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ChemistryEdu Logo Chemical Equilibrium | Equilibrium Constant#

Equilibrium Constant (Kc)#

  • The ratio of rate constants of forward and backward reaction is known as equilibrium constant.
\[Reactants ⇌ Products\]
\[K_c = {K_f \over K_b}\]

Here, Kf = Rate of forward reaction and Kb = Rate of backward reaction

Example

\[n_1A + n_2B ⇌ n_3C + n_4D\]

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, rf = Rate of forward reaction and rb = Rate of backward reaction

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

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

Equilibrium Constant (Kp)#

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

Example

\[n_1A_{(g)} + n_2B_{(g)} ⇌ n_3C_{(g)} + n_4D_{(g)}\]
\[ K_p = {{{p_C}^{n_3} \times {p_D}^{n_4}} \over {{p_A}^{n_1} \times {p_B}^{n_2}}}\]

Here, pC = Partial pressure of C

pD = Partial pressure of D

pA = Partial pressure of A

pB = 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.

\[CaCO_{3_{(s)}} ⇌ CaO_{(s)} + CO_{2_{(g)}}\]

So, we can define both Kc and Kp for this reaction.

\[K_p = p_{CO_2}\]
\[K_c = [CO_2]\]

Example 2

\[N_{2_{(g)}} + 3H_{2_{(g)}} ⇌ 2NH_{3_{(g)}}\]

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

\[K_c = {{[NH_3]^2} \over {[N_2]\ [H_2]^3}}\]
\[K_p = {{p^2_{(NH_3)}} \over {p_{N_2}\ p^3_{H_2}}}\]

Example 3

The reaction given below is a homogeneous liquid phase reaction.

CH3COOC2H5(l) + H2O(l) ⇌ CH3COOH(l) + C2H5OH(l)

In this case, we can define only Kc.

\[K_c = {{[CH_3COOH]\ [C_2H_5OH]} \over {[CH_3COOC_2H_5]\ [H_2O]}}\]

Relation between Kp and Kc#

  • Let us consider a gaseous phase reaction:
\[n_1A_{(g)} + n_2B_{(g)} ⇌ n_3C_{(g)} + n_4D_{(g)}\]
\[K_p = {{p^{n_3}_C \times p^{n_4}_D} \over {p^{n_1}_A \times p^{n_2}_B}}\]
  • According to ideal gas equation, we can write:
\[PV = nRT\]
\[P = {n \over V}RT\]
\[P = CRT\]
  • By applying P = CRT in the above expression of Kp, we can write that:
\[K_p = {{(C_CRT)^{n_3} (C_DRT)^{n_4}} \over {(C_ART)^{n_1} (C_BRT)^{n_2}}}\]
\[K_p = {{(C_C)^{n_3} (C_D)^{n_4}} \over {(C_A)^{n_1} (C_B)^{n_2}}} \times {RT}^{(n_3+n_4)-(n_1+n_2)}\]
\[K_p = K_c \times (RT)^{Δn_g}\]
  • Here, Δng = Total no. of gaseous moles of products - Total no. of gaseous moles of reactants.

Example

\[N_{2_{(g)}} + 3H_{2_{(g)}} ⇌ 2NH_{3_{(g)}}\]
\[Δn_g = 2 - (3+1) = -2\]
\[K_p = K_c \times (RT)^{-2}\]
  • The units of Kp and Kc depend on Δng. The unit of Kp is (atm)Δng and that of Kc is (mol/L)Δng

Characteristics of Equilibrium Constant#

  • When an equilibrium reaction is reversed, its equilibrium constant (Kp or Kc) 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 Kx.