The knowledge of equilibrium constant helps us to
1. Predict the direction in which the net reaction will take place
2. Predict the extent of the reaction and
3. Calculate the equilibrium concentrations of the reactants and products.
It is to be noted that these constants do not provide any information regarding the rates of the forward or reverse reactions.

(i) Ripening of fruits and vegetables in few days.
(ii) Tarnishing of silver in few months.
(iii) Rusting of iron slowly.

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.

2NO (g) + O₂(g) ⇌ 2NO₂(g)

\(K_p={(p_{NO_2})^2\over (P_{NO})^2(P_{o_2})}={0.96\times 0.96\over 0.04\times 0.04\times 0.52}\)
K_p = 1.017 x 10³.

\( \therefore\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=0.4 \mathrm{~mol} \mathrm{dm}^{-3} ; \quad\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=0.6 \mathrm{~mol} \mathrm{dm}^{-3} ; \quad\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=0.6 \mathrm{~mol}^{-3} \)
\(\mathrm{K}_{c}=\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=\frac{0.6 \times 0.6}{0.4}\)
K_c = 0.9 mol dm^-3

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

If a system at equilibrium is disturbed, then the system shifts itself in a direction that nullifies the effect of that disturbance.

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:
\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\Delta n_{g}=0 \therefore K_{p}=K_{c}(R T) \Delta n_{g}\)
\(K_{p}=K_{c}(R T)^{\circ} \)
\(K_{p}=K_{c} \text {. }\)

N₂O₂(g) \(\rightleftharpoons \) 2NO₂(g)
Kc = 0.21 at 373 K. The concentrations N₂O₄ and NO₂ are found to be 0.125 mol dm^-3 and 0.5 mol dm^-3 respectively at a given time. From the above information we can predict the direction of reaction as follows.
\(Q={[NO_2]^2\over [N_2O_4]}={0.5\times 0.5\over 0.125}=2\)
The Q value is greater than K_c. Hence, the reaction will proceed in the reverse direction until the Q value reaches 0.21.

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.

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 \(log\cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta H }{ 2.303R } \left[ \cfrac { 1 }{ { T }_{ 1 } } -\cfrac { 1 }{ { T }_{ 2 } } \right] \)

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

2 NO(g) + O₂ (g) ⇌ 2NO₂(g)

\(K_p={P^2_{NO_2}\over P^2_{NO_2}.Po_2}\)
\(={0.96\times 0.96\over 0.04\times 0.04\times 0.52}\)
= 11.07 x 10² (atm)^-1
K_eq = 41.6 x 10² M^-1.

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 I₂react together to form 2x moles of HI.
\({ H }_{ 2\left( g \right) }+{ I }_{ 2\left( g \right) }\rightleftharpoons { 2HI }_{ \left( g \right) }\)

Applying law of mass action,
\({ K }_{ C }=\cfrac { { \left[ HI \right] }^{ 2 } }{ { \left[ H \right] }_{ 2 }\left[ { I }_{ 2 } \right] } \)
= \(\cfrac { \left( \cfrac { 2x }{ V } \right) ^{ 2 } }{ \left( \cfrac { a-x }{ V } \right) \left( \cfrac { b-x }{ v } \right) } =\cfrac { { 4x }^{ 2 } }{ \left( a-x \right) \left( b-x \right) } \)
The equilibrium constant Kp can also be calcu•lated as follows:
We know }he \rer~tionship between the K_c and K_p
Here the \(\Delta n_{ g }\)=n_p -n_r = 2 - 2 =0
Hence K = K_c ;\({ K }_{ p }=\cfrac { { 4x }^{ 2 } }{ \left( a-x \right) \left( b-x \right) } \)

Given \(m_{N_2}\) = 28 g \(m_{H_2}\) = 6g
V = 1 L
\((n_{N_2})_{initial}={28\over 28}=1\ mol\)
\((n_{H_2})_{initial}={6\over 2}=3\ mol\)
N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)

\([NH_3]=({17\over 17})=1\ mol=1\ mol\)
Weight of N₂ = (no. of moles of N₂) × molar mass of N₂
= 0.5 x 28 = 14 g
Weight of H₂ = (no. of moles of H₂) × molar mass of H₂
= 1.5 x 2 = 3 g

Let us consider the formation of ammonia in which, 'a' moles nitrogen and 'b' moles hydrogen gas are allowed to react in a container of volume V. Let 'x' moles of nitrogen react with 3x moles of hydrogen to give 2x moles of ammonia.
\({ N }_{ 2 }\left( g \right) +3{ H }_{ 2 }\left( g \right) \rightleftharpoons { 2NH }_{ 3 }\left( g \right) \)

Applying law of mass action
\({ K }_{ C }=\cfrac { \left[ { NH }_{ 3 } \right] ^{ 2 } }{ \left[ { N }_{ 2 } \right] \left[ { H }_{ 2 } \right] ^{ 3 } } \)
= \(\cfrac { \left( \cfrac { 2x }{ V } \right) ^{ 2 } }{ \left( \cfrac { a-x }{ V } \right) \left( \cfrac { b-3x }{ V } \right) ^{ 3 } } \)
= \(\cfrac { \left( \cfrac { 4x }{ V } \right) ^{ 2 } }{ \left( \cfrac { a-x }{ V } \right) \left( \cfrac { b-3x }{ V } \right) ^{ 3 } } \)
\({ K }_{ C }=\cfrac { 4{ x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 2 } } \)
The equilibrium constant K_p can also be calculated as follows:
\({ K }_{ p }={ K }_{ C }\left( RT \right) ^{ \left( \Delta { n }_{ g } \right) }\)
\(\Delta \)n_g =n_p - n_r = 2 - 4 = -2
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \left( RT \right) ^{ -2 }\)
Total number of moles at equilibrium,
n = a - x + b - 3x + 2x = a + b - 2x
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { PV }{ n } \right] ^{ -2 }\)
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { n }{ PV } \right] ^{ 2 }\)
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { a+b-2x }{ PV } \right] ^{ 2 }\)
\({ K }_{ p }=\cfrac { 4{ x }^{ 2 }\left( a+b\quad -2x \right) ^{ 2 } }{ { P }^{ 2 }\left( a-x \right) \left( b-3x \right) ^{ 3 } } \)