Spontaneous Processes

Thermodynamics is concerned with the question: can a

reaction occur?
First Law of Thermodynamics: energy is conserved.

CHAPTER 3 Any process that occurs without outside intervention is

spontaneous.
When two eggs are dropped they spontaneously break.
SECOND LAW OF THERMODYNAMICS
The reverse reaction is not spontaneous.
We can conclude that a spontaneous process has a
direction.

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Spontaneous Processes Spontaneous Processes

A process that is spontaneous in one direction is not
spontaneous in the opposite direction.
The direction of a spontaneous process can depend on
temperature:
Ice turning to water is spontaneous at T > 0C, Water
turning to ice is spontaneous at T < 0C.

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 Reversible and Irreversible
Spontaneous Processes Processes
Reversible and Irreversible Processes A reversible process is one that can go back and forth
When 1 mol of water is frozen at 1 atm at 0C to form 1 mol between states along the same path.
of ice, q = Hvap of heat is removed. ◦ heat flow between two objects that have almost
similar temperature can be reversed by having
To reverse the process, q = Hvap must be added to the 1 infinitesimal change in temperature.
mol of ice at 0C and 1 atm to form 1 mol of water at 0C.
Therefore, converting between 1 mol of ice and 1 mol of A irreversible process is one that cannot simply be
water at 0C is a reversible process. reversed to restore the system and its surroundings to
their original states.
Allowing 1 mol of ice to warm is an irreversible process. To
◦ Heat can not flow from the colder object to the
get the reverse process to occur, the water temperature
must be lowered to 0C. hotter one.

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Reversible and Irreversible Reversible and Irreversible

Processes Processes
Irreversible process Chemical systems in equilibrium are reversible.
In any spontaneous process, the path between
reactants and products is irreversible.
Thermodynamics gives us the direction of a process. It
cannot predict the speed at which the process will
occur.

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The Spontaneous Expansion of The
a Gas Spontaneous
Consider an initial state: two flasks connected by a Expansion of
closed stopcock. One flask is evacuated and the other
contains 1 atm of gas.
a Gas
The final state: two flasks connected by an open
stopcock. Each flask contains gas at 0.5 atm.
The expansion of the gas is isothermal (i.e. constant
temperature). Therefore the gas does no work and
heat is not transferred.

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The Spontaneous Expansion of The Spontaneous Expansion of

a Gas a Gas
Consider the simple case where there are two gas When there are many molecules, it is much more
molecules in the flasks. probable that the molecules will distribute among to
the two flasks than all remain in only one flask.
Before the stopcock is open, both gas molecules will be
in one flask.
Once the stopcock is open, there is a higher probability
that one molecule will be in each flask that both
molecules being in the same flask.

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Second Law of Entropy and the
Second LawLaw
Second of
Thermodynamics Thermodynamics
of Thermodynamics
Entropy
Entropy, S, is a measure of the disorder of a system.
Any irreversible process results in an overall increase in
entropy, whereas a reversible process results in no Spontaneous reactions proceed to lower energy or
overall change in entropy higher entropy.
In ice, the molecules are very well ordered because of
the H-bonds.
Therefore, ice has a low entropy.

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Entropy and the

Second LawLaw
Second of
Thermodynamics
of Thermodynamics
Entropy
As ice melts, the intermolecular forces are broken
(requires energy), but the order is interrupted (so
entropy increases).
Water is more random than ice, so ice spontaneously
melts at room temperature.

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Entropy and the
Second LawLaw
Second of Entropy
Thermodynamics
of Thermodynamics
Entropy
There is a balance between energy and entropy
considerations.
When an ionic solid is placed in water two things
happen:
◦ the water organizes into hydrates about the ions (so
the entropy decreases), and
◦ the ions in the crystal dissociate (the hydrated ions
are less ordered than the crystal, so the entropy
increases).

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Entropy and the

Second LawLaw
Second of Entropy and the
Second LawLaw
Second of
Thermodynamics
of Thermodynamics Thermodynamics
of Thermodynamics
Entropy Entropy
Generally, when an increase in entropy in one process Suppose a system changes reversibly between state 1
is associated with a decrease in entropy in another, the and state 2. Then, the change in entropy is given by
increase in entropy dominates. q
Entropy is a state function. Ssys = rev (constant T )
T
For a system, S = Sfinal - Sinitial. ◦ at constant T where qrev is the amount of heat added
If S > 0 the randomness increases, if S < 0 the order reversibly to the system. (Example: a phase change
increases. occurs at constant T with the reversible addition of
heat.)

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Entropy and the
Second LawLaw
Second of Entropy and the
Second LawLaw
Second of
Thermodynamics
of Thermodynamics Thermodynamics
of Thermodynamics
.
The Second Law of Thermodynamics The Second Law of Thermodynamics
The second law of thermodynamics explains why For a reversible process: Suniv. = S syst. + S surr. = 0
spontaneous processes have a direction.
For a irreversible process: Suniv. = S syst. + S surr.  0
In any spontaneous process, the entropy of the universe
increases. Note: the second law states that the entropy of the
Suniv. = S syst. + S surr. universe must increase in a spontaneous process. It is
possible for the entropy of a system to decrease as long
the change in entropy of the universe is the sum of the as the entropy of the surroundings increases.
change in entropy of the system and the change in entropy
of the surroundings. For an isolated system, Ssys = 0 for a reversible process
and Ssys > 0 for a spontaneous process.
Entropy is not conserved: Suniv is increasing

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The Molecular Interpretation

of Entropy
A gas is less ordered than a liquid that is less ordered
than a solid.
Any process that increases the number of gas
molecules leads to an increase in entropy.
When NO(g) reacts with O2(g) to form NO2(g), the total
number of gas molecules decreases, and the entropy
decreases.

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The Molecular Interpretation The Molecular Interpretation
of Entropy of Entropy
There are three atomic modes of motion: Energy is required to get a molecule to translate,
◦ translation (the moving of a molecule from one point vibrate or rotate.
in space to another), The more energy stored in translation, vibration and
◦ vibration (the shortening and lengthening of bonds, rotation, the greater the degrees of freedom and the
including the change in bond angles), higher the entropy.
◦ rotation (the spinning of a molecule about some In a perfect crystal at 0 K there is no translation,
axis). rotation or vibration of molecules. Therefore, this is a
state of perfect order.

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The Molecular Interpretation The Molecular Interpretation

of Entropy of Entropy
Third Law of Thermodynamics: the entropy of a perfect
crystal at 0 K is zero.
Entropy changes dramatically at a phase change.
As we heat a substance from absolute zero, the entropy
must increase.
If there are two different solid state forms of a
substance, then the entropy increases at the solid state
phase change.

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 The Molecular Interpretation
of Entropy
Boiling corresponds to a much greater change in
entropy than melting.
Entropy will increase when
◦ liquids or solutions are formed from solids,
◦ gases are formed from solids or liquids,
◦ the number of gas molecules increase,
◦ the temperature is increased.

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The Molecular
Entropy Interpretation
Changes in Chemical
of Entropy
Reactions
Absolute entropy can be determined from complicated
measurements.
Standard molar entropy, S°: entropy of a substance in
its standard state. Similar in concept to H°.
Units: J/mol-K. Note units of H: kJ/mol.
Standard molar entropies of elements are not zero.
For a chemical reaction which produces n moles of
products from m moles of reactants:
S  =  nS (products ) −  mS (reactants )

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Gibbs Free Energy Gibbs Free Energy
For a spontaneous reaction the entropy of the universe There are three important conditions:
must increase. ◦ If G < 0 then the forward reaction is spontaneous.
Reactions with large negative H values are spontaneous. ◦ If G = 0 then reaction is at equilibrium and no net
reaction will occur.
How to we balance S and H to predict whether a reaction ◦ If G > 0 then the forward reaction is not
is spontaneous? spontaneous. If G > 0, work must be supplied from
Gibbs free energy, G, of a state is the surroundings to drive the reaction.
G = H − TS For a reaction the free energy of the reactants
decreases to a minimum (equilibrium) and then
For a process occurring at constant temperature increases to the free energy of the products.
G = H − TS

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Gibbs Free Energy Gibbs Free Energy

Consider the formation of ammonia from nitrogen and
hydrogen:
N2(g) + 3H2(g) 2NH3(g)
Initially ammonia will be produced spontaneously (Q <
Keq).
After some time, the ammonia will spontaneously react
to form N2 and H2 (Q > Keq).
At equilibrium, ∆G = 0 and Q = Keq.

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 Gibbs Free Energy
Standard Free-Energy Changes
We can tabulate standard free-energies of formation, Gf
(c.f. standard enthalpies of formation).
Standard states are: pure solid, pure liquid, 1 atm (gas), 1 M
concentration (solution), and G = 0 for elements.
G for a process is given by
G =  nG f (products ) −  mG f (reactants )
The quantity G for a reaction tells us whether a mixture
of substances will spontaneously react to produce more
reactants (G > 0) or products (G < 0).

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Free Energy and Temperature Free Energy and Temperature

Focus on G = H - TS:
◦ If H < 0 and S > 0, then G is always negative.
◦ If H > 0 and S < 0, then G is always positive.
(That is, the reverse of 1.)
◦ If H < 0 and S < 0, then G is negative at low
temperatures.
◦ If H > 0 and S > 0, then G is negative at high
temperatures.
Even though a reaction has a negative G it may occur
too slowly to be observed.

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 Free Energy and The Free Energy and The
Equilibrium Constant Equilibrium vConstant
Constant
Recall that G and K (equilibrium constant) apply to At equilibrium, Q = K and G = 0, so
standard conditions. G = G + RT ln Q
Recall that G and Q (equilibrium quotient) apply to 0 = G  + RT ln K eq
any conditions.
It is useful to determine whether substances under any
G  = − RT ln K eq
conditions will react: From the above we can conclude:
G = G + RT ln Q ◦ If G < 0, then K > 1.
◦ If G = 0, then K = 1.
◦ If G > 0, then K < 1.

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You place the substance A(g) in a container. Consider the following reaction under standard
conditions to produce the substance B(g):

For this reaction as written, the equilibrium constant is a very large, positive number.
a) When A(g) reacts to give B(g), does the standard free energy (G°) of the reaction
change as the reaction proceeds or does it remain constant? Explain.
b) When A(g) reacts to give B(g), does the free energy (G) of the reaction change as the
reaction proceeds, or does it remain constant? Explain.
c) Is this reaction spontaneous? How do you know?
d) When the reaction reaches equilibrium, is the following statement true: ∆G° = ∆G = 0?
if not, what can you say about the values of ∆G° and ∆G when equilibrium has been
reached?
e) When the reaction has reached equilibrium, what can you say about the composition
of the reaction mixture? is it mostly A(g), is it mostly B(g), or is it something close to
equal amounts of A(g) and B(g)?

Now consider running the reaction in reverse: B(g) → A(g). For the reaction as written, what
can you say about ∆G°, ∆G, the equilibrium constant, and the composition of the reaction
mixture at equilibrium? Also, is the reaction spontaneous in this direction?

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 Based on the following reaction, oxidation of SO2(g) to Reaction A is given below as:
SO3(g) can produced sulphuric acid. Analyse whether
this reaction is possible to be carried out. Provide
justification to your answer. A→B+C H=10.5 kJ/mol, S = 30 J/K.mol;

Given ΔH˚ SO2(g) = -296.9 kJ/mol, ΔH˚ O2(g) = 0 kJ/mol, Predict if reaction A is spontaneous at 25C. If reaction
and ΔH˚ SO3(g) = -395.2 kJ/mol, ∆S˚ SO2 (g) = 248.12 A is not spontaneous at 25C, estimate the minimum
J/K.mol, ∆S˚ O2 (g) = 205.03 J/K.mol, ∆S˚ SO3 (g) = temperature (in C) for this reaction to become
256.72 J/K.mol spontaneous.

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