Chemical Equilibria
By Curtis Bustos
In theory, nearly all
chemical reactions are reversible. However, if the reversibility is relatively small, we can treat it as negligible.
Key terms:
- Equilibrium
– chemical and biological definitions are acknowledged below.
-
Steady
State (aka Dynamic Equilibrium)– used interchangeably with homeostasis. However,
homeostasis refers to entire systems while steady states refer to specific chemical reactions or mechanisms inside a system such as glycolysis, sodium channels, proton gradients, etc. Without a steady
state, homeostasis would not be possible.
-
Homeostasis
- When an organism maintains a relatively constant internal environment within a system (cells, animals, plants, etc.) despite
changes to its external environment. Homeostasis does
not imply equilibrium!
-
ΔG –
Delta G represents a “change in free energy (G).” ΔG =
ΔH –TΔS. When ΔG = < 0, the reaction is spontaneous; when ΔG = > 0,
the reaction is non-spontaneous.
- Keq or K – equilibrium constant. A ratio of
molar concentrations of products to reactants in
a chemical reaction at equilibrium.
-
Exergonic
– a spontaneous change in free energy (G).
-
Endergonic
– a non-spontaneous change in free energy (G).
- Equilibrium (chemical perspective)
– when ΔG = 0 and no further net-change in free energy (G) is occurring.
What does it mean when ΔG = 0?????
When ΔG = 0 in a chemical reaction, the rate of a forward reaction is equal to the rate of a reverse reaction. A graphical representation may help:
The graphical representation above shows I
as a forward reaction (A à
B+C exergonic) and II as a reverse reaction (B+C
à A endergonic). If the forward reaction has a ΔG = -50 kJ/mol (spontaneous)
and the reverse reaction has a ΔG = +50 kJ/mol (non-spontaneous), then -50 and +50
will cancel out and result in a net ΔG = 0… this is called equilibrium. At this point, the ratio of molar concentrations
of products to reactants and the rate of the forward and reverse reaction
will remain unchanged over time.
- Equilibrium (biological perspective)
– A biological organism is DEAD if ΔG = 0.
What is the meaning of death and how is ΔG = 0 related?
To answer this question, we need to think
about how to define “life.” While this may prove to be a difficult task,
Nivaldo J. Tro mentions a description that suggests non-living
things are in equilibrium with their environment. In contrast, living things are not in equilibrium or are in
disequilibrium with their environment. This makes sense as animals
physiologically maintain a stable internal-environment by homeostasis and are not in equilibrium
with their always-changing external-environment. Therefore, dead organisms do not maintain homeostasis and as a result are in equilibrium with their environment where ΔG = 0.
How does a chemical reaction reach equilibrium? A chemical reaction can go back and forth from its reactants
to products and products to reactants until the rate of the forward reaction is equal to the rate of the reverse reaction. Therefore, the rate of the forward and reverse reaction will remain unchanged over time.This can also be represented graphically by referring to a typical
chemical-reaction
(A + B ⇌ C + D):
Additionally, when a chemical reaction reaches equilibrium, the products
and reactants will maintain their respective molar concentrations and will not change with time:
If these molar concentrations change due to external
factors, then a new Keq may need to be established as described by Le Châtelier’s principle.
Equilibrium does not suggest that molar
concentrations of products and reactants are equal to each other!
Equilibrium does not suggest that molar concentrations of products and reactants are equal to each other!
Equilibrium does not suggest that molar concentrations of products and reactants are equal to each other!
Equilibrium does not suggest that molar concentrations of products and reactants are equal to each other!
Equilibrium does not suggest that molar concentrations of products and reactants are equal to each other!
Now we turn our attention to an equilibrium constant or Keq. Keq can be explained mathematically as a measurement
of molar concentration of products over molar concentration of reactants at
equilibrium. For example:
When molar concentrations are
greater in the products than they are in the reactants, products will be
favored. A+B ⥂ C+D. These double-arrows imply products are favored at
equilibrium.
When molar concentrations are
greater in the reactants than they are in the products, reactants will be
favored. A+B ⥃ C+D. These double-arrows imply reactants are favored at equilibrium.
Quantitatively (K = Keq) :
K ≈ 1 neither
products or reactants in a chemical process are favored at equilibrium.
K >> 1 products are favored in a chemical process at
equilibrium. The molar concentration of products (numerator) outweigh the molar
concentration of reactants (denominator).
K << 1 reactants are favored in a chemical process at
equilibrium. The molar concentration of reactants (denominator) outweigh the molar
concentration of products (numerator).
By the way, there is a direct relationship
between Keq or K and ∆G standard (ΔG°): ∆G = ΔG° + RT lnKeq
*NOTE - ∆G = 0 at equilibrium; ΔG° may or may not = 0 at equilibrium
*NOTE - ∆G = 0 at equilibrium; ΔG° may or may not = 0 at equilibrium
Keq or K
(units are canceled out)
|
ΔG° (kJ/mol)
|
Products or
reactants are favored?
|
10-3
|
17.1
|
Reactants
|
10-2
|
11.4
|
Reactants
|
10-1
|
5.7
|
Reactants
|
1
|
0.0
|
Neutral
|
101
|
-5.7
|
Products
|
102
|
-11.4
|
Products
|
103
|
-17.1
|
Products
|
Moreover, -∆G = spontaneous and products will be favored at equilibrium. Whereas +∆G = non-spontaneous and reactants will be favored at equilibrium. This makes sense as chemical-reactions are inclined to move toward a lower free-energy state for stability. As illustrated below, products on the -∆G graph and reactants on the +∆G graph are both favoring their lower free-energy states. Therefore, any system is more stable at its lowest free-energy state and chemical reactions will respond accordingly to reach equilibrium.
*Concept check – from the graph below, estimate whether the products or reactants will be favored? Will this chemical-reaction be spontaneous or non-spontaneous? Draw a Gibbs free energy diagram to support your understanding.
Answer – at equilibrium, the products have a greater molar concentration than the reactants. This would suggest that the products are favored. We can also use
Keq =
[products]/[reactants]. If products are favored, then K >>1 (magnitude of K will vary with concentration). This information also
suggests a -∆G˚' at equilibrium.
Conceptual challenge cont.:
In a chemical reaction at equilibrium A ⇌ B + C, if the forward reaction-process is exergonic and the reverse reaction-process is endergonic, how can the reaction-rates for both the forward and reverse reaction be equal if the activation-energy of the forward reaction is lower than the activation-energy of the reverse reaction?
Answer - the activation-energy required from A ⇌ B + C is lower than the activation-energy of B + C ⇌ A. From this, it would be conceivable to think that the rate of A ⇌ B + C would be much faster than B + C ⇌ A. However, as mentioned earlier, B + C have a much higher molar-concentration than A. With a higher molar-concentration, the rate of chemical collisions in B + C will be much greater than the rate of chemical collisions from A. This will compensate for the higher activation-energy required to reach the transition-state in B + C ⇌ A. Therefore, the forward and reverse reactions will have equal reaction rates...EQUILIBRIUM.
Conceptual challenge cont.:
In a chemical reaction at equilibrium A ⇌ B + C, if the forward reaction-process is exergonic and the reverse reaction-process is endergonic, how can the reaction-rates for both the forward and reverse reaction be equal if the activation-energy of the forward reaction is lower than the activation-energy of the reverse reaction?
Answer - the activation-energy required from A ⇌ B + C is lower than the activation-energy of B + C ⇌ A. From this, it would be conceivable to think that the rate of A ⇌ B + C would be much faster than B + C ⇌ A. However, as mentioned earlier, B + C have a much higher molar-concentration than A. With a higher molar-concentration, the rate of chemical collisions in B + C will be much greater than the rate of chemical collisions from A. This will compensate for the higher activation-energy required to reach the transition-state in B + C ⇌ A. Therefore, the forward and reverse reactions will have equal reaction rates...EQUILIBRIUM.