Download Gibbs Free Energy and chemical equilibrium

1/21/14
OCN 623: Thermodynamics,
Chemical Equilibrium, and Gibbs
Free Energy
or how to predict chemical reactions
without doing experiments
Introduction
•  We want to answer these questions:
•  Will this reaction go?
•  If so, how far can it proceed?
We will do this by using thermodynamics.
•  This lecture will be restricted to a small subset of
thermodynamics...
1
1/21/14
Outline
• 
Definitions
• 
Thermodynamics basics
• 
Chemical Equilibrium
• 
Free Energy
• 
Aquatic redox reactions in the environment
Definitions
•  Extensive properties -depend on the amount of
material
–  e.g. # of moles, mass or volume of material
–  examples in chemical thermodynamics:
–  G -- Gibbs free energy
–  H -- enthalpy
•  Intensive properties - do not depend on quantity or
mass
e.g.
–  temperature
–  pressure
–  density
–  refractive index
2
1/21/14
Reversible and irreversible
processes
•  Reversible process occurs under equilibrium conditions
e.g. gas expanding against a piston Net result - system & surroundings back to initial state
P
p
p = P + ∂p reversible
p = P + ∆p irreversible
Reversible Process
•  No friction or other energy dissipation
•  System can return to its original state
•  Very few processes are really reversible in
practice
–  e.g.
Daniell Cell Zn + CuSO4 = ZnS04 + Cu
with balancing external emf
3
1/21/14
Daniell Cell Zn + CuSO4 = ZnS04 + Cu
with balancing external emf
-cell voltage is a measure of:
ability of oxidant to gain
electrons,
ability of reductant to lose
electrons.
Zn oxidation, Cu reduction
Zn(s) = Zn2+(aq) + 2e-
Cu2+(aq) +2e- = Cu(s)
Compression of gases
Vaporization of liquids
Reversibility is a concept used for comparison
Will get more of this next lecture, covering redox potential
Spontaneous processes
•  Occur without external assistance
–  e.g.
- expansion of a gas from region of high
pressure to low pressure
- diffusion of a solute in a solvent
Are irreversible by definition -- they occur at a finite rate
4
1/21/14
Thermodynamics: The Basics
•  Nature of energy: W = F x d
capacity to do work or transfer heat
•  Goals of (re) learning thermodynamics
–  To compute the equilibrium of a system
–  To determine how far a given system is from
equilibrium
–  To evaluate the energy available from a given reaction
1st law of thermodynamics
•  Energy cannot be created or destroyed
–  when one kind of energy disappears it is transformed into an
equivalent quantity of another kind
∆U = q-w
dU = dq - dw
dU = change in internal energy of the system
dq = heat absorbed by the system +
dw = work done by the system +
Energy is conserved
5
1/21/14
•  U is a thermodynamic function
dU depends only on the initial and final
states of the system-not the path taken
q and w are NOT thermodynamic
functions
Internal energy of system is increased by
gaining heat (q)
Internal energy of system is decreased when
work is done by the system
dU = dq - dw
wrev. > wirrev.
• 
Work done by system during irreversible
process is less than could be obtained if
process was reversible
e.g. if piston moved at finite rate,
friction dissipates energy 6
1/21/14
Chemical thermodynamics
Heat
•  Heat (q) is measured in terms of:
amount of substance x molar heat capacity x temp rise
•  For phase change
q = moles of substance x molar latent heat
•  q is the heat of reaction
•  Molar heat capacity = Heat energy per mole required to
raise temperature of 6.022x1023 molecules 1oC
WORK
(1) Work of expansion = P ∆V
where P = pressure and ∆V is change in volume
at constant volume ∆V = 0, therefore w =0
no work is done
7
1/21/14
(2) Electrical work – done on charged
species
•  Occurs in electrical cells e.g.
Zn + Cu2+ = Zn2+ + Cu
•  Electrical energy = I * E * t E is the emf (voltage), t = time, I = current flowing
It = zF for 1 mol
z = # of electrons transferred,
F (Faraday)= 96,490 Coulombs/mol
Electrical energy = zEF
•  System at constant volume
(all non-gas reactions are at constant volume)
P∆V = w= 0
∆U = qv qv= heat at constant volume •  System at constant pressure
(all reactions open to atmosphere)
∆U = qp - P∆V
Ideal gas law P∆V = ∆nRT, (n = amount of substance)
R is the gas constant = 8.314 Joules mol-1 K-1
therefore ∆U = qp - ∆nRT
rearranging qp = ∆U + ∆nRT
8
1/21/14
• 
qp is called the enthalpy H change in enthalpy ∆H = ∆U + P∆V
in absolute terms H = U + PV •  H is a thermodynamic property,
is defined in terms of thermodynamic
functions: U, P and V
For an infinitesimal change at constant pressure:
dH = dU + PdV
•  ∆Hf o298 is the heat of formation of 1 mole of a compound from its
elements at 298oK
•  ∆H is neg = exothermic reaction •  ∆H is pos = endothermic reaction
(seen from system perspective)
•  ∆H is proportional to amount of material
is an extensive property
H2(g) + 1/2O2(g) --> H2O(g)
This reaction releases 241.8 kJ of heat, therefore heat of formation is:
-241.8 kJ
9
1/21/14
•  ∆H is equal in magnitude but opposite in
sign for the reverse reaction, because it is a
thermodynamic quantity
∆Hproducts - ∆Hreactants
∆H for reaction is same regardless of number of
steps between reactants and products
∆H = ∆U + ∆nRT
•  Can use component reactions to calculate enthalpy
change for unknown reaction
Reaction Enthalpy
H2 + 0.5 O2 = H2O ∆H = -285.8 kJ mol-1 (a)
C + O2 = CO2
∆H = -393.3 kJ mol-1 (b)
C2H6 + 3.5 O2 = 2 CO2 + 3H2O ∆H = -1559.8 kJ mol-1 (c) can use above to calculate ∆H for ethane formation
2*b + 3*a - c
2C + 2O2 + 3H2 + 1.5 O2 - C2H6 - 3.5 O2 = 2CO2 + 3H2O - 2
CO2 - 3H2O
canceling yields
2C + 3 H2 = C2H6 ∆H = 2* -393.3 + 3* -285.8 - (-1559.8)
= -84.2 kJ mol-1
10
1/21/14
•  But enthalpy change alone is insufficient to
allow prediction of likelihood of reaction
•  Entropy change is also needed
2nd law of thermodynamics
dS = dqrev
T
change in entropy = amount of reversible heat
absorbed by system divided by temperature
Entropy of an isolated system which is not in equilibrium will tend to
increase over time
(a) Entropy is the degree of disorder of the
system
chance of finding something in a fixed volume- higher
pressure, greater chance, liquid lower entropy than gas,
solid lower than liquid
(b) Change in entropy measures capacity for
spontaneous change
diffusion of a solute from high concentration to low
concentration, or a gas from high pressure to low pressure
•  A system undergoing spontaneous change
is moving to a greater degree of disorder 11
1/21/14
•  Entropy is an extensive property
•  Units heat/temperature = Joules K-1 mol-1
for solids dq = moles x molar heat capacity x
temperature rise dq = CmdT
Cm = molar heat capacity and dT = temperature change
molar entropy change:
∆Sm = from T1 to T2 ∫ CmdT
T
∆Sm = Cm ln T2
T1
•  For phase change
∆S = L
T
where L is the latent heat of fusion
e.g. for water L = 6 kJ mol-1 ∆S = 6000 = 22.0 J K-1mol-1 for melting ice
273.1
•  reversible reaction the net change in entropy
is zero
•  irreversible reaction there is a net increase
in entropy
12
1/21/14
•  Since nearly all reactions are irreversible,
entropy is increasing -- called times arrow
•  always increasing --cannot run time
backwards
Third law of thermodynamics
The entropy of a perfectly crystalline material
at the absolute zero (-273.1˚ C) is zero
•  entropy allows prediction of the feasibility
of reactions
Why study chemical equilibrium?
• 
Is a geochemical system at chemical equilibrium?
• 
If not, what reactions are most likely to occur?
• 
Solubility -
• 
Redox -
• 
Complexation -
• 
Carbonate system -
13
1/21/14
Why study chemical equilibrium?
• 
Is a geochemical system at chemical equilibrium?
• 
If not, what reactions are most likely to occur?
• 
Solubility - diatoms in surface seawater
• 
Redox - organic matter oxidation
• 
Complexation - iron speciation & plankton growth
• 
Carbonate system - CaCO3 in marine sediments
Equilibrium vs. kinetics
• 
Equilibrium calculations give the energy available from
a reaction & the direction it will proceed
- Usually simpler & require less info than those for
kinetics
• 
Often a good approximation for many systems
• 
In some cases, equilibrium calculations can predict
kinetic rate constants
• 
HOWEVER, most natural water reactions are not at
equilibrium and knowledge of kinetics is often required
- Steady state, states of dynamic equilibrium, reaction
rates, are topics outside of the scope of this course,
but will be covered in 643 in Fall.
14
1/21/14
Chemical Equilibrium - 1
Consider a reversible reaction taking place at constant
temperature:
aA + bB
cC + dD
The reactants A and B combine to form products C and D.
The concentrations of A and B decrease until they reach
values that do not change with time:
Chemical Equilibrium – 2
•  The time-invariant concentrations of reactants and
products are called equilibrium concentrations.
•  The ratio of these concentrations (or activities –
active concentrations) is characteristic for each
reaction, and is called the equilibrium constant, K:
{C}c {D}d
K≡
{ A}a {B}b
•  Note that at equilibrium, the forward and reverse
reactions proceed at the same, stable rate.
15
1/21/14
Free Energy – 1
•  A criterion for equilibrium is that the total free energy
(Gibbs free energy, Gr) of the reaction is at a minimum:
•  If we add more reactant or more product, the reaction will
proceed spontaneously (without external help) as long as the
value for Gr decreases.
•  Thus, a reaction in the direction of decreasing Gr is
spontaneous. A reaction in the direction of increasing Gr is not
spontaneous, and will not occur in a closed system.
Free Energy - 2
As any reaction proceeds an incremental amount, the
change in Gr can be calculated as:
&
#
&
#
ΔGr = $ ∑υi G fi !
− $ ∑υi G fi !
% i
" products % i
" reactants
where νi is the stoichiometric coefficient (a,b,c,d) and Gfi is the free
energy of formation per mole.
1.  If ΔG < 0, (i.e., ΔG is negative and thus Gr decreases as the
reaction proceeds), then the reaction proceeds spontaneously as
written.
2.  If ΔG > 0, (i.e., ΔG is positive and thus Gr increases as the reaction
proceeds), then the reaction proceeds spontaneously in the
opposite direction as written.
3.  If ΔG= 0, (i.e., ΔG is at a minimum), then the reaction is at
equilibirium and will not proceeds spontaneously in either direction.
16
1/21/14
Free Energy - 3
Values for ΔG for a reaction give us a powerful tool to predict if a
reaction is possible. We calculate in-situ ΔGr using this equation:
ΔG r = ΔG r ° + RT ln
In-situ
Std. state
{C}c {D}d
{ A}a {B}b
where
&
#
&
#
ΔGr ° = $ ∑υi G° fi !
− $ ∑υi G° fi !
% i
" products % i
" reactants
(The the superscript zero (°) indicates standard state: 25°C (298°K),1 atm
pressure, and activity = 1.)
Gfi° is the standard-state free energy of formation per mole of species i.
{ } = activity (active concentration)
R = the ideal gas constant = 1.987 cal °K-1 mol-1 = 8.31 J °K-1 mol-1
T = °K
Free Energy - 4
Standard free energy of formation (ΔG°fi):
•  G°fi = 0 at standard state for all pure elements (solid
reference).
•  G°fi = 0 for H+ at a concentration of 1 mole/liter at standard
state (solution reference).
•  Allows the measure of the energy change involved in
forming compounds at standard state from their
component elements at standard state.
•  Measured values are listed in tables. Units are:
•  kJ/mol (SI units)
•  kcal/mol
17
1/21/14
Table is included in reading
(handout)
See complete table in handout.
18
1/21/14
Sample Calculation #1
Consider the dissolution of CaCO3 (calcite) in aqueous solution:
CaCO3 → Ca2+ + CO32Does the reaction proceed spontaneously as written?
Calculate the free energy of reaction at standard state (products and
reactants at activity = 1, P = 1 atm, T = 25°C):
&
#
&
#
ΔG°r = $ ∑υi G° fi !
− $ ∑υi G° fi !
% i
" products % i
" reactants
ΔG°r = (1(-132) + 1(-126) – (1(-270)) kcal/mol
=
=
(-258 + 270) kcal/mol
+12 kcal/mol
Therefore, at standard state, the reaction spontaneously proceeds in the
opposite direction to what is written.
19
1/21/14
Activities – How to Calculate
Activity of water = 1
Equilibrium at In-Situ Conditions - 1
We have already seem that:
ΔG = ΔG ° + RT ln
In-situ
Std. state
{C}c {D}d
{ A}a {B}b
We can define a reaction quotient, Q, using in-situ conditions:
Q≡
Thus:
{C}c {D}d
{ A}a {B}b
ΔG = ΔG° + RT ln Q
In the case of equilibrium, Q = K and ΔG = 0:
Thus:
0 = ΔG° + RT ln K
ΔG° = -RT ln K
Substituting (for the general case): ΔG = -RT ln K + RT ln Q = RT ln Q/K
20
1/21/14
Equilibrium at In-Situ Conditions - 2
ΔG = RT ln Q/K
This allows us to develop a set of criteria to determine in which
direction a reaction will proceed under non-standard conditions.
This is because Q/K will determine the sign of ΔG:
1.  If Q/K < 1, then ΔG is negative, and the reaction is
spontaneous as written.
2.  If Q/K = 1, then ΔG = 0 and the system is at equilibrium.
3.  If Q/K > 1, then ΔG is positive, and the reaction is
spontaneous in the opposite direction as written.
Sample Calculation #2
Again consider the dissolution of CaCO3 (calcite):
CaCO3 → Ca2+ + CO32In the surface ocean, does the reaction proceed spontaneously as written?
Use ΔG° = -RT ln K
From Sample Calculation #1, we know: ΔG° = +12 kcal/mole
Thus:
cal & 1 kcal #
12 kcal
mol = − 1.9987 ° K mol $
cal !(298° K )ln K
% 1000
"
(
)
(all units cancel out)
ln K = -20.27
K = 1.58 x 10-9 (“equilibrium constant”)
21
1/21/14
Sample Calculation #2 (cont.)
Surface seawater: {CaCO3} ≡ 1
[Ca2+] = 0.01 mol/L; γ = 0.28; {Ca2+} = 0.0028
[CO32-] = 45 µmol/L; γ = 0.21; {CO32-} = 9.45 x 10-6
{Ca }{CO }= (2.8x10 )(9.45x10 )= 2.65x10
2−
2+
Q=
−3
−6
3
{CaCO3 }
−8
1
Q 2.65x10 −8
=
= 17.6
K 1.58x10 −9
Therefore:
•  The reaction goes in the opposite direction as written (Q/K > 1)
•  CaCO3 precipitates in the surface ocean
•  Surface seawater is supersaturated with respect to calcite
Note: No information is given on the kinetics of the reaction!
Temperature and Pressure Effects - 1
The amount of heat that is released or taken up by a reaction is
called the enthalpy change (ΔH°).
Similarly to ΔG°, we can calculate ΔH° using tabulated data:
&
#
&
#
ΔH r ° = $ ∑υi H ° fi !
− $ ∑υi H ° fi !
% i
" products % i
" reactants
where νi is the stoichiometric coefficient (a,b,c,d), and H°fi is the enthalpy of
formation of species i (kcal/mol or kJ/mol) at standard state.
The van’t Hoff equation gives the temperature dependence of K:
ΔH r °
' ∂ ln K $
%
" =
2
∂
T
&
# P RT
For a small change in T, ΔH° does not change much, so we can
integrate directly:
& K # ΔH r ° & 1
1 #
ln$$ T * !! =
−
$
!
R % 298 T * "
% K 298 "
where T* is the temperature of interest.
22
1/21/14
Temperature and Pressure Effects - 2
The change in the partial molar volume (ΔV°) during a reaction at
standard state is also calculated using tabulated data:
&
#
&
#
ΔVr ° = $ ∑υiV ° fi !
− $ ∑υiV ° fi !
% i
" products % i
" reactants
where νi is the stoichiometric coefficient (a,b,c,d), and V°fi is the partial molar
volume of formation of species i (cm3/mol) at standard state.
The pressure dependence of K is also known:
ΔVr °
' ∂ ln K $
%
" =
RT
& ∂P #T
Again integrating directly:
( K % ΔVr °
(P * −1atm )
ln&& P* ## =
' K1atm $ RT
where P* is the pressure of interest (in atm).
Temperature and Pressure Effects - 3
See the following example in the handout for an example of
temperature-correcting the equilibrium constant:
23
1/21/14
•  For exothermic reactions lnK increases as 1/T increases,
(T decreases), favouring product formation
•  For endothermic reactions lnK increases as 1/T decreases
(T increases) favouring product formation
Exothermic
lnK
Endothermic
1/T
…putting it all together in terms of redox…
24
1/21/14
Redox Reactions in Nature
•  Example: net reaction for aerobic oxidation of organic matter:
CH2O + O2 → CO2 + H2O
•  In this case, oxygen is the electron acceptor – the half-reaction is:
O2 + 4H+ + 4e- → 2H2O
•  Different organisms use different electron acceptors, depending on
availability due to local redox potential
•  The more oxidizing the environment, the higher the energy yield of the
OM oxidation (the more negative is ΔG, the Gibbs free energy)
•  The higher the energy yield, the greater the benefit to organisms
that harvest the energy
•  In general:
–  There is a temporal and spatial sequence of energy harvest
during organic matter oxidation
–  Sequence is from the use of high-yield electron acceptors to
the use of low-yield electron acceptors
25
1/21/14
Why is organic matter an electron donor?
Z-scheme for photosynthetic electron transport
Falkowski and Raven (1997)
ADP→ATP
Energy Scale
Ferredoxin
ADP→ATP
to Krebs cycle
and carbohydrate
formation
photooxidation
of water
energy from sun converted to C-C, energy rich, chemical bonds
26
1/21/14
Environmentally Important Organic Matter
Oxidation Reactions Eh (V)
ΔG
+0.812
-29.9
+0.747
-28.4
+0.526
-23.3
-0.047
-10.1
-0.221
-5.9
-0.244
-5.6
Reduction of O2
+
-
O2 + 4H +4e --> 2H2O
Reduction of NO3-
+
-
2NO3 + 6H + 6e --> N2 + 3H2O
Reduction of Mn (IV)
MnO2 + 4H+ + 2e- --> Mn2+ +2H2O
Reduction of Fe (III)
+
2+
Fe(OH)3 + 3H + e --> Fe +3H2O
Reduction of
2-
SO42+
-
SO4 + 10H + 8e --> H2S + 4H2O
Reduction of CO2
CO2 + 8H+ + 8e- --> CH4 + 2H2O
DECREASING ENERGY YIELD
Reducing Half-reaction
The Next Lecture: Thursday January 23
“Redox & pE-pH”
27
1/21/14
Thermodynamics and kinetics
Lecture 2. Energy and life
Catalysis
Catalysis = Speeding reactions without getting used in the process.
K = equilibrium constant
k = rate constant
Energy
Exergonic direction
Endergonic direction
kf
kb
Energy
Catalysis (Cat) = the catalyzed reactions proceed faster toward equilibrium (K).
Catalyst
Energy
Energy
Kinetic inhibition may have been used by life to create an energy capacitor
George Cody citing Everet Shock, OLEB, 1990.
28