Chemical equilibrium
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CHEMICAL EQUILIBRIUM. Dr. Saleha Shamsudin. What is chemical equilibrium?. A dynamic state in which the rates of forward (f) and reverse (r) reaction are identical. Chemical Equilibrium – A Dynamic Equilibrium.

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Dr. Saleha Shamsudin

What is chemical equilibrium?

  • A dynamic state in which the rates of forward (f) and reverse (r) reaction are identical.

Chemical Equilibrium – A Dynamic Equilibrium

  • Upon addition of reactants and/or products, they react until a constant amount of reactants and products are present = equilibrium.

  • Equilibrium is dynamic since product is constantly made (forward reaction), but at the same rate it is consumed (reverse reaction).

1. Chemical Reactions

1.1 Introduction

  • Chemical reactions

    Involves chemical change.

    Chemical change: the atoms rearrange themselves in such a way that new substances are formed.

    Components decrease in quantity are called reactants and those increase in quantity are called products.

    The chemical reaction is:

    aA + bB ↔ cC + dD

The rate of reaction = constant x concentration each species raised to the power of the number of molecules participating in the reaction.

Forward reaction:

kf= rate constant

[A] and [B]= molar concentrations of A and B

Forward and reverse rates are equal

kf[A]a[B]b = kb[C]c[D]d

Molar equilibrium constant, K,

kb[C]c[D]d = kf = K

[A]a[B]b kb

  • Rearranging these equation gives molar equilibrium constant (K) :

    [C]c[D]d = kf = K

    [A]a[B]b kb

    When the two rates become equal, the system is in a state of equilibrium.

    Forward and reverse rates are equal

    kf[A]a[B]b = kb[C]c[D]d

Type of equilibria

  • We can write equilibrium constants for many types of chemical processes. The equilibria may represent :

    • Dissociation (acid/base, solubility)

    • Formation of products (complexes)

    • Reaction (redox)

    • Distribution between 2 phases (water and non-aqueous solvent)

Types of equilibria

Equilibrium constant of reverse reaction:

aA + bB  cC + dD

cC + dD  aA + bB

Equilibrium constants for dissociating and associating species

  • For weak electrolyte

  • Slightly soluble substances

  • Example:

    1) AB  A + B

  • Equilibrium constant written as:

    [A][B] = Keq


  • Stepwise dissociation

    A2B  A + AB K1 = [A][AB]


    AB  A + B K2 = [A][B]


    Overall dissociation:

    A2B  2A + B Keq = [A]2[B]


3) Heterogeneous equilibria

  • Involving more than one physial phase, example, solids, water and pure liquids.

  • Concentrations of any solid reactants and products are omitted from the equi. cons. expression.

  • Molar concentrations of water (or of any liquid reactant or product) is omitted from the equi. cons. expression.

E.g. Write out the equilibrium expression for K using the reaction below:

N2(g) + 3H2(g)  2NH3(g) Kc = ?

Calculating Equilibrium Constants

Try this example:

An aqueous solution of ethanol and acetic acid, each with a concentration of 0.810 M, is heated to 100oC. At equilibrium, the acetic acid concentration is 0.748 M. Calculate K at 100oC for the reaction

C2H5OH(aq) + CH3CO2H(aq)  CH3CO2C2H3(aq) + H2O(l)

ethanol acetic acid ethyl acetate

  • Write the equilibrium constant expressions in terms of concentrations:

  • Calculate K at 100oC for the reaction:

  • The equilibrium constant expressions is:

  • K = [CH3CO2C2H3] notice that liquid

  • [C2H5OH][CH3CO2H] water does not

  • 2. The calculation of K is: appear in equi.

  • = 0.062 = 0.11 expressions.

  • (0.748)(0.748)

Calculating an Equilibrium from an Equilibrium Constant

Problem: K= 55.64 for H2(g) + I2(g)  2HI(g)

Has been determined at 425oC. If 1.00 mol

Each of H2 and I2 are placed in 0.5 L flask at 425oC, what are the concentrations of H2, I2 and HI when equilibrium has been achieved?


  • Write the equi. Cons. Expression.

  • Set up an ICE table

  • Assume x is equal to the quantity of H2 or I2

  • consumed in the reaction

  • Substitute the equi. Con. Into K expression.

    55.64 = (2x)2 = (2x)2

    (2.0 – x)(2.0 – x) (2.0 – x)2

    55.64 = 7.459 = 2x

    2.0 – x

    x = 1.58

    H = I = 2.0 – x = 0.42 M

    HI = 2x = 3.16 M

Calculating Equilibrium Concentrations

  • Using initial concentrations, stoichiometry and Kc, equilibrium concentrations of all components can be determined.

    E.g.2 [H2]o = [I2]o = 24 mM, were mixed and heated to 490oC in a container. Calculate equil. composition. Given that Kc = 46.

    Solution: set up an equilibrium table and solve for unknown after substitution into equilibrium expression.

    H2(g) +I2(g)2HI(g)

  • Equilibrium concentrations of each component are in last row. Substitute into equilibrium expression and solve for x.

  • It may be necessary to rearrange so that quadratic equation can be used.

  • .

  • Rearrange to ax2 + bx + c = 0; determine a, b, c and substitute into quadratic equation:

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