CHEMISTRY 2000
Topic #3: Thermochemistry and Electrochemistry –
What Makes Reactions Go?
Spring 2012
Dr. Susan Lait
Free Energy Change and Equilibrium

We saw that the free energy change for a reaction under any
conditions (rG) can be calculated from the free energy change
of the same reaction under standard conditions (rGº):
 r Gm   r Gmo  RT ln Q
where Q is the reaction quotient:
product of activities of products
Q
product of activities of reactants

We also know that:



The forward reaction is spontaneous if ____________
The reverse reaction is spontaneous if ____________.
What if rG = 0?
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Free Energy Change and Equilibrium

When rG = 0, we can say that:
Which rearranges to give:

But what is Q at equilibrium?

So, we can say that:
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Free Energy Change and Equilibrium

If we know the equilibium constant for a reaction, we can
calculate its standard molar free energy change:
 r Gmo  RT ln K

If we know the standard molar free energy change for a reaction,
we can therefore calculate its equilibrium constant:
Note that these equations refer to rGº (for standard conditions)
NOT to rG (which is ZERO at equilibrium!)
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Q vs. K
Q
product of activities of products
product of activities of reactants
and K 
product of activities of products
product of activities of reactants
So, what’s the difference?

Consider a very generic equilibrium reaction:
reactants

When does Q = K?

When is Q < K?

When is Q > K?
products
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Free Energy Change and Equilibrium

Now, consider a real reaction, the dimerization of NO2:
2 NO2(g)
N2O4(g)
Calculate the equilibrium constant for this reaction at 25 ºC.
kJ
mol
kJ
 fG(N2O4( g ) )  97.89 6
mol
 fG(NO2( g ) )  51.31
Free Energy Change and Equilibrium

Is the dimerization of NO2 spontaneous at 25 ºC if the partial
pressure of NO2 is 0.4 bar and the partial pressure of N2O4 is
1.8 bar?
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Free Energy Change and Equilibrium: Acids


The acid ionization constant, Ka, is the equilibrium constant for
the reaction in which an acid ionizes. As an equilibrium
constant (and therefore based on activities), Ka has no units.
K a  6.6  104
HF(aq)
e.g.
H+(aq) + F-(aq)
In CHEM 1000, we used pKa as a measure of strength of an
acid. An acid with a low pKa was _______________ than an
acid with a high pKa. The pKa value for an acid comes from its
acid ionization constant:
pK a   log K a

So, we can also say that an acid will be _________________
than another acid whose Ka value is smaller. (Even a “big” Ka
value is pretty small. Ka values range from about 10-50 to 1010.)
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Free Energy Change and Equilibrium: Acids

Calculate the standard molar free energy of formation for HF(aq).
HF(aq)
H+(aq) + F-(aq)
K a  6.6  104
 fG(H(aq ) )  0
kJ
mol
kJ
 fG(F( aq ) )  278.8 9
mol
Free Energy Change and Equilibrium:Solubility


The solubility product, Ksp, is the equilibrium constant for the
reaction in which an ionic compound dissolves in water. Like all
other equilibrium constants, Ksp has no units.
K s p  ???
e.g.
PbI2(s)
Pb2+(aq) + 2 I-(aq)
Calculate the solubility of lead(II) iodide in water at 25 ºC.
Solubility is reported in #moles of solid soluble in 1L water.
kJ
mol
kJ
 fG(I(aq ) )  51.57
mol
kJ
 fG(PbI2( s ) )  173.6 10
mol
 fG(Pb(2aq ) )  24.43
Free Energy Change and Equilibrium:Solubility
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Free Energy Change and Equilibrium:Solubility


The problem with simple Ksp calculations is that they almost
never give good numbers. Why not? Most of the time, there
are multiple equilibria that must be considered, but the Ksp
expression only accounts for one.
When PbI2 is dissolved in water, two complex ions also form:
Pb2+(aq) + 4 I-(aq)
Pb2+(aq) + 3 OH-(aq)

[PbI4]2-(aq)
K f  3.0  104
[Pb(OH)3]-(aq)
K f  3.8  1014
In the case of PbI2, the “simple” Ksp calculation gives almost the
same answer as a proper calculation factoring in both complex
ions because the Kf values for both complex ions are relatively
small (compared to some Kf values in the 1030 range). In many
systems, however, the complex ion formation is significant
enough that it must be accounted for or the Ksp calculation will
give very poor values.
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Free Energy Change and Equilibrium:Solubility

Complex ions are not the only factor complicating solubility
calculations. What equilibria would you have to consider if you
wanted to calculate the solubility of sodium carbonate in water?
How would you increase the solubility of sodium carbonate in
water?
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