Chapter 7.5 - Homogeneous Equilibrium

In the previous section, we saw the equilibrium constant K_c. In this section, we will see homogeneous equilibrium

◼  In a homogeneous system, all the reactants and products are in the same phase
Let us see some examples:
• In the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), all the reactants and products are in the gaseous phase
• In the reaction CH₃COOC₂H₅(aq) + 3H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq), all the reactants and products are in the liquid phase
• In the reaction Fe³⁺(aq) + SCN^-(aq) ⇌ FeSCN²⁺(aq), all the reactants and products are in the liquid phase

◼  When the system is homogeneous, the 'equilibrium constant' will have some peculiarities
• We will now see a homogeneous system in which all the reactants and products are in the gaseous phase

• In the discussions so far in this chapter, we have seen a number of systems which are gaseous
• In those systems,we calculated K_c using the molar concentrations
• But if the system consists of gases only, it is more convenient to use partial pressures
• Let us see how it is done. It can be explained in 13 steps:
1. We have the ideal gas equation: pV = nRT
• This equation can be rearranged as: p = (ⁿ⁄_V)RT
2. In this equation, all quantities are expressed in SI units:
   ♦ Pressure p is expressed in pascal (pascal is another word for N m^-2)
   ♦ Volume V is expressed in m³
   ♦ n is the number of moles
   ♦ Temperature T is expressed in kelvin
◼  So the value of R will also be in SI units. We know that, the value is:
8.314 pa m³ K^-1 mol^-1 (Details here)
3. So in SI units, (ⁿ⁄_V) will be in mol m^-3
• But we know that, for calculating K_c, we want (n/V) in mol L^-1
• So in our present case, we need to use L instead of m³
4. When we use L for volume, corresponding changes should be made for p
• That is., we need to use bar instead of pascal
5. When we use L and bar, the value of R will change
• We know that:
   ♦ 1 pascal = 10^-5 bar
   ♦ 1 m³ = 1000 L
• So the new value of R can be calculated as follows:
8.314 pa m³ K^-1 mol^-1
= 8.314 ( × 10^-5 bar) ( × 10³ L) K^-1 mol^-1
= 0.08314 bar L K^-1 mol^-1
6. Thus we can write:
• For calculating K_c using the equation p = (ⁿ⁄_V)RT, we must use the following units:
   ♦ p must be in bars
   ♦ V must be in L
   ♦ T must be in kelvin
   ♦ Value of R must be 0.08314 bar L K^-1 mol^-1
   ♦ n is a number 
7. Consider the term ⁿ⁄_V:
• It is 'number of moles per unit volume'
   ♦ So it is concentration
• Recall that, we express concentration by using square brackets
   ♦ So ⁿ⁄_V is same as: [gas]
8. So the equation in (6) becomes:
p = [gas]RT
• Where:
   ♦ p is in bar
   ♦ [gas] is in mol L^-1
   ♦ R is 0.08314 bar L K^-1 mol^-1
   ♦ T is in kelvin
9. At equilibrium, there will be more than one gas in the system
• If the system is inside a closed container, V will be a constant
• If the temperature T is also a constant, we can write:
Partial pressure exerted by any one gas will be proportional to it's n
• That is.,
   ♦ Partial pressure exerted by any one gas
   ♦ will be proportional to
   ♦ it's concentration
• So we can write: p_gas ∝ [gas]
• Where:
   ♦ p_gas is the partial pressure exerted by the gas
   ♦ [gas] is the concentration of the gas
10. At equilibrium, [gas] will be a constant. So p_gas will also be a constant
• So we can use the p_gas values also to find a constant of equilibrium
• The equilibrium constant calculated in this way is denoted as k_p
• So we can write:
For the general reaction a A(g) + b B(g) ⇌ c C(g) + d D(g), we can use:
Eq.7.2: $\mathbf\small{\rm{K_p=\frac{[p_C]^c[p_D]^d}{[p_A]^a[p_B]^b}}}$
11. Let us apply this to one of our familiar reactions: H₂(g) + I₂(g) ⇌ 2HI(g)
We get: $\mathbf\small{\rm{K_p=\frac{(p_{HI})^2}{(p_{H_2})(p_{I_2})}}}$
• But we have:
   ♦ p_HI = [HI(g)]RT
   ♦ p_H2 = [H₂(g)]RT
   ♦ p_I2 = [I₂(g)]RT
• So the result in (10) becomes:
$\mathbf\small{\rm{K_p=\frac{([HI(g)]RT)^2}{([H_2(g)]RT)([I_2(g)]RT)}=\frac{[HI(g)]^2}{[H_2(g)][I_2(g)]}=K_c}}$
◼  So for this reaction, K_p = K_c
12. Consider another reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
We have: $\mathbf\small{\rm{K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}}}$
⇒ $\mathbf\small{\rm{K_p=\frac{([NH_3(g)]RT)^2}{([N_2(g)]RT)([H_2(g)]RT)^3}}}$
⇒ $\mathbf\small{\rm{K_p=\frac{[NH_3(g)]^2}{[N_2(g)][H_2(g)]^3 }\times \frac{(RT)^2}{(RT)(RT)^3}}}$
⇒ $\mathbf\small{\rm{K_p=K_c\times (RT)^{-2}}}$
For this reaction, Kp is not equal to K_c. A factor of (RT)^-2 is also present
13. Based on the above discussion, we can write the general case:
For the reaction a A(g) + b B(g) ⇌ c C(g) + d D(g), we get:
$\mathbf\small{\rm{K_p=\frac{(p_{C})^c (p_{D})^d}{(p_{A})^a (p_{B})^b}}}$
⇒ $\mathbf\small{\rm{K_p=\frac{([C]RT)^c \;([D]RT)^d}{([A]RT)^a \; ([B]RT)^b}}}$
⇒ $\mathbf\small{\rm{K_p=\frac{[C]^c \;[D]^d}{[A]^a \; [B]^b}\times \frac{(RT)^c \;(RT)^d}{(RT)^a \; (RT)^b}}}$
⇒ $\mathbf\small{\rm{K_p=K_c\times (RT)^{(c+d)-(a+b)}}}$
Thus we get Eq.7.3: $\mathbf\small{\rm{K_p=K_c\times (RT)^{\Delta n}}}$
    ♦ Where Δn = (c+d) - (a+b)
• In other words, Δn is obtained as follows:
    ♦ Subtract the number of moles of gaseous reactants
    ♦ From number of moles of gaseous products

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Units of Equilibrium constant

◼ First we will see the units of K_c. It can be written in 4 steps:
1. We have: $\mathbf\small{\rm{K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}}}$
2. In terms of units, we can write:
Units of K_c = $\mathbf\small{\rm{\frac{[mol \, L^{-1}]^c[mol \, L^{-1}]^d}{[mol \, L^{-1}]^a[mol \, L^{-1}]^b}}}$
⇒ Units of K_c = [mol L^-1]^(c+d)-(a+b)
3. So it is clear that, the units will depend on the values of a, b, c and d
• Let us see some possibilities:
(i) If (c+d) = (a+b), we get:
    ♦ Units of K_c = [mol L^-1]⁰ = 1
    ♦ In this case we say that: K_c has no units
(ii) If (c+d) - (a+b) = 1, we get:
    ♦ Units of K_c = [mol L^-1]¹ = [mol L^-1]
    ♦ In this case we say that: units of K_c is mol L^-1
(iii) If (c+d) - (a+b) = -1, we get:
    ♦ Units of K_c = [mol L^-1]^-1
    ♦ In this case we say that: units of K_c is (mol L^-1)^-1
4. So there are numerous possibilities for the units
It depends on the values of a, b, c and d

◼ Next we will see the units of K_p. It can be written in 4 steps:
1. We have: $\mathbf\small{\rm{K_p=\frac{[p_C]^c[p_D]^d}{[p_A]^a[p_B]^b}}}$
2. In terms of units, we can write:
Units of K_p = $\mathbf\small{\rm{\frac{[bar]^c[bar]^d}{[bar]^a[bar]^b}}}$
⇒ Units of K_p = [bar]^(c+d)-(a+b)
3. So it is clear that, the units will depend on the values of a, b, c and d
• Let us see some possibilities:
(i) If (c+d) = (a+b), we get:
    ♦ Units of K_p = [bar]⁰ = 1
    ♦ In this case we say that: K_p has no units
(ii) If (c+d) - (a+b) = 1, we get:
    ♦ Units of K_p = [bar]¹ = [bar]
    ♦ In this case we say that: units of K_p is bar
(iii) If (c+d) - (a+b) = -1, we get:
    ♦ Units of K_p = [bar]^-1
    ♦ In this case we say that, units of K_c is bar^-1
4. So there are numerous possibilities for the units
It depends on the values of a, b, c and d

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We will now see some solved examples

Solved example 7.5
PCl₅ , PCl₃ and Cl₂ are at equilibrium at 500 K and having concentration 1.59 M
PCl₃ , 1.59 M Cl₂ and 1.41 M PCl₅ . Calculate K_c for the reaction:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Solution:
• In this problem we do not have to find K_p. We need K_c only
1. We have the general formula:
K_c for the reaction a A + b B ⇌ c C + d D is given by:
Eq.7.1: $\mathbf\small{\rm{K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}}}$
2. In our present case, the balanced equation is: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
• So we can write:
[A] = [PCl₅] = 1.41 M; a = 1
[B] = [PCl₃] = 1.59 M; c = 1
[C] = [Cl₂] = 1.59 M; d = 1
3. Substituting the above values in (1), we get:
$\mathbf\small{\rm{K_c=\frac{(1.59)^1 \times (1.59)^1}{(1.41)^1}}}$ = 1.79

Solved example 7.6
The value of K_c = 4.24 at 800K for the reaction, CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) Calculate equilibrium concentrations of CO₂, H₂, CO and H₂O at 800 K, if only CO and H₂O are present initially at concentrations of 0.10 M each.
Solution:
1. Let at equilibrium, x moles each of CO₂ and H₂ be present
• From the balanced equation, it is clear that, if x mol each of CO₂ and H₂ are formed, the same x mol each of CO and H₂O would be consumed
• So the concetrations of CO and H₂ at equilibrium would be (0.10 - x) mol each
2. For this reaction, K_c can be obtained as: $\mathbf\small{\rm{K_c=\frac{[CO_2]^1[H_2]^1}{[CO]^1[H2O]^1}}}$
• Substituting the values from (1), we get: $\mathbf\small{\rm{4.24=\frac{x^1 \times x^1}{(0.1-x)^1 \times (0.1-x)^1}=\frac{x^2}{(0.1-x)^2}}}$
3. Solving this quadratic equation, we get: x = 0.067 or 0.194
• 0.194 is not acceptable. The reason can be explained in 2 steps:
(i) If 0.194 M each of CO₂ and H₂ is formed, it would mean that, 0.194 M each of CO and H₂O are consumed
(ii) But only 0.10 M CO and H₂ were available. So 0.194 M must be discarded
4. So we can write:
• The concentrations at equilibrium are:
[CO₂] = [H₂] = x = 0.067
[CO] = [H₂O] = (0.1 - 0.067) = 0.033

Solved example 7.7
For the equilibrium, 2NOCl(g) ⇌ 2NO(g) + Cl₂ (g) the value of the equilibrium constant, K c is 3.75 × 10^-6 at 1069 K. Calculate the K_p for the reaction at this temperature?
Solution:
1. We have Eq.7.3: $\mathbf\small{\rm{K_p=K_c\times (RT)^{\Delta n}}}$
• In our present case, Δn = (2+1) - 2 = 1
2. Substituting the values, we get:
K_p = 3.75 × 10^-6 × (0.0831 × 1069)¹ = 3.33 × 10^-4

Solved example 7.8
At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms
I₂(g) ⇌ 2I (g)
Calculate K_p for the equilibrium.
Solution:
1. Let the total number of moles be n
• 40% of n will constitute I atoms
• So number of moles of I atoms = 0.4n
2. So number of moles of I₂ molecules = (n - 0.4n) = 0.6n
3. Mole fractions:
    ♦ Mole fraction of I atoms = ^0.4n⁄_n = 0.4
    ♦ Mole fraction of I₂ molecules = ^0.6n⁄_n = 0.6
4. Partial pressures (See fig.5.19 of section 5.4)
• Partial pressure of I atoms
= Total pressure  × Mole fraction of I atoms
= 105 × 0.4 = 0.4 × 105 pa = 0.4 bar
• Partial pressure of I₂ molecules
= Total pressure × Mole fraction of I₂ molecules
= 105 × 0.6 = 0.6 × 105 pa = 0.6 bar
5. We have:
Eq.7.2: $\mathbf\small{\rm{K_p=\frac{[p_C]^c[p_D]^d}{[p_A]^a[p_B]^b}}}$
Substituting the values, we get:
$\mathbf\small{\rm{K_p=\frac{0.4^2}{0.6^1}}}$ = 0.267 bar

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Link to some more solved examples is given below:

Solved example 7.9 to Solved example 7.20

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• In the next section, we will see heterogeneous equilibrium


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