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The Concept of Equilibrium

Few physical and chemical changes proceed only in one direction. Fire is one chemical change that does. Once burned, a piece of paper cannot be restored.

The evaporation of water in a closed system is a reversible physical change that establishes a dynamic equilibrium. At equilibrium, when water seems to ceases evaporating, water is still evaporating but recondensing at the same rate.

There is no net change, (therefore equilibrium)but molecules are continually shuttling between the liquid and vapor state (dynamic). Many chemical systems act in this manner where product is continually being formed and then back-reacting to form the reactants again.

The Equilibrium Constant

Weak acid dissociation in water is an example of a dynamic equilibrium. Acid molecules dissociate to anion and hydrogen ion, but the molecular acid reforms, keeping the acid concentration relatively high.

HA  ⇔ H⁺¹  + A^-1

A value called the equilibrium constant can be assigned to the chemical system by measuring, at equilibrium, the concentrations of all the species.

The equilibrium constant, K can be taken as the product of the product molar concentrations divided by the reactant molar concentration.

• K_eq  = [H⁺¹] · [A^-1]
•                    [HA]

The usefulness of K_eq is that remains constant for the system at one particular temperature. Its value does not change when any of the reactant or product concentrations are altered. The other concentrations adjust so that K_eq remains the same.

Ex: What happens to the system if [H⁺¹] is changed? What happens if [HA] is changed? What happens if the temperature is changed?

In general, all chemical equilibrium systems obey the law of mass action. For the general equilibrium equation

aA + bB ⇔ cC + dD

the equilibrium constant will be given by

•         K = [C]^c [D]^d 
•                [A]^-a [B]^-b

The law of mass action can be extended to systems of any number or reactants and products. If K is much larger than unity, the reaction is said to lie on the right. If K is much smaller that one, then the equilibrium lies to the left.

As an example of the use of the exponents, note that if A is the same species as B and if D is not formed, in other words

1A + 1A ⇔ cC + 0D or 2A ⇔ [C]^c

then the equilibrium expression becomes

K = [C]^c [D]⁰/[A]^-1 [A]^-1  = [C]^c/[A]²

 Equilibrium Constant Expressions

 Substances in solution will have their concentrations expressed in moles/liter or M. Gases require an expression to convert between M and pressure. Solids and liquids in heterogeneous systems will have their concentration expressed simply as unity, since their effective concentration will not change as long as some condensed phase remains.

Homogeneous Equilibria

All reactants and products are in the same liquid or gas phase.

A  ⇔ B

A can be (g) or (l) and B can be (g) or (l)

Concepts of K_c based on molar concentrations in gas or in solution and K_p, based on partial pressures of gaseous species.

K_{c =} [B]/[A]

whether A and B are in solution or in gas phase

K_{p =} P_B/P_A

In general K_c ≠ K_p unless the number of moles of gas does not change during the reaction.

If the equilibrium expression does not contain different powers of [A] or [B] in the numerator or denominator, the ration of [B]/[A] is the same as that of P_B/P_A. If there are different powers of [B] or A in the expression, then the ratio changes.

Heterogeneous Equilibria

If a solid or liquid is part of a chemical equilibrium, its “concentration” is taken as unity and may be eliminated from the equilibrium expression.

NH₃(g) + HCl(g)  ⇔  NH₄Cl(s)

K_eq  = [NH₄Cl] / [NH₃]·[HCl]

Since NH₄Cl is a solid, its effective concentration in the reaction does not change as long as some solid is present. Since [NH₄Cl] is constant, it can be eliminated from the right side of the equation and incorporated into the K_eq constant, which them becomes

K_eq  = 1/[NH₃]·[HCl] 

if the molar concentrations of the gases will be expressed,

or

K_p  = 1/P _NH3 · P _HCl   

 if the gas pressures will be expressed.

Example: How will [NH₃] vary in the system as [HCl] increases or decreases?

Multiple Equilibria

The K_eq values from two or more successive reactions are simply multiplied together if the reactions can be added together to make one net reaction.

For A  ⇔B  K_eq = [A]/[B] and

For B ⇔ C, K_eq = [B]/[C]

Adding the two equations together gives

                 A ⇔ C with K_eq = [A]/[C]

The same K_eq expression would have been obtained by multiplying together the two first K values. Applications include stepwise dissociation of polyprotic acids such as H₃PO₄

The allover K = K₁ · K₂ · K₃ for the equation

               H₃PO₄ ⇔ PO₄^-3  + 3H⁺¹

The Form of K and The Equilibrium Constant

The K for an equilibrium reaction is the reciprocal of the K for the reverse reaction. Therefore it is important to specify what reaction the K meant to represent.

For A ⇔ B + C, K_f  = [B][C]/[A]

For B + C ⇔ A, K_r  = [A]/[B][C]

Therefore K_{f x} K_r  = [A][B][C]/[A][B][C] = 1

Remember that the reactant and product concentrations in the K_eq expression are raised to the power equal to their coefficients. What happens if the equation is written in more than one way?

Will the K_eq values be different or the same for the two ways of writing the same equation? Consider both as representing the same system. Set [H₂] = 0.020M, [I ₂] = 0.030M and [HI] = 0.010M. Calculate both K_eq values.

K₁  = [H₂][I₂] /[HI]²  = (0.020M)(0.030M)/0.010M)²  

                                     = 6.0

K2  = [H2]^½ [I2]^½ /[HI] = (0.020M)^½ (0.030M)^½ /(0.010M)¹ 

                                      = 2.4 (or 6^½)

                                  K₁ = (K₂)²

Both K_eq values will give the same reactant and product concentrations when applied to the corresponding equations.