Transcript Unit 9 Notes

AP
CHEMISTRY:
CHAPTERS 6
AND 17
NOTES
THERMODYNAMICS
6.1 THE NATURE
OF ENERGY
6.1 The Nature of Energy
• Energy in terms of chemical
potential energy and how
that energy can change form
to accomplish work.
– Chemical Potential energy is energy
stored in bonds
• Thermochemistry = involves
heat and energy transfer.
ENERGY AND
WORK
• Energy- the capacity to do work or to
produce heat
• Work- force acting over a distance
–Work = Force x distance
– It involves a transfer of energy
The 1 st LAW OF THERMODYNAMICS:
LAW OF CONSERVATION OF ENERGY
• 1st Law of Thermodynamics- also known as
the Law of Conservation of Energy.
• States that energy can be converted from
one form to another but it can be neither
created nor destroyed.
– The total amount of energy in the universe is
constant.
Energy can be
classified in two
ways:
• Potential energy- energy due to position or
composition (included chemical potential energy)
• Kinetic energy- energy due to the motion of an
object
– Kinetic energy is dependent on the mass and velocity of an
object
– KE = ½ mv2
– m = mass in kg
– v = velocity in m/s
– units are J, since J = (kg.m2)/s2
Heat
• Heat- (q) involves a transfer of energy
between two objects due to a temperature
difference.
– Heat always moves from warmer matter to
cooler matter.
SYSTEM VS
SURROUNDINGS
• Describe where heat moves using the terms
system and surroundings.
• chemical reaction = system
• surroundings = everything else
– including things like the container the reaction
occurs in, the room it sits in, etc.
TEMPERATURE
• Temperature- a property that reflects
random motions of the particles of a
particular substance
• Different from heat, which again is the
transfer of energy between 2 objects due to
a difference in temperature
EXOTHERMIC
• reaction which releases heat
• Energy flows out of the system
• Potential energy is changed to
thermal energy
• Products have lower potential energy
than reactants
ENDOTHERMIC
• Reaction which absorbs heat
• energy flows into the system
• thermal energy is changed into
potential energy
• products have higher PE than
reactants
Equation vs. Graph
Is this reaction exothermic or endothermic?
Chemical energy
C H 4 + 2O 2  C O 2 + 2H 2 O + H eat
C H 4 + 2O 2
CO 2 + 2 H 2O
Is this reaction exothermic or endothermic?
Potential energy
C H 4 + 2O 2  C O 2 + 2H 2 O + H eat
C H 4 + 2O 2
Heat
CO 2 + 2 H 2O
THE VALUE OF “E”
• Internal energy (E) of a system is the sum of the
kinetic and potential energies of all the particles in a
system.
• Thermodynamic quantities always consist of a
number and a sign (+ or -).
• The sign represents the system’s point of view.
(Engineers use the surrounding’s point of view)
– Exothermic -E (systems energy is decreasing)
– Endothermic +E (systems energy is increasing)
CALCULATING THE ENERGY
E = q + w
IN A SYSTEM



E is the change in the system’s internal
energy
q represents heat
w represents work usually in J or kJ
• Example: Calculate E if q = -50 kJ
and w = +35kJ.
E = q + w
= -50 kJ + 35 kJ
= -15 kJ
Work on
Gases
• For a gas that expands or is compressed, work can
be calculated by:
w = - PV
• w = work, units: L.atm
• P = pressure, units: atm
• V = volume, units L
Example: Calculate the work if the
volume of a gas is increased from
15 ml to 2.0 L at a constant
pressure of 1.5 atm.
• w = -PV
• w = -1.5 atm (1.985L)
• w = -3.0 L . atm
Expanding a gas: work is negative due to work
flowing out of the system.
Compressing a gas: work is positive due to work
flow into the system
HEAT ENERGY
Q = m x c x ΔT
• Energy = Mass(m) x Specific Heat (c) x Change in
Temperature (ΔT = Final – Initial)
• Specific Heat: Heat required to raise the temperature by 1
degree Celcius.
• Mass: Grams
• Specific Heat: J/(g·⁰C),
• Temp: ⁰C
UNITS
• Energy content of food is measured in calories or
kilocalories (1000 calories).
• 0.2390 calories = 1 J or
1 calorie = 4.184 J
• Conversions can also be made to kJ and kcal (Cal)
• Conversions to Kilo are decimal place changes
• kilo  unit : 3 right
• unit  kilo : 3 left
PRACTICE PROBLEM
If the temperature of 34.4 g of ethanol increases from
25 °C to 78.8 °C, how much heat has been absorbed
by the ethanol? The specific heat of ethanol is 2.44
J/(g·°C)
If 335 g of water at 65.5 °C loses 9750 J of heat, what is
the final temperature of the water? Liquid water has a
specific heat of 4.18 J/(g×°C).
ANOTHER PHASE DIAGRAM…TEMP & ENERGY
THE GREATER THE MATERIAL'S INTERNAL ENERGY, THE HIGHER THE
TEMPERATURE OF THAT MATERIAL .
HEAT IS THE ENERGY FLOW BET WEEN OBJECTS OF DIFFERENT TEMPERATURE.
HEAT AND TEMPERATURE ARE NOT THE SAME.
HEAT ENERGY & PHASE
CHANGES
• Heat of Vaporization
– Boiling: Liquid to Steam
– Heat required to change 1g of a liquid at its normal
evaporation point to a gas at the same temperature
• Heat of Fusion
– Melting: Solid to liquid
– Heat required to change 1g of a solid at its normal
melting point to a liquid at the same temperature
PHASE CHANGE OF WATER
Heat constants (use 3 sig figs)
I. Specific Heats “C” (energy to change the temp while in each phase)
– liquid (water) Cℓ = 1.00 cal /goC
– gas (steam)
Cg = 0.48 cal /goC
– solid (ice)
Cs = 0.50 cal /goC
II. Changing Phases (energy to change the phase but no temp change)
– heat of fusion
Hf = 80.0 cal / g
– heat of vaporization Hv = 540. cal / g
• III. Formulas
– q = m (∆T) Cp
• while in a phase
– m Hf or v
• formula during phase change
PRACTICE PROBLEM
• Calculate the amount of heat necessary to change 16.4 g of water
from -48.0oC to 112oC
PHASE CHANGE WITH
SOLUTIONS OTHER THAN
WATER
• Values will be given to you
• Remember freezing and boiling point temperatures will vary
• Can be embedded in the question or given to you in a chart.
6.2 ENTHALPY
AND
CALORIMETRY
LET’S TALK LAB…
• Solve for the calorimeter constant and write this on your cup. Be
sure to hang on to your cup!
• Solve for q=mCΔT for the solid you used. The mass used is the solid
and the liquid mass. The specific heat used is for water.
• Discuss your procedures for Friday with your lab partner. You
determine how you will calculate the amount of heat lost or gained
by an entire cold pack.
LAB DATA
Compound
Name
Heat of Solution
Group 1
% Error
1. Na2CO3
-20.8
28.4%
2. NaC2H3O2 101 (-17.4 legit)
680%
3. CaCl2
-90.026
28.2%
4. NaCl
29.9
6.98%
5. NH4NO3
28.294
11.4%
6. LiCl
-18.145
4.9%
LAB DATA
Compoun
d Name
Heat of
Solution
(J/g)
Heat of
Solution
(kJ/mol)
% Error
1. Na2CO3
-1471
-31.3
11.4
2.
NaC2H3O2
-877.8
-3.54
79.6
3. CaCl2
-4901
-543.86
31.6
4. NaCl
-160
-9.33
340
5. NH4NO3
-3120
-25.8
11.6
6. LiCl
-4759.3
-40.35
8.96
Enthalpy
• Enthalpy (H) concerns the heat
energy in a system.
• H = q at constant pressure
only
– Reactions that do not involve
gases or where moles of gases
do not change are considered
“at constant pressure”.
• At constant pressure, the
terms heat of reaction and
change in enthalpy are used
interchangeably.
• H = E + PV
– E is internal energy
– P is pressure
– V is volume
Enthalpy Change in a
System
• The change in the enthalpy of a
system can be calculated using:
H= ΣH products – ΣH reactants
• For an exothermic reaction, H is
negative
• For an endothermic reaction, H is
positive
EX AMPLE: Sulfur dioxide reacts with oxygen. Write the balanced
equation for the reaction of one mole of sulfur dioxide,
calculate the enthalpy value for the reaction, then draw an
energy diagram for this reaction based on your answer. Be sure
to label reactants, products, and the enthalpy of the system.
• Balanced Equation
• Enthalpy Value
H= ΣH products - ΣH reactants
H= [-395.7kJ/mol] - [-296.8kJ/mol + ½ 0] = -98.9 kJ/mol
• Energy Diagram
THERMODYNAMIC EQUATIONS
• For many reactions, a value is given alongside a
balanced equation called a thermochemical
equation. This value should be associated with
the moles of each substance given in the problem.
– 1/8 S8(s) + O2(g)  SO2(g)
H = -296.8 kJ
– As you can see, 1/8 mole of sulfur would release this
amount of energy while one mole of sulfur would
release eight times that amount of energy.
Example: For the reaction
2Na + 2H2O  2NaOH + H2 , H = -368 kJ
Calculate the heat change that occurs when 3.5
g of Na reacts with excess water.
• 3.5g Na 1 mol Na
•
-368 kJ
=
22.99g Na 2 mol Na
• H = -28 kJ (or say 28 kJ are released)
CALORIMETRY
Heat Capacity (C)
C = heat absorbed
Increase in temp.
C = J/g°C or
J/mol°C
• Calorimetry- the science of
measuring heat flow in a
chemical reaction.
– It is based on observing the
temperature change when a body
absorbs or discharges heat.
– The instrument used to measure
this change is the calorimeter.
Here are some
helpful tables
that are found in
the back of your
text book.
CALORIMETRY
q = mCT
J = (J/g°C )(g)(°C)
• Constant pressure calorimetry- pressure remains
constant during the process
– Constant pressure calorimetry uses a set up called a
coffee cup calorimeter
• The primary reaction to calculate heat changes in a
system is the “Mcat” equation.
– Energy released as heat = (heat capacity) (mass of
solution ) (increase in temp)
Example: A coffee cup calorimeter contains 150 g H 2 O
at 24.6 o c. A 110 g block of molybdenum is heated to
100 o c and then placed in the water in the calorimeter.
The contents of the calorimeter come to a temperature
of 28.0 o c. What is the heat capacity per gram of
molybdenum?
• qwater = mCΔT and qwater = qMo
• So… mwCw ΔTw = mMoCMo ΔTMo
• (150g)(4.18J/goC)(3.4oC) = (110g)
(CMo)(72oC)
• CMo = 0.27J/goC
100.0 ml of 0.100M silver nitrate is mixed with 100.0 ml
of 0.200M sodium chloride. Both solutions start at
room temperature (25.0°C) and, once combined into a
coffee cup calorimeter, the final temperature reading
after mixing is 26.8°c. Find the heat of reaction in
kj/mol of silver nitrate.
• AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
• q = mCΔT = (200.0g)(4.18 J/g°C)(1.8°C)
• q = 1500 J or 1.5 kJ
• (0.1000L)(0.100M AgNO3) = 0.0100 mol AgNO3 in the
reaction
• 1.5kJ/0.0100mol = 150 kJ/mol
EXTENSIVE VS INTENSIVE
PROPERTIES • Let’s Practice!
• In terms of calorimetry,
we can describe certain
properties of the
reaction as an:
I
– Extensive property- this
depends on the amount
of substance (ex. Heat
of reaction)
– Intensive propertydoesn’t depend on the
amount of substance
(ex. Temperature)
• The number of calories of energy
you derive from eating a banana
• The number of calories of energy
made available to your body
when you consume 10.0 g of
sugar
• The density of your blood
• The mass of iron present in
blood
• The electrical resistance of a
piece of 22-gauge copper wire.
• The melting point of copper
wire.
E
I
I
E
I
I
CALORIMETRY CAN ALSO BE DONE
IN A CLOSED, RIGID CONTAINER.
• This is called constant volume
calorimetry
– Ex. Flashbulb in a camera or a bomb
calorimeter
– No work can be done since the volume
doesn’t change
– Heat evolved = T x heat capacity of
calorimeter (energy required to change
the temp 1oC)
6.3 HESS’S L AW
HESS’S LAW
• Hess’s Law- States that the change in enthalpy
from products to reactants, or H, is the same
whether the reaction occurs in one step or in
several steps.
– H is not dependent on the reaction pathway.
– The sum of the H for each step equals the H for
the total reaction.
– If a reaction is reversed, the sign of H is reversed.
– If the coefficients in a reaction are multiplied by an
integer, the value of H is multiplied by the same
integer.
HESS’S LAW APPLIED
N2 + O2  2NO
2NO + O2  2NO2
N2 + 2O2  2NO2
Example: given the following reactions and
their respective enthalpy changes, calculate
H for the reaction: 2C + H 2  C 2H 2.
C2H2 + 5/2 O2  2CO2 + H2O
H = -1299.6 kJ/mol C2H2
C + O2  CO2
H = -393.5 kJ/mol C
H2 + ½ O2  H2O
H = -285.9 kJ/mol H2
• 2C + 2O2  2 CO2
H = 2(-393.5) kJ
• H2 + ½ O2  H2O
H = -285.9 kJ
• 2CO2 + H2O  C2H2 + 5/2 O2 H = +1299.6 kJ
• 2C + H2 C2H2
 H = 226.7kJ
Example: the heat of combustion of C to CO 2 is 393.5 kJ/mol of CO 2 , whereas that for
combustion of CO to CO 2 is -283.0 kJ/mol of CO 2 .
Calculate the heat of combustion of C to CO.
• First, write the equations given, as well as your goal
equation.
• Goal: C + ½O2  CO
• Given:
C + O2  CO2
H = -393.5 kJ
CO + ½ O2  CO2
H = -283.0 kJ
• Now, rearrange to find the goal equation!
• C + O2  CO2
H= -393.5 kJ
• CO2  CO + 1/2 O2
H= +283.0 kJ
• C + 1/2O2  CO
H= -110.5 kJ
6.4 STANDARD
ENTHALPIES OF
FORMATION
ΔH F°
• Standard enthalpy of formation (𝛥𝐻𝑓0 )
• The change in enthalpy that accompanies the
formation of one mole of a compound from its
elements with all substances in their standard
states at 25oC.
• The degree sign on a thermodynamics function
indicates that the process it represents has been
carried out at standard state conditions.
STANDARD STATE CONDITIONS
• for gases, pressure is 1 atm
• for a substance in solution, the concentration is 1 M
• for a pure substance in a condensed state (liquid or solid), the
standard state is the pure liquid or solid.
• for an element, the standard state is the form in which the
element exists under conditions of 1 atm and 25oC.
• Values of 𝛥𝐻𝑓0 are found in Appendix 4 and at the end of
these notes
• 𝛥𝐻𝑓0 reaction = 𝛥𝐻𝑓0 products - 𝛥𝐻𝑓0 reactants
Example: The standard enthalpy
change for the reaction
CaCO 3(s)  CaO + CO 2(g) is 178.1 kJ.
Calculate the 𝛥𝐻𝑓0 for CaCO 3(s).
• Solve for the heat of a single compound by using
𝛥𝐻𝑓0 reaction = 𝛥𝐻𝑓0 products - 𝛥𝐻𝑓0 reactants
• 178.1 kJ = [-635 kJ + -393.5] - [x]
• X = -1207 kJ
6.5 Present Sources of
Energy
• In this section, we will discuss some sources of
energy, including fossil fuels, and their effects on
the environment.
PETROLEUM
• Petroleum- A thick, dark liquid composed of
hydrocarbons chains of 5-25 carbons
– Refining petroleum involves the process of pyrolytic
cracking, or distilling the fractions of petroleum from
the main sample based on their molecular mass and
boiling point.
GASOLINE
• Gasoline
C5-C12
– Gasoline, when first used in car engines, caused a
dramatic knocking sound and was thus treated with
tetraethyl lead, (C2H5)4Pb, an antiknock agent. This
introduced lead into the atmosphere as the fuel was
spent and increased the amount of ingested lead in the
human and animal populations until 1960 when
“leaded” gas was finally phased out.
– kerosene & jet fuel C10-C18
– heating and lubricating oil and diesel fuel
– asphalt >C25
C15-C25
NATURAL GAS
• Natural gas- This
substance is usually
found alongside
petroleum and is
composed mostly of
methane. It also
contains ethane,
propane, and butane.
Coal
• Coal- formed from the remains of
plants buried under pressure for
many years. Cellulose, CH2Ox,
gradually loses its H and O.
• Coal develops through 4 stages:
–
–
–
–
lignite (least valuable)
subbituminous
bituminous (high sulfur)
anthracite (most valuable)
• Coal provides 20% of our energy in
the U.S.
CO 2 and the
Environment
• Effects of CO2 on the Climate
• Greenhouse effect - H2O and CO2 molecules in the
atmosphere reflect IR radiation and send it back to earth
thus raising the earth’s temperature.
• The CO2 concentration has increased by about 16% in the
past 100 years because of increase in the use of fossil fuels.
NEW
ENERGY
SOURCES
• Coal gasification- treating coal with oxygen and steam at high
temperatures to break down many of the C-C bonds and form C-O
and C-H bonds.
• The products are syngas (CO + H2) and methane gas. Syngas may be
converted to methanol.
– CO(g) + 2H2(g)  CH3OH(l)
NEW ENERGY SOURCES
• Hydrogen as a Fuel
• H2 (g) + ½O2(g)  H2O(l)
• Ho = -286 kJ
– ~2.5 times the energy of
natural gas
– 3 problems: production
(too expensive), transport
(too volatile), and storage
(large volume, decomposes
to H atoms on metal
surfaces, makes metal
brittle-forms metal
hydrides)
OTHER ALTERNATIVES:
• oil shale
• ethanol
• gasohol
• seed oil
(sunflower)
CHAPTER 17:
Spontaneity, Entropy
and Free Energy
17.1 & 17.2
S P O N TA N E O U S
PROCESSES AND
ENTROPY AND THE
SECOND L AW OF
THERMODYNAMICS
st
1
Law of
Thermodynamics
• The first Law of Thermodynamics states that energy is
neither created nor destroyed; it is constant in the
universe. We can measure energy changes in chemical
reactions to help determine exactly what is happening and
how it is occurring.
• Keep in mind that thermodynamics deals with the
reactants and products while kinetics deals with how
reactants become products.
SPONTANEOUS… WHAT??
• Spontaneous processoccurs without outside
intervention (ex. Rusting)
– This may be fast or slow
(CO2 sublimes at room
temperature vs. iron
rusting when in the
presence of oxygen)
ENTROPY
High
Entropy!
Low
Entropy!
• Entropy (S)- a measure of randomness or disorder
– This is associated with probability. (there are more ways for
something to be disorganized than organized)
– Entropy increases going from a solid to a liquid to a gas and when
solutions are formed.
– Entropy increases in a reaction when more atoms or molecules are
formed.
– Entropy increases with increasing temperature.
ENTROPY INCREASES…
• SLG
• More molecules
produced
• Temperature
increases
ND
2
LAW OF
THERMODYNAMICS
• 2nd Law of
Thermodynamics- In any
spontaneous process there
is always an increase in the
entropy of the universe.
The energy of the universe
is constant but the entropy
of the universe is
increasing.
17.3 & 17.4
ENTROPY AND
FREE ENERGY
FREE ENERGY
G = H -TS
• Free energy (G)- the amount of energy
available to do work.
– Free energy change is a measure of the
spontaneity of a reaction. It is the maximum
work available from the system.
• A spontaneous reaction carried out as
constant temperature and pressure has a
negative G. For example, when ice melts
H is positive (endothermic), S is positive
and G = 0 at 0˚C.
G = H -TS
S
H
G
Spontaneous?
Yes, always
Yes, at high T
Yes, at low T
No, never
17.5 ENTROPY
CHANGES IN
CHEMICAL
REACTIONS
RD
3
LAW OF
THERMODYNAMICS
• Third Law of Thermodynamics- The entropy of a
perfect crystal at 0 K is zero.
• Any substance that is not at 0 Kelvin must have a
value for entropy! This sets entropy apart for
enthalpy. Enthalpy values are just changes, while
entropy values are absolute and cannot drop
below zero.
Ex. Given the following standard
molar entropies, calculate S o for
the reaction:
2Al(s) + 3MgO(s)  3Mg(s) + Al 2O 3(s)
• Mg(s) = 33.0 J/K
Al2O3(s) = 51.0 J/K
• Al(s) = 28.0 J/K
MgO(s) = 27.0 J/K
• Soreaction = Soproducts - Soreactants
• So = [3(33.0J/K) + 51.0J/K] – [2(28.0J/K) + 3(27.0J/K)]
• So = 13.0 J/Kmol.rxn
17.6 FREE
ENERGY AND
CHEMICAL
REACTIONS
The Standard Free Energy
Change
• Standard free energy change (G°) -The change
in free energy that occurs if the reactants in their
standard states are converted to products in their
standard states.
– G° =Gfoproducts - Gforeactants at standard
conditions
– Gfo for a free element in its standard state is zero.
EX. GIVEN THE EQUATION N 2O 4(G)
 2NO 2(G) AND THE FOLLOWING
DATA, CALCULATE G O.
• Gfo for N2O4(g) = 97.82 kJ/mol,
• Gfo for NO2(g) = 51.30 kJ/mol
• Go = [2(51.30kJ/mol)] – [ 97.82 kJ/mol]
• = 4.78 kJ/mol rxn
THE GIBBS-HELMHOLTZ
EQUATION
• The Gibbs-Helmholtz equation works like
the Gibbs Free Energy equation, just at
standard state conditions.
– Go =Ho -TSo
– (When working this, change the units for S to
kJ).
Ex. For the given reaction and the
following information, calculate G o
at 25°C.
2PbO(s) + 2SO 2(g)  2PbS(s) + 3O 2(g)
PbO (s)
SO2 (s)
PbS (s)
O2 (g)
Ho (kJ/mol)
-218.0
-297.0
-100.0
?
So (J/mol.K)
70.0
248.0
91.0
205.0
Ho = [2(-100.0) + 3(0)]- [2(-218.0) + 2(-297.0)]
= 830.0 kJ/mol rxn
So = [2(91.0) + 3(205.0)] – [2(70.0)+2(248.0)]
= 161.0 J/K.mol rxn = 0.1610 kJ/mol
Go = Ho - T So
Go= 830.0 kJ/mol -298K(0.1610 kJ/mol)
= 782 kJ/mol rxn
17.7 THE
DEPENDENCE OF
FREE ENERGY
ON PRESSURE
ENTROPY AND PRESSURE/
VOLUME
• In an ideal gas, enthalpy does not
depend on pressure. Entropy,
however, does depend on pressure
because it depends upon volume.
– Gases in large volumes have greater
entropy than in a small volume.
– Gases at a low pressure have greater
entropy than at a high pressure.
• Because entropy depends on
pressure, G of an ideal gas
depends on its pressure.
17.7 The Dependence of Free
Energy on Pressure
 If we incorporate PV = nRT with the equation
for free energy, we end up with an equation to
calculate free energy in relation to temperature
and pressure variables.
G = G° + RT ln (Q)
• Q = reaction quotient (partial pressure of
products/reactants raised to the power of their
coefficients)-only pressures of gases are included.
• T = temperature in Kelvin
• R = gas constant 8.3145 J/K.mol
• G° = free energy change at 1 atm (be sure to
change to Joules!)
Ex. Calculate G at 298K for the
following reaction if the reaction mixture
consists of 1.0 atm N 2, 3.0 atm H 2, and 1.0
atm NH 3.
• N2(g) + 3H2(g)  2NH3(g)
Go = -33.32 kJ/mol
• G = Go + RT lnQ
• G = -33,320J/mol + 8.314J/Kmol(298K) ln
• G = -41,500J/mol rxn or
• -41.5 kJ/mol rxn
(1.0𝑎𝑡𝑚)2
(1.0 𝑎𝑡𝑚)1 (3.0𝑎𝑡𝑚)3
17.8 FREE
ENERGY AND
EQUILIBRIUM
FREE ENERGY AND
EQUILIBRIUM
• The equilibrium point in
terms of kinetics occurs
when the forward and
reverse reactions were
occurring at an equal rate.
• In terms of free energy, the
equilibrium point occurs at
the lowest value of free
energy available to the
reaction system.
FREE ENERGY AND
EQUILIBRIUM
• These two definitions are the same!
– G = Gproducts - Greactants = 0
• If a process has just shifted from nonspontaneous to
spontaneous, then at the point where it changes, the value
for G is zero. If G is zero, then Ho = TSo.
Ex. Given for the reaction Hg(l)  Hg(g)
that H o = 61.3 kJ/mol and S o = 100.0
J/mol . K, calculate the temperature of the
normal boiling point of Hg.
• Go =Ho -T So
• G = 0 at phase change
• 0 = 61.3 –T(0.1000)
• T = 613K or 340oC
THE “RAT LINK” EQUATION
• We can utilize the previous two equations (G = Gproducts Greactants = 0) and (G = Go + RT ln (Q)) to form an
equation that describes the relationship between free
energy and the value of the equilibrium constant.
• Go = -RT ln(K) (the “rat link” equation)
– When Go = 0, free energy of reactants and products are equal
when all components are in their standard states. During a phase
change, G = 0.
– When Go < 0, Go products < Go reactants The reaction is not at
equilibrium, K > 1 since pressure of products is > 1 and the
pressure of reactants is < 1.
– When Go > 0, Go reactants < Go products The reaction is not at
equilibrium, K < 1 since pressure of products is < 1 and the
pressure of reactants is > 1.
Ex. Calculate the approximate standard
free energy for the ionization of
hydrofluoric acid, HF (K a = 1.0 x 10 -3), at
25 o C.
• Go = -RT ln K
• Go = -8.314(298)ln (1.0 x 10-3)
• = 1.7 x 104J or 17 kJ
FREE ENERGY AND
EQUILIBRIUM
• We can use G =Go +
RT ln(Q) to calculate the
direction that a reaction
will shift to reach
equilibrium.
– Free energy is energy
available to do useful
work. Wmax = G