Equilibria Types & Relation Between K, Q & G - Chemistry Class 11 - NEET

Table of Contents
1. Types of Equilibria
2. Homogeneous Equilibria
3. Heterogeneous Equilibria
4. Applications of the Equilibrium Constant
5. Equilibrium Constant, Reaction Quotient and Gibbs Free Energy
6. Relationship between \(K_p\) and \(K_c\)
7. Summary
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Types of Equilibria

• We encounter so many reactions around us in our day-to-day lives; the rusting of iron, the burning of paper, the souring of curd, the formation of ozone, etc. Many of these reactions involve the presence of components in different phases, such as solid iron coming in contact with gaseous oxygen to form solid iron oxide, which we know as rust. 
• Similarly, gaseous hydrogen combines with gaseous oxygen to form liquid water. Dealing with such reactions is a tedious task. If the components are in the same phase, the interaction of the components can be understood easily, while when the components are in different phases, the interaction gets more complicated.

• Homogeneous equilibria: all reacting species are in the same phase (for example, all gases or all aqueous species).
• Heterogeneous equilibria: reacting species are present in more than one phase (for example, solids in contact with gases or liquids).

Homogeneous Equilibria

Consider the gaseous equilibrium

\( \mathrm{N_2 (g)} + 3\mathrm{H_2 (g)} \rightleftharpoons 2\mathrm{NH_3 (g)} \)

For a general balanced reaction

\( a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D} \)

the equilibrium constant expressed in terms of molar concentrations is

\( K_c = \dfrac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}} \)

and the equilibrium constant expressed in terms of partial pressures (for gases) is

\( K_p = \dfrac{(p_{\mathrm{C}})^{c}(p_{\mathrm{D}})^{d}}{(p_{\mathrm{A}})^{a}(p_{\mathrm{B}})^{b}} \)

When gases behave ideally, the molar concentration and partial pressure are related by the ideal gas equation.

\( PV = nRT \)
Where P is the pressure of the system, V is the volume of the system, n is the number of moles of components present in the system, R is the universal gas constant and T is the temperature of the system.

From this, molar concentration \(c = \dfrac{n}{V}\) can be written as

\( c = \dfrac{P}{RT} \)

Using this relation for each gaseous species and substituting into the expression for \(K_c\), we obtain a relation between \(K_p\) and \(K_c\).

\( K_p = K_c\,(RT)^{\Delta n} \)

Here, \(\Delta n\) is the change in the number of moles of gas during the reaction, defined as

\( \Delta n = \) (moles of gaseous products) - (moles of gaseous reactants).

Heterogeneous Equilibria

A heterogeneous equilibrium involves species in more than one phase. In writing equilibrium expressions for such systems we use the fact that the activity of a pure solid or a pure liquid is effectively constant (conventionally taken as unity) at a given temperature. Therefore pure solids and pure liquids do not appear explicitly in the equilibrium expression.

• Example: take a container with ice and water at a temperature that allows the existence of both phases simultaneously, such that, both ice and water are present in a state of equilibrium. This state is termed a heterogeneous equilibrium.
  In terms of its equation, this can be written as:
  \( \mathrm{H_2O (s)} \rightleftharpoons \mathrm{H_2O (l)} \)
•  Example : we can consider an aqueous solution of a solid such as calcium hydroxide. We notice that the solid calcium hydroxide is in equilibrium with its saturated solution.
  Writing the equilibrium constant for heterogeneous reactions is different from that of the homogeneous reactions. 
• Example: dissolution/precipitation or thermal decomposition involving solids and gases
  Consider the thermal decomposition of calcium carbonate:
  \( \mathrm{CaCO_3 (s)} \rightleftharpoons \mathrm{CaO (s)} + \mathrm{CO_2 (g)} \)
  Since \(\mathrm{CaCO_3}\) and \(\mathrm{CaO}\) are pure solids, their activities are constant and do not appear in the equilibrium expression. Thus the equilibrium constant (in terms of partial pressure of CO₂) is
  In terms of K_p, we can write
  \( K_p = p_{\mathrm{CO_2}} \)

This means that at a given temperature there is a fixed partial pressure of CO₂ in equilibrium with the solid phases \(\mathrm{CaCO_3}\) and \(\mathrm{CaO}\).

Applications of the Equilibrium Constant

Equilibrium constant and extent of reaction

The magnitude of the equilibrium constant gives a clear idea of how far a reaction proceeds toward products under standard conditions.

• Large \(K\) (≫ 10³): the forward reaction is strongly favoured; at equilibrium product concentrations are much larger than those of reactants. Example: \( \mathrm{H_2 + Br_2 \rightleftharpoons 2HBr},\; K_c = 5.4\times 10^{18} \).
  H₂(g) + Cl₂(g) ⇌ 2HCl(g) ⇒ Kc = 4 × 10³¹
  H₂(g) + 12O₂(g) ⇌ H₂O(g) ⇒ Kc = 2.4 × 10⁴⁷
• Intermediate \(K\) (≈ 10⁻³ to 10³): reactants and products are present in comparable amounts at equilibrium. Example: \( \mathrm{Fe^{3+} + SCN^- \rightleftharpoons [Fe(SCN)]^{2+}},\; K_c = 138 \) at 298 K.
  H₂ (g) + I₂ (g) ⇌ 2HI (g) ⇒ K_c = 57 at 700 K. 
• Small \(K\) (≪ 10⁻³): the reverse reaction is favouredi.e. concentration of reactants is much larger than that of products i.e. the reaction proceeds to a very small extent in the forward direction. Example: \( \mathrm{N_2 + O_2 \rightleftharpoons 2NO},\; K_c = 4.8\times 10^{-31} \) at 298 K.
  H₂O (g) ⇌ H₂ (g) + (1/2) O₂ (g) ⇒ K_c = 4.1 × 10^-48 

Reaction quotient and direction of spontaneous change

To predict the direction in which a non-equilibrium mixture will move, we define the reaction quotient \(Q\). For a balanced reaction \(a\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}\) :

\( Q_c = \dfrac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}} \)

\( Q_p = \dfrac{(p_{\mathrm{C}})^{c}(p_{\mathrm{D}})^{d}}{(p_{\mathrm{A}})^{a}(p_{\mathrm{B}})^{b}} \)

• If \(Q = K\), the system is at equilibrium.
• If \(Q < k\), the reaction will proceed in the forward direction until 
• \(Q\) increases to \(K\).
• If \(Q > K\), the reaction will proceed in the reverse direction until \(Q\) decreases to \(K\).

Calculating equilibrium concentrations 

The degree of dissociation of an equilibrium involving gas can be calculated by knowing the equilibrium constant and the concentration of the gaseous reactant/product. Consider the thermal decomposition (dissociation) of ammonia:

\( 2\mathrm{NH_3 (g)} \rightleftharpoons \mathrm{N_2 (g)} + 3\mathrm{H_2 (g)} \)

Let the initial number of moles of \(\mathrm{NH_3}\) be \(C\). Define \(\alpha\) such that the extent of the reaction produces \(C\alpha\) moles of \(\mathrm{N_2}\). By stoichiometry, the amount of \(\mathrm{NH_3}\) consumed is \(2C\alpha\) and the amount of \(\mathrm{H_2}\) produced is \(3C\alpha\).

Moles at equilibrium are then:

\( \text{moles } \mathrm{NH_3} = C(1 - 2\alpha) \)

\( \text{moles } \mathrm{N_2} = C\alpha \)

\( \text{moles } \mathrm{H_2} = 3C\alpha \)

Total moles at equilibrium:

\( n_{\text{total}} = C(1 - 2\alpha) + C\alpha + 3C\alpha = C(1 + 2\alpha) \)

Partial pressures (using Dalton's law):

\( p_{\mathrm{NH_3}} = \dfrac{C(1-2\alpha)}{C(1+2\alpha)}P_t = \dfrac{1-2\alpha}{1+2\alpha}\,P_t \)

\( p_{\mathrm{N_2}} = \dfrac{C\alpha}{C(1+2\alpha)}P_t = \dfrac{\alpha}{1+2\alpha}\,P_t \)

\( p_{\mathrm{H_2}} = \dfrac{3C\alpha}{C(1+2\alpha)}P_t = \dfrac{3\alpha}{1+2\alpha}\,P_t \)

Now write the expression for \(K_p\):

\( K_p = \dfrac{p_{\mathrm{N_2}}(p_{\mathrm{H_2}})^{3}}{(p_{\mathrm{NH_3}})^{2}} \)

Substitute the partial pressures and simplify:

\( K_p = \dfrac{\dfrac{\alpha}{1+2\alpha}P_t \left(\dfrac{3\alpha}{1+2\alpha}P_t\right)^3}{\left(\dfrac{1-2\alpha}{1+2\alpha}P_t\right)^2} \)

\( K_p = \dfrac{27\alpha^4 P_t^{\,4}}{(1+2\alpha)^4} \times \dfrac{(1+2\alpha)^2}{(1-2\alpha)^2 P_t^{\,2}} \)

\( K_p = \dfrac{27\alpha^4 P_t^{\,2}}{(1+2\alpha)^2(1-2\alpha)^2} \)

This relation can be used to determine \(\alpha\) if \(K_p\) and \(P_t\) are known. For small \(\alpha\) (small dissociation), the expression may be approximated by \(K_p \approx 27\alpha^4 P_t^{\,2}\).

Using vapour density to find degree of dissociation

For ideal gases, pV = nRT = [ω / M] × RT
M = [ωRT] / VP = ρ [RT] / P = ρ[RTV] / RTn = ρ V / n = 2 × Vapour Density
Vapour density = ρ × V / 2n = α × 1 / n
At equilibrium V and ρ are constant and Vapour Density is α × 1 / n
Vapour density measurements can also be used to determine the degree of dissociation for gases that change molecular weight on dissociation. The vapour density (VD) is proportional to the average molar mass (molecular weight) of the gaseous mixture. If \(M\) is the initial molecular mass and \(m\) is the average molecular mass at equilibrium, then

\( \dfrac{\text{VD}_{\text{initial}}}{\text{VD}_{\text{equilibrium}}} = \dfrac{M}{m} = \dfrac{\text{moles at equilibrium}}{\text{moles at start}} \)

Example

For a specific reaction, using the relation between moles at equilibrium and the initial moles one can derive \(\alpha\) in terms of vapour densities or molecular masses; for example, if the ratio \(D/d\) of initial to equilibrium vapour density equals \(1+\alpha\), then

\( \alpha = \dfrac{D}{d} - 1 \)

Knowing the vapour densities (or molecular masses) before and after dissociation allows calculation of \(\alpha\).

Equilibrium Constant, Reaction Quotient and Gibbs Free Energy

The spontaneity of a reaction under given non-standard conditions is determined by the Gibbs free energy change \(\Delta G\). The free energy change at any instant is related to the reaction quotient \(Q\) by the equation

\( \Delta G = \Delta G^\circ + RT\ln Q \)

Here, \(\Delta G^\circ\) is the standard free energy change (all species at 1 mol L⁻¹ or 1 bar for gases), \(R\) is the gas constant and \(T\) is the absolute temperature in kelvin.

At equilibrium the system satisfies \(Q = K\) and \(\Delta G = 0\). Substituting these values gives the fundamental relation between \(\Delta G^\circ\) and the equilibrium constant \(K\):

\( \Delta G^\circ = -RT\ln K \)

Equilibrium Constant vs Reaction Quotient

• If \(\Delta G < 0\)(equivalently \(Q < K\))the reaction will proceed to form products.
• If \(\Delta G = 0\) (equivalently \(Q = K\)) the system is at equilibrium; no net change occurs.
• If \(\Delta G > 0\) (equivalently \(Q > K\)) the reaction proceeds in the reverse direction until equilibrium is re-established.

Relationship Between ΔG^o & K

• Free energy, G, denotes the self-intrinsic electrostatic potential energy of a system. This means that in any molecule if we calculate the total electrostatic potential energy of all the charges due to all the other charges, we get what is called the free energy of the molecule. It tells about the stability of a molecule with respect to another molecule. The lesser the free energy of a molecule more stable it is. 
• Every reaction proceeds with a decrease in free energy. The free energy change in a process is expressed by ΔG. If it is negative, it means that the product has lesser G than reactants, so the reaction goes forward. If it is positive the reaction goes reverse and if it is zero the reaction is at equilibrium.
• ΔG is the free energy change at any given concentration of reactants and products. If all the reactants and products are taken at a concentration of 1 mole per liter, the free energy change of the reaction is called ΔG^o ( standard free energy change ).
• One must understand that ΔG^o is not the free energy change at equilibrium. It is a free energy change when all the reactants and products are at a concentration of 1 mole/L. ΔG^o is related to K (equilibrium constant ) by the relation, ΔG^o= -RT ln K.
• K may either be K_c or K_p. Accordingly, we get . The units of ΔG^o depend only on R.
• T is always in Kelvin, and if R is in Joules, ΔG^o will be in joules, and if R is calories then ΔG^o will be in calories.

Relationship between \(K_p\) and \(K_c\) 

Using the relation \(c_i = p_i/RT\) (for each gaseous species \(i\)), substitution into the expression for \(K_c\) leads to

\( K_c = \dfrac{(p_{\mathrm{C}}/RT)^{c}(p_{\mathrm{D}}/RT)^{d}}{(p_{\mathrm{A}}/RT)^{a}(p_{\mathrm{B}}/RT)^{b}} \)

Simplification yields

\( K_p = K_c\,(RT)^{\Delta n} \)

with \(\Delta n\) as defined earlier. This relation is frequently used to convert between \(K_c\) and \(K_p\) for gaseous equilibria.

Summary

Chemical equilibria are classified as homogeneous or heterogeneous depending on the phases involved. Equilibrium constants \(K_c\) and \(K_p\) quantify the composition at equilibrium. The reaction quotient \(Q\) is used to predict the direction in which a reaction will proceed. Gibbs free energy links thermodynamics and equilibrium through \(\Delta G = \Delta G^\circ + RT\ln Q\) and at equilibrium \(\Delta G^\circ = -RT\ln K\). For gaseous systems the relation \(K_p = K_c\,(RT)^{\Delta n}\) converts between pressure- and concentration-based equilibrium constants. Practical calculations frequently use these relations to find equilibrium compositions, degrees of dissociation and to predict the effect of changing conditions.