### Free Energy I

```UNIT 5
Thermodynamics
Free Energy
Electrochemistry
Internal Energy
FIRST LAW OF THERMODYNAMICS: ΔE = q + w
The signs of
thermodynamic variables
are important.
• ΔE > 0 means the internal
energy of the system
increased.
• q > 0 means heat was
transferred into the system.
• w > 0 means work was
done on the system.
w>0
q>0
ΔE
State Functions
Functions with a fixed value for a given
P, T, V, and component composition are
called state functions.
E is a STATE FUNCTION.
Enthalpy
Enthalpy is a state function that accounts for
heat flow in chemical changes occurring at
constant pressure.
The enthalpy H is defined in terms of the internal
energy, pressure, and volume of a system:
H = E + PV
When pressure is constant,
ΔH = ΔE + PΔV
H is a STATE FUNCTION.
Enthalpy
When a chemical reaction is performed
under conditions of constant pressure, the
enthalpy can be used to determine whether
the reaction is exothermic or endothermic:
ΔH = qp
Summary
1. Work and heat are ways to transfer
energy.
2. First Law of Thermodynamics:
“The energy of the universe is
conserved.”
The first law shown in terms of the system:
ΔE = q + w
Preview
 For chemical reactions carried out at constant
temperature and pressure,
 the spontaneity of the reaction is given by the
Gibbs Free Energy G
ΔG = ΔH - TΔS
 where ΔS is the change
system.
in entropy of the
Entropy
Entropy (S) is a state function that is related to
the degree of randomness of a system.
S is a STATE FUNCTION.
Entropy - FYI
On a molecular level, the entropy is
defined as follows:
S = k ln W
where k is Boltzmann constant (the gas
constant R divided by Avogadro’s number)
and W is the number of available
microstates. We will concentrate on
changes in S, not on its absolute value.
Standard Molar Entropies S°
THE STANDARD MOLAR ENTROPY S° is for the
substance in its standard state.
The standard state of any substance is its pure form at
1 bar pressure and the temperature of interest (often
25°C).
1. Unlike Δf H°, S° of an element in its standard state is
NOT zero at 298 K.
2. S° for a gas is greater than S° for a liquid or solid.
3. S° generally increases with increasing molar mass.
4. S° generally increases with increasing numbers of
atoms in the formula of a substance.
Comparing Entropies
Which of each pair has greater entropy?
a) 1 mol of NaCl(s) at 25°C or 1 mol of HCl(g) at 25°C?
1 mol of HCl(g) at 25°C
b) 2 mol of HCl(g) at 25°C or 1 mol of HCl(g) at 25°C?
2 mol of HCl(g) at 25°C
c) 1 mol of HCl(g) at 25°C or 1 mol of Ar(g) at 25°C?
1 mol of HCl(g) at 25°C
d) 1 mol of N2(g) at 25°C or 1 mol of N2(g) at 84 K?
1 mol of N2(g) at 25°C
Predicting ΔS for a Reaction
Entropy generally increases for the following processes:
1. Liquids or solutions formed from solids.
2. Gases formed from liquids or solids.
3. An increase in the number of moles of gas as the result of a
reaction.
Predict the entropy change in each of the following
isothermal reactions.
a) CaCO3(s)  CaO(s) + CO2(g)
ΔS>0
b) N2(g) + 3H2(g)  2NH3(g)
ΔS<0
c) N2(g) + O2 (g)  2NO(g)
ΔS≈0
Using S° Data to Find ΔS° for a
Reaction
This is done in exactly the same way we
calculated heats of reaction ΔrxnH°.
ΔrxnS° = ΣmS° (products) – ΣnS°( reactants)
(m and n represent the stoichiometric coefficients)
We get the values for S°(25°C) from Appendix C.
ΔS of a Reaction
2H2(g) + O2(g)  2H2O(l)
ΔS° = S°final – S°initial
ΔS°(25°C) = ?
This is the same as
ΔS° = S°products – S°reactants
ΔS°(25°C) = 2S°(H2O(l)) - 2S°(H2(g)) - S°(O2(g))
ΔS°(25°C) = 2(69.91) - 2(130.58) - 205.0
ΔS°(25°C) = -326.3
Units are not

.

Note: The only time entropy is
zero is for a perfect crystal at 0
K.
Entropy is an Extensive Property

2H2(g) + O2(g)  2H2O(l) ΔS°(25°C) = -326.3
Burning two moles of H2 in O2 (under conditions of
constant P) would decrease the entropy of the system
by 326.3 J/K for every mole of O2 or every two moles
of water or H2.
How much does the entropy change when 0.275 g of H2 is
ignited with O2 in a constant-pressure system?
0.275 g H2(g) 1 mol H2
-326.3 J/K = -22.3
2.016 g H2

2 mol H2
ΔS = -22.3

The randomness of the system decreases.
Second Law of Thermodynamics
For any process, the entropy of the universe
either increases or does not change.
ΔSuniv ≥ 0
ΔSuniv = ΔSsys + ΔSsurr
The entropy of the system is the one
in which we are most interested.
First Law of Thermodynamics
For any process, the energy of the universe
does not change. It is conserved.
ΔEuniv = ΔEsys + ΔEsurr = 0
Second Law of Thermodynamics
For any process, the entropy of the universe
either increases or does not change. It is
NOT conserved!
ΔSuniv = ΔSsys + ΔSsurr ≥ 0
Second Law of Thermodynamics
 The second law of thermodynamics says the
 This expresses the notion that there is an
inherent direction in which processes occur.
For the universe OR for an isolated system (a
system which does not exchange energy or
matter with its surroundings),
ΔS = 0 for a reversible process, and
ΔS > 0 for a spontaneous, irreversible process.
Reversible Processes
 A reversible process is a special way in
which the state of a system can change. In a
reversible process, the system and
surroundings can be restored to their
original state by exactly reversing the
change.
 Reversible processes tend to be small, slow
changes made to a system at equilibrium.
Reversible Processes
Reversible Process - Example #1
Ice melting at 0°C and 1 bar. Ice and water are in
equilibrium under these conditions. Adding heat slowly
to ice at 0°C will cause it to melt. The process can be
undone and the ice reformed by slowly removing the
same amount of heat.
+ heat
+ heat
ice, 0°C
water, 0°C
- heat
- heat
Entropy - FYI
An often-used definition of entropy is in
terms of heat:

≡

where dqrev is the amount of heat going
into the system for a reversible process.
(To get ΔS, the equation must be integrated.)
ΔS for a Phase Change
This definition means that the change in
entropy ΔS for a phase change (which is an
isothermal process) is
∆
∆ =
=

For melting, ∆ =
For boiling, ∆ =
∆

∆

Irreversible Processes
An irreversible process is one in which the
system and surroundings cannot be restored to
their original state by exactly reversing the
change.
 dropping a vase and breaking it
 reacting hydrogen and oxygen to form
water
 burning a match
Irreversible = Spontaneous
 Spontaneous chemical
reactions are irreversible.
 Systems not at equilibrium
tend to undergo spontaneous,
irreversible changes.
 A reaction being spontaneous is NOT
the same as a reaction being fast!
Spontaneous reactions can be slow.
Spontaneous Processes
A chemical reaction carried out at
constant temperature and pressure is
spontaneous if the change in its Gibbs
Free Energy is less than 0.
ΔG < 0: process is spontaneous
ΔG = 0: process is at equilibrium
ΔG > 0: process is nonspontaneous
(but the reverse reaction is spontaneous)
Gibbs Free Energy
G ≡ H - TS
The change in Gibbs free energy is what we
want to know:
ΔG = ΔH – Δ(TS)
For a process that takes place at constant
T and P (like a lot of open beaker lab
reactions)
ΔG = ΔH – TΔS
Gibbs Free Energy
If only P-V work is performed, the change
in Gibbs Free Energy ΔG is the maximum
work* that can be obtained from a
chemical reaction at constant temperature
T.
ΔG = ΔH – TΔS
*It is the maximum P-V work for an ideal gas, actually.
More generally true is that ΔG = wnon-PV for constant T
and P.
Gibbs Free Energy
There two ways to obtain ΔG:
1.
ΔrxnG = ΣmG (products) – ΣnG( reactants)
(m and n represent the stoichiometric coefficients)
2.
ΔG = ΔH – TΔS
Standard Gibbs Free Energy G°
The standard Gibbs Free Energy G°
applies to substances in their
standard states at a specified T.
There are two ways to calculate ΔG°.
1.
ΔrxnG° = ΣmG° (products) – ΣnG°( reactants)
(m and n represent the stoichiometric coefficients)
2.
ΔG° = ΔH° – TΔS°
Standard Gibbs Free Energy of
Formation ΔfG°
These are the values that are tabulated
in Appendix C.
The same rules that apply to ΔfH° apply
to ΔfG°…
Free Energies of Formation
• The free energy of formation applies to the formation
of a compound from its elements.
**The state of each element MUST be specified.**
• THE STANDARD FREE ENERGY OF FORMATION is
designated ΔfG° and is for the formation of 1 mole of
the compound in its standard state from its elements in
their standard states.
• The standard state of any substance is its pure form
at 1 bar pressure and the temperature of interest (often
25°C).
• ΔfG° for an element in its standard state is 0 at all
temperatures.
Calculating ΔG° for a Reaction
Calculate ΔG° for the combustion of methane
using both methods.
1. ΔrxnG° = ΣmΔfG° (products) – ΣnΔfG° ( reactants)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Look up ΔfG° in Appendix C.
ΔrxnG° = ?
Calculating ΔG° for a Reaction
Calculate ΔG° for the combustion of methane
using both methods.
2.
ΔG° = ΔH° – TΔS°
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Look up ΔfH° and S° in Appendix C.
ΔrxnG° = ?
Calculating ΔG° for a Reaction
Calculate ΔG° for the combustion of methane
using both methods.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
 Is the reaction spontaneous under standard
conditions at 25°C?
 What drives the reaction at 25°C?
 Is there a T at which the reaction is no
longer spontaneous under standard
conditions?
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