Chapter 7.6 - Heterogeneous Equilibrium

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

◼ Equilibrium in a 'system having more than one phase' is called heterogeneous equilibrium
Let us see an example. It can be written in 3 steps:
1. We have seen that, if water is placed inside a closed container, some water molecules will escape from the liquid mass and form a gaseous mass above the water surface
2. We have also seen that, there will be an equilibrium between the liquid water and gaseous water. We represent this equilibrium as: H₂O(l) ⇌ H₂O(g)
3. We see that in this equilibrium, there is a gaseous phase and a liquid phase. So it is a heterogeneous equilibrium

Another example:
• In the equilibrium: CaCO₃(s) ⇌ CaO(s) + CO₂(g),
CaCO₃ and CaO are in the solid phase but CO₂ is in the gaseous phase
• So it is a heterogeneous equilibrium

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• Let us see how we write the equilibrium constant for heterogeneous equilibrium. It can be written in 11 steps:
1. For the above reaction, the equilibrium constant can be written as: $\mathbf\small{\rm{K_c=\frac{[CaO][CO_2]}{[CaCO_3]}}}$
2. But there is no meaning in writing the concentration of CaCO₃ and CaO
• That is., there is no meaning in writing [CaCO₃] and [CaO]
• The following steps from (3) to (6) will give the reason
(3) At equilibrium, the CaO will be existing as solid pieces
• What ever quantities of CaO is present, will be confined in those solid pieces
(4) We can write ‘concentration’ only if the calcium oxide is dispersed as 'separate CaO molecules' into the volume of the container
• In such a situation, we obtain concentration by dividing the mass (number of moles) by the volume
5. But in our present case, the CaO is not occupying the volume of the container (volume of the system)
• So we cannot write concentration of CaO
6. Similarly, we cannot write concentration of CaCO₃
7. It is assumed that, the concentration of a pure solid is a constant. So they are not included in the equation for equilibrium
8. Thus we can write:
Equilibrium constant of the reaction CaCO₃(s) ⇌ CaO (s) + CO₂, is given by:
K_c = [CO₂]
• Also we get: K_p = p_CO2
9. We can site the same argument for pure liquids. It can be written in 2 steps:
(i) If there is a pure liquid in the system, it will be occupying a certain volume at the bottom of the container
• It’s molecules are not spread out into the volume of the container
(ii) So if any pure liquid is present at equilibrium, we need not consider it’s concentration while calculating the equilibrium constant
10. However, if instead of pure liquid, we are given a solution, the concentration do play a role
• Because, the solute is spread out into the volume of the solution
11. There is an important point to note:
(i) In the reaction CaCO₃(s) ⇌ CaO (s) + CO₂, we said that, we ignore [CaCO₃] and [CaO]
◼ But adding more solid pieces of CaCO₃ and/or CaO will change the equilibrium. Then why do we say that, [CaCO₃] ans [CaO] can be ignored?
• The answer can be written in 3 steps:
(i) When we increase the quantity of solid CaCO₃ and/or solid CaO, the quantity of CO₂ (g) changes
• That means [CO₂] changes
(ii) When [CO₂] changes, the equilibrium also changes
(iii) So even though we ignore [CaCO₃] and [CaO], we are indirectly taking them into account because, K_c is based on [CO₂]

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Let us see two more examples where pure solids or liquids are present:
• The equilibrium constant for the reaction
Ni(s) + 4CO(g) ⇌ Ni(CO)₄(g)
can be obtained as: $\mathbf\small{\rm{K_c=\frac{[Ni(CO)_4]}{[CO]^4}}}$
• The equilibrium constant for the reaction
Ag₂O(s) + 2HNO₃(aq) ⇌ 2AgNO₃ (aq) +H₂O(l)
can be obtained as: $\mathbf\small{\rm{K_c=\frac{[AgNO_3]^2}{[HNO_3]^2}}}$

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Now we will see some solved examples:
Solved example 7.21
Write the expression for the equilibrium constant, K_c for each of the following reactions:
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
(ii) 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g)
(iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq)
(iv) Fe³⁺(aq) + 3OH(aq) ⇌ Fe(OH)₃(s)
(v) I₂(s) + 5F₂ ⇌ 2IF₅
Solution:
(i) $\mathbf\small{\rm{K_c=\frac{[NO]^2 [Cl_2]}{[NOCl]^2}}}$
(ii) $\mathbf\small{\rm{K_c=[NO_2]^4[O_2]}}$
(iii) $\mathbf\small{\rm{K_c=\frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5]}}}$
(iv) $\mathbf\small{\rm{K_c=\frac{1}{[Fe^{3+}][OH^-]^3}}}$
(v) $\mathbf\small{\rm{K_c=\frac{[IF_5]^2}{[F_2]^5}}}$

Solved example 7.22
Find out the value of K_c for the following reaction:
(ii) CaCO₃ (s) ⇌ CaO(s) + CO₂ (g); K_p = 167 at 1073 K
Solution:
1. We have Eq.7.3: $\mathbf\small{\rm{K_p=K_c\times (RT)^{\Delta n}}}$
• In our present case, Δn = 1 - 0 = 0
2. Substituting the values, we get:
167 = K_c × (0.0831 × 1073)1
⇒ K_c = 1.90

Solved example 7.23
The value of K_p for the reaction, CO₂ (g) + C (s) ⇌ 2CO (g) is 3.0 at 1000 K. If initially P_CO2 = 0.48 bar  and P_CO = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO₂
Solution:
1. Let at equilibrium, x moles each of CO be present
• From the balanced equation, it is clear that, if x mol of CO is formed, x/2 mol CO₂ would be consumed
• So the concentration of CO₂ at equilibrium would be (0.48 - x) mol
• We need not consider the equilibrium concentration of C because, it is in the solid state
2. For this reaction, K_p can be obtained as: $\mathbf\small{\rm{K_p=\frac{p_{CO}^2}{p_{CO_2}^1}}}$
• Substituting the values from (1), we get: $\mathbf\small{\rm{3.0=\frac{x^2}{(0.48-\frac{x}{2})^1}}}$
3. Solving this quadratic equation, we get: x = 0.6651 or -2.165
• -2.165 is not acceptable because it is negative
4. So we can write:
• The partial pressures equilibrium are:
   ♦ p_CO = x = 0.66 bar
   ♦ p_CO2 = (0.48 - x/2) = 0.15 bar

Solved example 7.24
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass
C(s) + CO₂(g) ⇌ 2CO(g)
Calculate K_c for this reaction at the above temperature
Solution:
1. Mass at equilibrium is given
• Let m be the total mass of the gaseous mixture
   ♦ Then mass of CO = 0.9055m
   ♦ So mass of CO₂ = (m - 0.9055m) = 0.0945m
2. One mole of CO has a mass of 28 grams
⇒ One gram CO = ¹⁄₂₈ mol
⇒ Number of moles in 0.9055m = ^0.9055m⁄₂₈ moles
3. One mole of CO₂ has a mass of 44 grams
⇒ One gram CO₂ = ¹⁄₄₄ mol
⇒ Number of moles in 0.0945m = ^0.0945m⁄₄₄ moles
4. So total number of moles in the gaseous mixture
= (^0.9055m⁄₂₈ + ^0.0945m⁄₄₄) = 0.0345m
5. So we get:
• Mole fraction of CO = $\mathbf\small{\rm{\frac{\frac{0.9055m}{28}}{0.0345m}}}$ = 0.9374  
• Mole fraction of CO₂ = $\mathbf\small{\rm{\frac{\frac{0.0945m}{44}}{0.0345m}}}$= 0.0622
6. Thus we get:
• Partial pressure of CO = Total pressure × mole fraction of CO
= 1 atm  × 0.9374 = 0.9374 atm
• Partial pressure of CO₂ = Total pressure × mole fraction of CO₂
= 1 atm  × 0.0622 = 0.0622 atm
7. Now we can calculate K_p:
$\mathbf\small{\rm{K_p=\frac{(p_{CO})^2}{(p_{CO_2})}=\frac{0.9374}{0.0622}}}$ = 14.105
8. Using Kp, we can calculate K_c
• We have: $\mathbf\small{\rm{K_p=K_c\times (RT)^{\Delta n}}}$
• Substituting the values, we get:
14.105 = K_c (0.0831 × 1127)¹
⇒ K_c = 0.150

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• In the next section, we will see the applications of equilibrium constants


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