Download Chemistry and the material world

Chemistry and the material world
123.102
Unit 4, Lecture 4
Matthias Lein
Gibbs free energy
Gibbs free energy to predict the direction of a chemical process.
Exergonic and endergonic reactions.
Temperature dependence of the Gibbs free energy.
The chemical equilibrium.
The law of mass action.
Reaction quotients.
Gibbs free energy and equilibrium constants.
Equilibrium temperature.
Consider the reaction:
N2(g) + 3 H2(g) → 2 NH3(g)
Let us calculate ΔH° and ΔS° for the following reaction to decide in which
direction each of these factors will drive the reaction.
ΔHf°(N2) = 0 kJ/mol, ΔHf°(H2) = 0 kJ/mol, ΔHf°(NH3) = -46.11 kJ/mol
ΔSf°(N2) = 191.61 J/molK, ΔSf°(H2) = 130.68 J/molK, ΔSf°(NH3) = 192.54 J/molK
ΔH° = -92.22 kJ
ΔS° = -198.78 J/K
Enthalpy favours the reaction. It is exothermic, heat is released.
Entropy dis-favours the reaction. Order is increased, the number of particles
decreases.
The questions remains:
Will the reaction occur spontaneously?
The Gibbs free energy G is defined as:
G = H -TS
a measure that connects the two previously treated thermodynamic
functions H, the enthalpy and S, the entropy.
ΔG < 0
ΔG = 0
ΔG > 0
favored reaction (Spontaneous)
Neither the forward nor the reverse reaction prevails (Equilibrium)
disfavored reaction (Non-spontaneous)
Josiah Willard Gibbs
United States,
1839 – 1903
Just like reactions are called exothermic and endothermic depending
on the heat produced (ΔH), reactions are also called:
ΔG < 0
exergonic (Spontaneous)
ΔG > 0
endergonic (Non-spontaneous)
Wether a reaction is exergonic or endergonic depends on the reaction
enthalpy, the reaction entropy and the temperature at which the
reaction is carried out.
Consider the reaction (at 25° C):
N2(g) + 3 H2(g) → 2 NH3(g)
Let us calculate ΔH° and ΔS° for the following reaction to decide in which
direction each of these factors will drive the reaction.
ΔHf°(N2) = 0 kJ/mol, ΔHf°(H2) = 0 kJ/mol, ΔHf°(NH3) = -46.11 kJ/mol
ΔSf°(N2) = 191.61 J/molK, ΔSf°(H2) = 130.68 J/molK, ΔSf°(NH3) = 192.54 J/molK
ΔH° = -92.22 kJ
ΔS° = -198.78 J/K
ΔG = ΔH° -TΔS°
ΔG = -92,220 J - (298.15 K × -198.75 J/K)
ΔG = -32.96kJ
Answer: at 25° C the reaction occurs spontaneously (it is exergonic)
How about the same reaction at 500° C?
N2(g) + 3 H2(g) → 2 NH3(g)
ΔH° = -92.22 kJ
ΔS° = -198.78 J/K
ΔG = ΔH° -TΔS°
ΔG = -92,220 J - (773.15 K × -198.75 J/K)
ΔG = 61.4kJ
Answer:
at 500° C the reaction does not occur spontaneously
(it is endergonic)
The entropic term becomes more and more important as the temperature
rises.
ΔG = ΔH° -TΔS°
The sign of ΔG tells us the direction in which a reaction will proceed to reach
an equilibrium and the magnitude of ΔG tells us how far from equilibrium the
reaction still is (ΔG = 0 means equilibrium).
Thus, as the reaction progresses, the magnitude of ΔG will become smaller
and smaller until an equilibrium is reached. The chemical reaction is then
finished.
ΔG = ΔH° -TΔS°
When we calculate ΔG values for reactions, we are comparing G values for
the reactants and products. Therefore, the calculated ΔG value for the
reaction only represents the true ΔG value for the very start of the reaction.
Over the course of the reaction the magnitude of ΔG will go to zero.
Equilibrium:
N2O4(g) → 2 NO2(g)
colourless
brown
Equilibrium:
N2O4(g) → 2 NO2(g)
Equilibrium:
N2O4(g) → 2 NO2(g)
Equilibrium:
N2O4(g) → 2 NO2(g)
Law of mass action (Kc is the equilibrium constant)
This equation only describes the equilibrium.
Equilibrium:
N2O4(g) → 2 NO2(g)
Systems that are not at equilibrium
can be described by a similar
Qc
expression.
Qc is called the reaction quotient
Remember that Qc and Kc depend on temperature and pressure.
Equilibrium:
N2O4(g) → 2 NO2(g)
The connection between the equilibrium constant
and the Gibbs free energy:
ΔG° = -RT ln K
So, to calculate ΔG for a given reaction quotient:
ΔG = ΔG° +RT ln Q
(note that the expression becomes zero for Q = K)
What does the magnitude of Kc mean?
K = 1 means that the reaction quotient is balanced. There is approximately the
same amount of educt and product in the reaction mixture.
K > 1 means that that the numerator in the reaction quotient is larger. There is
more product in the reaction mixture.
K < 1 means that the denominator in the reaction quotient is larger. There is
more educt in the reaction mixture.
For example:
2 SO2(g) + O2(g) → 2 SO3(g)
ΔG° = -1.40⋄102 kJ/mol
What is the value of the equilibrium constant K?
We know:
ΔG°
=
-RT ln K
(R = 8.314 J/molK, T = 298K)
ln K
=
-ΔG° / RT
=
-(-1.40⋄102 kJ/mol) / (8.314 J/molK × 298K)
=
56.5
=
e56.5
So:
K
=
3⋄1024
The equilibrium temperature
We know that at equilibrium:
ΔG = 0
which means that
ΔH – TΔS = 0
That means that if we know two quantities out of ΔH, ΔS and T, we can
calculate the value of the third at the equilibrium.
For example:
Br2(l) → Br2(g)
ΔH° = 30.9 kJ/mol, ΔS° = 93.2 J/molK
At what temperature is the two phases of liquid and gaseous Br2in
equilibrium?
At equilibrium:
ΔG = ΔH – TΔS = 0
so
ΔH = TΔS
and
T = ΔH / ΔS
T = 30900 J/mol / 93.2J/molK = 332 K
332 K = 59° C
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Today we covered:
●
Gibbs free energy to predict the direction of a chemical process.
Exergonic (ΔG < 0) and endergonic (ΔG > 0) reactions.
Temperature dependence of the Gibbs free energy.
The chemical equilibrium.
The law of mass action.
Reaction quotients.
Gibbs free energy and equilibrium constants (ΔG° = -RT ln K).
Equilibrium temperature (ΔH = TΔS).
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