Transcript International Baccalaureate Chemistry

International
Baccalaureate
Chemistry
Topic 5 - Energetics
Thermochemistry
*
• Thermochemistry is the study of
energy changes associated with
chemical reactions.
– Studies the amount of energy in a
chemical reaction
– Energy is measured in Joules (J)
• Energy evolved or absorbed in a *
chemical reaction has nothing to do
with the rate of the reaction (how fast
or slow the reaction takes place.
Potential Energy
Energy an object possesses by virtue of its
position or chemical composition.
Kinetic Energy
Energy an object possesses by virtue of its
motion.
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KE =  mv2
2
Energy
• The ability to do work or transfer
heat.
– Work: Energy used to cause an object
that has mass to move.
– Heat: Energy used to cause the
temperature of an object to rise.
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Enthalpy
• Also called heat content
• Energy stored by the reactants and
products
• Units kJ mol-1
• Bonds are broken and made in chemical
reactions but the energy absorbed in
breaking bonds is almost never exactly
the same as the energy that is released in
making new bonds.
– Bonds broken = required energy
– Bonds formed = released energy
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Enthalpy Change
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• All reactions have a change in
potential energy of the bonds.
• This change in potential energy is
known as an enthalpy change.
• Enthalpy changes can only be
measured for chemical reactions, not
for state changes of substances.
• Given the symbol ΔH
Endothermic and
Exothermic
• A process is
endothermic
when H is
positive.
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Endothermicity and
Exothermicity
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• A process is
endothermic
when H is
positive.
• A process is
exothermic when
H is negative.
Endothermic Reactions
• Chemical reaction where the total enthalpy of the reactants is
less than the total enthalpy of the products
• Positive H values
• Heat energy is absorbed from surroundings
• They get cooler or external heat must be added to make the
reaction work.
• Will not occur as a spontaneous reaction.
• Bonds broken are stronger than bonds formed
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Exothermic Reactions
•
•
•
•
•
•
Chemicals lose energy as products are formed.
Negative H values
Heat energy is lost and surroundings get warmer.
Most spontaneous reactions are exothermic.
Products are more stable than the reactants.
Bonds made are stronger than bonds broken.
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Summary
Type of
Reaction
Heat
Energy
Change
Temperature
Change
Relative
Enthalpies
Sign of
ΔH
Exothermic
Heat
energy
evolved
Becomes
Hotter
Hp < Hr
Negative
Endothermic
Heat
energy
absorbed
Becomes
colder
Hp > Hr
Positive
Enthalpies of Reaction
This quantity, H, is called the enthalpy of
reaction, or the heat of reaction.
Enthalpies of Reaction
The change in
enthalpy, H, is the
enthalpy of the
products minus the
enthalpy of the
reactants:
H = Hproducts −
Hreactants
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Enthalpy Conditions
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• By definition, an enthalpy change must
occur at constant pressure.
• Thermochemical standard conditions have
been defined as a temperature of 25°C, a
pressure of 101.3 kPa, and all solutions
have a concentration of 1 mol dm-3.
• Thermochemical quantities that relate to
standard conditions are indicated with a
standard sign (θ).
• ΔH θ – Standard thermochemical
conditions.
Types of Enthalpies
• ΔHrxn – Heat produced in a chemical reaction.
• ΔHcomb – Heat produced by a combustion
reaction.
• ΔHneut – Heat produced in a neutralization
reaction.
• ΔHsol – Heat produced when a substance
dissolves.
• ΔHfus – Heat produced when a substance melts.
• ΔHvap – Heat produced when a substance
vaporizes.
• ΔHsub – Heat produced when a substance
sublimes.
Calculation of Enthalpy
Changes
• Temperature:
– A measure of the average kinetic energy
of the molecules.
• Heat:
*
– The amount of energy exchanged due to
a temperature difference between two
substances.
• An exchange of heat causes a
change in temperature.
Calorimetry
Since we cannot
know the exact
enthalpy of the
reactants and
products, we
measure H
through
calorimetry, the
measurement of
heat flow.
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Heat Capacity and Specific
Heat
• The amount of energy required to raise
the temperature of a substance by 1 K
(1C) is its heat capacity.
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• We define specific heat capacity (or
simply specific heat) as the amount of
energy required to raise the temperature
of 1 g of a substance by 1 K.
Heat Capacity and Specific
Heat
Specific heat, then, is
heat transferred
Specific heat =
mass  temperature change
c=
q
m  T
q = mC  T
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Constant Pressure
Calorimetry
By carrying out a
reaction in aqueous
solution in a simple
calorimeter such as
this one, one can
indirectly measure
the heat change for
the system by
measuring the heat
change for the water
in the calorimeter.
Constant Pressure
Calorimetry
Because the specific
heat for water is well
known (4.184 J/g K),
we can measure H
for the reaction with
this equation:
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q = mC  T
Example
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• Calculate the heat that would be
required an aluminum cooking pan
whose mass is 400 grams, from 20oC
to 200oC. The specific heat of
aluminum is 0.902 J g-1 oC-1.
• Solution
• q = mCT
•
= (400 g) (0.902 J g-1 oC-1)(200oC – 20oC)
•
= 64,944 J
Example
•
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What is the final temperature when 50 grams of
water at 20oC is added to 80 grams water at 60oC?
Assume that the loss of heat to the surroundings is
negligible. The specific heat of water is 4.184 J g-1
oC-1
Solution:
q (Cold) = -q (hot)
Let T = final temperature
mCT= -mCT
(50 g) (4.184 J g-1 oC-1)(T- 20oC) = -(80 g) (4.184 J g-1 oC-1)(T-60oC)
(50 g)(T- 20oC) = -(80 g)(T-60oC)
50T -1000 = – 80T + 4800
130T = 5800
T = 44.6 oC
Bomb Calorimetry
Reactions can be
carried out in a
sealed “bomb,”
such as this one,
and measure the
heat absorbed by
the water.
Bomb Calorimetry
• Because the
volume in the
bomb calorimeter
is constant, what is
measured is really
the change in
internal energy,
E, not H.
• For most reactions,
the difference is
very small.
First Law of
Thermodynamics
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• Energy is neither created nor destroyed.
• In other words, the total energy of the universe is
a constant; if the system loses energy, it must be
gained by the surroundings, and vice versa.
Use Fig. 5.5
Hess’s Law
*
 H is well known for many
reactions, and it can be
inconvenient to measure H for
every reaction in which we are
interested.
• However, we can estimate H
using H values that are published
and the properties of enthalpy.
Hess’s Law
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Hess’s law states
that “If a reaction
is carried out in a
series of steps, H
for the overall
reaction will be
equal to the sum of
the enthalpy
changes for the
individual steps.”
Hess’s Law
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Because H is a
state function, the
total enthalpy
change depends only
on the initial state of
the reactants and
the final state of the
products.
Standard Enthalpies of
Formation
Standard enthalpies of formation, Hθf, are
measured under standard conditions (25°C
and 1.00 atm pressure).
Hess’s Law Calculations
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• If a chemical reaction involves two or
more steps, the overall change in enthalpy
equals the sum of the enthalpy changes of
the individual steps.
• If a reaction is reversed, then the sign of
H must be reversed.
• If a reaction is multiplied by an integer,
then H must also be multiplied by the
integer.
• Attention must be paid to physical states.
Example 1:
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• Calculate the heat of combustion (H comb)
of C(s) to CO(g)
Given:
H (kJ)
C(s) + O2(g) → CO2(g)
-393.5
CO(g) + ½O2(g) → CO2(g
-283
C(s) + ½O2(g) → CO(g)
????
(-110.5 kJ)
Example 2
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• Calculate ΔH for the reaction:
2C(s) + H2(g) → C2H2(g)
Given:
ΔH (kJ)
C2H2(g) +5/2O2(g) → 2CO2(g) + H2O(l)
C(s) + O2(g) → CO2(g)
H2(g) + ½O2(g) → H2O(l)
-1299.6
-393.5
-285.9
(+226.7 kJ)
Example 3
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• Given:
ΔH(kJ)
NO(g) + O3(g) → NO2(g) + O2(g)
O3(g) → 3/2 O2(g)
O2(g) → 2O(g)
-198.9
-142.3
+495.0
• Find ΔH for the reaction:
NO(g) + O(g) →
NO2(g)
(-304.1 kJ)
Try This!
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• Calculate ΔH for the synthesis of diborane from its elements
according to the equation:
Given:
2B(s) + 3H2(g) → B2H6(g)
2B(s) + 3/2O2(g) → B2O3
B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g)
H2(g) + ½O2(g) → H2O(l)
H2O(l) → H2O(g)
ΔH(kJ)
-1273
-2035
-286
+44
(+36 kJ)
Bond Enthalpies
• Bond enthalpy is the amount of energy required to convert
a gaseous molecule into gaseous atoms.
HCl(g) → H(g) + Cl(g)
note: Cl2 and H2 are not formed.
• Limitations:
– Can only be used if all the reactants and products are
gaseous
– Values have been obtained from a number of similar
compounds and therefore, will vary slightly in different
compounds.
• Enthalpy changes (ΔH) can be calculated using bond
enthalpies.
• Bond formation: negative (exothermic)
• Bond breaking: positive (endothermic)
• ΔH = H bonds broken – H bonds formed
Example 1
• C2H4(g) + H2(g)
Bonds Broken (kJ mol-1)
C=C
612
4(C-H)
4(412)
H-H
436
Total:
2696 kJ
→
C2H6(g)
Bonds Formed(kJ mol-1)
C-C
348
6(C-H)
6(412)
2820 kJ
ΔH = 2696 - 2820 = -125 kJ mol-1
Example 2
• Estimate the ΔH for the following reaction from bond
energies.
C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(g)
Bonds Broken (kJ mol-1)
6 C-H
6(413) = 2478
7/2 O=O 7/2(495) = 1732.5
1 C-C
1(347) = 347
Total = 4557.5
Bonds Formed (kJ mol-1)
4 C=O 4(745) = 2980
6 O-H 6(467) = 2802
Total = 5782
ΔH = 4557.5 – 5782 = -1224.5 kJ mol-1