Chapter 19 Chemical Equilibrium by img20336

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									                    Chemical Thermodynamics
•   Spontaneous Processes

•   Entropy and the Second Law of Thermodynamics

•   The Molecular Interpretation of Entropy

•   Entropy Changes in Chemical Reactions

•   Gibbs Free Energy

•   Free Energy and Temperature

•   Free Energy and the Equilibrium Constant




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Spontaneous Processes

First Law of Thermodynamics stated:
Energy can not be created nor destroyed, it can be converted from one form to
another or transferred from a system to surroundings (or vice versa)




A reaction is spontaneous if; once it starts, it
proceeds on its own, regardless of how fast the
reaction occurs.

Reactions that are spontaneous in one direction
are non-spontaneous in the reverse direction



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Natural change occurs in one direction only, for example
the rusting of iron, melting of ice above 0 oC, a ball
rolling downhill etc.

A non spontaneous process can only be brought about
by doing work, it won’t occur unless some external
action is continually applied, e.g. pushing the ball back
up the hill




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To know if a reaction is going to be spontaneous or not we need to
distinguish between reversible and irreversible reactions




A (chemical) system at equilibrium is reversible, an
infinitesimal change in the system (reactants or
products) can reverse the direction

Reversible processes are those that can reverse
direction whenever there is an infinitesimal
change in the property of the system

 A reversible process is one in which:




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A chemical system that is not at equilibrium is irreversible, the reaction will
proceed spontaneously until equilibrium is reached.




In this case the system and surroundings are not BOTH returned to their
original condition – IRREVERSIBLE process. I had to do work to return the
system (gas) to its original volume.




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Entropy

Spontaneous reactions often give out heat ( exothermic ) but this isn’t always
the case , so enthalpy alone is unable to account for the direction of
spontaneous change.
Entropy is the other thermodynamic quantity that accounts for the direction of
spontaneous change.
Entropy is nature’s driving force to move to a condition of maximum
randomness or disorder.




Entropy is given the symbol S
Entropy is a state function, the
change in entropy only depends on
the initial and final states of the
system
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For an isothermal process (constant T), e.g. a phase change

e.g. Phase Changes                   ΔS = qrev / T      (absolute Temp.)

Recall: qrev is the heat exchanged at constant pressure, the rev implies that
the heat must be transferred reversibly (in infinitesimally small amounts)




Ethanol boils at 78 oC. The molar enthalpy of vaporization is 38.56 kJ/mol.
What is the change in entropy in the system when 68.3 g of ethanol at 1 atm is
vaporized at the normal boiling point?

Because S is a state function we can use the final and initial states to
calculate ΔS for any isothermal process, not just those that are
reversible, we imagine the process as reversible

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Second Law of Thermodynamics

Unlike energy, entropy is NOT conserved:
The Universe wants to become more and more disordered, it wants its
total entropy to increase continually!




Reversible process ΔSuniv = ΔSsys + ΔSsurr = 0
Irreversible process ΔSuniv = ΔSsys + ΔSsurr > 0

The rusting of a car is a spontaneous process. During this process
the entropy of the system (the car) decreases. What conclusion can
be made about the Entropy of the surroundings
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A Molecular interpretation of Entropy
Why is greater randomness the driving consideration for spontaneity


The relationship of Entropy to this disorder (randomness) or probability
was studied by Boltzmann.
Boltzmann was thinking about a mole of gas particles in random motion
and then considered what state they would be in if you froze them all and
took a picture. Their positions at that time he called a microstate
To figure out the seemingly impossible problem of relating Entropy to how
many different microstates there could possibly be in a system Boltzmann
came up with a very simple equation

                       k is Boltzmann’s constant 1.38 x 10-23J/K (the gas constant
                       R, divided by Avogadro’s number) and W is the number of
                       microstates associated with a particular state.


What would be some of the motions of a water molecule?
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∆S = k ln Wfinal - k ln Wfinal

Boltzmann’s equation demonstrates:
Any change in a system that increases the number of microstates leads to
an increase in the entropy
Therefore any factors that can cause an increase in the number of
microstates (motions and energies of the molecules) will increase the
entropy

These Factors are:




Solid is sublimed to a vapour, molecules
have more degrees of motion, and a
greater range of Kinetic Energies


Increase temperature further, Energies increase further
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The more ‘stuff’ you have in your house, the messier it is!

For example dissociating (NH4)2Cr2O7(s) to Cr2O3(s), N2(g) and 4H2O(g)




Vaporizing water, the molecules spread out into a larger volume and have a
greater range of motional energies (microstates)


Predict whether the Entropy change will be positive or negative for the
following processes:
CO2(s) → CO2(g)
2SO2(g) + O2(g) → 2SO3(g)
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So, if increasing the temperature increases the number of microstates
could we decrease the temperature to a point where all motion stops and
we have a single microstate?
The Third Law Of Thermodynamics




At 0K, all the units in the lattice have no thermal energy, no motion, 1 Microstate
S = k lnW W = 1 lnW = 0
Therefore: S = 0
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What happens to Entropy as we continue to heat




        Why is the Entropy of phase changes following the order
                            Ssolid < Sliquid < Sgas
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Entropy Changes in Chemical Reactions

We can measure ΔH of a reaction by using calorimetry, however there is
no such easy method for determining ΔS for a reaction

Absolute values for Entropies, S, can be obtained



Standard Molar Entropy So is defined for pure
substances at 1 atm pressure and 298K



                                    Why?




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The Entropy change for a reaction (system) can be calculated from the
Standard Molar Entropies


Using Standard Molar Entropies, calculate the Entropy Change ΔSo for
the reaction:
N2 (g) + 3H2(g) → 2NH3 (g)         does the answer make sense? Why?

Entropy Changes in the Surroundings
The surroundings act as a big heat sink, the entropy change of the
surroundings depends only on the heat gained or lost:




For a constant pressure reaction (usual) then qp (heat exchanged at
constant pressure) = ΔH, so the entropy change of the surroundings can
be written as

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Because Souniv is positive (increases) for any spontaneous reaction we can
put together the equations for calculating ΔSsys with ΔSsurr to predict
whether a reaction will be spontaneous


What would be the entropy change (ΔSouniv) for the following reaction:

CO(g) + 2H2(g) → CH3OH (l) at 298 K          ΔHosys = -128.1 kJ

Use you answer to predict whether the reaction would be spontaneous


Prediction of Spontaneity depends on

The sign of ΔH and ΔS

The magnitude of ΔH, ΔS and the temperature (in KELVIN)

A spontaneous reaction can be exothermic or endothermic

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Gibbs Free Energy
JW Gibbs came up with a new state function G, that connected entropy
and enthalpy to predict whether a reaction occurring at a constant
temperature would be spontaneous




From the definition of ΔSuniv we can relate the state function G to spontaneity:
ΔSuniv = ΔSsys + ΔSsurr    = ΔSsys + (-ΔHsys/T)

Multiply both sides by: –T to give:




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Standard Free Energy

The standard free energy change ΔGo is defined as the change in the free
energy when reactants in their standard states are converted to products
in their standard states:


  Conventions used in establishing    Standard free energies
  Solid                               Pure solid
  Liquid                              Pure liquid
  Gas                                 1 bar pressure
  solution                            1M
  elements                            ∆Gof in standard state = 0




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Standard Free energy change can be calculated from the standard enthalpy change
and the standard entropy change.




ΔHo and ΔSo don’t change (significantly) with temperature, but ΔGo does depend on
temperature and might even change sign as the temperature is changed.


Calculate the standard free energy change ΔGo for this reaction at 25 oC
and 500 oC. What do the calculated values tell you?
                          Given that ΔHo = -92.38 kJ and ΔSo = -198.4 J/K

N2(g) +     3H2(g)          2NH3(g)




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Standard Free Energy of Formation

The standard Gibbs free energy of formation is the change in free energy when
of 1 mole of substance is formed from its component elements in their most
stable (standard) form at 1 atm pressure and 298K.
From these values we can calculate the standard free energy change of
reactions at 298K
G is a State Function, so:




  Calculate the standard Free energy change ΔGo for the reaction of glass
  being etched by hydrogen fluoride at 298 K. Use the answer to predict if
  the reaction happens spontaneously
  SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(g)
  ΔGfo SiO2 = -856 kJ/mol    ΔGfo HF = -273 kJ/mol
  ΔGfo SiF4 = -1573 kJ/mol   ΔGfo H2O = -224 kJ/mol

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Free Energy and Temperature

The entropy term (TΔS) is dependant on the absolute temperature and as
such the change in free energy will also vary with temperature.

As we saw, it is possible that the temperature could change the sign of ΔG and
a reaction that is non spontaneous at one temperature could become
spontaneous at another temperature, for example:




The decomposition of sodium bicarbonate is non spontaneous at 298K . The enthalpy of the
reaction is 135.4 kJ and the Entropy of reaction is 333 J/K
Calculate the temperature above which the reaction would become spontaneous assuming ΔH and
ΔS do not change with temperature
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Recall: that when ΔG = 0 the reaction is at equilibrium and therefore:




When ΔG = 0 slight adjustments in temperature can change the direction of
spontaneity (see previous NaHCO3 example)

Recall: a substance at its melting point and boiling point exists in equilibrium:

 ΔH = ΔS and so: |ΔH| = |TΔS| (isothermal)
 T
The normal boiling point is the temperature at which a pure liquid is
in equilibrium with its vapour at 1 atm (bar) pressure. Given that ∆Ho
and ∆So for CCl4 at 298K are 32.6 kJ and 95.0 J/K, calculate the normal
boiling point of CCl4 in units of oC

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Free Energy and the Equilibrium Constant

We can use the value of ΔGo to calculate the value of ∆G under non standard
conditions.

For any chemical process the relationship between the standard free energy
change and the free energy change under any other set of conditions at a given
temperature is:



The expression for Q is the same as the Equilibrium Constant [products]n
                                                               [reactants]m
Where m and n are the stoichiometric coefficients.
But for Q, the reactants and products need not be in equilibrium.
Therefore under non standard conditions we can calculate ΔG by
calculating Q and using our value of ∆Go at the given temperature.

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Calculate ΔG for this reaction at 298 K when Q = 9.3 x 10-3. Use the
previously calculated value of ΔGo of -33.3 kJ
N2(g) +    3H2(g)           2NH3(g)

At equilibrium ΔG = 0 and Q = K (the equilibrium constant) therefore:




If ΔGo is negative then ln K must be positive, i.e. K > 1. The larger the value of K
the more negative the value of ΔGo which means products will be favoured in
the equilibrium mixture. (For a given temperature).

Use the value of Go at 25 oC to calculate the equilibrium constant K at 25 oC

                    N2(g) + 3H2(g)        2NH3(g)

And predict whether products or reactants will be favoured in the
equilibrium mixture at this temperature
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Driving non-spontaneous reactions
This is done by coupling the non-spontaneous reaction with a
spontaneous reaction for example:
Cu2S(s) → 2Cu(s) + S(s)                ΔGo = +86.2kJ
S(s) + O2(g) → SO2(g)                  ΔGo = -300.4kJ

What would be the standard free energy change ΔGo for the overall
coupled reaction?




Note: The 10th edition of Brown, US version of the course textbook uses 1 atm as
the standard state and the 11th edition uses 1 bar. 1 bar is universally accepted.
As these values are (to a close approximation) the same, these notes use bar and
atm interchangeably. Atm is more commonly used as we have become
accustomed to using this value in Gas Laws and thermochemistry.



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