CHEMISTRY A LEVEL(FORM SIX) NOTES – PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)

PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM

Chemical reactions which take place in both directions are called reversible reactions. These reactions do not proceed to completion but rather to dynamic equilibrium between products and reactants.

As a result, the concentration of reactants and products becomes constant (but not equal).

Reactants products

At equilibrium, the rate of forward reaction is equal to the rate of backward reaction.

Note: The concentration of all species in the equilibrium remains constant since both opposing reactions proceed at the same rate.

The concentration of reactants decreases with time while those of products increase with time until equilibrium is reached where the concentrations are constant (not equal).

The equilibrium has been attained with a high concentration of products than reactants; therefore, equilibrium lies on the product side.

The equilibrium has been attained with a high concentration of reactants than products; therefore, equilibrium lies on the reactants side.

The rate of reaction depends on the concentration of reactants. As the concentration of reactants decreases, the rate of forward reaction decreases too. The rate of backward reaction increases since the concentration of products increases until equilibrium is attained.

Characteristics of chemical equilibrium

1. The equilibrium can be established or attained in a closed system (no part of the reactants or products is allowed to escape).
2. The equilibrium can be initiated from either side. The state of equilibrium of a reversible reaction can be approached whether we start from reactants or products.

Consider the reactions:

H₂(g) + I₂(g) ⇌ 2HI(g)

3. Constancy of concentration: When a chemical equilibrium is attained, concentration of various species in the reaction mixture becomes constant.
4. A catalyst cannot change the equilibrium point. When a catalyst is added to a system in equilibrium, it speeds up the rate of both forward and backward reactions to equal extent. Therefore, a catalyst cannot change the equilibrium point except that it is reached earlier.

Types of chemical equilibrium

1. Homogeneous equilibrium: This is equilibrium where reactants and products are in the same physical states, i.e., all solids, all liquids, or all gases.
   Example: H₂(g) + I₂(g) ⇌ 2HI(g)
   N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
2. Heterogeneous equilibrium: This is equilibrium when the reactants and products are in different physical states.
   CaCO₃(s) ⇌ CaO(s) + CO₂(g)
   3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)
3. Ionic equilibrium: This is an equilibrium which involves ions.
   

LAW OF MASS ACTION

The law relates the rates of reactions and the concentration of the reacting substances. The law states that, “the rate of a chemical reaction is directly proportional to the product of the molar concentration of the reactants at constant temperature, at any given time”.

The molar concentration means number of moles per litre and is also called active mass.

Consider the following simple reactions:

The rate of the reaction, R = K [A]^a [B]^b where [A] and [B] are the molar concentrations of the reactants A and B respectively and K is a constant of proportionality known as rate constant or velocity constant.

If the concentration of each of the reactants involved in the reaction is unity, i.e., the concentration of [A] = [B] = 1, thus the rate constant of a reaction at a given temperature may be defined as the rate of the reaction when the concentration of each of the reactants is unity.

Generally for a reaction:

Where a and b are stoichiometric coefficients or mole ratio of the reactants, A and B.

LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT

The law of mass action is applied to reversible reactions to derive a mathematical expression for equilibrium constant known as the law of chemical equilibrium.

Consider a simple chemical reaction (reversible):

A + B ⇌ C + D

The forward reaction is A + B ⇌ C + D

Rate of forward (R_f) = K_f [A][B]
K_f = Rate constant for forward reaction

Similarly for the backward reaction:

Rate of backward (R_b) = K_b [C][D]
K_b = Rate constant for backward reaction

At equilibrium both rates are equal:

R_forward = R_backward

K_f [A][B] = K_b [C][D]

=

K_e = ^K_f/_Kb = equilibrium constant

Generally:

For the reaction:

The equilibrium constant for this expression is given by:

K_e =

K_e is changed to K_c for equilibrium concentration (equilibrium constant).

K_c =

Where K_c is equilibrium concentration (equilibrium constant).

[C], [D] etc. are molar concentrations of species in mol per litre; a, b etc. are mole ratios or stoichiometric coefficients.

But for gaseous equilibrium we know that PV = nRT.

PV = nRT

=

Concentration of a gas is directly proportional to partial pressure; therefore, the equilibrium constant can be represented in terms of both concentration and partial pressure.

For the above equilibrium:

K_p =

Example 1.

1. N₂O₄(g) dissociates to give 2NO₂(g)

K_c =

K_p =

2. H₂(g) + I₂(g) ⇌ 2HI(g)

K_c =

K_p =

A large value of K_p or K_c means the equilibrium lies on the side of the product, and a small value means equilibrium lies on the side of the reactants. Thus, the equilibrium constant shows to what extent the reactants are converted to the products.

• If K_c is greater than 10³, products predominate over reactants; equilibrium proceeds nearly to completion.
• If K_c is less than 10^-3, reactants predominate over products; the reaction proceeds to a very small extent.
• If K_c is in the range of 10^-3 and 10³, appreciable concentrations of both reactants and products are present.

SOLIDS AND PURE LIQUIDS IN EQUILIBRIUM EXPRESSION

The concentration of a solid or pure liquid (but not a solution) is proportional to its density. Therefore, their densities are not affected by any gas pressure and remain constant; hence their concentration never appears in the equilibrium expression.

Example

1. What is the equilibrium expression for the following reactions?

i) 3Fe(g) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)

Solution:

K_c = OR K_p =

ii) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

K_c =

K_p =

iii) HCl(g) + LiH(s) ⇌ H₂(g) + LiCl(s)

K_c =

K_p =

iv) Cu(OH)₂(s) ⇌ Cu²⁺(aq) + 2OH^–(aq)

K_c = [Cu²⁺] [OH^–]²

v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)

K_c = [CO₂]

vi) [Cu(NH₃)₄]²⁺(aq) ⇌ Cu²⁺(aq) + 4NH₃(aq)

2. The equilibrium constant for the synthesis of HCl, HBr and HI are given below;

H₂(g) + Cl₂(g) ⇌ 2HCl(g) K_C = 10¹⁷

H₂(g) + Br₂(g) ⇌ 2HBr(g) K_C = 10⁹

H₂(g) + I₂(g) ⇌ 2HI(g) K_C = 10

a) What do the values of K_c tell you about the extent of each reaction?

b) Which of these reactions would you regard as complete conversion? Why?

Answers:

a) For all the 3 reactions, the equilibrium lies on the product side (i.e., RHS) and the extent of formation of HCl is greater than HBr which in turn is greater than HI.

b) For the 1^st and 2^nd reactions, it can be regarded as complete conversion because products predominate over reactants at equilibrium.

Characteristics of equilibrium constant

1. Equilibrium constant is applicable only when the concentrations of reactants and products have attained their equilibrium.
2. The value of equilibrium concentration is independent of the original concentration of reactants.
3. The equilibrium constant has a definite value for every reaction at a particular temperature.
4. For a reversible reaction, the equilibrium constant for the forward reaction is the inverse of the equilibrium constant for the backward reaction.
5. The value of equilibrium constant tells the extent to which reaction proceeds in the forward or reverse direction.
6. Equilibrium constant is independent of the presence of catalyst. This is because the catalyst affects the rate of forward and backward reaction equally.

Relationship between K_c and K_p for a gaseous equilibrium

For any given reaction, K_c or K_p is a function of the reaction itself and temperature.

Consider the following gaseous equilibrium:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

K_c = —————— (1)

K_p = ————– (2)

From PV = nRT

[X] = =

i.e., [A] = , [B] =

Substitute these concentrations in terms of partial pressures in equation (1):

K_c = = .

= K_p

But; (c + d) = total number of moles in the products.

(a + b) = total number of moles in the reactants.

(c + d) – (a + b) = The difference between moles of products and reactants.

So, (c + d) – (a + b) = Δn

Thus, the numerical value of K_p and K_c are equal when there is the same number of moles on products and reactants side numerically.

Example:

1. What is the relationship between K_p and K_c in the following reactions?

2H₂(g) + O₂(g) ⇌ 2H₂O(g)

From; K_p = K_c (RT)^Δn

= K_c (RT)^2-(1+2)

= K_c (RT)^-1

ii) Derive the relationship between K_p and K_c for the particular reaction:

2H₂(g) + O₂(g) ⇌ 2H₂O(g)

K_c = ………… (1)

K_p = ……….. (2)

From PV = nRT

P = RT

P = [X] RT

Substitute the partial pressure in equation (2):

K_p = K_c (RT)^2-(1+2) = K_c (RT)^-1

DETERMINATION OF EQUILIBRIUM CONSTANT

Consider the reaction:

CH₃CH₂OH + CH₃COOH ⇌ CH₃COOCH₂CH₃ + H₂O

Example:

1. An equilibrium system for the reaction between H₂ and I₂ to form HI at 670K in a 5L flask contains 0.4 moles of H₂, 0.4 moles of I₂ and 2.4 moles of HI. Calculate the equilibrium constant K_c.

H₂(g) + I₂(g) ⇌ 2HI(g)

At t=0: a b 0

Equilibrium: a-x b-x 2x

2x = 1.87 x 10^-3

x = 9.35 x 10^-4 mol/dm³

a = 1.0 x 10^-3

a-x = 1.0 x 10^-3 – 9.35 x 10^-4 = 6.5 x 10^-5 mol/dm³

b = 2.0 x 10^-3

b-x = 2.0 x 10^-3 – 9.35 x 10^-4 = 1.065 x 10^-3 mol/dm³

K_c = =

K_c = 50.51

2. A mixture of 1.0 x 10^-3 mol/dm³ H₂ and 2.0 x 10^-3 mol/dm³ I₂ are placed into a container at 450°C. After equilibrium was reached the HI concentration was found to be 1.87 x 10^-3. Calculate the equilibrium constant.

H₂(g) + I₂(g) ⇌ 2HI(g)

K_c = =

K_c = 3.95 ≈ 4 shown

b) From the same reaction above, 3 moles of acetic acid and 5 moles of ethanol reacted. Find the amounts which will be present at equilibrium.

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOCH₂CH₃(l) + H₂O(l)

t= 0 3 5 0 0

t = t (3-x) (5-x) x x

Equilibrium Concentration

K_c =

K_c = 4 =

X² = 60 – 32x + 4x²

3x² – 32x + 60 = 0

X = 8.24 or x = 2.43

Logically; x = 2.43

At equilibrium CH₃COOH = 3 – 2.43 = 0.57 moles

C₂H₅OH = 5 – 2.43 = 2.57 moles

4. For the reaction:

CO₂(g) + H₂(g) ⇌ CO(g) + H₂O(g)

The value of K_c at 552°C is 0.137. If 5 moles of CO₂, 5 moles of H₂, 1 mole of CO and 1 mole of H₂O are initially present, what is the actual concentration of CO₂, H₂, CO and H₂O at equilibrium?

At start; 5 1 1 1

At equilibrium 5-x 1-x 1+x 1+x

K_c = 0.137 =

0.137 =

3.425 – 1.37x + 0.137x² = x²

3.425 – 1.37x + 0.137x² – x² = 0

3.425 – 1.37x – 0.863x² = 0

x =

x = 1.349

COMBINING EQUILIBRIUM REACTIONS

1. Reversing an equilibrium reaction

Consider the reaction:

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) …………… (i)

K_c(i) = …………… (i)

Reversing the reaction:

PCl₃(g) + Cl₂(g) ⇌ PCl₅(g)

K_c(ii) = …………… (ii)

When the reaction equation is reversed, the equilibrium constant is reciprocated, i.e., from equation (i) and (ii):

K_c(ii) = 1 / K_c(i)