Transcript Chapter 5 Notes Thermochemistry

1411
Chapter 5
Thermochemistry
1
Energy
• Energy – the ability to do work or transfer
heat.
– Energy used to cause an object that has mass
to move is called work.
– Energy used to cause the temperature of an
object to rise is called heat.
• Thermodynamics – study of energy and its
transformations
– Thermochemistry – branch of thermodynamics
involving chemical reactions and changes in
heat
2
Kinetic Energy
Kinetic energy is energy an object possesses by
virtue of its motion
1
KE =  mv2
2
m = mass
v = velocity
Atoms and molecules can have mass and motion
and therefore possess kinetic energy!
3
Potential Energy
• Potential energy is energy an object possesses by
virtue of its position or chemical composition.
• Potential energy arises when a force operates on an
object
– Force – any kind of push or pull exerted on an object
• Ex: Gravity is the force exerted on the cyclist at the top of the hill.
The cyclist possesses potential energy.
PE = mgh
m = mass
g = gravitational constant, 9.8 m/s2
h = height of an object relative to
some reference height
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Potential Energy
• Potential energy is energy an object possesses by
virtue of its position or chemical composition.
• Many substances (ie, fuels) release energy when the
react.
• The chemical energy of these substances is due to the
potential energy stored in their chemical bonds.
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Units of Energy
• The SI unit of energy is the joule (J) (“jewel”).
kg m2
1 J = 1 
s2
• An older, non-SI unit is still in widespread use:
the calorie (cal).
1 cal = 4.184 J
– In nutrition, the Calorie (Cal) is used
1 Cal = 1000 cal = 1 kcal
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Definitions:
System and Surroundings
• The system includes the
molecules we want to study
(here, the hydrogen and
oxygen molecules).
• The surroundings are
everything else (here, the
cylinder and piston).
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Classifications of Systems
1. Open: matter and energy can be exchanged with
the surroundings
– Example: Boiling pot of water on stove with no lid –
heat enters the system from the stove and water
vapor escapes from the system into the surroundings.
2. Closed: energy, but not matter, can be exchanged
with the surroundings
– Example: Cylinder with movable piston containing
H2(g) and O2(g) – these form water but mass is not
lost. But energy is exchanged with surroundings in the
form of heat and work.
H2(g) + O2(g)  H2O(g) + energy
3. Isolated: neither energy nor matter are
exchanged with the surroundings
– Example: An insulated thermos containing hot coffee
is an approximation.
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Transferring Energy: Work
• Work – the energy used
to move an object over
some distance
• w=Fd
w = work
F = force
d = distance over
which the force
is exerted
• The object being moved is
the system
• Whatever is moving the
object is part of the
surroundings and is
performing work on the
system
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Transferring Energy: Heat
• Energy can also be transferred as
heat.
• Heat flows from warmer objects to
cooler objects.
• In this combustion reaction, the
chemical energy stored in the fuel
molecules is released as heat
(energy). The heat is transferred
from the hotter system to the
cooler surroundings.
– Substances involved in combustion
reaction = system
– Everything else = surroundings
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Example
A book with a mass of 590 g is raised from the floor to a height of 2.1 m. It then
falls to the floor.
a) What is the potential energy when the book is at its highest?
590 g * 1kg = 0.59 kg
1000 g
PE = mgh
PE = (0.59 kg)*(9.8 m/s2)*(2.1 m)
PE = 12.1422 kg*m2/s2
PE = 12 J
b)
How much work is done to raise the book? (Note: The force due to gravity is
F = m x g, where g is the gravitational constant, 9.8 m/s2.)
AND F = m * g
w=F*d
w =m*g*d
w = (0.59 kg)*(9.8 m/s2)*(2.1 m)
w = 12.1422 kg*m2/s2
w = 12 J
c)
Assuming all the work done in part (b) has been converted to kinetic energy,
what is the speed of the book at the instant before it hits the floor?
w = 12.1422 kg*m2/s2
w = KE
KE = ½mv2
w = ½mv2
v2 = 2w/m
v = √(2w/m)
v = √{2 * (12.1422 kg*m2/s2 ) / (0.59 kg)}
v = √{41.16 m2/s2}
v = 6.41561 m/s
v = 6.4 m/s
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Conversion of Energy
• Energy can be converted from one type to another.
• For example, the cyclist above has potential energy as she sits on top
of the hill.
• As she coasts down the hill, her potential energy is converted to
kinetic energy.
• At the bottom, all the potential energy she had at the top of the hill is
now kinetic energy.
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First Law of Thermodynamics
• 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
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Internal Energy
• The internal energy of a system is the sum of all kinetic and
potential energies of all components of the system; we call it E.
• By definition, the change in internal energy, E, is the final
energy of the system minus the initial energy of the system:
E = Efinal − Einitial
Initial state = reactants
Final state = products
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Changes in Internal Energy
• When energy is
exchanged between
the system and the
surroundings, it is
exchanged as either
heat (q) or work (w).
• That is, E = q + w.
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E, q, w, and Their Signs
“+” means “deposit”
“–” means “withdrawal”
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Example, E
• If a system does 45 J of work and receives 28 J
of heat, what is the value of E for this
change?
E = q + w
w = -45 J
q = +28 J
E = (+28 J) + (-45 J)
E = -17 J
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Exchange of Heat between System
and Surroundings
• When heat is absorbed by the system from the
surroundings, the process is endothermic. (“endo” = into)
Endothermic reaction:
Reactants + Heat  Products
System = reactants and products only
Surroundings = solvent and everything else
Temperature of reaction mixture decreases from 20 °C to -9°
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Exchange of Heat between System
and Surroundings
• When heat is released by the system into the
surroundings, the process is exothermic. (“exo” = out of)
Exothermic reaction:
Reactants  Products + Heat
System = reactants and products only
Surroundings = solvent and everything else
Temperature of reaction mixture increases by a LOT!
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State Functions
• Usually we have no way of knowing the exact quantity of E
– Finding that value is simply too complex a problem
• However, we do know that the internal energy of a system is
independent of the path by which the system achieved that state.
– In the system below, the water could have reached room temperature from either
direction.
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State Functions
• Therefore, internal energy is a state function.
• It depends only on the present state of the system, not on the
path by which the system arrived at that state.
• And so, E depends only on Einitial and Efinal.
Real-life analogy of a state function: Your checkbook balance!
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Work
Atmospheric
pressure
Atmospheric
pressure
Gas
Gas
• Gas is in a closed vessel
with a movable piston:
– The gas does work on the
surroundings by pushing
the piston against the
force of atmospheric
pressure pressing down
on it.
Pressure is constant
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Work
• pressure-volume work (P-V work) – the work
involved in the expansion or compression of gases
• For constant pressure scenarios (like the example
with pistons):
w = -PV
P = pressure
V = change in volume = Vfinal - Vinitial
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Enthalpy
• If a process takes place at constant pressure
(as the majority of processes we study do)
and the only work done is this pressurevolume work, we can account for heat flow
during the process by measuring the enthalpy
(H) of the system.
• Enthalpy is the internal energy plus the
product of pressure and volume:
H = E + PV
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Enthalpy
• When the system changes at constant pressure, the change
in enthalpy, H, is
H = (E + PV)
• This can be written
H = E + PV
• Since E = q + w and w = -PV, we can substitute these
into the enthalpy expression:
H = (q+w) − w
H = q
• So, at constant pressure, the change in enthalpy is the
heat gained or lost.
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Endothermicity and Exothermicity
Because H = q,
• A process is
endothermic when
H is positive (heat is
absorbed).
– because q is positive
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Endothermicity and Exothermicity
Because H = q,
• A process is
endothermic when
H is positive (heat is
absorbed).
– because q is positive
• A process is endothermic
when H is negative (heat is
released).
– because q is negative
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Enthalpy of Reaction
The change in enthalpy, H, of a chemical
reaction is the enthalpy of the products minus
the enthalpy of the reactants:
H = Hproducts − Hreactants
This quantity, H, is called the enthalpy of
reaction, or the heat of reaction.
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Truths about Enthalpy
1.
Enthalpy is an extensive property.
–
–
So the amount of reactants/products matters
H for the following reaction is only true when 1 mol CH4 reacts
with 2 mol O2
CH4 (g) + 2O2(g)  CO2(g) + 2H2O(l)
–
2.
If twice as much of each (2 mol CH4 and 4 mol O2) reacts, twice as
much heat is released (-890 kJ * 2 = -1780 kJ)
H for a reaction in the reverse reaction is equal in size, but
opposite in sign, to H for the forward reaction.
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
3.
H = -890 kJ
H = 890 kJ
H for a reaction depends on the state of the products and the
state of the reactants.
CH4 (g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ
H2O gas formed instead of liquid
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Examples
1) A gas is confined to a cylinder under constant atmospheric
pressure. When 378 J of heat is added to the gas, it
expands and does 56 J of work on the surroundings. What
are the values of H and E for this process?
2) For the reaction,
2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(g) H = -1199 kJ
a)
b)
c)
d)
Is heat absorbed or released?
Is the reaction endothermic or exothermic?
What is H for the reverse reaction?
Calculate the amount of heat transferred when 46.0 g of
methanol combusts.
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Examples
1) A gas is confined to a cylinder under constant atmospheric
pressure. When 378 J of heat is added to the gas, it
expands and does 56 J of work on the surroundings. What
are the values of H and E for this process?
q = +378 J
w = -56 J
q = H = 378 J
2) For the reaction,
E = q + w
E = (+378 J) + (-56 J)
E = 322 J
2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(g) H = -1199 kJ
a)
b)
c)
d)
Is heat absorbed or released? released
Is the reaction endothermic or exothermic? exothermic
What is H for the reverse reaction? H = 1199 kJ
Calculate the amount of heat transferred when 46.0 g of
methanol combusts.
46.0 g CH3OH
x
1 mol CH3OH
-1199 kJ
x
32.042 g CH3OH
2 mol CH3OH
= -860.65 kJ  -861 kJ
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Calorimetry
Calorimeter – measures heat flow
Since we cannot know
the exact enthalpy of
the reactants and
products, we measure
H through
calorimetry, the
measurement of the
temperature change
the heat flow
produces.
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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 (C).
C=
q
T
Example of units:
J/K
• And we can specify the amount of substance:
– molar heat capacity (Cm) – heat capacity of 1 mole of substance
q
Cm = mol  T
Example of units:
J/mol-K
– specific heat (Cs) – heat capacity of 1 gram of substance
Cs =
q
m  T
Example of Units:
J/g-K
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Using Specific Heat
• We usually know Cs, but not the heat (q), so we rearrange the
equation to get
q = Cs  m  T
• Example:
If an iron key with a mass of 5.5 g changes in temperature
from 25.0 to 28.0 °C, how much energy (in joules) has it
absorbed? (The specific heat of Fe(s) is 0.45 J g-1 K-1.)
mass = 5.5 g
T = Tfinal – Tinitial
T = 28.0 °C – 25.0 °C
T = 3.0 °C = 3.0 K
Cs = 0.45 J/g-K
q = Cs  m  T
q = (0.45 J/g-K)  (5.5 g)  (3.0 K)
q = 7.425 J  7.4 J
We don’t have to convert to Kelvin because we only
care about the change in temperature, and 1 °C is
the same size as 1 K. So if we convert to Kelvin, we
get the same change:
T = Tfinal – Tinitial
T = (28.0°C + 273.15) – (25.0°C + 273.15)
T = 3.0 K
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Constant-Pressure Calorimetry
Simple “Coffee-cup” calorimeter
• Reaction done in aqueous solution
• Not sealed, so pressure is constant
• Heat change for the system is indirectly measured by measuring
the heat change for the water:
– Reactants & products = system
– Water, cup, and everything else = surroundings
– We assume calorimeter prevents solution from gaining or losing
heat
– Therefore, all the heat produced by reaction, qrxn, is entirely
absorbed by the solution (stays within calorimeter)
qsoln = -qrxn
qsoln = Cs(water)  msolution  T = -qrxn
•
If T > 0, then temperature of solution increased, so rxn “lost”
heat to surroundings, so surroundings “gained” heat (that’s why
temperature of solution rose)
– So qrxn < 0, reaction was exothermic
Cs(water) = 4.184 J/g-K
•
If T < 0, then temperature of solution decreased, so rxn “gained”
heat from surroundings, so surroundings “lost” heat (that’s why
temperature of solution decreased)
– So qrxn > 0, reaction was endothermic
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Example: Coffee-cup calorimeter
When a 3.88 g sample of solid ammonium nitrate dissolves in 60.0 g of water in a
coffee-cup calorimeter, the temperature drops from 23.0 °C to 18.4 °C.
a) Calculate ΔH (in kJ/mol NH4NO3) for the solution process
NH4NO3(s)  NH4+(aq) + NO3-(aq)
Assume that the specific heat of the solution is the same as that of pure water, 4.184
J/g-K.
We need to find ΔH (in kJ/mol) for the reaction
And ΔHrxn = qrxn
We know that qsoln = -qrxn (so let’s find qsoln)
qsoln = Cs(water)  msoln  T = -qrxn
msoln = 60.0 g + 3.88 g = 63.88 g
T = 18.4 °C – 23.0 °C = -4.6 °C = -4.6 K
qsoln = (4.184 J/g-K) * (63.88 g) * (-4.6 K) = -1229.46 J
The negative sign tells us that the solution “lost” 1229.46 J of heat to the reaction.
Since qsoln = -qrxn, then qrxn = +1229.46 J, meaning the reaction “gained” 1229.46 J of heat from the solution.
Now, the quantity 1229.46 J is how much heat is gained when 3.88 g NH4NO3 reacts, but the question asks for
the units of kJ/mol, meaning when 1 mole of NH4NO3 reacts. So we convert to kJ/mol:
1229.46 J
3.88 g NH4NO3
x
1 kJ
1000 J
x
80.043 g NH4NO3
1 mol NH4NO3
b) Is this process endothermic or exothermic?
= 25.354 kJ/mol  25 kJ/mol = ΔH
endothermic
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Bomb Calorimetry
• Reactions can be carried
out in a sealed “bomb”
such as this one.
• The heat absorbed (or
released) by the water is a
very good approximation
of the heat released in the
reaction, qrxn.
qrxn = -qcal
qcal = Ccal  T
qrxn = -Ccal  T
Heat capacity of the calorimeter
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Example: Bomb Calorimeter
Under constant-volume conditions the heat of combustion of benzoic acid
(C6H5COOH) is 26.38 kJ/g. A 1.640-g sample of benzoic acid is burned in a
bomb calorimeter. The temperature of the calorimeter increases from 22.25
°C to 27.20 °C.
a) What is the total heat capacity of the calorimeter?
b) A 1.320-g sample of a new organic substance is combusted in the
same calorimeter. The temperature of the calorimeter increases from
22.14 °C to 26.82 °C. What is the heat of combustion per gram of the
new substance?
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Example: Bomb Calorimeter
Under constant-volume conditions the heat of combustion of benzoic acid
(C6H5COOH) is 26.38 kJ/g. A 1.640-g sample of benzoic acid is burned in a
bomb calorimeter. The temperature of the calorimeter increases from 22.25
°C to 27.20 °C.
a) What is the total heat capacity of the calorimeter?
H = 26.38 kJ/g
1.640 g
x 26.38 kJ = 43.2632 kJ = qcal
1g
T = 27.20 °C – 22.25 °C = 4.95 °C
qcal
43.2632 kJ
Ccal =
=
= 8.74004 kJ/°C  8.74 kJ/°C
4.95°C
T
b) A 1.320-g sample of a new organic substance is combusted in the
same calorimeter. The temperature of the calorimeter increases from
22.14 °C to 26.82 °C. What is the heat of combustion per gram of the
new substance?
Ccal = 8.740 kJ/°C
T = 26.82 °C – 22.14 °C = 4.68 °C
qrxn = -Ccal  T
qrxn = -(8.740 kJ/°C) * (4.68 °C) = -40.903 kJ
H = -40.903 kJ = -30.987 kJ/g = -31.0 kJ/g
1.320 g
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Hess’s Law
• H is well known for many reactions, and
it is inconvenient to measure H for
every reaction in which we are
interested.
• However, we can estimate H using
published H values and the properties
of enthalpy.
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Hess’s Law
Hess’s law states that “[i]f 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.”
Example: Combustion of CH4(g) to form liquid water can be
thought of as two steps. The enthalpy change for the overall
process is the sum of the enthalpy changes for the 2 steps, so
we add the reactions like an algebraic equation:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
(add)
2H2O(g)  2H2O(l)
CH4(g) + 2O2(g) + 2H2O(g)  CO2(g) + 2H2O(g) + 2H2O(l)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = -890 kJ
H = -802 kJ
H = -88 kJ
H = -890 kJ
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Hess’s Law Problems
The goal is to manipulate the given equations so that we can add them to
look like the final equation of interest. So we use the Truths of Enthalpy from
earlier.
1. Work with one substance at a time by comparing how it looks in the
given equation to the desired equation. (Preferably, pick a substance that
only appears once in all the given equations.)
a.
Determine if its coefficient changes
•
b.
Determine if the substance is on the same side of the rxn arrow or not
•
2.
3.
If so, then multiply AN ENTIRE GIVEN EQUATION (and the associated H) by an appropriate
number so that coefficients now match.
If that changes, re-write the given equation, swapping the places of reactants and products.
Remember to multiply the associated H by -1 (ie change its sign).
Repeat until you can add up the equations to give the desired equation,
cancelling substances that appear on both sides of the arrow.
And finally (the point of the problem), add the manipulated H
quantities to give H for the desired equation.
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Example: Hess’s Law
Given the data
C(s) + O2(g)  CO2(g)
H = -393.5 kJ
CO(g) + ½O2(g)  CO2 (g)
H = -283.0 kJ
use Hess’s law to calculate H for the reaction
C(s) + ½O2(g)  CO(g)
H = -110.5 kJ
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Example: Hess’s Law
Given the data
N2(g) + O2(g)  2NO(g)
2NO(g) + O2(g)  2NO2(g)
2N2O(g)  2N2(g) + O2(g)
H = +180.7 kJ
H = -113.1 kJ
H = -163.2 kJ
use Hess’s law to calculate H for the reaction
N2O(g) + NO2(g)  3NO(g)
H = 155.7 kJ
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Enthalpies of Formation
An enthalpy of formation, Hf, is defined as
the enthalpy change for the reaction in which
a compound is made from its constituent
elements in their elemental forms.
Ex:
C(s,graphite) + O2(g) → CO2(g)
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Standard Enthalpies of Formation
Standard enthalpy of formation, Hf°, is the change in
enthalpy for the reaction that forms one mole of the
compound from its elements, with all the substances in
their standard states (25 °C and 1.00 atm pressure).
• Hf° = 0 for elements in their most stable elemental
form (ie C(graphite), H2(g), O2(g)
• The values of Hf° for other compounds can be found in
a table of known enthalpies of formation (Appendix C).
• Units: kJ/mol
elements (in their standard states)  1 mol compound (in standard state)
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Calculation of H
We can use Hess’s law in this way:
H =  nHf°products –  mHf° reactants
where n and m are the stoichiometric
coefficients.
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Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H =  nHf°products –  mHf° reactants
= [3Hf°(CO2) + 4Hf°(H2O)] – [1Hf°(C3H8) + 5Hf°(O2)]
= [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]
= (-2323.7 kJ) – (-103.85 kJ)
= -2219.9 kJ
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