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Th d iThermodynamics
6.1 The Nature of Energy Law of Conservation of Energy
Energy is converted among forms but never created or destroyed.
The energy if the universe is constant The energy if the universe is constant.
2 basic types of EnergyPotential Energy Kinetic Energy
Bond energy or stored energy
Due to position or composition
Attractive and repulsive forces
energy of motion
KE / 2p
lead to PE
PE = mgh
m = mass (kg)
g = gravity constant (m s‐2)
h = height (m)
units are kg m2 s–2 J
KE = 1/2mv2
m = mass (kg)
v = velocity (m s–1)
v2 = (m2 s–2)
units are kg m2 s–2 = J
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Units of Energy The SI unit of energy is the joule (J).
1 J = 1 kg m2
2
An older, non‐SI unit is still in widespread use: the calorie (cal).
1 cal = 4.184 J
© 2009, Prentice-Hall, Inc.
s2
Heat
Involves the transfer of energy due to temperature (random motion of particles) difference
Work
Force acting over a distanceForce acting over a distance
Pathway
Best described as the way energy transfer occurs (ie: what contributes to the total energy transferred)
Therefore, total energy is NOT dependent on the pathway but contributors such as work and heat DO depend on pathway
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State Function
Refers to the characteristics of a system’s current state (ie: how much energy a system has now.
Does not depend on pathway
Properties that depend on path are not state functions
State Function (cont.)
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.
Chemical Energy System
Specific part of the universe
Includes the molecules we want to study (here the want to study (here, the hydrogen and oxygen molecules).
Surroundings
Everything else (here, the cylinder and piston).
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Exothermic
Energy flowing out of the system
Forming of bonds is always exothermic
Endothermic
Energy flowing into the system
Breaking of bonds is always endothermic
Thermodynamics
The study of energy and its transformations
1st Law of Thermodynamics
Energy of the universe is conserved (cannot be created or destroyed)or 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.
Internal Energy (E) The sum of all kinetic and potential energies of all components of the system
Internal energy can change through the flow of work (w), heat (q), or both.
E = q + w
Sign of each quantity is always written from the system’s point of view
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Think:
Heat given off by system (exothermic) is negative (‐q)
Heat absorbed by system (endothermic) is positive (+q)
Work done by system on surroundings is negative (‐W)
Work done on system by surroundings is positive (+W)
Work Energy used to move an object over some distance is work.
Often expansion or compression of a gas
Energy transfer between a system and its surroundings
The development of the work quantity:
w= F x d
P = F/ area d = V/area
w = (P x area) x (V/area)
W = (‐)(PV)
W = (‐)(PV)
This is the magnitude of the work required to expand a gas (V) against a pressure (P)
Why negative?V and W must have
opposite signs.
( )( )‐V means the system is
compressing and therefore Work is (+)
(+)W = (‐)P(‐V)
+V means the system is expanding and therefore
Work is (‐)
(‐)W = (‐)P(+V)
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This means work has the units of L‐atm
1 L‐atm = 101.3 J
6.2 Enthalpy and Calorimetry Enthalpy (H)
Enthalpy is a state function.
It describes heat
H = E + PV Where,
E = internal energy
P = pressure
V = volume
At constant pressure, where only PV work is done, then the enthalpy is the same thing as the heat flow.
When a reaction is studied at constant pressure, the flow of heat is a measure of the change in enthalpy or the heat of the in enthalpy or the heat of the reaction.
H = Hprod – Hreact
+ H = endothermic
‐ H = exothermic
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Calorimetry
Science of measuring heat through an observation of temperature change
Heat capacity (C) = heat absorbed (q)
increase in temp (tincrease in temp (t
The amount of heat required to raise
temp of the system by 1 degree
units of J oC–1 or J K–1
Specific heat capacity (s)
The heat capacity of a 1 gram sample
Specific heat (s) = C = q
m mt units of J g–1 oC–1 or J g–1 K–1
A useful form of the specific heat equation is: A useful form of the specific heat equation is:
q = m s T
Molar heat capacity
The product of specific heat times the molar mass of a substance
units are J mol–1 K–1
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Constant‐Pressure Calorimetry Called a coffee cup calorimeter
Change in enthalpy equals heat
Energy released by the reaction: = energy absorbed by the solution
T = m s T Joules
Is an extensive property (depends on the amount of substance)
Can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.
Because the specific heat for water is well known (4.184 J/g‐K), we can measure H for the reaction with this equation:
Tq = m s T
A process carried out at constant pressure would mean that the volume of the gas would have to expand in order to keep the pressure constant
Therefore,
qp + w
But the gas is expanding soqp – PV
And that means,
qp = E + PV
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Now, we said enthalpy is defined as
H = E + PV
And that heat at constant pressure equals,
qp = E + PV
Resulting in,
qp = H
Constant Volume Calorimetry
Called a bomb calorimeter.
The heat capacity of the calorimeter is known and tested.
Remember that,
E = q + PV Since V = 0,
PV = 0 Meaning,
E = qv
Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water.
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6.3 Hess’s Law Enthalpy is a state function.
It is independent of the path.
Number of steps required for a rxn to take place is also NOT importantNOT important.
The magnitude of H is directly proportional to the quantities of reactants and products
Two rules
If the reaction is reversed the sign of H is changed If the reaction is multiplied, so is H
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.”
Practice…
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6.4 Standard Enthalpies of Formation Some reactions are not easily studied using calorimetry and for those we use Standard Enthalpies of Formation (Hf
o)
Hfo = the change in enthalpy that accompanies the
formation of 1 mole of a compound from its elements with all substances in their standard states.
Standard State (o)
Indicates that the corresponding process has been carried out under standard conditions
For a compound
If gas, assume it is at 1 atm
If a liquid or solid assume the sample is pure If a liquid or solid, assume the sample is pure
If a solution, assume the concentration is 1 M
For an element
Will be in the form that the element exists under 1 atm and 25oC
For example:
Oxygen = O2(g) at a pressure of 1 atm
Sodium = Na(s) Mercury = Hg(l)
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General Expression
Horeaction = np × Ho
f (products) – nr × Hof (reactants)
Use Hess’s law and choose any convenient pathway from reactants to products.
Keep in mind:
If rxn is reversed, the magnitude of H stays the same but sign changes.
Value of H for that rxn must be multiplied by same integer used to balance equation
Elements in standard state, Hof = 0
16.1 Spontaneous Process and Entropy Spontaneous
Occurs without outside interventionintervention
Thermo only requires knowledge of energy of reactant and product relative to one another
Kinetics requires knowledge of the pathway
Is this reaction spontaneous?
The characteristic common to spontaneous processes is an increase in the entropy (S) of the universe.
Entropy (S)
Measure of randomness
order disorder
low S high S
Describes the number of arrangements (positions and/or energy levels) that are available to a system and is closely associated with probability
Nature spontaneously proceeds toward states that have the highest probability of existing
Electron orbitals being described as areas of probability, by definition, the electron will be found there
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Positional Probability
The probability of occurrence of a particular arrangement (state) depends on the number of ways (microstates) in which that arrangement can be achieved
Positional probability increases in going from
solid liquid gas
Solution Formation
+S when 2 pure substances mix because there are many more microstates for the mixed condition than for the separated condition
When 2
substances come
together V = S
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16.2 Entropy and the Second Law of Thermodynamics Second Law of Thermodynamics
Any spontaneous process results in an increase in the entropy of the universeentropy of the universe
Entropy is NOT CONSERVED!!!!! (you can’t break even)
In fact, the entropy of the universe is ALWAYS increasing
Higher temp and lower pressure = increase in entropy (more random)
Suniverse = Ssystem + Ssurroundings
The sign here will tell us if a reaction occurs:
(+) spontaneous in the
These two must be considered together in order to arrive at the sign for Suniverse
direction written
(‐) spontaneous in the opposite direction
(0) system is in equilibrium
g universe
16.3 The Effect of Temperature on Spontaneity Most often when heat flows into a system, the positional entropy of the particles in the system INCREASES (Ssys)INCREASES (Ssys) They move around more and therefore assume more positions
When heat flows into a system it must flow out of its surroundings thus lowering the positional energy of the surrounding particles (Ssurr)
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With +Ssys and –Ssurr what will Suniv be??? It depends on how high or low the temperature is
To understand why this is so, we must discuss in more detail the factors that control entropy changes in the surroundingssurroundings
All things seek lower energy (higher stability)
In the system, this requires releasing energy to the surroundings thus causing Ssurr to increase.
The effect any given amount of energy transferred to the surroundings has, depends on the temperature of those surroundings to begin with
$50 to a millionaire or a chemistry student, where will there be a real effect?
50J added to an already high temp will not cause as large a percent change as if it were added to very low temps
The tendency for a system to lower it’s energy by releasing to surrounding is a stronger driving force at lower temperatures where the heat can have a real impact
There are 2 important characteristics of the entropy changes that occur in the surroundings:
The sign of Ssurr depends on the direction of the heat flow. Exo = heat flows into surroundings from system (+Ssurr) Endo = heat flows from surroundings into system (‐S ) Endo = heat flows from surroundings into system ( Ssurr)
The magnitude of Ssurr depends on the temperature. A given quantity of heat produces a much greater impact in the randomness of the surroundings at low temp than at high temp.
Ssurr depends directly on the quantity of heat transferred and inversely on temperature (tendency for the system to lower its energy becomes a driving force at lower temps)
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= =
These ideas are summarized by:
Exo process: S + J
Driving force provided by energy flow (heat)
Magnitude of the entropy change of the surroundings
Quantity heat (J)
Temperature (K)
Exo process: Ssurr = + J
K
Endo process: Ssurr = ‐ J
K
We can express Ssurr in terms of H for a process at Pc; heat flow = H
=‐
‐H because the sign for Ssurris opposite the heat flow. H
refers to the system and the eqnbelow expresses the properties
of the surroundings
If H is negative (exo relative to the system), then Ssurr will be positive
surrIf H is positive (endo relative to the system), then Ssurr will be negative.
Example Exercise Antimony is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores:
Sb2S3 (s) + 3 Fe (s) 2 Sb (s) + 3 FeS (s) H = ‐125 kJSb2S3 (s) + 3 Fe (s) 2 Sb (s) + 3 FeS (s) H 125 kJ
Carbon is used as the reducing agent for oxide ores:
Sb4O6 (s) + 6 C (s) 4 Sb (s) + 6 CO (g) H = 778 kJ
Calculate Ssurr for each of these reactions at 25oC and 1 atm.
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Ssys Ssurr Suniv Process Spontaneous?+ + + Yes
‐ ‐ ‐ No, reaction will occur in opposite direction
+ ‐ ? Yes, if Ssys has a larger magnitude than Ssurr
‐ + ? Yes, if Ssurr has larger magnitude than Ssys
16.4 Free Energy Free Energy (G)
Available energy not total energy
Portion of total energy to do stuff with
Negative refers to direction
(heat flow)
G’s relationship to spontaneity Divide both sides by –T
G = H + (‐T)S‐T ‐T ‐T
Remember that at Tc and Pc
Ssurr = ‐H
All quantities refer to the system
Ssurr HT
So we can write
G = Ssurr + S = SunivT
Which shows that
Suniv = G @ Tc and Pc
T
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This means that a process carried out at Tc, Pc wil be spontaneous (+ Suniv) when G is negative or where the free energy of the system is negative
The opportunity for less free energy results in spontaneity
2 ways to determine spontaneity:
Suniv – must increase and can be used for all processes
Gsys – must decrease and can only be used at constant T and P
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16.5 Entropy Changes in Chemical Reactions
N2 (g) + 3 H2 (g) 2 NH3 (g)
1 mole gas 3 moles gas 2 moles gasg 3 g g
4 total moles gas
For gaseous molecules, positional entropy is dominated by the relative number of molecules of gaseous reactants and products
The Third Law of Thermodynamics
The entropy of a perfect crystal at 0 K is zero.
The purpose here is to move from this statement to a determination of absolute entropy values
The entropy of a substance
increases with temperature
The entropy value of a
substance at any temp. can
be calculated by knowing
the temperature dependence
of entropy (So)
Because S is a state function:
ΔSoreaction = Σ np Soprod – Σ nr S
oreact
extensive process
Th l h l l h hi h i h The more complex the molecule, the higher is the standard entropy value
ΔSo H2O (g) > ΔSo H2 (g)
More rotational and vibrational motion due to
shape
Linear, diatomic is much more rigid
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Water
Hydrogen gas
16.6 Free Energy and Chemical Reactions ΔGo
Standard free energy change
Change in free energy that will occur if the reactants in Change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states
Not measured directly rather calculated from other measurable quantities
The more negative ΔG, the further to the right the equilibrium of a reaction will be
Calculating ΔGo
1) ΔGo = ΔHo ‐ T ΔSo @ Tc
2) Hess’s law for free energy (because ΔG is a state function, pathway is unimportant)function, pathway is unimportant)
3) ΔGo = Σ np Gfo(products) – Σ nr Gf
o(reactants)
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16.7 The Dependence of Free Energy on Pressure
At Tc and Pc, the system will proceed spontaneously in the direction that lowers it’s free energy
h l b ll h l f The equilibrium position will represent the lowest free energy available to a system
Total free energy of a system changes as a reaction proceeds
Pressure of gases and concentrations of species in solution change
Pressure of Gases
G = H ‐ T ΔS
Not pressure dependent
Depends on pressure because it depends on volume.
Slarge volume > Ssmall volume
Slow pressure > Shigh pressure
p p pPositional entropy increases with more space therefore positional entropy will also increase with less pressure (P = V)
We know that entropy and therefore free energy depend on pressure.
Based on arguments beyond the scope of this class, we can conclude that:
G = Go + RT ln(P)
Temp (K)
( )
Gas constant
Free energy of gas @ 1 atm
Free energy of gas at “P”
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For a reaction
ΔG = ΔGo + RT ln(Q)
Temp (K)
J/molrxnΔGo = Σ np Gf
o(products) – Σ nr Gf
o(reactants)
(Q)
8.3145 J/K molrxn
J/molrxn“one mole of the reaction” Reaction quotient
(Pprod)n
(Preactant)n (Preactant)
n
Where n = # of moles in reaction
The meaning of ΔG for a Chemical Reaction
A system reaches the lowest possible ΔG at equilibrium, which of course explains why the system “stops”
The ball will roll to point B since it has a lower potential energy than point A (Similar to how ice spontaneously melts to liquid water). There is no intermediate substance
The ball will not get to point B since there is lower potential energy at point C.
The value of ΔG for a given reaction system will tell us if a reaction is favored but it does not mean that the
system will proceed to pure products (if ΔG is ‐) or remain at pure reactants (if ΔG is +)
The system will spontaneously go to the equilibrium The system will spontaneously go to the equilibrium position, the lowest free energy available to it.
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16.8 Free Energy and Equilibrium Equilibrium
Lowest value of free energy of the system
Let’s consider the following hypothetical rxn:
A BA(g) B(g)
1 mole @ 2 atm
As A reacts to form B, the total free energy of the system changes, yielding the following results:
Free energy of A = GA = GoA ‐ RT ln(PA)
Free energy of B = GB = GoB ‐ RT ln(PB)
Total free energy of the system = G = GA + GB
As A B, GA because PA and GB because PB
The reaction proceeds right as long as Gsys decreases (as long as GB is less than GA)
When PeA and P
eB occur GA will equal GB and ΔG will be
zero and the relative pressures will remain constant (no zero and the relative pressures will remain constant (no more driving force to change A B or vice versa)
If the pressure represent equilibrium pressures, Q = K K allows us to make certain judgments about the extent of the chemical reaction.
If K > 1, the mixture contains mostly products.
If K < 1, the mixture contains mostly reactants.
Suppose the experiment is described by the figure
Minimum free energy is reached when 75% of A has been changed to B
Pressure of A is 0.25 times the original pressure since only ¼ moles f A l f ( )( ) of A are left (0.25)(2.0 atm) = 0.50
atm
Pressure of B is (0.75)(2.0 atm) = 1.5 atm
Use equilibrium pressure to calculate K for the rxn
PeB = 1.5 atm
PeA = 0.50 atm
K = = 3.0
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In summary, when substances undergo a chemical reaction, the reaction proceeds to the minimum free energy (equilibrium)
Gproducts = Greactants
ΔG = Gproducts – Greactants = 0
Quantitative relationship between G and K
ΔG = ΔGo + RT ln(Q)
and @ equilibrium ΔG = 0 and Q = K so
ΔG = 0 = ΔGo + RT ln(K)
0r ΔGo = ‐ RT ln(K)
When ΔGo = 0, the system is in equilibrium. The free energies are equal when prod and react are in standard states (1 atm) meaning K = 1.
When ΔGo < 0, Goprod < G
oreact.
The system will move right to reach equilibrium and K > 1 because the product pressure will be high at equilibrium.
When ΔGo > 0, Goprod > G
oreact.
The system will move left to reach equilibrium and K < 1 because the reactant pressure will be high at equilibrium.
The Temperature Dependence of K
ΔGo = ‐ RT ln(K) = ΔHo ‐ T ΔSo
therefore,
‐ RT ln(K) = ΔHo ‐ T ΔSo
Divide both sides by ‐ RT
‐ln(K) = ‐ ΔHo + ΔSo
RT R
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Rearrange to get,
‐ln(K) = ‐ ΔHo 1 + ΔSo
R T R
y = m x + b
which matches the equation of a line.
This means that if values of K for a given reaction are determined at various temperatures, the plot will be linear with slope and intercept
This assumes that both ΔHo and ΔSo are independent of T in the range tested
16.9 Free Energy and Work Remember,
At Tc, Pc the sign of the change in free energy determine spontaneity
‐ΔG = spontaneous +ΔG = nonspontaneous
Thermodynamic information tells scientists which reactions are worth pursuing effectively and economically Those with slow kinetics would require searching for the appropriate catalysts
Wmax = ΔG
The maximum possible useful work obtainable from a process at Tc, Pc is equal to the change in free energy
Explain why the function is called free energy
(‐) energy that is free to do work( ) energy that is free to do work
(+) energy needed to make a process occur
When ΔG is compared to actual work accomplished, the % efficiency of the reaction can be measured
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Reversible process
Process that allows the energy of a system and surroundings to remain exactly the same after a forward and reverse cycle
Irreversible process
Process that results a change in the free energy of the g gyuniverse (the universe is different after the cyclic process occurs)
All real processes fit here.
ie: More work will always be required to recharge a battery than the battery produces at discharge.
Even though battery returns to original state, the surroundings have not
