Derive the Relation Between Kp and Kc

Equilibrium (eq) constants are expressed using Kp and Kc notations. 

  • Kp is the equilibrium constant calculated by taking the pressure (p) of the gaseous reactants and products. Kp can be expressed in terms of atmospheric pressure. 
  • Kc is the equilibrium constant calculated by taking concentration(c) of the reactants and products. Kc can be expressed in terms of molarity. This article helps you understand the relation between Kp and Kc.

When the reaction happens in a closed system, there is a change in the concentration of reactant and product until ‘the rate of the forward reaction is equal to the rate of the backward reaction’ at equilibrium, at this point, there is no change in the concentration of reactants or products. 

For example, for the equilibrium reaction: A = B

Irrespective of the initial concentrations, the concentrations of reactants and products present at equilibrium are related through the Equilibrium Constant (Keq) as follows:

aA (g)  + bB (g) cC (g)  + dD(g)

 

Where,

a is the mole of reactant for A, 

b is the mole of reactant for B, 

c is the mole of reactant for C, 

d is the mole of reactant for D and 

g is referred to as a gaseous phase.

 

All the reactants and products are in the same phase,

Keq= Kc =  (The subscript “c” stands for “concentration”) equation (1)

 

If all reactants and products are ideal gases, partial pressures or concentrations can be used in the equilibrium constant expression.  In terms of partial pressures:

Keq = Kp (The subscript “p” stands for “pressure”.) equation (2)

The numerical value of the two equilibrium constants above will be different. As per ideal Gas Law, “the partial pressure of a gas is directly proportional to its molarity”.

P= = MRT (M is molarity and R = 0.0821 L-atm/mol-K).

pA=[A]RT

pB=[B]RT

pC=[C]RT

pD=[D]RT

Substituting the values in equation (2), we have

 

 

            [C]c(RT)c[D]d(RT)d​

Kp​  = —————————-

            [A]a(RT)a[B]b(RT)b

 

               [C]c  [D]d

⇒Kp​  = ————— ​(RT)[(c+d)−(a+b)]

                    [A]a  [B]b

⇒Kp​=Kc​(RT)Δng ​From equation (1)

 

By thus using this relationship,

Kp = Kc(RT)Δn(here “T” stands for “Temperature”.)

 

Where, 

Δn is the change in the number of moles of gaseous molecules. 

[That is Δn = product – reactants in mole this is only for gaseous molecule]

When the change in the number of moles of gas molecules is zero, that is Δn = 0

⇒Kp=Kc(RT)Δng From equation (1)

With an example 

Example  1

2H2O(g)  ⇋  2H2 (g) + O2(g)

 

          [pH2]2 (pO2)2​

Kp​  = ——————— p=CRT

            [pH2O]2

 

         ( [H2]2. RT)2​  ([O2].RT)      (RT)2​  (RT)

Kp​  = —————————–          —————- =RT

            ([H2O].RT)2                       (RT)2​ 

 

          [H2]2 [O2]

Kp​  = —————— RT

            [H2O]2

 

                      KC

⇒Kp​=Kc​ (RT)

 

Example  2

2H2F(g)  ⇋  2H2 (g) + F2(g)

 

          [pH2] (pF2)

Kp​  = ——————- p=CRT

            [pHF]2

 

         ( [H2]. RT)​  ([F2].RT)      (RT)​  (RT) 

Kp​  = —————————–          —————- =1

               ([HF].RT)2                       (RT)2​ 

 

          [H2] [F2]

Kp​  = —————— 1

            [HF]2

 

                      KC

⇒Kp​=Kc​

 

 

Example 1, If Δng = 0, then Kp=Kc.i.e., there is a change in the number of moles in gas molecules is zero

Example 2, If ΔΔng> 0, then Kp>Kc.i.e., there is a change in the number of moles in gas molecules that is positive.

Example 3, If ΔΔng< 0, then Kp<Kc. i.e., there is a change in the number of moles in gas molecules that is negative.

Note: when reactants are converted to products, the mathematical value of the equilibrium constant can be expressed as follows: 

  • If Keq>> 1 here, “products are favoured”.  
  • If Keq<< 1 here, “reactants are favoured”.   
  • If the equilibrium constant is very high (>1000), the reaction shall be near completion.
  • If the equilibrium constant is very less (<0.001) means less product shall present at equilibrium

Calculations

  1. Ammonia decomposes upon heating by the following reaction

         2NH3 (g)  ⇋  3H2 (g) + N2(g),

a.       If Kc =9.31×10-3 at 800 0C, what is Kp at this temperature?

b.      If a 2.50  L vessel contained 0.0185 mol of ammonia NH3 0.158mol of hydrogen gas, and 0.870 mol nitrogen gas at 298 K in an equilibrium mixture, calculate the value of Kc at 298 K?

c.       Calculate the value of Kp at 298 K?

 

Answers

a.       Kp = Kc(RT)Δn

    2NH3 (g)  ⇋  3H2 (g) +1 N2(g),

             3+1=4

b.        Kp = Kc(RT)2

800 0C +273.15 = 1073.15 K

Kp =9.31×10-3 . (0.08206 × 1073.15)2

Kp =72.1

 

c.        

            [H2]3 (N2)               n

Kc​  = ——————- C=  —–

            [NH3]2                   V

 

           0.158 mol    

CH2​  = —————- = 0.0632 M

            2.50 L           

 

           0.870 mol    

CN2​  = —————- = 0.346 M

            2.50 L    

 

           0.185 mol    

CNH3​  = —————- = 0.00740 M

            2.50 L           

  

           (0.0632)3 (0.346)

Kc  = ———————— = 1.60

              (0.00740)2

 

    Kp = Kc(RT)2

T=298K, R=0.08206

Kp = 1.60. (0.08206×298) 2 =597

 

Key points about equilibrium constants:

  1. Each reaction has its individual equilibrium constant that mainly depends on temperature. The equilibrium constant is independent of the initial concentration of reactants and products.
  2. The equilibrium constant has no units. Since the pressure or concentration is the actual ratio of the pressure or concentration to their standard values. For a solution, the standard concentration is 1M (one molar). For a Pressure standard pressure for gas is 1 atm (one atmosphere). 
  3. The pure solids and liquids concentrations are constant. For any pure substance, don’t change, and the ratio that appears in the equilibrium constant expression is always one.
  4. All the reactants and products should be at equilibrium, despite pure solids and pure liquids not appearing in an equilibrium constant expression.
  5. By altering the initial concentrations of reactants or products do not alter the equilibrium constant, the equilibrium concentrations of reactant and products varies but the equilibrium constant will be the same.
  6. The rate of equilibrium constant depends mainly on the identity of reactant, products and temperature; all reactions have a unique equilibrium constant at a fixed temperature.
  7. The equilibrium constant can’t be altered by adding a catalyst. A catalyst either increases or decreases the rate of reaction in the same factor.
  8. The equilibrium constant can be altered by varying temperature; if the temperature increases, it increases the equilibrium constant for an endothermic reaction; if the temperature decreases, it decreases the equilibrium constant for an exothermic reaction.
  9. A reaction in a reverse direction (reversible reaction) of the equilibrium constant is reciprocal to that of the equilibrium constant in the forward direction.
  10. The equilibrium constant for a reaction that has been multiplied by a number is the original equilibrium constant raised to a power equal to that number.
  11. The equilibrium constant for a net reaction produced by adding two or more steps is the product of the equilibrium constants for the individual steps.

Conclusion:

Knowing the relationship between Kp and Kc at different variable factors will help to know about the many nature of chemical reactions. Chemical reactions based on temperature, pH, ionization can be understood, numerous biological reactions, and physical transformations of molecules at the time of reactions. Many other important factors like ionization, factors affecting acid or base strength, etc., shall be derived. 

Plagiarism Report – 

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