 6 CO2 (g) 2 H2O
  (g)
• The change in heat, or DH, is the energy released
  for the reaction the way it was written!
• We only used .1964 grams of the chemical!
• The reaction calls for one mole of the chemical!
• 1 mole C6H4O2 (s) 108 grams
• So I set up a ratio
• -4,994.02 J/.1964 g X/108 g
• X -2,746,202.4 J -2746.2024 kJ
• So, DH -2700 kJ (2 significant figures)

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2. Using Hess Law
What if a reaction is too costly or dangerous to
conduct, but we still want to calculate its DH?
4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g)
We can use algebra to manipulate other reactions
to look like the desired reaction making sure
we change the energies as well!
This is called Hess Law
N2(g) O2(g) ? 2NO(g) ?H 180.6 kJ
N2(g) 3H2(g) ? 2NH3(g) ?H -91.8 kJ
2H2(g) O2(g) ? 2H2O(g) ?H -483.7 kJ
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Goal
4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g)
Using the following sets of reactions
N2(g) O2(g) ? 2NO(g) ?H 180.6 kJ
N2(g) 3H2(g) ? 2NH3(g) ?H -91.8 kJ
2H2(g) O2(g) ? 2H2O(g) ?H -483.7 kJ
4NH3 ? 2N2 6H2 ?H 183.6 kJ
NH3
Reverse and x 2
Any chemical in more than one reaction - skip
O2
2N2 2O2 ? 4NO ?H 361.2 kJ
NO
x2
H2O
x3
6H2 3O2 ? 6H2O ?H -1451.1 kJ
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Goal
4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g)
4NH3 ? 2N2 6H2 ?H 183.6 kJ
NH3
Reverse and x2
O2
NO
x2
2N2 2O2 ? 4NO ?H 361.2 kJ
H2O
x3
6H2 3O2 ? 6H2O ?H -1451.1 kJ
Cancel terms and take sum.
5O2
?H -906.3 kJ
6H2O
4NH3
?
4NO
Is the reaction endothermic or exothermic?
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Determine the heat of reaction for the
reaction C2H4(g) H2(g) ? C2H6(g) Use
the following reactions C2H4(g) 3O2(g)
? 2CO2(g) 2H2O(l) ?H -1401 kJ C2H6(g)
7/2O2(g) ? 2CO2(g) 3H2O(l) ?H -1550
kJ H2(g) 1/2O2(g) ? H2O(l) ?H -286
kJ
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Determine the heat of reaction for the
reaction Goal C2H4(g) H2(g) ?
C2H6(g) ?H ? Use the following
reactions C2H4(g) 3O2(g) ? 2CO2(g)
2H2O(l) ?H -1401 kJ C2H6(g) 7/2O2(g) ?
2CO2(g) 3H2O(l) ?H -1550 kJ H2(g)
1/2O2(g) ? H2O(l) ?H -286 kJ
C2H4(g) use 1 as is C2H4(g) 3O2(g) ?
2CO2(g) 2H2O(l) ?H -1401 kJ H2(g)
3 as is H2(g) 1/2O2(g) ?
H2O(l) ?H -286 kJ C2H6(g) rev 2
2CO2(g) 3H2O(l) ? C2H6(g) 7/2O2(g)
?H 1550 kJ
C2H4(g) H2(g) ? C2H6(g) ?H -137 kJ
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3. Using Heats of Formation Tables
A. Heat of Formation (?Hfº) the heat released
or absorbed when one mole of a substance is
formed from its elements.
EX H2(g) ½ O2(g) ? H2O(l) ?Hfº -289 kJ
The reactant elements are in their standard
state their most stable form at 25 ºC and 1
atm. This is usually indicated with a 0 by the
?Hf.
The ?Hfº of an element in its standard state is
zero. You Cannot make an element from elements!
The heat of formation for a substance is like
having its Potential energy it is a
measurement of how stable or Unstable it is!
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How do we use the table to figure out the ?H For
a reaction?
The ?H of a rxn is equal to the sum of the ?Hfºs
ofthe products minus the sum of the ?Hfºs of
the reactants.
(Each products or reactants ?Hfº must be
multiplied by its coefficient.)
?Hrxn S ?Hfº(products) S ?Hfº(reactants)
YOU MUST PRINT OFF THE TABLE FROM MY WEBSITE
IT WAS NOT INCLUDED IN YOUR PACKET!
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1. Calculate the ?H for 2 Na(s) 2 H2O(l) ?
2 NaOH(aq) H2(g) ?H ?
?Hrxn S ?Hfº(products) S ?Hfº(reactants) ?Hrx
n (2)(NaOH(aq)) (1)(H2(g)) (2)(Na(s))
(2)(H2O(l))
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1. 2 Na(s) 2 H2O(l) ? 2 NaOH(aq) H2(g)
?H ? ?Hrxn S ?Hfº(products) S
?Hfº(reactants) ?Hrxn (2)(NaOH(aq))
(1)(H2(g)) (2)(Na(s)) (2)(H2O(l)) ?Hrxn
(2)(-469.6 kJ) (1)(0 kJ) (2)(0 kJ)
(2)(-285.84 kJ) ?Hrxn -367.52 kJ Notice that
the table from my website has some values listed
twice and they are slightly different you
might get slightly differing answers based on the
values that you use that is fine!
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2. Calculate the ?H for C2H5OH (l)
3 O2(g) ? 2 CO2(g) 3 H2O(l) ?H ?
?H -1366.89 kJ