An Analogy for Chemical
Equilibrium,DG, and DG**°**

**(view Simulation of this
process)**

**We will consider the
apparatus below as a chemical reactor in which we can carry out the
reaction:**

**H _{2}O(left)
-> H_{2}O(right)**

**We will work with 90 cm ^{3}of
water, which is approximately 5.0 moles.**

**Note that the cylinder on
the right has a larger diameter than the cylinder on the left.**

**We will take the area of the
left cylinder as 10 cm ^{2}, and that of the right cylinder as
20cm^{2}.**

**We can start with all of
the water on either the left or the right.**

**When the stopcock is
opened,the outcome will be the same:**

**This is the nature of
Chemical Equilibrium -**

**whether you start with only
reactants or only products,**

**the system comes to
equilibrium to satisfy some inherent property.**

**The equilibrium condition
could have been determined by the relative volumes,**

**relative masses, relative
potential energy, or some other property of the system.**

**In this case, the
equilibrium is determined by the relative height of the liquid**

**in the two cylinders.**

(More precisely, the equilibrium is determined by equal pressures on both sides of the stopcock.)

**Irrespective of the
starting condition, water will move from the cylinder**

**with the higher level to the
cylinder with the lower level.**

**In a chemical system at
fixed temperature and pressure,**

**the reaction proceeds so as to
lower the Gibbs Free Energy (G) of the system.**

**The equilibrium condition
is determined by the minimum free energy of the system.**

**However, our description of
the equilibrium condition involves**

**an equilibrium constant (K)
which is related to**

**the relative amounts of the
reactants and products.**

**The situation is further
complicated by the fact that the value of the equilibrium constant**

**is determined by the Gibbs
free energy required (DG ^{o})**

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**Let's try to apply the ideas
of
chemical equilibrium to this very simple system.**

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**In this very simple case of a
"reaction",**

**H _{2}O(left)
-> H_{2}O(right)**

**there is no difference
between Gibbs Free Energy (G) and Potential Energy (E),**

**so we will work with the
simpler quantity, E.**

**The potential energy of an
object is the product of its mass (m),**

**the gravitational constant (g=
9.80 m/sec ^{2}), and the height of its center of gravity (h):**

**E = mgh .**

**Using mass values in grams
and height values in meters**

**gives the potential energy
in milliJoules (mJ).**

**For a column of water of
height H(meters), volume V (cm ^{3}), and Density (g/cm^{3}):**

**m = DV
; h = H/2**

**E = gDVH/2 .**

**The energy is different in
the left and right columns,**

**E _{L} = gDV_{L}H_{L}/2 ;
E_{R} = gDV_{R}H_{R}/2**

**E _{total} = E_{L}+
E_{R} = (gD/2)(V_{L}H_{L} + V_{R}H_{R})**

**While the transfer is most
easily considered in terms of volume,**

**Chemists prefer to work
with the Chemical Amount (moles):**

**moles = DV/M (M
=Molar Mass in grams) .**

**We can then describe the
energies on the left and right in molar terms:**

**(E _{m})_{L}
=gMH_{L}/2 ; (E_{m})_{R}=
gMH_{R}/2**

**The difference between
these two terms is called DE _{rx}:**

**Now we consider the changes
in these quantities as we allow the liquid**

**to flow from left to right
until it reaches equilibrium,**

**then we will force all of
the liquid to the right.**

**The equilibrium condition
of equal heights of the two columns**

**now has two additional
conditions:**

**1. The total energy
hasa minimum value at this point.**

**2. The energy/mole has the
same value in both columns.**

**The quantity DE _{rx }is related to the driving force of the
process**

**H _{2}O(left)
-> H_{2}O(right)**

**While chemical reactions
are much more complex than this simple transfer,**

**the equilibrium condition
is very similar:**

**1. The total Gibbs Free
Energy of the system has a minimum value at equilibrium.**

**2. The sums of the Partial
Molar Gibbs Free Energies (Chemical Potentials) are**

**the same for the Reactants
and the Products (each multiplied by its**

**stoichiometric constant).**

**A quantity DG _{rx }is defined as the difference between the**

**When this quantity is
negative, the reaction proceeds as it is written,**

**and if this quantity is
positive, the reverse reaction is favored.**

**One other point may be made
about this system:**

**The standard energy change (DE°) for the reaction**

**H _{2}O(left)
-> H_{2}O(right)**

**DE°=
- 3.97 mJ/mole**

**This represents the energy
(mechanical work) per mole of water**

**required to convert
the initial state (all on the left) to**

**the final state (all on
the right).**

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**For Chemical Reactions,
there is a DG°, which represents the**

**Chemical Work required to
convert the reactants**

**completely to the products
at constant T & P.**

**This quantity is not to be
confused with DG _{rx}, which represents the**

** If DG°is negative, the equilibrium condition
will
favor**

**products over reactants**

**and the opposite is true if DG° is positive.**

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