Reversible Equilibrium Reactions

From Kinetics to Equilibrium

Reversible Reaction of Higher Order

Law of Mass Action

The Hydrogen Ion in Solution

Strong and Weak Electrolytes

Biprotic Weak Acids

Triprotic Acids

Biprotic Acid with Salt/Hydroxide Precipitation

 


From Kinetics to Equilibrium

Reversible reactions are those in which the products can be converted back to the original reactants. In a system of reversible reactions, the apparent rate of the forward reaction will decrease as the reaction products accumulate, until eventually a state of dynamic equilibrium is established. At equilibrium, the forward and backward reactions proceed at equal rates.

A first-order reversible reaction can be represented by this simple mechanism

A B (rate constants = k1,k2)

where k1, k2 rate constants refer to the direct (from left to right) and inverse reaction.

The true rate of concentration change for either reactant is given by the difference in the rates in the opposite directions, each proportional to the concentration of the reacting compound. The differential equations for this mechanism are:

- d[A]/dt = k1•[A]- k2•[B]

- d[B]/dt = k2•[B]- k1•[A]

Instead of finding a mathematical solution for these equations, the usual iterative procedure as described below will be used.

 

dim A(100),B(100)

A(0) = 40 ' initial concentration of A

B(0) = 10 ' initial concentration of B

k1 = 0.04 ' reaction constant of direct reaction

k2 = 0.02 ' reaction constant of inverse reaction

t1 = 100 ' final time

for t = 1 to t1 ' reactions take place

A(t) = A(t-1) - A(t-1)*k1 + B(t-1)*k2

B(t) = B(t-1) + A(t-1)*k1 - B(t-1)*k2

next t

call DrawScreen



Fig.3.1 First-order irreversible reaction

When a reversible chemical reaction like the above reaches equilibrium, the rates of the forward (v1) and reverse (v2) reactions are exactly equal, and the net rate of reaction (Vnet) is zero (Fig 3.1). Under these conditions v1 = v2, and seeing that v1 = k1•[A] , v2 = k2•[B]

one obtains:

k1•[A] = k2•[B] k1/k2 = [B]/[A] = Keq (3.1)

If we are interested in the equilibrium concentrations, no matter how long it takes for the reaction to reach them, there is only one constant to be addressed, Keq (equilibrium constant)

Keq is the ratio between the two rate constants and, as it is independent from concentrations, only temperature affects it. By knowing the initial concentrations of reactants [A]0 and [B]0 and Keq, we are always able to calculate the final (equilibrium) concentrations. This can be done mathematically by solving first, second or even third degree equations, or by iterative procedure, as shown in the following.


Reversible Reaction of Higher Order

Let’s discuss this with a slightly more complex reversible reaction, with an order higher than one. For example, the reaction

A + B C + D

As soon as the substances A and B start to react in the reaction vessel, they produce C and D with a rate that can be expressed as

r1 = k1 • [A][B] (3.2)

where each of the quantities in brackets is the molar concentration at a given time.

Being an equilibrium reaction, as soon as C and D are formed, they start to react together (unless removed from the reaction vessel). The rate of this inverse reaction will increase with C and D concentrations:

r2 = k2 • [C][D] (3.3)

whilst the rate r1 will decrease due to the shortage of A and B. At a certain time, the molar concentration of reactants and products will reach such a value that the rate of direct and inverse reaction will match, so we can write

r1 = k1 • [A][B] = r2 = k2 • [C][D]

It is a dynamic equilibrium, as it is the case overall in chemistry, during which the quantities of A and B that disappear are balanced by those obtained by C and D.

By transforming the above equation, one obtains

    k1

     [C]•[D]

_____ = Keq =

____________ 

    k2

    [A]•[B]

where the constant Keq is a new constant (the ratio of the forward and reverse rate constants) called the equilibrium constant. Each of the quantities in brackets is the equilibrium concentration of the substance shown. At any given temperature, the value of Keq remains constant no matter whether you start with A and B, or C and D, and regardless of the proportions in which they are mixed. Keq varies with temperature because k1, and k2 vary with temperature, but not by exactly the same amount. This dependence on temperature is discussed at the end of this chapter.


Law of Mass Action

Equation 3.4 applies to any chemical reaction at equilibrium, no matter how many or how complicated the intermediate steps in going from reactants to products. It is worth stressing that as long as eq. xx-1 is valid, the concentrations therein are the equilibrium concentrations. Some reactions are very fast, so that equilibrium is reached after just a few seconds, others reach it after years or never.

If we consider the general case

aA + bB cC + cD

where a,b,c,d indicate the stoichiometric coefficients, the equilibrium constant can be written as


[C]c • [D]d

Keq =

____________ 


[A]a • [B]b

which is the general formulation of the law of mass action (Guldberg-Waage, 1864) that states: in a chemical system at equilibrium and constant temperature, the ratio between the product of the concentrations of chemicals formed (each elevated to its stoichiometric coefficient) and that of the reagents is a constant value.

It can be shown by thermodynamics that in the law of mass it is the use of coefficients that balances the reaction itself. They are not therefore to be confused with α,β coefficients of the reaction rate of an irreversible reaction.

The above expression of the law of mass action is particularly useful in solution. However, it can also describe equilibrium between two or more phases, as will be shown in the following four cases:

In gaseous reactions concentrations are usually substituted by partial pressures, giving the expression


P(C)c • P(D)d

Kp =

________________  


P(A)a • P(B)b

where P(A)...are the partial pressures (in atm.) of the reacting gases and it can be demonstrated that:

Kp = Keq(RT)(c+d-a-b)  

R being the universal gas constant and T the absolute temperature.

In the equilibrium systems considered above, all the substances are in a homogeneous phase (liquid, gaseous). However, chemical reactions may also occur in heterogeneous systems, with two phases. Let’s consider, as an example, a reaction between a solid (CaCO3) and its decomposition products. The temperature reaction is 900°C

CaCO3 (solid) CaO(solid) + CO2(gas)

As an initial prediction, we can write


P(CaO)•P(CO2)

Kp' =

____________________  


P(CaCO3)

but because vapour pressures of solids are very low, they can be disregarded, or more precisely, as they are constant values at a certain temperature they can be included in the equilibrium constant, thus obtaining a new Kp constant:

Kp' = a/b•P(CO2) Kp = Kp'•(b/a) =P(CO2) 

In summary: if heterogeneous equilibria concerns solids and gases, equilibrium expression only contains partial pressures of gases, each raised to the stoichiometric coefficients.


If we consider the solution of a sparingly soluble salt, for example CaCO3, once it has formed a saturated solution, an equilibrium exists between the solid salt and its ions in solution, as in

CaCO3 Ca++ + CO3- -

For this heterogeneous equilibrium, the mass action law is applicable (where sp stands for solubility product)


[Ca++]•[CO3 - -]

Keq =

___________________  


[CaCO3]

As solids always have the same concentration, it is included in Keq, thus obtaining

Ksp = [Ca++]•[CO3 - -]

For a general salt of composition MmAn ↔ mMb+ + nAa- the solubility product will be expressed as

Ksp = [Mb+] m • [Aa-] n  

As an example, let’s consider sodium phosphate Mg3(PO4)2, which is present with a variable concentration in seawater. It has a very low solubility product, so as to be a nearly-insoluble salt:

Mg3(PO4)2 3 Mg++ + 2 PO4 - - - Ksp = [Mg++] 3 [PO4 - - - ] 2 = 6.310 -26

Gas – liquid equilibria are also common. In ocean chemistry, one fundamental reaction is the dissolution of atmospheric carbon dioxide (CO2) in water which, in the first instance, forms carbonic acid (H2CO3). The mass action law for the dissolution reaction

CO2 + H2O ↔ H2CO3 states that


[H2CO3]

Keq =

____________________ 


P(CO2 )• [H2O]

However, molar concentration of water is constant (1000/18 in pure water) and is incorporated in Keq. The partial pressure of carbon dioxide is expressed in atm in the S.I. System. The term [H2CO3] comprises the concentration of CO2 dissolved in water or CO2 (aq) which is the first reaction product of gaseous CO2 with water.



The Hydrogen Ion in Solution


Chemical reactions in aqueous solutions (including the chemistry of life processes) very often depend on the concentration of hydrogen ions (H+) in the solution.

As we shall see, we may deal with hydrogen ion concentrations which range from over 1 M to less than 10-14 M. Consequently, it is convenient to express these concentrations on a logarithmic basis; for this purpose the terms "pH" and "pOH," have been introduced.

Hydrogen ion concentrations are represented by "pH" and hydroxide ion concentrations by "pOH", denoted by the relations:


pH = -log [H+]

pOH = -log [OH-]


In keeping with this usage, we also use pKw, = -log Kw

You may recall that log AB = log A + log B, therefore pH + pOH = pKw = 14

Water is a weak electrolyte, ionizing slightly and reversibly as H2O H+ + OH-

The H+ is hydrated, forming chiefly H3O+ ions. Just as we ignore the hydration of all the metal ions (for convenience in writing equations), we also often ignore the hydration of H+. We must always remember, however, that a bare proton (an H+) can never exist in solution by itself.

This dissociation reaction is always at equilibrium, with extremely rapid formation and recombination of H+ and OH-. Because it is always at equilibrium, the principles of chemical equilibrium discussed apply, and we can write the equilibrium constant expression.

This equilibrium is of such importance that K bears the special subscript w. In pure water at 25.0°C, the concentration of H+ is 1.0 ·10-7 M, that is, [H+] = 1.0·10-7 M. Because the dissociation provides equal numbers of H+ and OH- ions, it also follows that in pure water [OH-] =1.0 ·10-7 M. Knowing the equilibrium concentrations, we can evaluate Kw numerically: Kw = 1.0·10-7 x 1.0·10-7 = 1.0·10-14 (mol/L)2

This constant applies to all water solutions. It follows that if we add acid to water, and thereby increasing the [H+], there must be a corresponding decrease in [OH-], and vice versa. HCl is a strong acid that completely dissociates in water.

This means that in a 0.10 M HC1 solution [H+] = 0.10 M. Because [H+][OH-]= 1.0·10-14, it follows that [OH-] = 1.0·10-14 M, one millionth of the concentration in pure water. NaOH is a strong base, which is also completely dissociated in water. A 0.10 M NaOH solution will have [OH-] = 0.10 M, and an associated [H+] that is 1.0·10-13 M.



Strong and Weak Electrolytes

When we put a strong electrolyte (such as HCl) into solution, essentially all the molecules dissociate to ions; in this case, H+ and Cl-. But when we put a weak electrolyte into solution, such as acetic acid (CH3COOH), only a small fraction of the molecules dissociate. The equation is:

CH3COOH CH3COO- + H+

Because this reaction is at equilibrium, we can apply the mathematical expression

Kdiss = [H+][CH3COO- ]/[ CH3COOH]

The equilibrium constant for the ionization of a weak electrolyte is usually designated as Kdiss, which we call the ionization constant. Ionization constants are determined by experimental measurements of equilibrium concentrations. For example, to determine Kdiss for acetic acid, we prepare a solution of known concentration and, by any of several methods, measure the H+ concentration or the pH. The method most widely used today is measuring with a pH meter, which gives a direct dial reading for the pH.

However, in practical calculations on weak and very weak acids, self-dissociation of water must be considered, and we thereby obtain a system with two equations and two unknowns.

Kdiss = [H+][CH3COO- ]/[ CH3COOH]

Kw = [H+][OH-] = 1.0·10-14 at 25°C

The corresponding iterative procedure for hydrofluoric acid (a simple monoprotic acid found in the composition of seawater) is listed and commented on in the appendix (code001.bas).

 


Biprotic Weak Acids

In some cases, we find two or even three hydrogen atoms in the same molecule that are able to dissociate in aqueous solution. An important example in seawater is carbonic acid, H2CO3, which originates from CO2 (aq), i.e. the dissolved form of CO2 (gas). In this dissolved form, the water dipoles interact with the charges in the CO2 molecule without forming new chemical bonds. The concentration of H2CO3 is, however, much smaller (about 0.3%) than that of CO2 (aq). The sum of the two electrically neutral forms, true carbonic acid H2CO3 and aqueous carbon dioxide, which are chemically inseparable, is usually denoted as H2CO3* or simply H2CO3.

In the computer code listed in the appendix (code 002.bas), carbonic acid is used as an example. Three equilibria (including water self-dissociation) are solved simultaneously by means of an iterative procedure. The dissociation constants Ka1 and Ka2 refer to pure water at 25°C, and given that the reaction vessel is deemed as closed, it is not necessary to consider CO2 partial pressure at this stage.

The code is self-explanatory to some extent. Moreover, the program structure can simulate the addition of a strong base, as in NaOH, the example having a concentration of 0.03 M (mol/L).

 


Triprotic Acids

As an example of triprotic acid we shall consider phosphoric acid, being a constituent of seawater. It dissociates according to the following (the Keq refers to pure water at 25°).

The three dissociation constants of phosphoric acid, Ka1 = 7.5•10-3 (1st dissociation constant), Ka2 = 6.2•10-3 (2nd dissociation constant) and Ka3 = 4.4•10-13 (3rd dissociation constant) and Kw = 1•10-14 (ionic product of water) refer to pure water and 25°C.

In the simulation, the possible addition of a strong base (NaOH) and a strong acid (HCl) is considered, both assumed to be completely dissociated and therefore regarded as ions, as can be seen in the code003.bas listed in the appendix.

 


Biprotic Acid with Salt/Hydroxide Precipitation

In an even more complex system of reactions we deal with a biprotic acid (carbonic acid again) which interacts with calcium ions (Ca++) in solution giving (if the solution becomes supersaturated) a white precipitate of CaCO3. Moreover, magnesium ions (Mg++), whose concentration in seawater is five times greater than calcium ions, may form solid magnesium hydroxide (in this case). The latter is a hydroxide with a very low solubility (Ksp = 8.9 •10-12). Programs can simulate the addition to the reaction vessel of many chemicals, like magnesium and calcium chloride, sodium carbonate and bicarbonate, sodium hydroxide (strong base) and chloridric acid (a strong acid).

The different equilibria are of course solved with iterative procedures, in which pH values are varied in certain ranges, each time with a smaller step (the cycle uses the 'j' variable). The reaction container is deemed closed with respect to CO2 exchange with the surrounding atmosphere. Due to the many reactants, the code list takes up some pages, and it is listed in the appendix as code004.bas. In the example the starting composition for simulation is as follows (values in mol/liter):

Na2CO3 = 0.2 ' Soluble salt completely dissociated in Na+ and CO3-- ions

NaHCO3 = 0.1 ' Soluble salt completely dissociated in Na+ and HCO3- ions

CaCO3 = 0.0 ' Sparingly soluble salt. Solubility product Ksp needed

NaOH = 0.1 ' Strong electrolyte, completely dissociated in Na+ and OH- ions

HCl = 0.4 ' Strong electrolyte, completely dissociated in H+ and Cl- ions

CaCl2 = 0.6 ' Soluble salt completely dissociated in Ca++ and Cl- ions

MgCl2 = 0.2 ' Soluble salt completely dissociated in Mg++ and Cl- ions

H2CO3 = 0.005 ' Hydrated CO2