EQUILLIBRIUM at a glance

EQUILLIBRIUM is said to be established in a reversible reaction when the rate of forward reaction is equally to the rate of backward reaction.

 We can describe equilibrium here as a state where there is no observable change.

In this Chapter we will be looking at Equilibrium in Chemistry and try to explain it as best as we can

Types of Equilibrium

Equilibrium can be Static  and  Dynamic

I. Static equilibrium: - In static equilibrium there is basically no movement within the system and the substances involved are at the same level. An example of a static equilibrium is when two children on a seesaw are suspended in space.

You will observe that both children suspended in space have the same potential energy. But any slight disturbance and the equilibrium will be lost.

Ii. Dynamic equilibrium: - In dynamic equilibrium, there is actually movements, but the changes are not observable. For example, a child going up an escalator that is actually descending at the same speed as the child, the child would seem to remain at a particular spot even though he is moving up.

Dynamic equilibrium can be grouped into two

I. Physical equilibrium and 

Ii. Chemical equilibrium

I. Dynamic physical equilibrium: -In dynamic equilibrium no new substance is formed (just like a physical change). For example, when sodium chloride is dissolved in water to give a saturated solution, the dissolved crystals will crystallize out of the solution just as undissolved crystals are dissolving into the solution, with these forward and backward reactions occurring at the same rate, we say the saturated solution is in equilibrium. 

NaCl(s) ⇌ NaCl(aq)

II. Dynamic Chemical Equilibrium: - In this equilibrium system a new substance is formed (just like a chemical change).

An example is the reaction between Nitrogen and Hydrogen to yield Ammonia

N2(g) + 3H2(g) ⇌ 2NH3(g)

In dynamic equilibrium a new substance is formed.

Properties Of a System in Equilibrium

If a system is in equilibrium, then it will possess the following properties

I. The reaction must be a reversible reaction 

Ii. The rate of forward reaction must be equal to the rate of backward reaction 

III. ∆G must be equal to zero. (delta G is the Gibbs free energy)

IV. The equilibrium can be approached from either side of the reaction.

V. The reaction must occur in a closed system.

Factors affecting a system in equilibrium position of a reversible reaction

I. Temperature 

II. Concentration 

III. Pressure

IV. Catalyst

 

Le Chatelier's Principle: See laws and principles

The effect in equilibrium position brought about by a change in any of the factors mentioned above was predicted and may be explained by Le Chatelier's principle 

i. How Temperature Affects equilibrium position of a reversible reaction: - Given a reaction where the forward reaction is exothermic, then the backward reaction will be endothermic.  For such a reaction, an increase in temperature will shift the position of Equilibrium to the left, that is, it will favor the backward reaction on the other hand, a decrease in temperature will favor the forward reaction since it does not require much heat.

How Concentration Affects the equilibrium position of a reversible reaction: - When any one of the reactants in a reversible reaction is increased this will cause the equilibrium position to shift to the right, favouring the forward reaction.   This is because an equilibrium has already been established and adding more reactants will alter that equilibrium, and so according to Le Chatelier's principle the reactants will be quickly used up and be converted to product. 

If the products are removed from the system as they produced, this can also cause the equilibrium to shift to the right.

For example, let us consider the equilibrium reaction of the Haber process, that the reaction between H2 and N2 the produce NH3 

 

                    N2(g) + 3H2(g) ⇌ 2NH3(g) 

An increase in the concentration of the N2 or H2 will cause the equilibrium position to shift to the right, favouring the forward reaction.

How Pressure Affects the equilibrium position of a reversible reaction: - For Pressure to affect the equilibrium position of a reversible reaction, at least one of the reactants or products must be a gas and the total number of mols (vol.) of the reactants must be different from the total number of mols (vol.) of the product. 

For example, considering the two reactions 

  1. Fe(s) + 4H2O(g) ⇌ Fe2O3 + 4H2(g) 

                  4vol.                      4vol.

      

         2.  N2O4(g) ⇌ 2NO2(g)

              1vol.        2vol.

Pressure cannot affect the position of equilibrium of equation 1 because the volumes of both the gaseous reactant and gaseous product are the same (i.e. 4vol.)

However, in equation 2. the total volume of reactants is different from the volume of product, so an increase in pressure for such a reaction will lead to a decrease in volume (Boyle's law) hence causing the equilibrium to shift the left (the position that has a small volume) favouring the backward reaction. Similarly, a decrease in pressure will result to an increase volume and this will cause the equilibrium to shift to the right (area with larger volume) favouring the forward reaction.

Equilibrium Constant 

The equilibrium constant for a reaction in equilibrium, is the product of the molar concentrations (or pressures) of the products divided by the product of the reactants raised to the power of the coefficient of each reactant.

For a reaction 

            aA +bB ⇌ cC + dD

The equilibrium constant for the reaction will be 

                Kc = [C]^c[D]^d
                        [A]^a[B]^b

         

OBJECTIVE QUESTIONS

1. what will happen if more heat is applied to the following system in equilibrium

   

     X_2(g) + 3Y_2(g) ⇌ 2XY_3(g); ∆H= -xkjmol^-1

   

A. The yield of XY₃ will increase.

B. More of will XY₃ decompose 

C. more of X₂ will react 

D. The forward reaction will go to completion

2. What is the expression for the equilibrium constant (Kc) for the following reaction?  N_2(g) + O_2(g) ⇌ 2NO_(g)

A.      [NO]²
      [N₂] + [O₂]  

 

b.       [2NO]²
         [N₂]]O₂]

 

c.     [N2] + [O]
          2[NO]²

 

d.       [NO]²

        [N₂][O₂]

3. A reaction represented by the equation below  

     A_2(g) + B_(2(g) ⇌ 2AB_(g); ∆H = +X kjmol^-1             

which of the following statement about the system is correct

a. The forward reaction is exothermic 

b. The reaction goes to completion at equilibrium 

c. Pressure has no effect on the equilibrium mixture

d. At equilibrium increase in temperature favours the reverse reaction.

 

4. What will happen if more heat is applied to the following system at equilibrium?

                X_2(g) + 3Y_(2(g) ⇌ 2XY_(g); ∆H = -X kjmol^-1

a). The yield will increase of XY³ will increase.

b).  More of XY₃ will decompose 

c). . More of X₂ will react 

d).  The forward reaction will go to completion

5. The position of equilibrium ina reversible reaction is affected by 

a). Particle size of the reactants 

b). Change in concentration of the reactants 

c). Change in the size of the reaction vessel 

d). Vigorous stirring of the reaction mixture.

THEORY

1(a). When few drops of aqueous KSCN are added to a solution of iron (III) salt the following equilibrium is s_{The equilibrium mixture has a pale colour}

                                          ^{Yellow      colourless         deep red} 

(i). Explain what will happen if more KSCN were added to the equilibrium mixture 

(ii). Which of the ions in the equilibrium mixture forms an insoluble hydroxide with NAOH (aq)?   Write an equation for the reaction

(iii).  State two changes observed on adding NaOH to the equilibrium mixture.

(b)i.