Thermochemistry

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Determination of Specific Heat for a Metal
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3
Tf
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1
Ti
Lines 1 and 4 are horizontal, 2 is vertical, 3
follows the experimental points until intersects
2.
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Chapter 6 Energy and Chemical Reactions
RVCC Fall 2009 CHEM 103-12 General Chemistry I
Chemistry The Molecular Science, 3rd Ed. by
Moore, Stanitski, and Jurs
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Thermodynamics

• Thermodynamics heat changes, science of heat,
  work and the transformations of one to the other
• Thermochemistry the study of energy changes that
  occur during physical processes and chemical
  reactions

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Energy

• Energy the capacity to do work (w -Fd)
• Kinetic Energy is energy by virtue of motion.
  (mechanical, thermal, electrical)
• EK cT ½mv2
• Potential Energy is energy by virtue of position.
  (gravitational, electrostatic, chemical)
• EP magh

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Energy Units

• 1 Joule (J) ½ (2 kg) (1.0m/s)2
• 1match 2000J or 2kJ
• 1 cal 4.184 J
• 1 calorie the energy needed to raise the
  temperature
• of 1 gram of water from 14.5-15.5C
• 1 Cal (dietary calorie) 1000 cal 1 kcal

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Units of Energy
The label on this Australian packet of Equal
indicates that it supplies 16 kJ of nutritional
energy. How many American Calories are in
this packet of Equal?
1 cal 4.184 J 1Cal 1000 cal
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Energy

• The Law of Conservation of Energy (First Law of
  Thermodynamics) Energy may be converted from one
  form to another, but the total quantities of
  energy remain constant.

We will be studying the transfer of energy from
one form or one place to another. How is energy
transferred?
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Conservation of Energy 1st Law

Gravitational potential energy being converted
into kinetic energy.
heat
work
A single Fritos chip burns in O2
The chemical potential energy stored in food
(chemical bonds) being converted into thermal
(kinetic) energy.
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Work

• Work is one process that transfers energy to an
  object.
• For example
• separate a Na from a Cl- increases the
  potential energy of the object
• throwing a baseball increases the kinetic
  energy of an object

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Heat

• Heat is another process that transfers energy to
  an object.
• Heat refers to the energy transfer that occurs
  whenever two samples of matter at different
  temperatures (thermal or kinetic energy) are
  brought into contact.
• Energy always transfers from the hotter sample to
  the cooler sample until both are at the same
  temperature (in thermal equilibrium).

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Heat

• All matter consists of nanoscale particles that
  are constantly in motion (KMT Ch.1).
• Higher temperature ? faster motion of particles ?
  more thermal energy
• Raising the temperature of
• a sample increases the
• thermal (kinetic) energy
• of that sample.

So does CO2(g) and C12H22O11 (s) at room
temperature have the same kinetic energy?
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Specific Heat Capacity

• The specific heat capacity (c) is the heat
  required to raise the temperature of one gram of
  a substance by one degree Celsius.

q - heat absorbed or released c specific heat
for a substance DT change in temperature
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Specific Heat

Metals have low specific heats.
Water has high specific heat. (4.184 J/g?C
or 1.000 cal/g?C)
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Example 1

If a 10.0 g sample of each substance below has
250 J applied to it. Which substance will have
the greatest increase in temperature? iron (c
0.46 J/g?C) water (c 4.184
J/g?C) copper (c 0.39 J/g?C) aluminum (c
0.92 J/g?C)
? ? ? - constant
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Example 2 solve for c

• Calculate the specific heat of a substance if
  2.35 kJ of heat is needed to raise the
  temperature of 200. g of this substance from
  19.0ºC to 98.0ºC

? ? ? ?
? ? ?
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Example 3 solve for qloss

• How much energy will be loss when 500.0 g of iron
  cools from 55C to 22C?
• cFe 0.451 J g-1 C-1.
• How much energy will be loss when 500.0 g of gold
  cools from 55C to 22C?
• cAu 0.128 J g-1 C-1.

-7400J or -7.4kJ -2100J or 2.1kJ
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Example 4 solve for T
What will be the final temperature of a 50.00 g
silver ring at 37.0 ?C that gives off 25.0 J of
heat to its surroundings (c 0.235 J/g
?C)?
? ? ? ?
? ?
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Heat Transfer (gain loss)

• The hot steel bar transfers energy to the cool
  water until the two are at the same temperature
  (in thermal equilibrium).

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LAB! - Transfer of Kinetic Energy

• qloss,Fe qgain,H2O
• mFe cFe ?TFe mH2O cH2O ? TH2O

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Example 5 -qloss qgain , solve c

• A 19.6 g sample of metal was heated to 61.67 ºC
  and placed in 26.7 g of water in a coffee cup
  calorimeter. The temperature of water increased
  from 25.00 ºC to 30.00 ºC. What is the specific
  heat of the metal?

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Example 6 -qloss qgain , solve ?T

• A 200. g block of Fe heated in a bunsen
• burner flame is plunged into 1.00 kg of water
• (T 23.4 C) in an insulated container. The
  final equilibrium T of water and Fe is 33.4C.
  The specific heat of Fe is 0.448J/gC. What was
  the temperature of the flame?

heat lost by Fe heat gained by H2O
-(200.g)(0.448J/gC)(?Tm) (1000.g)(4.184
Jg-1C-1)(33.4- 23.4C) ?Tm (1000.g)(4.184
Jg-1C-1)(10.0C) -467C -(200.g)(0.448J/gC)
?Tm -467C33.4 C -Ti Ti
500.C
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From Applied Optics vol. 11, (14) 1972         
The temperature of Heaven can be rather
accurately computed. Our authority is Isaiah
3026,                     "Moreover, the light
of the Moon shall                     be as the
light of the Sun and the light                   
  of the Sun shall be sevenfold, as the
                    light of seven days."
         Thus Heaven receives from the Moon as
much radiation as we do from the Sun, and in
addition 77 (49) times as much as the Earth does
from the Sun, or 50 times in all. The light we
receive from the Moon is one 1/10,000 of the
light we receive from the Sun, so we can ignore
that ... The radiation falling on Heaven will
heat it to the point where i.e., Heaven loses 50
times as much heat as the Earth by radiation.
Using the Stefan-Boltzmann law for radiation,
(H/E)4 50 and the temperature of the earth
(300K), gives H (heaven) as 798K (525C). The
exact temperature of Hell cannot be computed ...
        However Revelations 218 says
                    "But the fearful, and
unbelieving ... shall                     have
their part in the lake which burneth
                    with fire and brimstone."
        A lake of molten brimstone (sulfur)
means that its temperature must be at or below
the boiling point, 444.6C. So Heaven (525C)
is hotter than Hell (445C)!
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Energy Transfer, ?H Enthalpy
q - heat, energy transfer (kinetic) that occurs
when two samples at different temperature are
brought into contact. H qp - enthalpy,
energy transfer (potential) that occurs when a
system changes physical state or undergoes a
chemical reaction.
stored in bonds
intermolecular bonds
intramolecular bonds
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Enthalpy, Heats of Reaction
As already stated

• An endothermic process is a chemical reaction or
  physical change in which heat is absorbed (qp is
  positive).
• An exothermic process is a chemical reaction or
  physical change in which heat is released (qp is
  negative).

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Phase Changes Endothermic and Exothermic

• Boiling

Endothermic qp lt 0 (-) net bond breaking

• Condensation

Exothermic qp gt 0 () net bond breaking
Which is worse? A burn from 100C water or 100C
steam?
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Changes of State, Enthalpy
Change Name value for H2O (J/g) solid ?
liq enthalpy of fusion (melting) 333 liq
? gas enthalpy of vaporization 2260
liq ? solid enthalpy of freezing -333 gas
? liq enthalpy of condensation
-2260 Note ?Hfusion - ?Hfreezing
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Energy Transfer, q and ?H
no change in temperature!
q ?H(qp) q and ?H
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Phase ChangesFreezing and Melting of Water
KE qmc?T
PE ?H
KE qmc?T
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Problem

• How much energy is used in converting 5.0g of ice
  at -50.C to water at 50.C?

?Hfusion 333 J/g cwater 4.184
J/gC cice 2.06 J/gC
Ice is melting. T remains at 0C
3200 J
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Enthalpy Change- Chemical Reactions
Endothermic
DH gt 0 ( )
net bond breaking
Exothermic
DH lt 0 ( - )
net bond making
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Bond Enthalpies

• During a chemical reaction
• Old bonds break requires E (endothermic)
• New bonds form releases E (exothermic)

Both typically occur net bond
breaking, -184 kJ/mol
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Thermochemical Equations

• A thermochemical equation is the chemical
  equation for a reaction in which the enthalpy of
  reaction for these molar amounts is written
  directly after the equation.

When 2 moles of ammonia are produced, the
reaction gives out (produces) 92.22 kJ of heat.
?Ho is the enthalpy change at the standard
pressure of 1 bar and a specified temperature.
When temperature is not specified, it is intended
to be 25oC
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WritingThermochemical Equations

• In a thermochemical equation it is important to
  note phase labels because the enthalpy change,
  DH, depends on the phase of the substances.
  

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WritingThermochemical Equations
Exothermic reaction
Can be written as
Heat is a product.
Endothermic reaction
Can be written as
Heat is a reactant.
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ManipulatingThermochemical Equations
Rule 1 When a thermochemical equation is
multiplied by any factor, the value of DH for the
new equation is obtained by multiplying the DH in
the original equation by that same factor.
Multiply by 1/2
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ManipulatingThermochemical Equations
Rule 2 When a chemical equation is reversed,
the value of DH is reversed in sign.

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Example 1
What is the enthalpy change for the following
reactions?
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Example 2

• Given the following thermochemical reaction
• N2(g) 3H2(g) ? 2NH3(g) DH -92.22kJ
• Write the thermochemical reaction for
• a.) Formation of 1 mol of ammonia gas
• b.) Decomposition of 6 moles of ammonia gas
• c.) Combination of 1 mole of hydrogen with a
  stoichiometric amount of nitrogen to produce
  ammonia gas

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• N2(g) 3H2(g) ? 2NH3(g) DH -92.22kJ

a.) Formation of 1 mol of ammonia gas
a.) 1/2 N2(g) 3/2 H2(g) ? NH3(g) DH
-46.11kJ
b.) Decomposition of 6 moles of ammonia gas
b.) 6NH3(g) ? 3N2(g) 9H2(g) DH 276.66kJ
c.) Combination of 1 mole of hydrogen with a
stoichiometric amount of nitrogen to produce
ammonia gas
c.) 1/3 N2(g) H2(g) ? 2/3 NH3(g) DH
-30.74kJ
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Heat StoichiometryInstant Cold Pack
?H 28.1kJ
ICYHOT? Instant Cold Pack contains 60.g of
NH4NO3. How many kJ of heat are absorbed by
the system (loss by the surroundings)?
2.1 kJ
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Heat StoichiometryInstant Heat Pack
?H -824 kJ
HEAT TREAT? Hand Warmer contains 2.5 g of Fe.
How many kJ of heat are released by the system
(gained by the surroundings)?
-18 kJ
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Enthalpy

• Enthalpy is a state function the enthalpy of a
  system depends only on its present state and not
  on the route by which it got there.

This allows us to apply laboratory measurements
to real-life situations. eg. The enthalpy change
of a chemical reaction is the same whether it
occurs in a lab or in your body.
Altitude is a state function.
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Hesss Law
?H for a reaction is the same whether it takes
place in a single step or several steps.

• If the equation for a reaction is the sum of the
  equations for two or more other reactions,
• then ?H for the 1st reaction must be the sum of
  the ?H values of the other reactions.

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Hesss Law

• Use Hesss Law to find ?H for unmeasured
  reactions.

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Hesss Law
Calculate ?H for the reaction 2C(graphite)
O2(g) ? 2CO(g)

• Rearrange

2C O2 CO2 2(-393.5)
-787.0 -12 CO O2 2 CO2 -1(-566.0) 566.0
or
Add, then cancel 2 C 2 O2 2 CO2 2 CO2 2
CO O2
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Hesss Law

• Determine ?H for the production of coal gas

Using
A B C
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Hesss Law

• Want 2 C(s) 2 H2O(g) CH4(g)
  CO2(g)

2 C 2 H2O ? CH4 CO2 ?H 15.3 kJ
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Heats of Reaction Calorimetry

• A calorimeter is a device used to measure the
  heat absorbed or evolved during a physical or
  chemical change.
• Whatever heat is produced during a chemical
  reaction, it is absorbed by the calorimeter.
• Whatever heat is absorbed by the chemical
  reaction, it will be supplied by the calorimeter
• Total heat, produced or absorbed remains inside
  the system

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Calorimetry

• When reactions take place in solution, it is
  easier to use a calorimeter that is open to the
  atmosphere.
• This allows for the direct measurement of ?H (q
  at constant pressure).
• An easy way to do this is with a coffee-cup
  calorimeter.

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Coffee Cup Calorimeter

• qsolution -?Hreaction
• The thermal transfer to the solution is equal to
  the thermal transfer from the reaction.
• The masses of other substances in the solution
  are so small compared to the mass of the solvent
  water that their heat capacities can usually be
  ignored (c ? 4.184 J/g?C)

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Example 1

• When a 13.0 g sample of NaOH (s) dissolves in
  400.0 g water in a coffee cup calorimeter, the
  temperature of the water changes from 22.6 to
  30.7 ?C. Calculate
• The heat transfer from system to surroundings.
• ?H for the reaction NaOH (s) ? Na (aq) OH-
  (aq)

qsolution -DHrxn
qsolution c m DT
qsolution 4.184 J/gºC 413.0g 8.1ºC 13997J
1.4 104J (14kJ)
-DHrxn 14kJ/13.0g NaOH
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Calorimetry

• When at least one of the reactants or products is
  a gas, a bomb calorimeter is used because it is
  closed to the atmosphere at constant volume.
• qreaction qbomb qwater0
• - qreaction qbomb qwater
• qreaction ?Ereaction

From the bomb calorimeter we evaluate ?Ereaction.
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Defining the System

• When analyzing thermodynamics, it is crucial to
  keep track of all energy transfers (work and
  heat) taking place
• It is practical to define the sample of matter
  that you are investigating as the system.
• (eg. chemicals in a flask, the coffee in your
  coffee cup,
• my textbook)
• Anything that can exchange energy with the system
  is defined as the surroundings. ( rest of the
  universe, or as much as needed)
• (e.g. the flask.perhaps the flask and this
  classroom,perhaps the flask and all of the
  building).

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System vs. Surroundings

• The total amount of energy must remain constant,
  so any energy transferred out of the system must
  be transferred into the surroundings, and vice
  versa.

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Thermodynamic Changes

• We will refer to ?E (change in energy) when
  keeping track of energy transfers.
• ?Esystem Efinal - Einitial
• The sign of ?E indicates the direction of
  transfer.
• ?Esystem gt 0 Energy transferred into the
  system.
• - Final E greater than initial E
• ?Esystem lt 0 Energy transferred out of the
  system.
• - Final E lower than initial E

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Thermodynamic Changes

• Indicating with q the heat and w the work, in
  most cases
• ?Esystem qsystem wsystem
• qsystem heat transferred to or from the
  system
• wsystem work done on or by the system
• q gt 0 heat absorbed by system
• q lt 0 heat given off by system
• w gt 0 work done on the system
• w lt 0 work done by the system

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Exercise 6.3

• It takes about 1.5kJ to raise the temperature of
  a can of Classic Coke from 25.0-26.0C. You put
  the can of Coke into a refrigerator to cool it
  from room temperature (25C) to 1C.
• What quantity of heat transfer is required?
• What is a reasonable choice of system?
• What constitutes the surroundings?
• What is the sign of ?E?