Chemical Equilibrium | Equilibrium Constant

Equilibrium Constant (K_c)

• The ratio of rate constants of forward and backward reaction is known as equilibrium constant.

$$Reactants\rightleftharpoons Products$$

$$K_c=\frac{K_f}{K_b}$$

Here, K_f = Rate of forward reaction and K_b = Rate of backward reaction

Example

$$n_{1}A+n_{2}B\rightleftharpoons n_{3}C+n_{4}D$$

From law of mass action, we can write:

$$r_f=K_f[A{]}^{n_1}[B{]}^{n_2}$$

$$r_b=K_b[C{]}^{n_3}[D{]}^{n_4}$$

where, r_f = Rate of forward reaction and r_b = Rate of backward reaction

At equilibrium, rate of forward reaction = rate of backward reaction.

$$K_f[A{]}^{n_1}[B{]}^{n_2}=K_b[C{]}^{n_3}[D{]}^{n_4}$$

$$\frac{K_f}{K_b}=\frac{[C{]}^{n_3}[D{]}^{n_4}}{[A{]}^{n_1}[B{]}^{n_2}}$$

$$K_c=\frac{[C{]}^{n_3}[D{]}^{n_4}}{[A{]}^{n_1}[B{]}^{n_2}}$$

Equilibrium Constant (K_p)

• There is another equilibrium constant, K_p defined for gaseous phase reactions.

Example

$$n_1{A}_{(g)}+n_2{B}_{(g)}\rightleftharpoons n_3{C}_{(g)}+n_4{D}_{(g)}$$

$$K_p=\frac{{p_C}^{n_3}\times {p_D}^{n_4}}{{p_A}^{n_1}\times {p_B}^{n_2}}$$

Here, p_C = Partial pressure of C

p_D = Partial pressure of D

p_A = Partial pressure of A

p_B = Partial pressure of B

Examples

Example 1

The reaction given below can be considered as a gaseous phase reaction because CO₂ is a gas and other substances are pure solids.

$$CaC{O}_{3_{(s)}}\rightleftharpoons Ca{O}_{(s)}+C{O}_{2_{(g)}}$$

So, we can define both K_c and K_p for this reaction.

$$K_p=p_{C{O}_2}$$

$$K_c=[C{O}_2]$$

Example 2

$$N_{2_{(g)}}+3H_{2_{(g)}}\rightleftharpoons 2N{H}_{3_{(g)}}$$

This is a gaseous phase reaction. So, we can define both K_c and K_p for this reaction.

$$K_c=\frac{[N{H}_3{]}^2}{[N_2][H_2{]}^3}$$

$$K_p=\frac{p_{(N{H}_3)}^2}{p_{N_2}{p}_{H_2}^3}$$

Example 3

The reaction given below is a homogeneous liquid phase reaction.

CH₃COOC₂H_5(l) + H₂O_(l) ⇌ CH₃COOH_(l) + C₂H₅OH_(l)

In this case, we can define only K_c.

$$K_c=\frac{[C{H}_{3}COOH][C_2{H}_{5}OH]}{[C{H}_{3}COO{C}_2{H}_5][H_{2}O]}$$

Relation between K_p and K_c

• Let us consider a gaseous phase reaction:

$$n_1{A}_{(g)}+n_2{B}_{(g)}\rightleftharpoons n_3{C}_{(g)}+n_4{D}_{(g)}$$

$$K_p=\frac{p_C^{n_3}\times p_D^{n_4}}{p_A^{n_1}\times p_B^{n_2}}$$

• According to ideal gas equation, we can write:

$$PV=nRT$$

$$P=\frac{n}{V}RT$$

$$P=CRT$$

• By applying P = CRT in the above expression of K_p, we can write that:

$$K_p=\frac{(C_{C}RT{)}^{n_3}(C_{D}RT{)}^{n_4}}{(C_{A}RT{)}^{n_1}(C_{B}RT{)}^{n_2}}$$

$$K_p=\frac{(C_C{)}^{n_3}(C_D{)}^{n_4}}{(C_A{)}^{n_1}(C_B{)}^{n_2}}\times {RT}^{(n_3+n_4)-(n_1+n_2)}$$

$$K_p=K_c\times (RT{)}^{\Delta n_g}$$

• Here, Δn_g = Total no. of gaseous moles of products - Total no. of gaseous moles of reactants.

Example

$$N_{2_{(g)}}+3H_{2_{(g)}}\rightleftharpoons 2N{H}_{3_{(g)}}$$

$$\Delta n_g=2-(3+1)=-2$$

$$K_p=K_c\times (RT{)}^{-2}$$

• The units of K_p and K_c depend on Δn_g. The unit of K_p is (atm)^Δn_g and that of K_c is (mol/L)^Δn_g

Characteristics of Equilibrium Constant

• When an equilibrium reaction is reversed, its equilibrium constant (K_p or K_c) gets inverted.
• When two reactions are added or subtracted, their equilibrium constants will get multiplied or divided respectively.
• When an equilibrium reaction with equilibrium constant K is multiplied by any number x, then equilibrium constant becomes K^x.