```RELATION BETWEEN KP AND KC
For a general equation,
aA + bB   cC + dD
Where, a , b, c and d are coefficients of the reacting substance
Kp = (PC)c (OD)d / (PA)a (PB)b
From gas equation
PV = nRT        =>  P = n/V RT
So,                       PA = [A]RT’ PB = [B]RT
PC = [C]RT; PD = [D]RT
Hence,           Kp = ([C]RT)c.([D]RT)d / ([A]RT)a / ([B]RT)b
or, Kp = [C]d[D]d / [A]a[B]b × (RT)[(c+d) – (a+b)]
Where Δng = {(c + d) – (a + b)} = change in the numbers of gaseous moles.
Hence,           Kp = Kc.(RT)Δng
When       Δn = 0; Kp = Kc
Δn > 0; Kp > Kc
Δn < 0;            Kp < Kc
Note: Similarly we can find equilibrium constant (Kx) interms of mole fraction and can find out its relation with Kp and Kc.
Illustration . At 27°C Kp value for reaction  is 0.1 atm, calculate its Kc value.
Solution:       KP = Kc(RT)Dn
Δn = 1
Kc = KP / RT = 0.1 / 0.82 × 300 = 4 × 10-3
Illustration . Solid NH4I dissociates according to the reaction at 400 K
NH4I(s)  NH3(g) + HI(g) ; Kp = 16 atm. In presence of catalyst HI dissociates in H2 and I2 as 2HI  H2 + I2. If partial pressure of H2 at this temp is 1 atm in the container when both the equilibrium exist simultaneously, calculate Kp value of second equilibrium (for the dissociation of HI).
Solution:      NH4I(s)  NH3 (g) + HI (g)            KP1 = 16
2HI  H2 +     I2
P - x            x/2 . x/2                    KP2 = ?
Here x/2 = 1 atm
x = 2 atm
P(P -2) = 16
P = 5.1
Illustration . A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on addition of graphite. What is the value of Kp if the total pressure at equilibrium is 0.8 atm?
Solution: CO2(g)      +    C(s)     ——>    2CO(g)
0.5                                    2n
or, 0.5-n                               0.5+n = 0.8
n = 0.3
Kp = (2×0.3)2 / (0.5–0.3) = 1.8 atm
Illustration . Given that the equilibrium constant for the reaction, H2(g) + I2(g)  2HI(g) is 50 at 700K. Calculate the equilibrium constant for the reaction.
HI(g)  1/2 H2(g) + 1/2 l2(g)
Solution: We have H2(g) + I2(g)  2HI(g)
Kc = [Hl]2 / [H2][l2] = 50
The new equilibria is
HI(g)  [H2]1/2[l2]1/2 / [Hl] = 1/√Kc
= 1/ √50 = √2 / 10 = 0.141
Illustration . One mole of N2 is mixed with 3 moles of H2 in a 4 litre container. If 25% of N2 is converted into NH3 by the following reaction
N2(g) + 3H2(g)  2NH3(g). Calculate Kc and Kp of the reaction. (Temperature = 227°C and R = 0.08231).
Solution: We have N2(g) + 3H2(g)  2NH3(g)
Percentage N2 reacted, 25%
x = 0.25
Now, (a – x) = 1 – 0.25 = 0.75
b – 3x = 3 – 0.75 = 2.25

= 1.48 × 10–5 L2 mol–2
Now, Kp = Kc . (RT)Δn = 1.48 × 10–5 × [0.0821] × (227 + 273)–2
1.48 × 10–5 / {0.082 × (500)}2
= 8.78 × 10–9```
Related Resources
Miscellaneous Exercises of Chemical Equilibrium

MISCELLANEOUS EXERCISES Exercise 1: For an...

Significance of Magnitude of Equilibrium Constant

SIGNIFICANCE OF THE MAGNITUDE OF EQUILIBRIUM...

Equilibrium in Physical and Chemical Process

EQUILIBRIUM IN PHYSICAL PROCESS The different...

Objective Questions of Chemical Equilibrium

Objective: Prob 1. Reaction between iron and steam...

Solution to Exercises

Solution Calculate the equilibrium pressure of...

Chemical Equilibrium Questions

Chemical Equilibrium Questions Prob 1. The...

Reaction Quotient

THE REACTION QUOTIENT ‘Q’ Consider the...

Law of Chemical Equilibrium

Law of Chemical Equilibrium According to this law,...

Le Chatelier Principle

LE CHATELIER’S PRINCIPLE “When an...

Solved Problems of Chemical Equilibrium

SOLVED PROBLEMS Subjective: Board Type Questions...

Introduction to Chemical Equilibrium

INTRODUCTION It is an experimental fact that most...

Characteristics of Equilibrium Constant

CHARACTERISTICS OF EQUILIBRIUM CONSTANT (K c ) (i)...

Effect of Catalyst on Equilibrium

EFFECT OF CATALYST ON EQUILIBRIUM Since the...

Law of Mass Action

LAW OF MASS ACTION Guldberg and Waage established...