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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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chemical equilibrium

CHEMICAL EQUILIBRIUM

Dr. Saleha Shamsudin

what is chemical equilibrium
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
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. 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

slide6

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

slide7
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
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)
slide10

Equilibrium constant of reverse reaction:

aA + bB  cC + dD

cC + dD  aA + bB

equilibrium constants for dissociating and associating species
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

[AB]

slide12
Stepwise dissociation

A2B  A + AB K1 = [A][AB]

[A2B]

AB  A + B K2 = [A][B]

[AB]

Overall dissociation:

A2B  2A + B Keq = [A]2[B]

[A2B]

slide13
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.
slide14

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

N2(g) + 3H2(g)  2NH3(g) Kc = ?

slide15

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:
slide16

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)
slide17

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?

slide18
Solution:
  • 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
slide19
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
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)

slide21

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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