(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}

Let us consider the melting of ice in a closed container at 273 K. In the process the total number of water molecules leaving from and returning to the solid phase at any instant are equal.
If some ice-cubes and water are placed in a thermos flask (at 273K and 1 atm pressure), then there will be no change in the mass of ice and water. At equilibrium
Rate of melting of ice = Rate of freezing of water
\({ { H }_{ 2 }O\left( S \right) }\rightleftharpoons { H }_{ 2 }O\left( 1 \right) \)
The temperature at which the solid and liquid phases of a substance are at equilibrium is called the melting point or freezing point of that substance.

In reversible processes, the rate of two opposing reactions equals at a particular stage. At this stage the concentration of reactants and products do not change with time. This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate.

(i) See-saw
There are different types of equilibrium. For example, if two persons with same weight sit on opposite sides of a see-saw at equal distance I from the fulcrum, then the see-saw will be stationary and straight and it is said to be in equilibrium.
(ii) Tug of war
Anotlier example of a state of equilibrium is the game of "tug-of-war." In this game a rope is pulled taut between two teams. There may be a situation when both the teams are pulling the rope with equal force and the rope is not moving in either direction. This state is said to be in equilibrium

The value of equilibrium constant changes with change in temperature. If K₁and K₂ are equilibrium constants at temperatures
T₁ and T₂ m-Heat of reaction at constant pressure. Then,
\(log\quad { K }_{ 2 }-log{ K }_{ 1 }=\cfrac { -1 }{ 2.303 } \left[ \cfrac { 1 }{ { T }_{ 2 } } -\cfrac { 1 }{ { T }_{ 1 } } \right] \Delta H\)
or \(log\cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta H }{ 2.303R } \left[ \cfrac { 1 }{ { T }_{ 1 } } -\cfrac { 1 }{ { T }_{ 2 } } \right] \)

\({ K }_{ c }=\cfrac { \left[ { NH }_{ 3 } \right] ^{ 2 } }{ \left[ { N }_{ 2 } \right] \left[ { H }_{ 2 } \right] ^{ 3 } } =\cfrac { 1.2\times { 10 }^{ -2 } }{ 1.5\times { 10 }^{ -2 }\times \left( 3\times { 10 }^{ -2 } \right) ^{ 3 } } \)

At any instant, the rate of a chemical reaction, at a given temperature is directly proportional to the product of the active masses of the reactants at that instant.
Rate of the reaction \(\alpha \) [Reactant]^x

\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\mathrm{K}_{\mathrm{c}}=\frac{4 x^{2}}{(a-x)(b-x)}\)
This expression doesn't involve, V. So, increase or decrease of volume will not affect the equilibrium and hence the yield of the product.

The equilibrium constant is a constant and it is for equilibrium condition. But 'Q', the reaction quotient is not a constant as it is for non - equilibrium condition. 'Q' is the ratio of the product of active masses of a reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants, under non - equilibrium conditions.
i) If Q = K_c; it is equilibrium
ii) If Q > K_c.; the reaction will proceed in reverse direction
iii) If Q < K_c ; the reaction will proceed in forward direction

This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate. No macroscopic change is observed.

This equation gives the quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy change (\(\triangle\)G^O) and equilibrium constant is
\(\Delta { G }^{ 0 }=-RTln\ K\)    ...(1)
We know that
\(\Delta { G }^{ 0 }=\Delta { H }^{ 0 }-T\Delta { S }^{ 0 }\)
Substituting (1) in equation (2)
\(-RTln\ K\ =\Delta { H }^{ 0 }-T\Delta { s }^{ 0 }\)
Rearranging
In \(K=\cfrac { -\Delta H^{ 0 } }{ RT } +\cfrac { { \Delta S }^{ 0 } }{ R } \)     ...(3)
Differentiating equation (3) with respect to temperature
\(\cfrac { d\left( In\quad K \right) }{ dT } =\cfrac { \Delta { H }^{ 0 } }{ { RT }^{ 2 } } \)       ...(4)
Equation 4 is known as differential form of Van't Hoff equation.
On integrating the equation 4, between T₁ and T₂ with their respective equilibrium constants K₁ and K_2.
\(\int _{ { k }_{ 1 } }^{ { K }_{ 2 } }{ d\left( In\ K \right) =\cfrac { \Delta { H }^{ 0 } }{ R } \int _{ { T }_{ 2 } }^{ { { T }_{ 2 } } }{ \cfrac { dT }{ { T }^{ 2 } } } } \)
\(\left[ In\quad K \right] _{ { K }_{ 1 } }^{ { K }_{ 2 } }=\cfrac { \Delta { H }^{ 0 } }{ R } \left[ -\cfrac { 1 }{ T } \right] ^{ { T }_{ 2 } }_{ { T }_{ 1 } }\)
\(In\quad { K }_{ 2 }-In\quad { K }_{ 1 }=\cfrac { \Delta { H }^{ 0 } }{ R } -\left[ \cfrac { 1 }{ { T }_{ 2 } } +\cfrac { 1 }{ { T }_{ 2 } } \right] \)
\(In\quad \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta { H }^{ 0 } }{ R } \left[ \cfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 }{ T }_{ 1 } } \right] \)
\(log\quad \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta { H }^{ 0 } }{ 2.303R } \left[ \cfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 }{ T }_{ 1 } } \right] \)  ...(5)
Equation (5) is known as integrated form of Van't Hoff equation.

Addition of an inert gas to a reaction at equilibrium, at constant volume has no effect.

If we know the value of k_c and Q, the reaction quotient, we can predict the direction of a reaction
If Q = Kc; the reaction is in equilibrium state.
If Q > Kc: the reaction will proceed in the reverse direction i.e., formation of reactants.
If Q < Kc: the reaction will proceed in the forward direction i.e., formation of products.

i) \(K_{p}=K_{c}(R T)^{\Delta n_{g}}\)
K_p = Equilibrium constant in term of partial Pressures.
K_c = Equilibrium constant in term of concentration.
R = Gas constant; T = Temperature
\(\Delta \mathrm{n}_{\mathrm{g}}\) = Difference between the sum of number of moles of products and the sum of number of moles of reactants in gas phases.
ii) Synthesis of HI:
\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\Delta n_{g}=0 \therefore K_{p}=K_{c}(R T) \Delta n_{g}\)
\(K_{p}=K_{c}(R T)^{\circ} \)
\(K_{p}=K_{c} \text {. }\)

\(K_c={[NH_3]^4[O_2]^5\over [NO]^4[H_2O]^6}\)
\(4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \rightleftharpoons 4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})}\)