Download Thermodynamics The First Law Work, Heat, Energy

The Basic Concepts: the system
Thermodynamics
The First Law
Work, Heat, Energy
System
Matter
System
Matter
Energy
Energy
Surroundings
Thermodynamics is the study of
the transformations of energy.
Surroundings
(b) closed
(a) open
System
Matter
Energy
Oxtoby, Chapter 10 (10.1-10.4)
Surroundings
(c) isolated
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Summer Session 2005 Chem 31
The Basic Concepts: state and path fns
A state function, X, is a property that depends only on the current state
of the system, and not on how it was prepared.
Changes in a state function depend only on the start and end
points of an experiment
The Conservation of Energy:
Energy can neither be created not destroyed
(experimental observation)
DX = Xfinal – Xinitial
e.g. the duration of this lecture depends only when I start and
when I finish
A path function, Y, is a property that depends on the history of the
system.
e.g. the boredom/interest factor for this lecture depends on a lot
more than just my first and last sentence
fl Energy is a state function
The principle of conservation of energy can be used to assess
the energy changes that accompany physical and chemical
processes.
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Work, Heat and Energy
ICE +
SOLID
C
+
+
heat
heat
O2
Melting
Liquid water
Work, w, is done when an object is moved against an opposing force
Liquid
CO2 +
heat
Work is a path function
(how strong is gravity?)
Expanding gas pushes out piston
against (e.g, atmospheric) pressure
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1
Work, Heat and Energy
Heat, q
The Energy, E of a system is its capacity to do work.
When the energy of a system changes as a result of a temperature
difference between it and its surroundings, it is said that energy has
been transferred as Heat, q
When spring unwinds,
work will be done
Processes that release energy as heat are exothermic;
Increasing its capacity to work
A
B + heat
Combustion reactions
When piston moves out
against higher pressure,
work will be done
Processes that absorb energy as heat are endothermic;
A + heat
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Hence, for an isolated system:
The First Law
In thermodynamics, the “total” energy of a system is called the
internal energy, E.
∆E = w + q = 0 + 0 = 0
Energy
Surroundings
∆E = Ef - Ei
The change in internal energy ∆E is the sum of work done on a
system and the energy transferred as heat to a system according to;
∆Euniverse = 0
System
Matter
Experiments measure the change in energy between start and finish:
and a closed system:
∆E = w + q
System
Matter
Heat must be a path function: the system can gain/lose energy either
as heat or as work.
Energy
Surroundings
SI unit for work, heat and energy is the Joule (or J/mol, kJ/mol, …)
the change in internal energy of
a closed system is equal to the
energy that passed through its
boundary — either as heat or as
work.
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Work, w
In general:
B
Melting, Vaporization of Water
Work and pressure
w=F. d
work
Pressure Volume Work:
distance
External pressure, p
Force along path
Area A
More specifically, for thermodynamics:
w=- F. d
∆h
w = - Fext . d
w = - Fext ∆ h
We focus on the force the system has to push against
w = - pext A ∆ h
Work is done on system by the surroundings
w = - pext ∆ V
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Work, and pressure
+/-? The Sign Convention
w = - pext ∆ V
Work done by the system on its surroundings is “lost” by the
system (a negative w).
Work done on the system by the surrounding increases the
internal energy (positive w) – energy is stored in the system
For pext = 0
Expansion ∆ V > 0
Compression ∆ V < 0
w < 0, hence system
performs work on
surroundings
w > 0, hence
surroundings perform
work on system
A positive heat flow (positive q) corresponds to heat gained
by the system.
Hence, a heat loss is to be understood as a negative gain
(negative q).
Thermal Equilibrium is established when two bodies have
the same temperature.
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Example:
Recall:
Calculate the work done when 25g of iron reacts with Hydrochloric Acid
in (a) a closed cell, (b) in a open beaker at 25oC?
(a) ∆ V = 0
∆E = w + q
&
w=0
w = - pext ∆V
If the system is kept at constant volume:
(b)
Fes + 2 HCl(aq)
w = - pext ∆ V
∆E = qv
FeCl2 (aq) + H2(g)
∆ V = Vf-Vi =Vgas
pV = nRT
n= m/M
25g mol
55.845 g
w = - pext ∆ V = - pext Vgas = - n R T
w=
- 0.45mol x 8.314 J x 298 K
K mol
= 0.45 mol
=
- 1.2 kJ
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Examples
Calorimetry
Adiabatic Bomb Calorimeter
We measure ∆E usually using Calorimetry (the measurement
of amounts of heat flowing into or out of a system and the
accompanying temperature changes)
Adiabatic means no heat is transferred
from calorimeter to surroundings
The change in temperature, ∆T, of the
calorimeter is proportional to the heat
that the reaction releases or absorbs.
q = Ccal ∆T
Calorimeter Constant
Calibrated using a process of known energy output
(eg, burning of a substance of known mass) or
from an electrical current, I, of known potential, V
q=VIt
(a) When 100 g of Napthalene was burned in a bomb calorimeter
releasing heat of 3.91 kJ, the temperature rose by 3.01 K. What is the
calorimeter constant?
q = Ccal ∆T
3.91 kJ = Ccal µ 3.01 K
Ccal = 1.30 kJ / K
(b) 10.0 A from a 12 V battery were passed through the heater of a
calorimeter for 150 s. The temperature rose by 5 K. What is the
calorimeter constant?
q = (10.0 A) µ (12 V) µ (150 s)=1.8 µ 104 A V s = 18 kJ
18 kJ = Ccal µ 5 K
Ccal = 3.6 kJ / K
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Heat Capacity
Heat Capacity
Generally: C =
q
∆T
Units: J K-1, cal K-1 1 cal = 4.184 J
Molar Heat Capacity (J K-1 mol-1 or cal K-1 mol-1)
q
C
Heat capacity of a sample divided by the
c=
=
n
chemical amount of substance, n
∆T n
The heat capacity C of a substance is the heat required to change its
temperature by one Kelvin, and has units of energy per Kelvin.
The heat capacity is an extensive variable: the quantity is proportional to
the amount of matter present.
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Heat Capacity
Example:
measured at constant pressure
measured at constant volume
(eg, conduct experiments in
closed container)
qp
Cp = ∆T
cp =
qp
=
∆T n
Cp
n
Specific Heat Capacity (in J K-1 kg-1 or cal K-1 kg-1)
q
C
Heat capacity of a sample divided by the
cs = ∆T m = m
mass of substance, m
c = M cs
molar
qv
Cv = ∆T
qv
cv =
=
∆T n
Cv
n
If reaction contains only solids and/or liquids, cp ≈ cv
A new calorimeter is filled with 200g of room temperature
water. Adding 1212 J of heat to the contents, raises the
interior temperature by 1.42 K. Calculate the calorimeter
constant given that the molar heat capacity of water, cp(H2O)
is 75.38 J K-1 mol-1.
1212 J = qwater+ qcalorimeter = Ccal ∆T + cp(H2O) n(H2O) ∆T
= Ccal ∆T + cp(H2O) m(H2O) ∆T
M(H2O)
Ccal =
1212 J
- 75.38 J K-1 mol-1 200 g 1.42 K
1.42 K 18 g mol-1
1.42 K
= 16 J K-1
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Recall:
∆E = w + q
&
w = - pext ∆V
If the system is kept at constant volume:
Enthalpy
∆E = qv
But if it is not kept at constant volume?
∆E = qv
Another type of energy, the ENTHALPY, is useful in these
circumstances
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The Enthalpy, H
Example: Relating ∆H and ∆U
For a system kept at constant pressure:
∆H = qp
where ∆H = Hf - Hi
The Enthalpy, like the internal Energy, is a state function
The Enthalpy is related to the Energy by:
H = U + pV
∆H = ∆(U+pV)=Hf - Hi = (U+pV)a - (U+pV)c
= Ua - Uc + (pV)a -(pV)c
as p = const.
= ∆ U + p ∆V
m
Va = ρ a
∆H- ∆U = p ∆V
a
H = U + pV = U + nRT
Va = 17.0 cm3
Vc = 18.5 cm3
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Enthalpy of Reaction - ∆Hr
The Reaction Enthalpy, ∆Hr
A
B
HA
HB
∆Hr = Hproducts - Hreactants
In General:
∆Hr = Σ ν Hproducts - Σ ν Hreactants
Enthalpy of Reaction: ∆Hr = HB - HA
Example 2)
Ma = Mc = 100 g/mol
ma = mc = 50.0 g
∆H- ∆U = 0.152 J
for an ideal gas
Example 1)
The internal energy change when 0.500 mol calcite (ρc =2.71 g cm-3)
converts to aragonite (ρa =2.93 gcm-3) is 0.100 kJ. Calculate the
difference between the changes in enthalpy and in internal energy at a
pressure of 1 atm. [calcite and aragonite are different forms of CaCO3]
Sum over all standard molar enthalpies taking
into account their stochiometric factors, ν
2C + D
2 HC HD
E +3F
HE 3 HF
Enthalpy of Reaction: ∆Hr = HE + 3 HF - HD - 2 HC
Example:
2 C2H6 + 7 O2
ν= 2
ν= 7
4 CO2
ν= 4
+
6 H2O
ν= 6
∆Hr = 6H(H2O) + 4H(CO2) - 7H(O2) - 2H(C2H6)
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Standard Enthalpies
Enthalpies of phase change
Enthalpy are normally tabulated for substances in their standard state;
Denotes standard condition
these are called the standard enthalpy, Ho.
Enthalpy changes also occur when a substance melts/freezes or
condenses/evaporates
The standard state of a substance at a specified temperature is its pure
form at 1 atm (should now be 1 bar).
Fusion (melting):
H2O(s)
kJ
H2O(l) ∆Hofus(273)=+6.0 mol
For dissolved species, the standard state is the concentration of 1M
under a pressure of 1 atm at a specified temperature.
Vaporisation (boiling): H O
2 (l)
kJ
H2O(g) ∆Hovap(373)=+40.7 mol
Standard enthalpies at T = 298 K are often denoted Hʅ
endothermic!
Hence, the standard enthalpy of reaction is
∆Hro
=Σν
Ho
Sublimation
products -
Σν
Ho
reactants
H2O(s)
kJ
H2O(g) ∆Hosub(298.15)=+46.7 mol
(direct conversion from solid to gas)
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Rules for enthalpy changes (1)
Rules for enthalpy changes (2)
A change in enthalpy is independent of the path taken
between two states:
Final state
Initial state
Enthalpy is a state function fl can derive some rules
∆Hovap(373 K)= + 40.7 kJ mol–1
(1)
H2O(l)
vaporisation
condensation
Sublimation
H2O(g)
solid
∆Hofus
gas
∆Hovap
gas
∆Ho(A→B) = - ∆Ho(B→A)
∆Hosub = ∆Hofus +
the enthalpy of a forward process differs from that of the backwards
process only in sign.
∆Hovap
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One could also have written …
Hess’s Law
H2O(s)
H2O(l)
∆Hofus = +6.0 kJ mol–1
H2O(l)
H2O(g)
∆Hovap = +40.7 kJ mol–1
Overall: H2O(s)
H2O(g)
∆Hosub = +6.0 + 40.7 kJ mol–1
= + 46.7 kJ mol–1
Because enthalpy is a state function, these rules apply to every type of
reaction or change
The enthalpy of an overall reaction is the sum of the
enthalpies of the individual reactions into which the
reaction can be divided.
Example: Calculate the enthalpy of combustion of benzene (C6H6)
from its enthalpy of hydrogenation (-205 kJ/mol) to
cyclohexane, and the enthalpy of combustion of
cyclohexane (∆Ho (C6H12) = –3920 kJ/mol)). The enthalpy
of the combustion for H2 is -286 kJ/mol.
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Standard Molar Enthalpies of Formation, ∆Hof
Enthalpy of Combustion of Benzene
We want to know: C6H6 + 7.5 O2
6 CO2
We do know:
C6 H6 +
3 H2
C6H12
C6H12 +
9 O2
6 CO2 + 6 H2O
C6 H6 +
H2O(g)
∆Hosub
solid
∆Hocond(373 K)= - 40.7 kJ mol–1
In general:
H2O(s)
+ 3 H2 O
-205 kJ/mol
3 H2 O
3H2 + 1.5 O2
7.5 O2
6 CO2 + 3 H2O
-3920 kJ/mol
-(3 µ -286) kJ/mol
-3268
kJ/mol
The enthalpy of formation is the enthalpy change when a compound is
formed from its elements, and those elements are in their most stable
form under the prevailing conditions.
When the prevailing conditions are the standard state, this is called the
standard enthalpy of formation, ∆Hof
H2(g) + 0.5 O2(g)
6 C(s, graphite) + 3 H2(g)
H2O(l)
∆Hof = - 286 kJ/mol
C6 H6(l) ∆Hof = + 49 kJ/mol
The standard enthalpies of elements in their reference states are zero
at all temperatures (graphite is the reference state of carbon!).
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Bond enthalpies
A–B(g)
Applications of bond enthalpies
A(g) + B(g)
∆Hrxn = energy of the A–B bond
But bond enthalpies are affected by the neighbouring bonds
CH4(g)
CH3(g)
+ H(g)
∆H = 439 kJ mol–1
C2H6(g)
C2H5(g) + H(g)
∆H = 410 kJ mol–1
CHCl3 (g)
CCl3(g
∆H = 380 kJ mol–1
)
+ H(g)
fl usually tabulate average bond enthalpies (determined from A–B
bond enthalpies in many different A–B containing molecules)
Average bond enthalpies can be used to estimate the enthalpy of a
compound: just count the number and type of bonds involved.
Using average bond enthalpies, estimate the enthalpy change for
C3H8(g) + 5 O2(g)
3 CO2(g)
+ 4 H2O(g)
Step 1: what bonds are in the reactants (draw Lewis structures)
C3H8(g)
H
H C
H
H H
C C H
H H
1 µ (8 C-H bonds),
1 µ (2 C-C bonds)
O2
O
O
5 µ (1 O=O double bond)
nb: angles are not 90o
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Step 2: what bonds are in the products (draw Lewis structures)
4 H2O
3 CO2
O C O
3 µ (2 µ CO double bonds)
H
O
H
4 µ (2 µ OH single bonds)
Step 3: calculate the net breakage / formation of bonds: energy is
absorbed in breaking reactant bonds, and released it in making them
DHrxn º 8H(C–H) + 2H(C–C) + 5H(O=O)
– 6H(C=O) – 8H(O–H)
Finally…
DHrxn º 8 µ 412 + 2 µ 348 + 5 µ 497
– 6 µ 743 – 8 µ 463) kJ mol–1
= –1685 kJ mol–1
Value from calorimetry experiments:
∆Hco = -2220 kJ/mol
Bond enthalpies appear to give a 30% error!
• Works better when not all bonds are being broken (1st, 2nd, 3rd and
4th C–H bonds in methane have very different strengths)
• better to use thermochemical groups bonded to at least two other
atoms rather than individual bonds
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Spontaneous Changes:
Why do some processes happen spontaneously?
•Why does a hot body get cooler (rather than hotter) when
surrounded by a cooler medium?
Entropy
•Why does a gas expand into all available volume of a container
rather than contract?
The driving force for spontaneous change (change that happens
without intervention — doing work or heating) is described in the
second law of Thermodynamics
Chapter 11
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No process is possible in which the sole result is the absorption of heat
from a reservoir and its complete conversion into work.
Energy is not accumulated in ball
and thermal motion is not directional
Kinetic energy converted
into thermal motion
Entropy, S, and the second law
These spontaneous changes happen because they increase the
randomness with which energy is spread through an isolated system
The Entropy, S, a thermodynamic state function, is a measure of
“molecular disorder”, or “freedom of movement” molecules have, and
helps us to define the direction of spontaneous change
The Entropy of an isolated system increases in the course of a
spontaneous change
∆Ssystem + ∆Ssurroundings =
∆Stotal > 0
Hence, in a spontaneous process: ∆Suniverse > 0
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Entropy and Equilibrium
Entropy and Heat
Nothing changes when a system is at equilibrium, including the
entropy of the system
So, for any process
∆Ssystem + ∆Ssurroundings = ∆Suniverse r 0
Entropy measures dispersal of energy in a system
Heat changes kinetic energy of molecules, i.e. disperses energy by
increasing the velocities of all the molecules
fl heat and entropy are related?
But, equilibrium is dynamic at a microscopic level
For a reversible process
reactants Ý products
This introduces the idea that changes can be reversible, i.e. a change
can be made, and then exactly undone (“reversed”), so
∆Suniverse = 0
DS = qrev / T
= DH / T (at constant pressure)
For an irreversible process
DS > qrev / T
for a reversible process
In practice, reversible changes are an idealised limit in which changes happen
infinitely slowly via a series of imperceptible shifts in the equilibrium
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Entropy and Disorder
But …
Water freezes (spontaneously) on a cold night!
The Entropy of the System
Entropy
increases ??
Solid
Gas
Liquid
Liquid (water molecules
able to move)
Entropy of system increases
Solid (molecules fixed in
crystal)
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Spontaneous Freezing
Entropy of Phase Changes
System at transition temperature
Water freezing is an exothermic process
H2O(l)
∆H = - 6.88 kJ/mol
H2O(s)
So heat is put into the surroundings
–∆ Hsystem = + ∆ Hsurroundings= qp,surroundings b ∆ Ssurroundings
decreases
Solid
increases
∆Ssystem + ∆Ssurroundings =
∆Suniverse > 0
Entropy is not the final determinant of spontaneity — see tomorrow
∆SSysfus =
similarly, for vaporization
∆Hfus
Liquid
TFus
∆SSysvap =
∆Hvap
Tvap
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Entropy of Reaction
Trends in entropy
Increasing Temperature
As entropy is a state function, the entropy of reaction is defined (by
analogy to Hess’ law for the enthalpy) as the difference between the
molar entropies for the pure, separated products and the pure,
separated reactants:
∆Sr = Σ ν Soproducts - Σ ν Soreactants
Where o as before indicates that all substances are in their standard
states at the specified temperature.
rotation
translation
vibration
More modes to distribute energy become available
ENTROPY INCREASES
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Entropy and Disorder: the microscopic view
The Statistical Entropy
Let’s say you are tossing a coin with your mates to check who will
have to go to the next lecture to take notes (if handouts are not
provided). You decide on head and your mate on tails. As you toss the
coin for the first time, the likelihood that of head:tail is 50:50, or
1/2:1/2.
Tail
Head
# of combinations
1
1
Increase in
Entropy
Example: mixing of two gases (pure O2 and pure N2)
To quantify S, see microstate considerations below
The chance that you win heads twice is ¼, as it is to win tails twice;
the mixed result has a probability of ½
#
Tail/Tail
Tail/Head or Head/Tail
Head
1
2
1
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9
Similarly for particles in boxes
For three throw
#
T/T/T
TTH,THT,HHT
1
3
HHT,HHT,THH
H/H/H
1
3
Imagine you have four balls and can put them into two containers (T or
H) then you can distribute them as follows;
Most likely outcome
T/T/T/T
TTTH,TTHT,
THTT, HTTT
# 1
4
TTTH,TTHT,
THTT, HTTT
T/T/T/T
…. and for four throws
#
TTHH, THHT, THTH,
HHTT, HTHT, HTTH
HHHT,HHTH,
HTHH, THHH
6
1
TTHH,THHT,HTTH,
THTH, HTHT,HTTH
4
HHHT,HHTH,
HTHH, THHH
6
H/H/H/H
1
4
H/H/H/H
4
1
T
H
T
H
T
H
T
H
T
H
Even split most favored
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The Boltzmann Formula
Third Law of Thermodynamics
The entropy change for any equilibrium between pure substance
becomes zero as the temperature approaches absolute zero
S = k ln W
The Entropy of any crystalline substance approaches zero as
T | 0 K.
W = weight of most probable configuration of the system!
k = Boltzmann constant = R/N0 = 1.308630 x 10-23 J K-1
Walther Nernst 1864-1941
Nobel Laureate in chemistry in recognition of his
work in thermochemistry
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The third law
Let’s remember when processes are
spontaneous
S = k ln W
As T | 0 K, every atoms is located at the perfect crystal
location, so W = 1
But …
•Removing every defect is difficult (if not impossible)
∆Suni > 0
Spontaneous
∆Suni = 0
Equilibrium; also true for idealised
changes that are so slow they never
disturb equilibrium (reversible)
∆Suni < 0
Does not happen (reverse reaction
will be spontaneous)
•QM ensures atoms still have zero-point energy, and so some
entropy persists (see last week)
fl can’t actually achieve absolute zero
Temperatures < 0.000001 K have been achieved
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10
The Gibbs Free Energy, G
∆Suni = ∆Ssystem + ∆Ssurroundings
Requires knowledge of both system and surroundings
Gibbs Free Energy, G
At constant pressure and temperature, and for reversible changes:
T∆Ssurroundings = qp,rev = –∆Hsystem
Define a new form of the energy function
G = H – TS
Called the Gibbs free energy.
G is a state function
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The Gibbs Free Energy: criterion for spontaneity
At constant T: ∆Gsys = ∆ (Hsys - Tsys Ssys) = ∆Hsys - T ∆Ssys
∆G = ∆H - T ∆S
But also:
∆Hsys
= - ∆Ssurr
T
∆Gsys = - T ∆Ssurr - T ∆Ssys
= - T (∆Ssurr + ∆Ssys)
= - T (∆Suni) < 0 for spontaneous pr.
and as ∆Suni > 0
Resulting in:
Criterion for Spontaneity
∆Suni > 0
Spontaneous
∆Gsys < 0
∆Suni = 0
Equilibrium
∆Gsys = 0
∆Suni < 0
Not spontaneous
(reverse reaction is
spontaneous)
∆Gsys > 0
N.B. left hand describes the whole universe, but right hand is just the
system. We can drive the system in the wrong direction (hence can
have reactions where DG is positive, but cannot force the universe to
show negative entropy
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Examples
∆G = ∆H - T ∆S
∆G < 0 for spontaneous
Remember ∆Gsys < 0 for spontaneous processes
2) An endothermic process where entropy decreases
∆G = ∆H - T ∆S
1) An exothermic process which also increase entropy
∆H = + ve (>0)
∆H = - ve (<0)
∆S = + ve ; -T ∆S = - ve (<0)
∆S = - ve ; -T ∆S = + ve (>0)
∆G = -ve (<0)
spontaneous
∆G = +ve (>0)
NOT spontaneous
(reverse reaction would be
spontaneous as G is a state
function)
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11
∆G = ∆H - T ∆S
∆G < 0 for spontaneous
∆G = ∆H - T ∆S
∆G < 0 for spontaneous
3) An exothermic process in which entropy decreases (freeze water)
Depends on relative
magnitudes of ∆H and -T
∆S
∆H = - ve (<0)
∆S = - ve ; -T ∆S = + ve (>0)
4) An endothermic process in which entropy increases (melting ice)
∆S = + ve ; -T ∆S = - ve (<0)
|∆H| < |T ∆S|
|∆H| = |T ∆S|
|∆H| > |T ∆S|
|∆H| < |T ∆S|
∆G > 0
∆G = 0
∆G < 0
∆G < 0
∆G = 0
Not spontaneous
Equilibrium
Spontaneous
Equilibrium
Spontaneous
For fixed ∆H and ∆S , spontaneity depends on T
Depends on relative
magnitudes of ∆H and -T
∆S
∆H = + ve (>0)
|∆H| = |T ∆S|
|∆H| > |T ∆S|
∆G > 0
Not spontaneous
For fixed ∆H and ∆S , depends on T
Stanford Chemistry
Stanford Chemistry
Summer Session 2005 Chem 31
Summer Session 2005 Chem 31
Driving forces in chemistry
Gibbs Energy for Phase Transitions
Two driving forces underpin Chemistry:
P = 1atm
¾ Systems tend to a state of minimum enthalpy
H2O(s)
H2O(l)
¾ Systems tend to a state of maximum entropy
H2O(l)
H2O(s)
The Gibbs free energy expresses the balance between these two
driving forces
DG = DH – T DS
(b 0 ?)
∆G = ∆H - T ∆S
kJ
∆Hofus(273.15)= + 6.0 mol
kJ
∆Hofrze(273.15)= – 6.0 mol
∆Hfrze, 273.15
∆Sfrze, 273.15
=
Tfrze=273.15
∆G273.15 = ∆H273.15 - Tfrze=273.15
Equilibrium ∆S273.15 = 0
Stanford Chemistry
Stanford Chemistry
Summer Session 2005 Chem 31
Summer Session 2005 Chem 31
If ∆H and ∆S are independent of temperature (a good, but not perfect,
approximation), then:
T < Tfrze
T > Tfrze
As the Gibbs energy is a state function, the (standard) Gibbs energy
of reaction is defined (in analogy to Hess’ law for the enthalpy) as:
∆G = ∆H - T ∆S
∆G < 0
Spontaneous Freezing
∆G = ∆H - T ∆S
∆G > 0
Gibbs Energy for Chemical Reaction
Spontaneous Melting of Ice
Clearly, the spontaneity of freezing (negative ∆S and negative ∆H)
depends on a balance of the entropy and enthalpy
∆Gro = Σ ν Goproducts - Σ ν Goreactants
Where o as before indicates that all substances are in their standard
states at the specified temperature.
∆Gro = ∆ Hro - T ∆ Sro
Stanford Chemistry
Stanford Chemistry
Summer Session 2005 Chem 31
Summer Session 2005 Chem 31
12
Standard Molar Gibbs Energy of Formation
Example:
By analogy with the definition of the standard molar Enthalpy of
formation we define
Pt (NH3)2I2, cis
Pt (NH3)2I2, trans
∆Gfo = ∆ Hfo - T ∆ Sfo
as the standard molar Gibbs energy of formation when 1 mole of a
substance forms in a standard state at a specified temperature from the
most stable forms of its constituent elements in standard states at the
same temperature.
∆Gfo (298.15K)
∆Hfo(298.15K)
cis
-286.6 kJ/mol
trans
-316.9 kJ/mol
-130.2 kJ/mol
-161.5 kJ/mol
Calculate the standard entropy,
Pt(s) +
2 NH3,(g)
Stanford Chemistry
Stanford Chemistry
Summer Session 2005 Chem 31
Gibbs Free Energy & Equilibrium Constants
Pt(NH3)2I2 (s)
cis
trans
∆ Sfo = (-316.9+161.5) kJ/mol
298.15K
= -0.5212 kJ/(K mol)
= -0.5246 kJ/(K mol)
∆ Sfo = So(Pt(NH3)2I2) - So(Pt(s)) - 2 So(NH3,(g)) - So(I2,(s))
-0.5425 kJ/(K mol)
So(Pt(NH3)2I2)cis = 17.9 J/(K mol)
Now remember, at
equilibrium, ∆Gr = 0
and Q = K
G
So(Pt(NH3)2I2)trans = 21.3 J/(K mol)
cCg +
bBg
dDg
For the “standard” reaction all reactants and products in standard
states, but in general they are related by the reaction quotient.
∆Gr = ∆Gro + RT lnQ
(PC)c (PD)d
N.B. both ∆G and Q can be used to
determine the direction of change:
∆G | 0
Q | K
Q=
(PA)a (PB)b
Stanford Chemistry
Stanford Chemistry
Summer Session 2005 Chem 31
Temperature and the Equilibrium Constant
∆Gr = ∆Gro + RT lnQ
∆Gro = - RT lnK
Assume ∆S and ∆ Hro don’t change with temperature!
∆Gr = - RT lnK + RT lnQ
= RT ln Q/K
RT2 ln K 2 = ∆H − T2 ∆S
RT1 ln K1 = ∆H − T1∆S
⎛K ⎞
∆H − T1∆S ∆H − T2 ∆S
ln ⎜ 1 ⎟ = ln K1 − ln K 2 =
−
RT1
RT2
⎝ K2 ⎠
1852-1911
Spontaneous
to products
reactants
aAg +
Summer Session 2005 Chem 31
Equilibrium
∆G=0 Q=K
∆G<0
Q<K
of both compounds at 298.15 K.
Summer Session 2005 Chem 31
∆ Sfo = ∆ Hfo - ∆ Gfo
T
+ I2,(s)
∆ Sfo = -286.6 + 130.2 kJ/mol
298.15K
So,
Nobel Prize in
Chemistry, 1901
The van’t Hoff Equation
∆G>0
Q>K
Not spontaneous to
reactants
ln
∆Hro
K2
K1 =
R
( 1T
1
1
T2
)
products
Stanford Chemistry
Stanford Chemistry
Summer Session 2005 Chem 31
Summer Session 2005 Chem 31
13
Similarly, for liquid-vapour equilibrium
The Clausius-Clapeyron Equation
ln
∆Hvapo 1
P2
=
(T
P1
R
1
1
)
T2
assuming ∆S & ∆ Hro independent of temperature
This expression is particularly useful when you want to determine the
boiling points of substances at different pressures.
Example:
Water boils at 373.15K at an atmospheric pressure of 1.0 atm.
Calculate the boiling temperature of water on top of Mt Everest
(assuming a pressure of about 0.36 atm). The enthalpy of
vaporization of water can be assumed to be 40.7 kJ/mol.
∆Hvapo 1
ln P2 =
( T1
P1
R
8.314 J ln
0.36 atm
1 atm
K mol 40.7 103 J mol-1
1
T2
P1 =1.0 atm P2 = 0.36atm
)
T1 = 373.15 K
∆Hvapo = 40.7 kJ/mol
1
373.15 K
=
1
T2
=
1
346 K
Stanford Chemistry
Stanford Chemistry
Summer Session 2005 Chem 31
Summer Session 2005 Chem 31
14