Lecture 4

Thermodynamic considerations

Pease read Ch2, Ch3 (White)

• Chemical Thermodynamics
– Into and The Phase Rule
• Enthalpy
• Entropy
• The Gibbs Function
• ∆G and K

GG325 L4, F2013

Chemical Thermodynamics

Thermodynamics is the study of the energetics of physical

and chemical transformations of matter.

The energetics of a "system " are described with

physical (e.g., temperature = T, pressure = P)

and
chemical (e.g., composition = X1, X2, X3,…) quantities.

A system can contain more than one chemical component

(e.g., H2O) and more than one physiochemical phase (ice,
water vapor, liquid water).
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Definitions
• system - a collection of matter in an identifiable place and
condition. A system can contain more than one chemical
component and more than one physiochemical phase.
• A component is a chemical entity (e.g., H2O) that can be used
to describe compositional variation in a system.
Normally, we choose the minimum number of components (C) to describe the
system: C = n – r , where n is the number of chemical species and r is the
number of reactions that can occur between these species. A component need
not physically exist in a system (e.g., plagioclase can be described as a solid-
solution mixture of albite and anorthite components, but neither exists in a
plagioclase grain of intermediate composition, like labradorite or andesine).

• A phase is a physical form of that chemical

e.g., for H2O: ice, water vapor, liquid water
for SiO2 quartz, chert, cristobalite.

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The Phase Rule:

This provides a basis for thermodynamically describing a system by telling us the
number of system variables (P, T, composition X1, X2, X3, etc.) we must define for
a unique set of conditions to apply

F = C - P +2
(Sometimes written as f = c -  +2)
F = degrees of freedom (we want this to be
a small number)
C = components (chemical constituents)
P = phases (i.e., minerals)
Some Examples:
a). A univariant system...
(see figure at right)

H2O (l) ↔ H2O (v) (=steam)

P=2 (aqueous and vapor)
C=1 (water)
F=1 (T or P)

i.e., if we know T, then we can determine P,

or vice-versa
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 Some More Examples:
b). A simple system... dissolved silica
SiO2 (s) + H2O (l) ↔ H4SiO4 (aq)

P=2 (aqueous and solid)

C=2 (SiO2 and water)
F=2 (e.g., T & P)

c). Another simple system...”salt” water

NaCl (s) + H2O (l) ↔ Na+ (aq) + Cl- (aq)

P=2 (aqueous and solid)

C=3 (water, sodium ions, chloride ions)
F=3 (e.g., T & P and either Na+ or Cl-)

d) The forsterite-fayalite-melt system...

Mg2SiO4 + Fe22SiO4 form the solid solution mineral olivine, so they count as
one phase.

P = 2 (melt + olivine)
C = 2 (forsterite + fayalite)
F = 2 (e.g., T & P)
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In a complex natural system,

such as sea water, or magma (melt and crystals),

there are significantly more components.

for instance, other minerals with additional chemical elements,

gasses such as water and carbon dioxide, etc..)

• We USUALLY restrict our system composition to the major

components and use them to approximate the overall system
conditions.

•Trace constituents may be used to refine our approximation.

Trace elements are present in such fleeting quantities that they

don't dramatically alter the energetics of the system or its
phases, so they do not contribute to F.
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Chemical Thermodynamics

► Please refresh your memory on:

enthalpy (H)
entropy (S)
Gibbs function (G)
and their relationships to measures of chemical equilibrium
(and chemical equilibrium constants, K).
for instance, the last slide of last lecture, inserted after this slide for convenience

In all cases the standard state condition (annotated by "°" next

to the variable) is for STP (298.15°K = 25°C, 1 atm)

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Energy and Chemical Equilibrium

Recall from your prior chemistry classes that the total system energy available for
chemical work is known as Gibbs Free Energy, ∆G.
► ΔGreaction = 0 at chemical equilibrium, (the total system energy is no longer
changing.
ΔGreaction is directly related to the equilibrium constant for a chemical reaction:
ΔGo = -RT lnK (the superscript  indicates conditions of STP)
For aA +bB ↔ cC + dD, K= Cc Dd …so ΔGo = -RT ln Cc Dd
Aa Bb Aa Bb
R is the Universal Gas Constant (8.314 J = 0.08206
L-atm = 82.06 cm3-atm = 1.987 calories all per mol oK)

large K (products favored) = negative ΔGo

Small K (reactants favored) = positive ΔGo
►What if we are not at chemical equilibrium?
ΔG = ΔGo + RT lnQ remember, AT EQUILIBRIUM (Q=K) and ΔG = 0

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Some Thermodynamic Quantities and Relationships
The laws of thermodynamics and the mathematical
relationships that derive from them describe infinitesimal
changes to system variables ("variables of state") using
differential calculus – see the White text for these derivations.
• In this class we will only “break down” the energetics of reactions part of the way.
• You will not have to derive thermodynamic expressions from first principles and
partial differential equations.
• The "finite" change versions of various equations will be sufficient (e.g., ∆H, ∆G)
for our applications.

Variables such as G, H and S are not measurable as absolute

values. Instead, G, H and S are compared to reference
materials in the standard state for which arbitrary values
(typically 0) are defined.

The variable (G, H or S) is then defined as a ∆ (change) relative to that

(i.e., ∆G, ∆H or ∆S).
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Enthalpy – H
Enthalpy is a measure of energy in the form of heat (which itself
is a measure of molecular vibration or motion).

Enthalpy is important for understanding the energy consumed or

released in changes of state of a system.

The relationships between

heat (Q) and

work (W = -P∆V) to
internal energy (U, e.g., ∆U = Q + W) and
enthalpy (H, e.g., H = U + PV)

are details we are going to skip over here

(see equations in the reading if you’re interested in the details).
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Enthalpy – H

∆H = ∆U + V∆P + P∆V, or ∆Q + ∆W + V∆P + P∆V

which reduces to
∆H = ∆Q + V∆P

It is therefore useful to remember that at constant pressure :

∆H = Q
(the Enthalpy change of a process is the heat gained or lost in the process)

∆H = Cp ∆T (where Cp = heat capacity)

(the Enthalpy change of a process is the heat capacity times the temperature change)

and…

∆H = T∆S + V∆P at constant temperature

(the Enthalpy change of a process is relates to changes in entropy, which we define
momentarily, and pressure)
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Enthalpy – H
In General, the enthalpy change of a chemical reaction at STP is
∆H°reaction = Σ∆Hf°(products)- Σ∆Hf°(reactants)

For A +B ↔ C + D
∆H°reaction = [ ∆Hf° (C) + ∆Hf° (D) ] - [ ∆Hf° (A) + ∆Hf° (B) ]

The enthalpy of a chemical reaction reflects changes in 3 basic

types of system heat:

1. heat that is internal to chemical bonds between different

atom pairs
2. heat associated with molecular ordering in solids versus
liquids versus gasses.
3. heat associated with molecular motion (kinetic energy)
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Enthalpy – H 1. "Bond Heat"
This energy associated with chemical bonds. The Heat of
Formation Hf of a compound is the heat input or given off when
making a molecule from its constituent elements.

We define Hf°=0 for the constituent elements in the form that is

stable at STP (i.e., for C, the STP form is graphite, and Hf°=0)

So C (graphite) ↔ C (diamond) must involve a transfer of heat

We can learn what is ∆Hf° (standard heat of formation) of

diamond relative to this from a table of thermodynamic data:

∆Hf° of C(diamond) = -453 cal/mol (= -1.897 kJ/mol,

using 1 calorie = 0.0041868 kJ)
453 calories or -1.897 kJ/mol of heat must be added to a mole of graphite to
transform it into diamond at STP.
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Enthalpy – H 1. "Bond Heat"

If two reactions can be mathematically combined to make a
third, we can also combine their ∆Hf° to calculate ∆Hf° of the
new reaction.

For example, we might find these two reactions in a table:

1. C (graphite) + O2 (g) ↔ CO2 ∆Hf°= 94.051 kcal/mol
2. C (diamond) + O2 (g) ↔ CO2 ∆Hf°= 94.504 kcal/mol

if we reverse the order of equation 2 and change the sign of

∆Hf°, we can add the two to get
C (graphite) ↔ C (diamond)

∆Hf° = (94.051 - 94.504) kcal/mol

∆Hf° = -0.453 kcal/mol = -1.897 kJ/mol

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Enthalpy – H 2. "phase change heat"
Heat of Fusion ∆Hfus is the heat absorbed by a solid as it melts.
-∆Hfus is the heat needed to be removed to resolidify

Heat of Vaporization ∆Hvap is the heat absorbed by a liquid as it

vaporizes. -∆Hvap is the heat needed to be removed to condense

+ heat

- heat

3. other "kinetic energy heat"

In the absence of phase changes and
chemical reactions, heat is gained or
Snowflake photos by Wilson Bentley
circa 1902 lost by temperature changes.
http://en.wikipedia.org/wiki/Wilson_Bentley
∆H = Cp ∆T
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Entropy - S
Entropy is a measure of system disorder.

In general, systems tend to evolve toward less ordered

conditions, reflected in the amount of heat absorbed by a system
from its surroundings during any physiochemical change at a
given temperature.
∆S = Qr/T in other words, Qr= T∆S
also, at constant P, ∆S = Cp/∆T (where Cp = heat capacity)

The standard state for any compound S° is determined relative

to S° of H+ = 0 at STP.

The calculation of ∆S°reaction is the same as that for H:

∆S°reaction = Σ∆Sf° (products)- Σ∆Sf° (reactants)
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"Free Energy", or the Gibbs Function – G
Total system energy available for chemical work is known as
Gibbs Free Energy, ∆G.

► ∆Greaction = 0 at chemical equilibrium, meaning that

the net chemical work is no longer occurring.

Like other variables of state, ∆G is additive:

∆G °reaction = Σ∆Gf° (products) - Σ∆Gf° (reactants)

Since ∆G = ∆G° + RT lnQ

(where Q is the reaction coefficient, not heat)

and since AT EQUILIBRIUM Q=K and ∆G = 0, we derive:

∆G° = -RT lnK

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"Free Energy", or the Gibbs Function – G

In closed systems, G is related to enthalpy H and entropy S by:

G = H-TS.

For isothermal (constant T) conditions:

∆G = ∆H - T∆S and ∆Gf° = ∆Hf° - T∆Sf°

In other words, the free energy change in a closed system

reflects the heat and the degree of disorder.

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"Free Energy", or the Gibbs Function – G
another useful general relationship which applies to
processes or reactions where T and/or P are changing is
∆G = V∆P - S∆T

Consider a P-T phase diagram like this one

At any boundary the two phases will be in
equilibrium, so that ∆G =0. Changes in
volume (∆V) and entropy (∆S) associated
with small changes in P and T can be written as:
d∆G = ∆VdP – ∆SdT = 0
Rearranging, we arrive at the famous Clapeyron Equation
dT = ∆V and because S = H/T, dT = T∆V
dP ∆S dP ∆H
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"Free Energy", or the Gibbs Function – G

Let’s solve a problem using the relationship of G and K
What is Keq for the precipitation reaction of the
mineral malachite from water?

2Cu2+(aq) + CO32-(aq) + 2 OH-(aq) ↔ Cu2(OH)2CO3 (s)

"copper 2+ "carbonate "hydroxyl "malachite"
cation" anion" anion"

First, notice how the equation is balanced:

reactants products
material: Cu 2 2
CO3 1 1
OH 2 2
charge: Cu 4 (2x2) --
CO3 -2 --
OH -2 (-1x2) --
Cu2(OH)2CO3 -- 0
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"Free Energy", or the Gibbs Function – G
Next, get ∆G°reaction = Σ∆Gf°(products) - Σ∆Gf°(reactants) using
values for ∆Gf° from a compilation of thermodynamic data.
A word on units:
Always watch units when solving problems using thermodynamic equations.
•R is the Universal Gas Constant (8.314 J = 0.08206
•L-atm = 82.06 cm3-atm = 1.987 calories all per mol oK).
•R should be in the same system of units as the thermodynamic data you are using
Some books give R in non-standard units (cm3-atm/mol-oK) or (cal/mol-oK). Others use
the SI units: use J/mol-oK and J/mol, respectively.

2Cu2+(aq) + CO32-(aq) + 2 OH-(aq) ↔ Cu2(OH)2CO3 (s)

∆G°reaction=-216.44-(2·15.50+ 2·-37.60 -126.22) kcal/mol
∆G°reaction=-46.02kcal/mol = -46,020 cal/mol
From ∆Go = -RT lnK, we get ∆Go/(RT) = -lnK
-46,020/(298.15·1.987) = -lnK K = e46,020/(298.15·1.987)
K= 5.45 x 1033 or about 1033.7
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