Transcript Thermochemistry

Thermochemistry
Thermochemistry**

The study of the changes in heat
energy that accompany chemical
reactions and physical changes.
As a group

On your index card Define:
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Heat
Temperature
Thermal Equilibrium
Heat**
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a form of energy (properly termed
internal energy)
depends on the amt of motion
increases as the particles move
faster
total KE of the particles

ex. rub hands together (quickly), slide
down rope quickly (rope burn)
Heat vs. Temperature
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How do you detect internal energy?
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temperature allows us to detect heat
HEAT ≠ TEMPERATURE
(are related)
Temperature**
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measure of the average KE of
molecules
increases the faster the molecules
are moving
thermometer is instrument
Mercury
red alcohol
temp scales o F, oC and K
Thermal Equilibrium
Thermal Equilibrium**


state in which two
bodies in contact
with each other
have identical
temperatures
basis for measuring
temperature with
thermometers
Heat
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
**Heat cannot be
measured directly –
only indirectly by  in
temp.
a change in
temperature indicates
the transfer of energy
between substances by
heat.
Heat

the transfer of energy between objects at
different temps.
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an increase in temp. indicates the addition of energy
a decrease in temp. indicates the removal of energy
**Symbol Q
**Joule(SI) or calorie (common for food) = unit
to measure heat
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How is the energy of food listed? Note the unit for
cal= C
1 Calorie = 1000 calories = 1kcal
1cal = 4.186J
calorimeter is instrument used to “measure” heat
Heat

Calorimetry is used to
determine the heat
released or absorbed in
a chemical reaction.
The calorimeters
shown here can
determine the heat of
a solution reaction at
constant (atmospheric)
pressure.
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Does the addition of heat insure an
increase in temperature? No, a phase
change is possible. If heat is
removed does that insure that the
temperature decreases? No, a phase
change is possible.
Questions to consider?

Why would food be measured in “C”
calories?
Questions to consider?
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Why does your mother make you keep
the thermometer in you mouth for at
least 3 mins.?.
I need a volunteer…
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Heat
Temperature
Thermal equilibrium
Questions to consider?
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Why do we put
warm drinks into
ice?
How does the ice
cool the drinks?
Ice
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Predict the
temperature of the
ice?
Is this possible?
Why?
Ouch!!
Ice
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Make a prediction
as to the
temperature that
this ice will melt.
(ie. If placed on a
hot plate will it
begin melting
immediately?)
Internal Energy Changes

When a
substance is
heated, the
energy of its
particles is
increased.
Internal Energy Changes
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If the potential energy
changes the physical
state of the substance
will change.
if Potential Energy increases: sl, lg, or sg
Internal Energy Changes

If the kinetic
energy increases
the temperature
of the substance
increases.
Changes of State

The changes of state from solid to
liquid and liquid to solid takes place at
the same temperature
for water
Melting point
 Freezing point
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__?__
__?__
Changes of State

The changes of state from liquid to
gas and gas to liquid take place at the
same temperature
for water
Boiling point
__?__
 Condensing point
__?__
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Changes of State
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The amount of heat needed for the
change depends on the particular
substance.
Q = m(Hf)
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m=mass;
Hf= heat of fusion
Solid/liquid
Q = m(Hv)
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m=mass
Hv= heat of vaporization
Liquid/Gas
Questions to Consider?
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When food was stored in cellars,
during the winter, people would often
place an open barrel of water in the
cellar alongside their produce.
Explain why this was done and why it
would be effective.
Phase Change Equations**
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Q = m(Hf)
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Hf = 334J/g for water**
Q = m(Hv)
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Hv = 2260J/g for water**
Make sure units cancel !!!
Sample problems
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Determine the energy change involved
in converting 16.2 grams of ice to
liquid water, both at 0oC.
Q = (16.2g)(334J/g)
= 5410 J
energy is absorbed
Sample problems
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Determine the energy change involved in
converting 5.8 grams of water to steam,
both at 100oC.
Q = (5.8g)(2260J/g)
= 1.3E4J
energy is absorbed
Why does it require so much more energy to go to a
gas?
Sample problems
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Determine the energy change involved
to:
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Convert 98.2 grams of water to ice, both
at 0oC.
Convert 52.6 grams of steam to water,
both at 100oC.
Sample problems
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Determine the energy change involved
to:
 Convert 98.2 grams of water to ice
at 0oC.
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Q=(98.2g)(334J/g) = 32800J
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energy is released = -32800J
Sample problems
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Determine the energy change involved
to:
 Convert 52.6 grams of steam to
water at 100oC.
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Q = (52.6g)(2260J/g)=1.19E5J
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Energy is released = -1.19E5J
Phase change graph
When temperature does
change
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Frequently when energy is added to a
substance the temperature does
increase.
Questions to consider?
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Can you give me an example of two
things which are exposed to the same
energy, yet have different
temperatures?
Changes in Temperature
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**Specific heat capacity – energy
required to Δ the temp of 1g of that sub
by 1oC
Relates heat, mass, and temp Δ
***Q = mCΔT***
Equation applies to both subs that absorb
energy and those that lose energy
When temp increases ΔT and Q are
positive
Temp decrease ΔT and Q are neg.
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Note = ice, water, and steam have different
specific heat capacities
Sample problems
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Hypothermia can occur if the body temperature
drops to 35.0°C, although people have been known
to survive much lower temperatures. On January
19, 1985, 2-year-old Michael Trode was found in
the snow near his Milwaukee home with a body
temperature of 16.0°C. If Michael's mass was
10.0 kg, how much heat did his body lose, assuming
his normal body temperature was 37.0°C?
(Happily, Michael survived!)
Chuman body =3.47 J/g°C
Sample problems
Q= (10 000g)(3.47 J/g°C)(16.0°C - 37.0°C )
= - 728700 J
= - 7.29 E5 J
Sample problems
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Determine the energy required (in
kilojoules) when cooling 456.2 grams
of water at 89.2 °C to a final
temperature of 5.9 °C
Sample problems
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Determine the energy required (in
kilojoules) when cooling 456.2 grams
of water at 89.2 °C to a final
temperature of 5.9 °C
Q =(456.2g)(4.18J/g°C)(5.9°C-89.2°C)
= - 158846.1 J
= - 1.59 E5 J
Sample problem
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Determine the energy released
when converting 500.0 g of ice at
-25.0 °C to steam at 110.0 °C.
Sample problem: Steps
1.
2.
3.
4.
5.
Heat solid to melting point.
Melt solid.
Heat liquid to boiling point.
Boil liquid.
Heat gas to required temperature.
Sample problem: Steps
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Q = (500.0g)(2.05J/goC)(0- -25.0oC)
25625J
Q = (500.0g)(334J/g)
167000J
Q =(500.0g)(4.18J/goC)(100.0oC-0oC)
209000J = 2.090E5
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Q =(500.0g)(2260J/g)
1130000J
Q =(500.0g)(2.02J/goC)(110.0oC-100.0oC)
10100J
Sample problem: Steps
Q = 25625J + 167000J + 209000J +
1130000J + 10100J
Q = 1541725J =1542000J
(thousands is least significant)
Determining specific heat
capacity
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If a hot sub. is placed in insulated container of
cool water, energy conservation requires that
the energy the sub gives up must equal the
energy absorbed by the water.
energy absorbed by water = energy released
by the substance
Cw*mw*ΔTw = Cx*mx*ΔTx
energy gained is positive, energy released is
negative
Sample problem
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Emily is testing her baby's bath water and
finds that it is too cold, so she adds some
hot water from a kettle on the stove. If
Emily adds 2.00 kg of water at 80.0°C to
20.0 kg of bath water at 27.0°C, what is
the final temperature of the bath water?
Sample problem
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Emily is testing her baby's bath water and
finds that it is too cold, so she adds some
hot water from a kettle on the stove. If
Emily adds 2.00 kg of water at 80.0°C to
20.0 kg of bath water at 27.0°C, what is
the final temperature of the bath water?
(2000g)(4.18J/g°C)(Tf-80.0°C) =
(20000g)(4.18J/g°C)(Tf -27.0°C)
Sample problem
- (2000g)(4.18J/g°C)(Tf-80.0°C)=(20000g)(4.18J/g°C)(Tf -27.0°C)
-(2000g)(4.18J/g°C)(Tf-80.0°C)=(20000g)(4.18J/g°C)(Tf -27.0°C)
-2000gTf – (-160000goC) = 20000gTf – 540000goC
+2000gTf
+540000goC +2000gTf +540000goC
700000goC = 22000gTf
22000g = 22000g
31.8 oC = Tf
Pressure and phase change
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Not only can temperature changes
cause phase changes, but pressure
changes can also cause phase changes.
Triple point graph
Heat of Reaction
Heat of Reaction


the quantity of heat released or
absorbed during a chemical reaction.
difference between stored energy of
the reactants and products.
Heat of Reaction
CH4
+ 2O2
 CO2 +
2H2O
H
H–C-H + 2O=O  O=C=O +2H-O
H
H
4 (C-H)
2(O=O)2(C=O) 4(H-O)
Heat of Reaction
H
H–C-H +
H
2O=O  O=C=O +2H-O
H
4 (C-H) + 2(O=O) = energy to break bonds (+)
2(C=O) + 4(H-O) = energy to make bonds(-)
Heat of Reaction
H
H–C-H +
H
2O=O  O=C=O +2H-O
H
4(C-H) + 2(O=O) =4(414kJ) + 2(499kJ)= 2654kJ
2(C=O) + 4(H-O) =2(799kJ) + 4(460kJ)= - 3438kJ
The sum of these two values is –784kJ.
Heat of reaction
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While bond energy calculations give
an estimate of heats of reaction; it is
possible to measure the energy
change during a reaction in the lab.
Heat of reaction
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**Heat that is absorbed or released during a
chemical rxn at const pressure is represented
by ΔH
**H = enthalpy – the heat content of a system
at const press.
Δ H=-q (per mole)
Heat of reaction
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Only changes in enthalpy can be measured.
Therefore, Δ H is the amount of heat
absorbed or lost by a system during a
process at constant pressure.
Δ H = Hproducts - Hreactants
Heat of reaction
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A thermometer
can be used to
follow heat flow
during the
reaction.
Heat of reaction
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**If the temperature increases – heat is
flowing from the system into the
thermometerExothermic**
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Δ Hrxn is negative (-)
q is positive(+)
Heat of reaction
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**If the temperature decrease – heat is
flowing into the system from the
thermometer  Endothermic**
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Hrxn is positive (+)
q is negative (-)
Thermochemical equations
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equation that includes the quantity of
heat released or absorbed during the
reaction as written.
ex: 2H2(g) + O2(g)  2H2O(g) + 483.6kJ
More common notation:
2H2(g) + O2(g)  2H2O(g) Δ H=- 483.6kJ
Thermochemical equations
2H2(g) + O2(g)  2H2O(g) + 483.6kJ
2H2(g) + O2(g)  2H2O(g) Δ H=- 483.6kJ
Note: these are exothermic. Whether
written as a product or as (-) Δ H.
Key points with
Thermochemical equations
1.
2.
3.
4.
Coefficient is moles not molecules.
physical state of each substance
required.
Energy is directly proportional to
number of moles present.
Δ H not significantly influenced by
temperature.
Molar Heat of formation
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heat released or absorbed when one
mole of a compounds is formed by
combination of its elements.
symbol, Δ Hof
Enthalpy
standard state, 1atm and 25oC (room
temperature.
symbol, o (means standard state)
Molar Heat of formation
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Large (-) ΔH  stable compound
Elements  ΔH = 0
Ex: CO2 ΔH = -393.5kJ/mol  CO2 is
more stable than the elements C and O2
indivuidually.
Appendix A=14
If heat of formation is (+) or slightly (-)
the compound is unstable (explosive).
Heat of Combustion

The heat released by the complete
combustion of one mole of a substance.
ΔHc
Appendix A=5
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Can be written as a thermochemical equation
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C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) + 2220kJ
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
ΔHc= 2220kJ
Questions to Consider?

Comment on the stability of each of
the substances listed.
H (kJ/mol)
Al2O3(s)
CaCO3(s)
NO(g)
O3(g)
-1676.0
-1206.92
90.29
142.7
Potential Energy Diagram
Hess’s law

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The overall enthalpy change in a
reaction is equal to the sum of
enthalpy changes for the individual
steps in the process.
this is the formal definition that
explains why
Δ H = Hproducts - Hreactants
Hess’s law example

C(s) + 2H2(g)  CH4(g)

given:
C(s) + O2(g)  CO2(g)
H2(g) + ½ O2(g)  H2O(l)
Δ Hof = ?
Δ Hoc=-393.5kJ/mol
Δ Hoc=-285.8kJ/mol
CH4(g) + 2O2(g)  CO2(g) + 2 H2O(l)
Δ Hoc=-890.8
Hess’s law problem rules
1.
2.
If a reaction needs to be reversed, the
sign of Δ Ho= is also reversed.
Multiply the coefficients of the known
equation so that when added together
they give the desired thermochemcial
equation.
Hess’s law example

C(s) + 2H2(g)  CH4(g)
Δ Hof = ?
Notice: methane is on the “wrong” side so it
needs to be reversed.
CO2(g) + 2 H2O(l)  CH4(g) + 2O2(g) Δ Hoc=+890.8
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Hess’s law example
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C(s) + 2H2(g)  CH4(g)
Δ Hof = ?
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Notice: hydrogen needs a coefficient of 2
2H2(g) + 1 O2(g)  2 H2O(l) Δ Hoc=2(-285.8kJ/mol)
Hess’s law example

C(s) + 2H2(g)  CH4(g)

given:
Δ Ho f = ?
C(s) + O2(g)  CO2(g)
Δ Hoc=-393.5kJ/mol
2H2(g) + 1 O2(g)  2 H2O(l) Δ Hoc=-571.6kJ/mol
CO2(g) + 2 H2O(l)  CH4(g) + 2O2(g) Δ Hoc=+890.8
Hess’s law example

C(s) + 2H2(g)  CH4(g)

given:
Δ Ho f = ?
C(s) + O2(g)  CO2(g)
Δ Hoc=-393.5kJ/mol
2H2(g) + 1 O2(g)  2 H2O(l)
Δ Hoc=-571.6kJ/mol
CO2(g) + 2 H2O(l)  CH4(g) + 2O2(g) Δ Hoc=+890.8
Hess’s law example

C(s) + 2H2(g)  CH4(g)

given:
Δ Ho f = ?
C(s) + O2(g)  CO2(g)
Δ Hoc=-393.5kJ/mol
2H2(g) + 1 O2(g)  2 H2O(l) Δ Hoc=-571.6kJ/mol
CO2(g) + 2 H2O(l)  CH4(g) + 2O2(g) Δ Hoc=+890.8
Hess’s law example

C(s) + 2H2(g)  CH4(g)

given:
Δ Ho f = ?
C(s) + O2(g)  CO2(g)
Δ Hoc=-393.5kJ/mol
2H2(g) + 1 O2(g)  2 H2O(l) Δ Hoc=-571.6kJ/mol
CO2(g) + 2 H2O(l)  CH4(g) + 2O2(g) Δ Hoc=+890.8
Hess’s law example

C(s) + 2H2(g)  CH4(g)
Δ Hof = ?
given:
C(s) + O2(g)  CO2(g)
Δ Hoc=-393.5kJ/mol
2H2(g) + 1 O2(g)  2 H2O(l)
Δ Hoc=571.6kJ/mol
CO2(g) + 2 H2O(l)  CH4(g) + 2O2(g) Δ Hoc=+890.8
C(s) + 2H2(g)
 CH4(g)
Δ Hof= -74.3kJ/mol
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Hess’s law example #2

NO(g) + ½ O2(g)  NO2(g)
given:
½ N2(g) + ½ O2(g) NO(g)
kJ/mol
½ N2(g) + O2(g) NO2(g)
Δ Hof = ?

Δ Hoc=90.25
Δ Hoc=33.2 kJ/mol