Download (s) + H 2 (g) - Gordon State College

CHEMISTRY
Chapter 6
Thermochemistry
Prentice Hall © 2003
Chapter 5
Thermodynamics
- is the study of energy and the transformations it
undergoes in chemical reactions.
Units:
Joules (J)
calorie = 4.184 J
calorie – the amount of E needed to raise the
temperature of 1 gram of water 1 oC
Prentice Hall © 2003
Chapter 5
Energy
• Examples of energy:
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•
•
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Potential E
Kinetic E
Work – E needed to move an object against a force
Heat – E transferred from hot to cold objects
- capacity to do work or transfer heat
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Chapter 5
Today’s Topics
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4 Thermodynamic Functions
Definition of State Function
Internal Energy
Point of Views – System, Surroundings Universe
Sign Conventions
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Chapter 5
4 Thermodynamic Functions
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E
H
S
G
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Internal Energy
Enthalpy
Entropy
Gibb’s Free Energy
Prentice Hall © 2003
Chapter 5
What is a state function?
• A state function is a property that depends on the
present condition and not on how the change occurs.
• Derived from calculating the change:
D = [Final value – initial value]
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Chapter 5
4 State Functions
• Therefore:
•
DE =
Ef -Ei
DH =
Hf - Hi
DS =
Sf - Si
DG =
Gf - Gi
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Chapter 5
Internal Energy (E)
• Internal Energy - is the sum of the kinetic and
potential energy of a system
• Is the sum of heat (q) and work (w)
DE = Efinal – Einitial
DE = q + w
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Chapter 5
Point of Views
• System - the reaction we are studying
• Surroundings – anything else besides the reaction
For example: the container, you, etc….
• Universe – system + surroundings
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Chapter 5
q = heat and w = work
• If q is (+), system gaining heat from surroundings
(endo)
• If q is (-), system giving up heat to the surroundings
(exo)
• If w is (+), system is the recipient of work from the
surroundings. (in short, surroundings is doing work
on the system.
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Chapter 5
Thermodynamic Functions
• Have value, unit and magnitude.
• The sign of DE depends on the magnitude of q and w.
• Knowing the value of DE does not tell us which
variable is larger, q or w.
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Chapter 5
Problem 1
• Calculate the change in internal energy of the system
for a process in which the system absorbs 140 J of
heat from the surroundings and does 85 J of work on
the surroundings.
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Chapter 5
Problem 2
• Consider the reaction of hydrogen and oxygen gases
to produce water. As the reaction occurs, the system
loses 1150 J of heat to the surroundings. The
expanding gas does 480 J of work on the
surroundings as it pushes against the atmosphere.
Calculate the change in the internal energy of the
system?
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Chapter 5
Problem 3
• Calculate DE and determine whether the process is
endothermic or exothermic.
• 1.) q = 1.62 kJ and w = -874 J
• 2.) The system releases 113 kJ of heat to the
surroundings and does 39 kJ of work.
• 3.) The system absorbs 77.5 kJ of heat while doing
63.5 kJ of work on the surroundings.
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Chapter 5
Thermodynamic Functions
• Have value, unit and magnitude.
• The sign of DE depends on the magnitude of q and w.
• Knowing the value of DE does not tell us which
variable is larger, q or w.
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Chapter 5
Enthalpy
• Enthalpy – accounts for heat flow in chemical
reactions that occur at constant P when nothing other
than P-V work are performed
DH = Hfinal - Hinitial
• If:
• Then:
H = E + PDV
DH = DE + PDV
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Chapter 5
Ways of Measuring DH
• Calorimetry
• Hess’s Law
• Heats of Formation (DHof)
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Chapter 5
Calorimetry
• - Is the measurement of heat flow
• Specific Heat capacity (C) - the amount of heat
needed to raise the temperature of 1 g of substance 1
oC.
• The greater the heat capacity, the greater the heat
required to produce a rise in temp.
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Chapter 5
Calorimetry Equation
• q = mCDT
• - qsystem =
• - qsystem =
qsurroundings
qwater + qcalorimeter
• - qsubstance = (mCDTsolution + mCDTcalorimeter)
• Simplifies to:
• - qsubstance = (mCDTsolution + CDTcalorimeter)
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Chapter 5
Calorimetry Problem
• A 30.0 gram sample of water at 280 K is mixed with
50.0 grams of water at 330 K. Calculate the final
temperature of the mixture assuming no heat loss to
the surroundings. The specific heat capacity of the
solution is 4.18 J/g-oC.
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Chapter 5
Calorimetry Problem
• A 46.2 gram sample of copper is heated to 95.4 oC
and then placed in a calorimeter containing 75.0
gram of water at 19.6 oC. The final temperature of
the metal and water is 21.8 oC. Calculate the specific
heat capacity of copper, assuming that all the heat
lost by the copper is gained by the water.
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Chapter 5
Calorimetry Problem
• A 15.0 gram sample of nickel metal is heated to 100.0
oC and dropped into 55.0 grams of water, initially at
23 oC. Assuming that no heat is lost to the
calorimeter, calculate the final temperature of the
nickel and water. The specific heat of nickel is 0.444
J/g-oC. The specific heat of water is 4.18 J/g-oC.
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Chapter 5
Problem 4
• The specific heat of water is 4.18 J/g-K.
• How much heat is needed to warm 250 g of water
from 22 oC to 98 oC?
• What is the molar heat capacity of water?
• Molar heat Capacity = C x molar mass
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Chapter 5
Problem 5
• Large beds of rocks are used in solar heated homes to
store heat. Assume that the specific heat of rocks is
0.82 J/g-K.
• A. Calculate the amount of heat absorbed by 50 kg.
of rocks if their temperature rose by 12.0 oC.
• B. What temperature change would these rocks
undergo if they emitted 450 kJ of heat?
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Chapter 5
• Work – E needed to move an object
against a force
• When P is constant, P-V work is given by
w = - PDV
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Chapter 5
Enthalpy
• Enthalpy of a reaction or heat of Reaction:
DH = Hproducts - Hreactants
• 1. sign of DH depends on the amount of reactant
consumed
• 2. DH sign is opposite for backwards reaction
• 3. DHrxn depends on the physical state of the
reactants and products.
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Chapter 5
Problem 1
• Given the reaction:
• 2H2 (g) + O2 (g)  2 H2O (g) DH = -483 kJ
• Calculate the DH value for:
• 2 H2O (g)  2H2 (g) + O2 (g)
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Chapter 5
Problem 2
• Given the reaction:
• 2H2 (g) + O2 (g)  2 H2O (g)
DH = -483 kJ
• How much heat is released when 10.5 grams of H2 is
burned in a constant-pressure system?
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Chapter 5
Ways of Measuring DH
• Calorimetry
• Hess’s Law
• Heats of Formation (DHof)
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Chapter 5
Hess’ Law
• If a reaction is carried out in steps, DH for the
reaction will equal the sum of the enthalpy changes
for the individual steps.
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Chapter 5
Problem 1
• Given:
• C (s) + O2 (g)
 CO2 (g)
• CO(g) + ½ O2 (g)  CO2 (g)
DH = -393.5 kJ
DH = -283.5 kJ
• Calculate the enthalpy of combustion for:
C(s) + ½ O2 (g)  CO (g)
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Chapter 5
Problem 2
• Given:
• C(graphite) + O2 (g)  CO2 (g)
• C(diamond) + O2 (g)  CO2 (g)
DH = -393.5 kJ
DH = -395.4 kJ
• Calculate the enthalpy of combustion for:
C(graphite)  C (diamond)
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Chapter 5
Problem 3
• Given:
• C2H2 (g) + 5/2 O2 (g)  2CO2 (g) + H2O (l) DH = -1299.6 kJ
• C (s) + O2 (g)  CO2 (g)
• H2 (g) + 1/2 O2 (g)  H2O (l)
DH = -395.4 kJ
DH = -285.8 kJ
• Calculate the enthalpy of combustion for:
2C (s) + H2 (g)  C2H2 (g)
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Chapter 5
Enthalpies of Formation
• - known as heat of formation
• - gives the energy needed for a compound to form
• Standard enthalpy - is the enthalpy change (DH)
when the reactants and products are in their
standard state, usually 1 atm and 25 oC - denoted by
DHo ( ex. DHof)
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Chapter 5
Standard Enthalpy of
Formation, (DHof)
• By definition:
• The standard enthalpy of formation ( ex. DHof) of the
most stable form of any element is ZERO because
there is no formation reaction needed when the
element is in its standard state.
• - important for diatomic molecules
• - need knowledge of standard states of compounds
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Chapter 5
Enthalpies of Formation
• If there is more than one state for a substance under
standard conditions, the more stable one is used.
• Standard enthalpy of formation of the most stable form of
an element is zero.
Using Enthalpies of Formation of Calculate
Enthalpies of Reaction
• We use Hess’ Law to calculate enthalpies of a reaction
from enthalpies of formation.
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Chapter 5
Enthalpies of Formation
Prentice Hall © 2003
Chapter 5
Enthalpies of Formation
Using Enthalpies of Formation of
Calculate Enthalpies of Reaction
• For a reaction
DH rxn   nDH f products    mDH f reactants 
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Chapter 5
Next Topic
DS = Entropy = disorder
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Chapter 5
Thermodynamic Question
• Can a process occur?
• Spontaneous or Non-spontaneous?
– Forward vs. Reverse reactions
– Example: Gas Expansion
• Reversible or irreversible?
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Chapter 5
Entropy
• Entropy, S, is a measure of the disorder of a system.
• Spontaneous reactions proceed to lower energy or
higher entropy.
• In ice, the molecules are very well ordered because of
the H-bonds.
• Therefore, ice has a low entropy.
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Chapter 5
• As ice melts, the intermolecular forces are broken
(requires energy), but the order is interrupted (so
entropy increases).
• Water is more random than ice, so ice spontaneously
melts at room temperature.
• Conclusion: The higher the entropy the more
spontaneous the reaction.
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Chapter 5
• Generally, when an increase in entropy in one process
is associated with a decrease in entropy in another,
the increase in entropy dominates.
• Entropy is a state function.
• For a system, DS = Sfinal - Sinitial.
• If DS > 0 the randomness increases, if DS < 0 the
order increases.
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Chapter 5
Entropy and the Second
Law of Thermodynamics
Entropy
• Suppose a system changes reversibly between state 1
and state 2. Then, the change in entropy is given by
qrev
DSsys 
(constant T )
T
– at constant T where qrev is the amount of heat added
reversibly to the system. (Example: a phase change occurs
at constant T with the reversible addition of heat.)
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Chapter 5
Entropy and the Second
Law of Thermodynamics
•
•
•
•
The Second Law of Thermodynamics
Spontaneous processes have a direction.
In any spontaneous process, the entropy of the
universe increases.
DSuniv = DSsys + DSsurr: the change in entropy of the
universe is the sum of the change in entropy of the
system and the change in entropy of the surroundings.
Entropy is not conserved: DSuniv is increasing.
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Chapter 5
Entropy and the Second
Law of Thermodynamics
The Second Law of Thermodynamics
• Reversible process: DSuniv = 0.
• Spontaneous process (i.e. irreversible): DSuniv > 0.
• DSsys for a spontaneous process can be less than 0 as
long as DSsurr > 0. This would make DSuniv still (+).
• For an isolated system, DSsys = 0 for a reversible
process and DSsys > 0 for a spontaneous process.
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Chapter 5
Third Law of
Thermodynamics
• The entropy of a perfect crystal at 0 K is zero.
• Entropy changes dramatically at a phase change.
• As we heat a substance from absolute zero, the
entropy must increase.
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Chapter 5
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Chapter 5
Entropy Changes in
Chemical Reactions
• Absolute entropy can be determined from complicated
measurements.
• Standard molar entropy, S: entropy of a substance in
its standard state. Similar in concept to DH.
• Units: J/mol-K. Note units of DH: kJ/mol.
• Standard molar entropies of elements (S ) are not 0.
• For a chemical reaction which produces n moles of
products from m moles of reactants:
DS    nS products    mS reactants 
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Chapter 5
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Chapter 5
Gibbs Free Energy
For a spontaneous reaction the entropy of the universe
must increase.
• Reactions with large negative DH values are
spontaneous.
• How do we correlate DS and DH to predict whether a
reaction is spontaneous?
• Gibbs free energy, G, of a state is
G  H  TS
• For a process occurring at constant temperature
DG  DH  TDS
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Chapter 5
Gibb’s Free Energy
• Is the capacity to do maximum useful
work.
• Heat decreases the amount of useful work
done.
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Chapter 5
WORK
• At constant P, wsys = - PDV
• At constant temperature, wsys = - TDS
• If DG = DH – TDS, for maximum useful
work, DH must be = 0.
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Chapter 5
Gibbs Free Energy
Three important conditions:
– If DG < 0 then the forward reaction is spontaneous.
– If DG = 0 then reaction is at equilibrium and no net reaction
will occur.
– If DG > 0 then the forward reaction is not spontaneous. If
DG > 0, work must be supplied from the surroundings to
drive the reaction.
• For a reaction the free energy of the reactants decreases to a
minimum (equilibrium) and then increases to the free energy
of the products.
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Chapter 5
Problem
• Correlate DS and DH to predict whether a reaction is
non-spontaneous or spontaneous. If spontaneous,
determine whether the reaction will be spontaneous
at all temperature, at high temperature, or at low
temperature.
• A. If DS is (-) and DH is (+).
• B. If DS is (-) and DH is (-).
• C. If DS is (+) and DH is (+).
• D. If DS is (+) and DH is (-).
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Chapter 5
Gibbs Free Energy
• Consider the formation of ammonia from N2 and H2.
N2(g) + 3H2(g)
2NH3(g)
• Initially ammonia will be produced spontaneously (Q
< Keq).
• After some time, the ammonia will spontaneously
react to form N2 and H2 (Q > Keq).
• At equilibrium, ∆G = 0 and Q = Keq.
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Chapter 5
Gibbs Free Energy
Standard Free-Energy Changes
• DGf
• Standard states are: pure solid, pure liquid, 1 atm (gas), 1
M concentration (solution), and DG = 0 for elements.
• DG for a process is given by
DG   nDG f products    mDG f reactants 
• The quantity DG for a reaction tells us whether a
mixture of substances will spontaneously react to produce
more reactants (DG > 0) or products (DG < 0).
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Chapter 5
Free Energy and
Temperature
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Chapter 5
Free Energy and The
Equilibrium Constant
• Recall that DG and K (equilibrium constant) apply to
standard conditions.
• Recall that DG and Q (equilibrium quotient) apply to any
conditions.
• It is useful to determine whether substances under any
conditions will react:
DG  DG  RT ln Q
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Chapter 5
Free Energy and The
Equilibrium Constant
• At equilibrium, Q = K and DG = 0, so
DG  DG  RT ln Q
0  DG  RT ln K eq
DG   RT ln K eq
• From the above we can conclude:
– If DG < 0, then K > 1.
– If DG = 0, then K = 1.
– If DG > 0, then K < 1.
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Chapter 5
Calculating DH, DS, DG
• Use Hess’ Law
• Standard Heats of Formation (DHo, DSo, DGo )
• Equations
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Chapter 5