The given reaction is: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
According to available data
N₂ = [3.0]; H₂ = [2.0]; NH₃ = [0.50]
\({ Q }_{ C }=\cfrac { \left[ { { NH }_{ 3 }\left( g \right) } \right] ^{ 2 } }{ \left[ { N }_{ 2 }\left( g \right) \right] \left[ { H }_{ 2 }\left( g \right) \right] ^{ 3 } } =\cfrac { \left[ 0.50 \right] ^{ 2 } }{ \left[ 3.0 \right] \left[ 2.0 \right] } =\cfrac { 0.25 }{ 24 } =0.0104\) 

We know that Kp = K, (RT) ^Δng
For the above reaction, Δn_g= (2 + 1) - 2 = 1
K_p = 3.75 x 10^-6(0.0831 x 1069) = 3.3 x 10^-4

(i) The value of equilibrium constant K_C tells us the extent of the reaction i.e., it indicates how far the reaction has proceeded towards product formation at a given temperature.
(ii) A large value of K_C indicates that the reaction reaches equilibrium with high product yield on the other hand, lower value of K_C indicates that the reaction reaches equilibrium with low product yield.
(iii) If K_C > 10³, the reaction proceeds nearly to completion.
(iv) If K; < 10 - 3 the reaction rarely proceeds.
(v) If the K_C is in the range 10^-3 to 10³, significant amount of both reactants and products are present at equilibrium.

(i) The units of Kp and Kc depend on the value of \(\triangle\)n_g 
(ii) If number of moles of reactants and products are equal (ie) \(\triangle\)ng = 0; Then K_p and K_c have no units.
(iii) If there is increase or decrease in the number of moles of the reaction, then
unit of K_p is (atmospherere)^{\(\triangle\)n}g
Unit of K_c is (mol per litre )^{\(\triangle\) ng}

(i) Heterogeneous equilibrium
(ii) Heterogeneous equilibrium
(ii) Homogeneous equilibrium
(lv) Homogeneous equilibrium.

(i) As the concentration of the products increases, more products collide and react in the backward direction.
(ii) As the rate of the reverse reaction increases, the rate of the forward reaction decreases.
(iii) Eventually the rate of both reactions becomes equal.

(i) Concentration of reactants decreases with time initially and concentration of products increases with time.
(ii) After sometime, equilibrium is reached i.e., concentration of reactants and products remains constant.

\({ I }_{ 2 }(S)\leftrightharpoons { I }_{ 2 }\left( g \right) \) 
Camphor (s) ⇌Camphor (g)
NH₄Cl(s)⇌ NH₄Cl(g)

(i) Consider a system in which the solid sublimes to vapour. e.g., 12(or) camphor.
(ii) When solid iodine is placed in a closed transparent vessel, after sometime, the vessel gets filled up with violet vapour due to sublimation of iodine.
(iii) Initially the intensity of the violet colour increases, after some time it decreases and finally it becomes constant as the following equilibrium is attained.
\({ I }_{ 2 }\left( s \right) \leftrightharpoons { NH }_{ 4 }Cl(g)\)

Solid in liquids:
1.When you add sugar to water at a particular temperature, it dissolves to form sugar solution. If you continue to which the added sugar remains as solid and the resulting solution is called a saturated solution. Here, as in the previous cases a dynamic equilibrium is established between the solute molecules in the solid phase and in the solution phase.
Sugar (Solid) In this process \(\rightleftharpoons \) Sugar (Solution)
Rate of dissolution of solute =Rate of crystallisation of solute
Gas in liquids:
1.When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecules in the gaseous state and those dissolved in the liquid.
2. In carbonated beverages the following equilibrium exist
\({ CO }_{ 2 }\left( g \right) \rightleftharpoons { CO }_{ 2 }\left( s \right) \)
3. Henry's law is used to explain such gas-solution equilibrium processes

Consider that 'a' moles of PCl , is taken in container of volume 'V'
Let x moles of PCl₅ be dissociated into x moles of PCl₃ and x moles of Cl₂

Applying law of mass action
\(\\ \\ { K }_{ C }=\cfrac { \left[ { PCl }_{ 3 } \right] \left[ { Cl }_{ 2 } \right] }{ \left[ { PCl }_{ 5 } \right] } =\cfrac { \left( \frac { x }{ V } \right) \left( \frac { x }{ V } \right) }{ \frac { a-x }{ V } } =\cfrac { { x }^{ 2 } }{ \left( a-x \right) V } \) 
K_p, calculation: K_P= K_C . RT^Δng
Δn_g = 2-1 = 1
We know that = PV = nRT
\(RT=\cfrac { PV }{ n } \) 
Where 'n' is the total number of moles at equilibrium
n = a - x + x + x = a + x
\({ LK }_{ P }=\cfrac { { x }^{ 2 } }{ \left( a-x \right) V } .\cfrac { PV }{ n } \)
\({ K }_{ P }=\cfrac { { x }^{ 2 }\times PV }{ \left( a-x \right) V(a+x) } \)
\({ K }_{ P }=\cfrac { { x }^{ 2 }P }{ \left( a-x \right) \left( a+x \right) } \)

Let us consider the formation of HI in which, 'a' moles of hydrogen and 'b' moles of iodine gas are allowed to react in a container of volume V. Let 'x' moles of each of H₂ and I₂ react together to form 2x moles of HI.
H₂(g) + I₂(g) ⇌ 2HI(g)

Applying law of mass action
\({ K }_{ C }=\cfrac { \left[ HI \right] ^{ 2 } }{ \left[ { H }_{ 2 } \right] \left[ I_{ 2 } \right] } \) 
\({ K }_{ C }=\cfrac { \left( \cfrac { 2x }{ V } \right) ^{ 2 } }{ \cfrac { \left( a-x \right) }{ V } \cfrac { \left( b-x \right) }{ V } } =\cfrac { { 4x }^{ 2 } }{ \left( a-x \right) \left( b-x \right) } \)
The equilibrium constant K_p can also be calculated as follows:
We know the relationship between the K_c and K_p
K_P, = K_C . RT^Δng
Here the
Δn_g = n_p - n_r = 2 - 2 = 0
Hence, K_P = K_C
\(\\ \\ \\ { K }_{ P }=\cfrac { { 4x }^{ 2 } }{ \left( a-x \right) \left( b-x \right) } \) 

The relation between \(\triangle\)G and Q is
\(\triangle\)G =\(\triangle\)G^o+ RT In Q
\(\triangle\)G = change in free energy as the reaction proceeds
\(\triangle\)G^o= standard free energy
Q reaction quotient
R gas constant
T absolute temperature in K
(i) Since, \(\triangle\)G^o = -RT in K
\(\therefore\) \(\triangle\)G^o = -RT in K + RT in Q;
\(\triangle G=RT \ in {Q\over K}\)
If Q < K,\(\triangle\)G will be negative and the reaction proceeds in the forward direction. If Q = K, \(\triangle\)G = 0 reaction is in equilibrium and there is no net reaction.
(ii)\(CO_{(g)}+3H_{2(g)}\rightleftharpoons CH_{4(g)}+H_2O_{(g)}\)
\(K_c={[CH_4][H_2O]\over[CO][H_2]^3}\)
On increasing pressure, volume decreases. If we doubled the pressure, volume will be halved but the molar concentrations will be doubled.
Then,
\(Q_c={2[CH_4].2[H_2O]\over 2[CO]\{2[H_2]\}^3}\)\(={1\over4}{[CH_4][H_2O]\over[CO][H_2]^3}={1\over4}K_c\)
Therefore, Q_c is less than K_c , so Q_c will tend to increase to re-establish equilibrium and the
reaction will go in forward direction.
\(CO_{(g)}+3H_{2(g)}\rightleftharpoons CH_{4(g)}+H_2O_{(g)}\)