Equilibria

According with our daily experiences, it results that mechanical systems tend to reach an equilibrium condition in which their energy is as low as possible. In fact, objects fall (to reach a minimum of potential energy), moving objects stops moving (to reach a minimum of kinetic energy).

When you raise a book from the floor (h₁) to your desktop (h₂), you spend some work which accumulates in the book as potential energy (see fig. A). Since

mgh₂ > mgh₁ (potential energy)

then you should expect that the book will fall on the floor to reach its condition of (relative) minimum energy. 

Fig. A	Fig. B	Fig. C

However, the book can stay on the table for years without reaching its condition of lower potential energy (on the floor). The book will "spontaneously" fall to the floor only if you supply it with the energy necessary to reach the border of the desktop. This energy is referred to as activation energy. Of course, when the same energy is supplied to the book on the floor, the book will not jump on the desktop.  Different processes require a different amount of activation energy. It should be evident that a paper sheet will require less activation energy to fall on the floor. A breeze can allow the sheet, but not the book, to reach the floor. Generally, the less the activation energy the higher the rate at which the spontaneous process will take place.

Consider now three differents states of equilibrium a book may have on a desktop (A, B, C in figure B). States A and C are both two states of stable equilibria, but, of course, state C is more stable than state A (more potential energy, higher center of gravity). In fact, by giving to A a little push (activation energy) it will reach the state of equilibrium C. The state A, for this reason, is said a state of metastable equilibrium. With some skill and in the absence of any disturbing force you can be able to reach state B which is a state of unstable equilibrium having a higher  potential energy (center of gravity) than A and C. From this state the book can reach either position A and C.

The differents states of equilibrium can be also represented by plotting potential energy versus the position of the center of gravity (Fig. C). It should be noted that stable states of equilibria are represented by minima in the curve (A and C) while unstable states of equilibria are represented by maxima in the curve (B). Furthermore let's note that stables states (A and C) are separated by an unstable state and that the difference of energy (B - A) represents the energy of activation. Eventually, for all the three states (A, B, C) the slope is zero, i.e., dU/dx = 0.

By summarizing, mechanical systems tend to reach a state of minimum energy provided that the necessary activation energy is supplied to them.

When the evaporation takes place in a closed evacuated container at constant temperature, we can observe that

only part of the liquid evaporates;

the pressure exerted by vapor molecules increases with time until to reach a constant value.

These observations can be explained by considering that vapor molecules can return to the liquid state (condensation) and the rate at which vapor molecules condense depends on pressure (the higher the pressure the higher the rate of condensation). At the beginning the rate of evaporation is greater than the rate of condensation and we observe a net formation of vapor but as the pressure increases the rate of condensation increases until to matches the rate of evaporation and no more vapor is formed. When the liquid-vapor equilibrium is reached we have that evaporation rate = condensation rate and the pressure exerted by vapor is constant (vapor pressure). Therefore, a liquid in a close container tends to reach an equilibrium in which part of the liquid is in the vapor state.

The process of evaporation requires energy to overcome the attractive forces of their neighbors molecules. Therefore, vapor molecules have higher energy than molecules in the liquid, it follows that a liquid-vapor system is not in a state of minimum energy. We have noted that at the liquid-vapor equilibrium, there are two conflicting forces, i.e., condensation and evaporation. Evaporation leads to a state of higher energy and of higher chaos (desorder or entropy) while condensation of the liquid leads to a state of minimum energy and minimum desorder (see entropy). Without the existence of these two conflicting forces all materials should eventually occur as vapors (maximum chaos) or solids (minimum energy).

Chemical equilibrium

Let's consider a generic chemical reaction involving two reactants (A and B) to give products C and D:

A + B = C + D

this equation states that compound A combines with compound B to gives compounds C and D (known as products)

It can be experimentally proved that the rate of the reaction is proportional to the concentration of reactants (A and B).

It can be also proved that the reverse reaction can occur

C + D = A + B

and, analogously to the reverse reaction, the rate at which the C and D are transformed into A and B is proportional to the concentration of C and D. 

This situation is similar to that encountered for liquid-vapor equilibrium.

At beginning, the concentration of reactants (A and B) is high and a net formation of C and D can be observed. As the concentration of products increases (and the concentration of reactants decreases), the rate of formation of A and B increases until the rate of the two opposite reactions will be the same. Therefore, at equilibrium, the concentration of reactants and products does not change with time.

Chemical reactions, as mechanical processes, require an energy of activation. As an example, consider a combustion reaction in which carbon atoms react with oxygen to give carbonic anhydride. It is well known that such reaction evolves heat thus indicating that the products of the combustion are more stables than carbon and oxygen. However, the carbon atoms presents in your book does not burn (by reacting with oxygen in the air) without the activation of the process by means of a lighter. Therefore, as in the case of mechanical equilibrium, the chemical transformation of substance A to substance C is achieved through the formation of an unstable state:

A ---> [B] ----> C

The state B is known as state transition or activated complex and the difference of energy (B - A) represents the energy of activation. In general, the higher [B] the higher the rate of the reaction.

It is well known that chemical reactions can be exothermic (evolving heat) or endothermic (absorbing heat) and  both kind of reactions can spontaneously take place. Therefore, the driving force determining a chemical equilibrium state is not a state of minimum energy otherwise only esothermic reactions should take places. As for liquid-vapor equilibrium, the chemical equilibrium is the results of two conflicting forces: the tendency of a system to reach a state of minimum energy and the tendency of a system to reach a state of maximum entropy (chaos, disorder).

Dynamic equilibria: LeChatelier principle

By comparing the force determining mechanical equilibria with those determining liquid-vapor and chemical equilibria an important difference results:

Mechanical systems tend to reach an equilibrium condition in which their energy is as low as possible.

Liquid-vapor and chemical equilibria are the results of two conflicting forces: one driving the system toward a maximum of disorder (entropy) and an opposite one driving the system toward a minimum of energy.

Equilibria which are the result of two conflicting forces, such as liquid-vapor and chemical reactions, are referred to as dynamic equilibria.

An interesting aspect of a dynamic equilibrium, is that it can be influenced by external factors. A simple principle to qualitatively predict the effect of an external factor on dynamic equilibria is LeChatelier's principle. This principle states that if a system at equilibrium is subjected to a disturbance or a stress, the system will react in such a way to minimize the effect of the disturbance. 

Let's consider a liquid, at constant temperature, in a container filled with a mobile piston. At the equilibrium evaporation rate = condensation rate and the pressure exerted by vapor is constant. If we move the piston up, we increase the volume of the container which results in a decrease of pressure in the container.

As previously observed, the condensation rate depends on pressure (the higher the pressure the higher the rate). Therefore the decrease in pressure leads to a decrease of condensation rate. On the other hand, the vapor pressure depends only on temperature and thus the change in pressure does not affect the evaporation rate. The consequence of these facts is that new molecules will enter the vapor state until a new equilibrium is reached at which the vapor pressure returns to its original value.

In the example described here, the decrease of pressure represents the stress while the variation of condensation rate represents the reaction of the system. When the pressure is decreased, the system minimizes the effect of the disturbance (stress), by allowing a net conversion of liquid into vapor which leads to an increase of pressure in the container. Consider now the effect of an increase in pressure (by moving the piston down). In this case we can expect that the condensation rate increases and the evaporation rate remain unchanged. Therefore, a net conversion of vapor into liquid occurs which leads to a decrease of pressure in the container. The conversion will continue until a new equilibrium is reached.

Chemical equilibrium

Chemical equilibrium. As for liquid-vapor equilibrium, the state of equilibrium of a chemical reaction is a compromise between two conflicting forces: the tendency of a system to reach a state of minimum energy and the tendency of a system to reach a state of maximum entropy (chaos, disorder). However, it should be pointed out that some reactions can be both energetically and entropically favored. In other words, the products of the reaction are both in a state of minimum energy and maximum entropy. Since no conflicting forces exist no equilibrium can exist. On these basis, chemicals reactions are distinguished in

irreversibles when no equilibrium exists between products and reactants and thus they are not affected by external factors.

reversibles when equilibrium exists between products and reactants and thus they are affected by external factors according to le Chatelier principle.

Reversible reactions. An example of reversible reaction is the dissociation of acetic acid in water

CH₃COOH = CH₃COO^- + H⁺

As previously stated, at the equilibrium the rate of dissociation of CH₃COOH matches the rate of association of CH₃COO^- and H⁺. Furthermore we remember that the rate of a reaction is proportional to the concentration of reactants.

When we add an external source of proton (the stress) to the system we have that the rate of association of acetate ions with protons will increases (the response of the system to the stress) thus leading to an increase of acetic acid in the system. Therefore, we can revert the dissociation of acetic acid by adding an external source of H⁺ or of acetate.

Irreversible reactions. An example of irreversible reaction is the dissolution of NaCl in water

NaCl --> Na⁺ + Cl^-

This reaction is both energetically and entropically favored and thus no equilibrium exists. The absence of equilibrium means that no association between the Na⁺ + Cl^- can take place and thus the addition of an external source of chloride ion (for example) can not revert the dissociation.

Eventually, let's remember that compounds like CH₃COOH and NaCl that ionize when dissolved in water to produce an electrically conductive medium are said electrolytes. CH₃COOH, which is partially dissociated (the equilibrium exists), is called weak electrolyte. On the contrary, sodium chloride which is fully dissociated (no equilibrium exists) is called strong electrolyte. It should be noted that double arrows (or  the = sign) are used for reversible reactions while a single arrow is used for irreversible reactions.

Let's consider a generic reaction such as

1. aA + bB = cC + dD

as previously stated at equilibrium the rate of the reaction of A with B matches the rate of reaction of C with B and the concentration of reactants and products does not change with time. It can be demonstrated (see thermodynamics) that at equilbrium equ. 2 holds:

[C]c [D]d 2. ----------- = K [A]a [B]b

[C]c [D]d = the concentration of products, each raised to a pover equal to its stochiometric coefficients. [A]a [B]b = concentration of reactants, each raised to a pover equal to its stochiometric coefficients. K = a constant, at constant temperature, known as equilbrium constant.

Please note that square brackets are generally used to indicate the concentrations (generally molarity) at the equilibrium.

The equilibrium constant for reactions in gaseous phase are more conveniently expressed in terms of partial pressure (generally in atm, at 25^oC) of reactants and products. As an example, for the reaction N₂O₄ = 2NO₂ we have

P²_(NO2)
3. K_p = --------- 
P_(N2O4)

where K_p is constant at constant temperature and pressure.

Equilibrium constant and nonideal solutions. If reactions take place in nonideal solutions the expressions for the equilibrim constant will still holds if concentrations are substituted with the activity (see solutions) of the species involved in the reaction, i.e.,

K = {C}^c {D}^d / {A}^a {B}^b.

This principle states that if a system at equilibrium is subjected to a disturbance or a stress, the system will react in such a way to minimize the effect of the disturbance. Let's consider acetic acid (a weak electrolyte) which in solution is partially dissociates in acetate and hydrogen ions:

1. CH3COOH CH3COO- + H+

[CH3COO-] [H+] Ka = ----------------- [CH3COOH]

If we add to the solution a source of H⁺ (e.g. HCl ) we have that [H⁺] increases and thus the ratio of acetate and hydrogen ion with acetic acid will be > K_a:

(CH₃COO^-) (H⁺)
2. ----------------- > K_a
(CH₃COOH)

In such situation the system is not at equilibrium and in order to return to a state of equilibrium, part of acetate and hydrogen ions must reassociate:

The addition of H⁺ is the stress or disturbance at which the system is subjected.

The reassociation of acetate and hydrogen ions is the reaction of the system to minimize the effects of the stress (disturbance).

This is the so called "common ion effect" of LeChatelier's principle.

Volume changes. Consider now a change of volume for a chemical system at equilibrium  (variation of pressure, dilution, etc.). Of course, the same variation of concentration for all the components of the system will be obtained. However two cases are possible:

1) reactions in which there is no variation of the numer of moles on the two side of the chemical equation:

2A = 2B

[B]2 K = ---- [A]2

2) Reactions involving a different number of chemical species on the two sides of the chemical equation:

2NO2 N2O4

[N2O4] K = ------- [NO2]2

In the first case, a change in volume (concentration) will influence at the same way numerator and denominator with not effect on the equilibrium. On the contrary in case 2, the concentration of NO₂ is raised to power 2 and dilution will increase the denominator more than numerator and thus the reaction will move towards the formation of N₂O₄ to return to a state of equilibrium.

Let's consider again the dissociation of acetic acid, considering that the [CH₃COO^-] = [H⁺] the constant can be written as

[H⁺]²
3. K_a = ---------
[CH₃COOH]

and thus dilution will increase numerator more than denominator and thus a reassociation of acetic acid will take place. In general when a reaction involves a different number of chemical species on the two sides of the chemical equation 

an increase in concentration  will move the equilibrium to the side containing the lesser number of species.

a diminution of concentration (an increase in volume) will move the equilibrium to the side containing the greater number of species.

Inert gases

An inert gas is one which is presents in a reaction system but which does not take part to the reaction, e.g., nitrogen is a gas inert during the synthesis of methanol (CO + H₂ = CH₃OH). Limiting our discussion to constant pressure processes, the addition of a inert gas leads to an increase of the total number of moles and thus to an increase of the volume of system according to the state equation of gas. Therefore according LeChatelier principle, the equilbrium will move toward the side containing the greater number of species (see above). This effect can be well understood if we express the equilibrium constant in terms of mole fraction and total pressure (Dalton's law). For a generic reaction (aA + bB = cC + dD), we have

(x_c P_t)^c (x_dP_t)^dx_c^c x_d^d
K_p = ---------------- = ------------(P_t)^c+d-a-b
(x_aP_t)^a (x_bP_t)^bx_a^a x_b^b

Remembering that the mole fraction is the moles of a component divided by the total number of moles we obtain:

n_c^c n_d^d(P_t)^c+d-a-b 
K_p = ------------ ------------ = K_n (P_t/N_t)^Dn
n_a^a n_b^b(N_t)^c+d-a-b 

where K_n = ratio of the number of moles and Dn = c+d-a-b.

Now let's consider the three possible cases, considering that the term (P_t/N_t) always decreases upon the addition of an inert gas (we are considering a constant pressure process):

1. Dn = 0 (no variation of the number of moles) and thus (P_t/N_t)^Dn = 1 and no effect can be observed after the addition of an inert gas.

2. Dn > 0 (positive variation of the number of moles) and thus (P_t/N_t)^Dn decreases. In order to mantain the value of K_p the value of  K_n should increase i.e., the reaction is favored.

3. Dn < 0 (negative variation of the number of moles) and thus (P_t/N_t)^Dn increases. In order to mantain the value of K_p the value of  K_n should decrease i.e., the reaction is not favored and the equilibrium will move to the left.

Of course, the reasoning above is valid only at constant pressure (for which K_p is constant). In a volume constant process V = P_t/RN_t and thus P_t/N_t is constant (no dilution occurs) and no influence on the equilibria is observed.

Inert gases can be also used to control the temperature reached by a reaction system. As an example, consider the combustion of carbon which is usually accomplished with air:

C + O₂ + 3.7N₂ = CO₂ + 3.7N₂, DH = -94000 cal/mole

If enough air is used, the complete conversion of C in CO₂ will liberate -94000 cal for mole of carbon. The heat liberated will produce an increase of the temperature of the reaction system. By admitting more air than the required air the same heat will be liberated but the temperature will result lower since the heat will be distributed on a greater number of molecules.

Influence of temperature

The effects of temperature on chemical equilibria can be also easily predicted by using LeChatelier's principle. 

When we increase or decrease the temperature of a chemical system at equilibrium, the heat we add to the system or subtract from the system represents the stress

If we increase the temperature, the reaction, in order to minimize the stress will move in the direction which absorbs heat.

if temperature is lowered the reaction will move in the direction that evolves heat.

Reactions 1 and 2 below both describe the formation of water from hydrogen and oxygen, the difference is just in the coefficients used to balance the reaction.

1. 2H2(g) + O2(g) = 2H2O(g)

[H2O]2 K1 = --------------     [H2]2 [O2]

2. H2(g) + 1/2 O2(g) = H2O(g)

[H2O] K2 = ----------------     [H2] [O2]1/2

It should be evident that the second reaction can be obtained by dividing by two the first one and K₂ is the square root of K₁. In general if a reaction is multiplied by a factor, the new equilibrium constant is obtained from the first raised to a power equal to the factor.

Let's consider a reaction and its reverse (3 and 4):

3. 2NO + O2 = 2NO2

[NO2]2 K1 = --------------    [NO]2 [O2]

4. 2NO2 = 2NO + O2

[NO]2 [O2] K2 = -----------    [NO2]2

It should be evident that

K₂ = 1/K₁

Consider reactions 5 to 7:

5. 2NO + O2 = 2NO2	[NO2]2 K1 = ----------------    [NO]2 [O2]
6. 2NO2 = N2O4	[N2O4] K2 = -------------     [NO2]2
7. 2NO + O2 = N2O4	[N2O4] K3 = --------------------      [NO]2 [O2]

It should be evident that

The last reaction is the sum of reaction 1 and 2. 

K₃ = K₁K₂

In general, when two or more chemical equations are added, 

the K for the resulting reaction is the product of the constants of the reactions added.