The Relationship Between Kp and Kc along with Examples and Solutions

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The Equilibrium Gas Constants

The equilibrium constants of the gaseous mixture are the Kc and Kp. In contrast, Kc depends on the molar concentration, whereas Qc is derived from the reaction rate. On the other hand, Kp is the partial pressure of the gas inside a closed system.

The forward reaction and the reverse reaction have the same rate when a reaction is at equilibrium. Although the forward and backward reactions are still occurring, the concentrations of the reaction components stay at equilibrium. To know better about the Kp and Kc Relation, let us study the topic in detail. 

Definition of the ratio of concentrations at equilibrium

Equilibrium constants are used to define the ratio of concentrations at equilibrium for a process at a certain temperature. The subscript c in Kc conveys that all concentrations are being indicated in terms of molar concentration. 

The following points must be considered during the calculation of Kp:

  1. The reaction has to be balanced, or else the stoichiometric coefficients and the equilibrium constant exponents will be incorrect.
  2. In this expression of equilibrium, pure liquids and solids have a concentration of One.
  3. Units are not mentioned while writing Kp. The pressure units are needed to be checked in the books during the solution of a Kp problem. It has the value of Kp, which depends on the units used for partial pressure.
  4. During the concentration of Kp, all the other partial pressures should have similar units.
  5. Reactions that have an inclusion of pure liquids and solids, Kp, can be written as their appearance is not observed in the equilibrium expression.

There are many steps to write down gas equilibrium constants. This can be used as the same process for finding Kc, Kp, Ksp, Q, etc.

The chemical reaction for an equilibrium constant is Kc. The reagent number is shown as molar concentration when c is used in Kc. 

For reactions like A B quals to AB, we can define it as the equilibrium constant Kc will be written as [AB]/[A][B]. The Brackets in the formula denote reagent concentrations that need to be given to computing Kc. For such situations, we will be calculating Kc not for one but two reactions. The first is the oxidation reaction among the components of carbon monoxide (CO) along with nitrogen (II) and oxide (NO). Meanwhile, the second one will be of the baking soda’s thermal decomposition.

There are various other ways to write the gas equilibrium constant

Even though the arrows point in the two directions  (⇌) in equilibrium equations, we usually assume that the left one is a reactant and the arrow pointing towards the right is a product.

The numerator (numerators) consists of the products. The reactants are always seen on the bottom of any given fraction which is commonly known as the denominator.

In the balanced chemical equation, the concentrations of the products and reactants are mainly grown to the power of their respective coefficient.

There are many examples of thermal decomposition:

Considering the thermal decomposition of NH4SH(s) :

NH4SH(s) ⇌ NH3(g) H2S(g)

This is also related to Ksp

Kc = [ NH3] plus [H2S]

[NH4SH]

As NH4SH is solid, we get : Kc = [NH3] plus [H2S]…………………………………….. [1]

Kc = [NH3] plus [H2S]

As Kp is same as Kc, so inplace of using [ ], Kp uses ( )

Therefore : Kp = ( NH3 ) plus (H2S )

                                    ( NH4SH) 

                       Kp = (NH3 ) plus (H2S ) ………………………………    ( 1 )

                       Kp = (NH3) plus ( H2S)  

Second Example

Hydrogen and Iodine

Considering the double replacement reaction of hydrogen and iodine gas :

                      H2(g) plus I2(g) ⇌ 2 HI(g)

                       Kc = [ HI ]2

                      [ H2 ] [ I2 ]

                      Kp = ( HI )2   

                       ( H2 ) (I2 )

Kc and Kp can also be defined as :

Kc is an equilibrium constant in terms of molar concentrations and is usually defined as :

                       Kc = [C]c [D]d / [A]a [B]b

The general reaction becomes, aA bB ⇌ cC dD

If Kc formed is large, then more products are formed, and a small Kc indicates that the reactants are favored by the reaction.

Now, Kp is an equilibrium constant concerning partial pressures and is usually defined as :

                    Kp = [C]c [D]d / [A]a [B]b

General reaction now becomes, aA bB ⇌ cC dD

So now, Homogeneous Equilibria: Reactants/products, they are all in a single-phase like :

                         A(g) B(g)⇌ C(g) D(g)

Heterogeneous Equilibria: Reactants / products in more phases than one, like :

               A(s) B(s)⇌ C(s) D(s)

Now, the relationship between the two equilibrium constants become:

                          Kp = Kc(RT)n

                         Kc = Kp / (RT)n

Equilibrium about equilibrium gas constants

Partial pressures or concentrations affect the value of k, being derived from ideal gas law (PV=nRT), the gas equilibrium constant relates to equilibrium (k).

Concentration of reaction denoted by Kc generally shown as:

                                c[C]c[D]c[A]c[B]

Partial pressure in reaction denoted by Kp is generally shown as:

                                  p(C)p(D)p(A)p(B)

From the above equations, we can conclude that the Kp = Kc(RT)n. However, the ideal gas law always helps in the derivation of this formula.

The alternate arrangement

The ideal gas law (PV= NRT) can also be arranged as :

 nL= PRT. as we know Kcin terms of Molarity MolesLitres

As concentration and Partial Pressure are directly proportional to each other, we get,

P = LRT

Here, the unit for pressure is Pascal(Pa), Atmosphere (atm)

Hence when Kc and Molarity are replaced, the equation turns out to be like Kp= Kc(RT)n

(RT)n = (RT)c(RT)d(RT)a(RT)b

likewise, nL = PRT can also be seen as   

Kc= Kp(RT)

K is also written in the same way as Kind 

Kp:

aA bB ⇌ cC dD

Therefore, K = [C]c[D]d[A]a [B]b

Questions

Problem 1

What will be the value of Kp if the Kc is 26 for a reaction, PCI5⇌PCI3 plus CI2 at 250

-0.57

-0.46 re-type answer is d

-0.83

-0.61

Problem 2

Out of the options given below, which of the following reactions show Kp is less than Kc?

  1. 2HI ⇌ H2 plus I2
  2. N2 plus O2 ⇌ 2NO
  3. N2O4 ⇌ 2NO2
  4. 2SO2 plus O2 ⇌ 2SO3

Answer: (d)

Problem 3

Which of the following is not the correct one?

  1. The large value of K indicates a large number of products
  2. The small value of K indicates a large number of reactants
  3. for K < 1 amount of reactants > amount of products
  4. none of the above

Answer: (d)

Conclusion for the equilibrium

The concentration of the solids never changes, as their density always remains the same. With the help of various derivates, it becomes easy to find a solution to our problems. The addition or removal of a liquid has an insignificant effect on the concentration of the system. The equilibrium constant is a constant.